Stoichiometric relationships 1 Flashcards

Question 1

Q

Simple molecular structures

Answer

A

Simple molecular structures consist of molecules not held to one another by formal bonds. The molecules themselves are constructed by means of covalent (shared electron pair) bonds between non-metal atoms
Molecular structures may normally be identified by their lack of metallic elements in the formula, however their are a few metal compounds that are covalently bonded and which consequently form simple molecular structures

Question 2

Q

Giant molecular strucutres

Answer

A

The whole network (lattice) is a construction of atoms all held together by formal covalent (shared electron pair) bonds. This effectively means that the structure is one giant molecule.

Question 3

Q

The smallest particle of each type of structure

Answer

A

atom, a molecule or an ion

Question 4

Q

The hydrogen standard

Answer

A

Hydrogen is the smallest atom and it was originally used as the standard by which all the other atoms were compared. It was assigned a value of 1 unit and other atoms masses calculated compared to hydrogen atoms.

The 1H isotope has a mass assigned a value of exactly 1 atomic mass unit. This was the original reference.

Question 5

Q

The carbon 12 standard

Answer

A

Nowadays the 12C isotope is used as a reference for comparison of relative atomic masses. This isotope has the assigned mass of 12.00000, all other atoms are measured relative to 12C.

Question 6

Q

Avogadros number

Answer

A

Avogadro’s number or constant is the number to which the mass of an atom must be multiplied to give a mass in grams numerically equal to its relative atomic mass.

Example:
Hydrogen has a relative atomic mass of 1 therefore 6.02 x 1023 hydrogen atoms have a mass of 1g

Question 7

Q

1 mole

Answer

A

The amount of any substance containing an Avogadro number of particles of that substance is called a mole.
1 mole of any substance has a mass equal to its relative mass expressed in grams

Question 8

Q

1 mole

Answer

A

The amount of any substance containing an Avogadro number of particles of that substance is called a mole.
1 mole of any substance has a mass equal to its relative mass expressed in grams

Example:
Example

1 mole of magnesium contains 6.02 x 1023 magnesium atoms

Question 9

Q

Simple atomic structurees

Answer

A

All of the atoms are held together by weak forces only. The substances are always gases at room temperature, liquifying and solidifying at temperatures far below zero celsius (0ºC). As the only known cases of simple atomic structure are the noble gases (Group 0), they are of little interest as regards moles calculations.

Question 10

Q

Simple molecular substances

Answer

A

1 mole of a simple molecular substance contains 1 mole of molecules of that substance = 6 x 1023 molecules

In the case of water, where 1 molecule is made up of two hydrogen atoms and 1 oxygen atom bonded together, it is clear that each molecule contains a total of three atoms. We can talk about these atoms as if they were not chemically bonded.

It is valid to say that 1 mole of water contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms.

Question 11

Q

Giant molecular structures

Answer

A

These structures have a giant lattice in which the atoms are all covalently bonded together, effectively making the entire structure 1 molecule. The atoms may be all the same, in the case of a giant molecular element such as diamond, or they may be different as in silicon dioxide.

Question 12

Q

Giant ionic structures

Answer

A

Ionic compounds are giant structures comprising a lattice of oppositely charged ions. The simplest ratio of the ions is called the formula unit. 1 mole of an ionic compound is understood to contain 1 mole of formula units of that substance.

Question 13

Q

Giant metallic structures

Answer

A

Metallic structures comprise a giant structure of metal atoms within which the valence electrons are delocalised. In terms of discussing the number of moles the structure can be considered to be made up of associated atoms.

Question 14

Q

Composition by mass

Answer

A

The ratio of element mass within a compound depends on both the relative masses of the atoms involved as well as the number of them present. The ratio of atoms in water is 2:1, hydrogen to oxygen respectively.

However, as the mass of a hydrogen atom is 1 and the mass of an oxygen atom is 16, then the ratio in terms of mass is:

mass of 2 atoms of hydrogen = 2 x 1 = 2
to mass of 1 atom of oxygen = 1 x 16 = 16
The mass ratio of hydrogen to oxygen in water is always, 2 : 16

Question 15

Q

Valency

Answer

A

The valency of an atom is the number of single chemical bonds that it can make (in the case of a covalently bonding substance) or the number of electrical charges that it carries (for an ion). Notice that once again the nature of the substance in question requires that the definitions be adapted appropriately. The concept of valence can be used to find the formula of a compound from the valencies of its constituent elements, or to find the valency of an elements within a compound of known formula.

Every atom within a substance is assigned a valency number that is either positive or negative. The total sum of all of the valencies within a formula unit is zero.
Example: Find the valency of the chlorine atoms in the compound Cl2O7

The oxygen has a valency of -2

7 oxygen atoms make a total of -2 x 7 = -14

The overall valency must cancel out, i.e sum of the valencies of oxygen + sum of the valencies of chlorine = 0

Therefore Cl2 = +14

Valency of chlorine in Cl2O7 = +14/2 = +7

Question 16

Q

Covalent compounds

Answer

A

A covalent bond is a shared pair of electrons. When non-metals bond to other non-metals they always do so by sharing electron pairs. The total number of bonds that an atom has is called its valency.

In the water molecule oxygen combines with two hydrogens and so has a valency of 2.
The hydrogen atoms are each sharing one pair of electrons - they have a valency of 1. The oxygen atom is sharing two pairs of electrons - it has a valency of 2.

The rules of valency in compounds say that the total valency of the hydrogens must equal the valency of the oxygen.

i.e. 1 + 1 = 2

If the most electronegative element is assigned a negative valency and the most electropositive element a positive valency then the sum of the atoms’ valencies must equal zero.

Hydrogen (electropositive) = +1

Oxygen (electronegative = -2

Sum: [2 x (+1)] + [1 x (-2)] = 0

Question 17

Q

Ionic compounds

Answer

A

Ionic substances are made up of giant ionic lattices. The simplest formula unit consists of the simplest ratio of oppositely charged ions. The total electrical charge MUST equal zero in a neutral compound.

The valency of an ion is the number of electrical charges that it carries. A sodium ion has a single positive charge - it has a valency of +1. An oxide ion (from oxygen) has a charge of two minus, it has a valency of -2.Example: For the compound formed from sodium and oxygen.
The same valency rules apply as for the covalent substances. The sum of positives must equal the sum of negatives.

Sodium has a valency +1

Oxide (from oxygen) has a valency of -2

In order for the sum to equal zero we must have two sodium ions for each oxide ion.

The formula = Na2O

Question 18

Q

Using valencies

Answer

A

Once the valencies of a few elements are known it becomes a simple matter to construct the formula of unknown compounds using the valency method. Remember that the sum of the valencies of all of the atoms in the compound must equal zero.

Where an atom may have either positive or negative valency, it is negative if it is the more electronegative element in the compound and positive if not.

Example: From the water molecule above we know that the valency of hydrogen is +1.

If the valency of nitrogen in ammonia is -3 then we can construct the formula of ammonia thus:

We need enough hydrogens to cancel out the -3 valency of nitrogen. Each hydrogen = +1 therefore we need three hydrogen atoms.

The formula of ammonia = NH3

Question 19

Q

Working with ions

Answer

A

When using valencies to work out the formula of an ion we have to remember the final charge on the ion must equal the sum of the valencies, taking into account whether the valency of each atom is negative or positive.

Example: Find the formula of the sulfate (2-) ion given that the valency of the sulfur atom is +VI and the valency of the oxygen atom is -II

Oxygen always has negative valencies (unless bonded to fluorine)

There is one sulfur atom with a valency of +6 and overall the ion has a valency of -2

Therefore +6 +(xO) = -2

Therefore (xO) = -2 -6 = -8

each O =-2 therfore there are four oxgen atoms in the ion

Formula of the sulfate ion = SO42-

Question 20

Q

ionic compounds

Answer

A

Ionic compounds have positive ions arranged in a giant lattice with negative ions. Every positive ion is surrounded by negaitive ions and every negative ion is surrounded by positive ions. The overall structure has no charge, therefore the number of positive charges is exactly cancelled out by the number of negative charges. This is also the case in the simplest formula unit.

Question 21

Q

covalent compounds

Answer

A

In covalent compounds all of the atoms in each molecule are held to one another by bonds comprising electron pairs. In the case of double bonds there are two electron pairs involved in the bond. There are no full charges in covalent molecules unlike their ionic counterparts, but there may be partial charges caused by dipoles between atoms having different electronegativities, such as oxygen and hydrogen.

Question 22

Q

hydrated compounds

Answer

A

Ionic compounds in which water molecules has been used to build the crystal lattice (water of crystallisation) are called ‘hydrated’ salts

When many substances are crystallised from aqueous solution, water molecules form part of the crystal lattice and become an integral part of the final crystal structure. These molecules are called ‘water of crystallisation’ and when the compound is weighed out they must be taken into account.

Example: Cobalt(II) chloride crystals contain two molecules of water for every cobalt ion.

The formula of the crystals must be written CoCl2.2H2O showing the two water molecules.

Any mass of cobalt chloride weighed out also contains the water molecules.

Example: Calculate the mass of copper sulfate pentahydrate, CuSO4.5H2O, that must be weighed out to prepare 1dm3 of 1 molar solution.

The solution contains 1 x 1 = 1 mole of solute.

Relative formula mass = 63.5 + 32 + (4 x 16) + [5 x (2 + 16)] = 249.5

Therefore 249.5g must be weighed out.

Question 23

Q

Efflorescence

Answer

A

Certain hydrated compounds lose some of their water of crystallisation when left in the open air. This is known as efflorescence. A case in point is that of sodium carbonate decahydrate Na2CO3.10H2O. The crystals develop a powdery layer on the outside as water of crystallisation is lost to the atmosphere. Salts that effloresce cannot be used as standards for accurate preparation of solutions, as the exact composition of the crystals cannot be known.

Question 24

Q

Hygroscopy

Answer

A

Some compounds, typically ionic salts, absorb water from the atmosphere and increase in mass on exposure to air. Once again, the exact composition of the compound cannot be known and such salts cannot be used as primary standards. Sodium nitrate behaves in this way and, as such, is unsuitable for use in gunpowder, potassium nitrate being preferred instead.

Question 25

Q

Deliquescence

Answer

A

This is a special case of hygroscopy, where the salt ends up dissolving in the water absorbed from the air. A pellet of sodium hydroxide will turn into a small pool of sodium hydroxide solution when left in the air for long enough.

Question 26

Q

Percentage composition

Answer

A

The percentage composition of a material is generally taken to mean the percentage by mass of the elements within a substance.

Example: Calcium carbonate - CaCO3

The relative formula mass = 40 + 12 + (3x16) = 100

Calcium makes up 40 relative mass unit parts of this formula, therefore the percentage by mass of calcium is 40/100 = 40%

Carbon makes up 12 relative mass unit parts of this formula, therefore the percentage by mass of carbon is 12/100 = 12%

Oxygen makes up 48 relative mass unit parts of this formula, therefore the percentage by mass of oxygen is 48/100 = 48%

Question 27

Q

Empirical formula

Answer

A

The empirical formula is the simplest ratio of atoms within a chemical compound.

Question 28

Q

Example: Find the empirical formula of the compound which has the following percentage by mass composition: Carbon 12.12%; Oxygen 16.16%; Chlorine 71.17%.

Answer

A

Divide each percentage composition by mass by the relative atomic mass.

Carbon 12.12/12 = 1.01, Oxygen 16.16/16 = 1.01, Chlorine 71.17/35.5 = 2.00

These are in the approximate integer ratio 1xC : 1xO : 2xCl

Thus the empirical formula = COCl2

Question 29

Q

Example: Find the empirical formula of the compound which has the following percentage by mass composition: Carbon 12.12%; Oxygen 16.16%; Chlorine 71.17%.

Answer

A

Divide each percentage composition by mass by the relative atomic mass.

Carbon 12.12/12 = 1.01, Oxygen 16.16/16 = 1.01, Chlorine 71.17/35.5 = 2.00

These are in the approximate integer ratio 1xC : 1xO : 2xCl

Thus the empirical formula = COCl2

This method is used to calculate the empirical formula of unknown substances. If we also know the relative formula mass, obtainable from experimental data, then the molecular formula can also be found.

Question 30

Q

Example: 1.6 g of an unknown organic substance X was burned in excess oxygen and the mass of carbon dioxide produced determined by absorption in concentrated KOH solution.

Answer

A

The mass of water was determined by first passing the gases produced from the combusion through anhydrous calcium choride.

The following results were obtained:

Mass of carbon dioxide = 4.889g
Mass of water obtained = 2.405g
Mass of carbon in 4.889g CO2 = 4.889 x 12/44 = 1.333g

Mass of hydrogen in 2.405g of H2O = 2.405 x 2/18 = 0.267g

percentage by mass of carbon = 1.333/1.6 = 0.833 x 100 = 83.3%

percentage by mass of hydrogen = 0.267/1.6 = 0.167 x 100 = 16.7%

empirical formula = carbon 83.3/12 : hydrogen 16.7/1

empirical formula = C 6.94 : H 16.7

This simplifies to give an empirical formula of C5H12

Question 31

Q

Relative atomic mass

Answer

A

(abbreviated Ar)

Relative atomic mass is used to define the relative mass of one atom of a metallic element, or non-metallic element, which may be considered to consist of unique atoms for the purposes of calculations.

Note: Although sulfur and phosphorus contain discrete molecules (S8 and P4, respectively), they are usually treated in calculations as monatomic.

Question 32

Q

Relative molecular mass

Answer

A

(abbreviated Mr)
Relative molecular mass is used to define the relative mass of one molecule of a covalent substance, whether element or compound.

Note :The halogens bromine and iodine are liquid and solid respectively, at room temperature and pressure.

Question 33

Q

Relative formula mass

Answer

A

(abbreviated Mr)
Relative formula mass is used to define the mass of the simplest ratio of particles in ionic and giant covalent compounds.

Question 34

Q

The mole

Answer

A

The amount of substance that contains an Avogadro number of particles (6.02 x 1023) is called 1 mole of that substance. Where the word amount is used it always refers to the number of moles.

metals and other elements
Moles = mass / relative atomic mass

Simple covalent substances
Moles = mass / relative molecular mass

Ionic and giant covalent compounds
Moles = mass / relative formula mass

All of these relationships can be summarised as an equation triangle

Question 35

Q

Example 1: Calculate the number of moles in 12 g of magnesium

Answer

A

Magnesium has a relative atomic mass = 24

Moles of magnesium in 12 g = 12/24 = 0.5 moles

Question 36

Q

Example 2: Calculate the amount represented by 15g of silicon dioxide

Answer

A

Silicon dioxide, SiO2, is a macromolecular structure with a relative formula mass = 60

Moles of silicon dioxide in 15 g = 15/60 = 0.25 moles

Question 37

Q

Stoichiometry

Answer

A

The stoichiometry is shown by the relative coefficients of the components appearing in the chemical reaction.

Example

4NH3 + 5O2 4NO + 6H2O
The stoichoimetry of the reaction tells us that 4 molecules of ammonia react with 5 molecules of oxygen and produce 4 molecules of nitrogen monoxide and 6 molecules of water

1 mole is equivalent to 6.02 x 1023 molecules therefore the stoichiometry also gives us the ratio of moles reacting.

Consider the reaction:

Fe + S FeS
The stoichiometry of the equation shows us that one atom of iron is needed to react with each atom of sulfur. Extending this idea we can see that the same number of iron and sulfur atoms are always needed for a complete reaction.

Therefore the moles of iron are always equal to the moles of sulfur in this reaction.

If we are told the mass of iron that we start with is 5.6g then we can calculate the mass of sulfur needed. The calculation proceeds via the number of moles. [Relative atomic mass of Fe=56, S=32]

Moles of iron = mass /RAM = 5.6/56 = 0.1 moles

Therefore moles of sulfur = 0.1 moles

RAM of sulfur = 32

Therefore mass of sulfur needed = moles x RAM = 0.1 x 32 = 3.2g

The procedure followed is:

Example: Calculate the mass of chlorine needed to completely react with 2.24g of iron [relative atomic masses Fe=56, Cl=35.5]

2Fe + 3Cl2 2FeCl3
Mass of iron = 2.24g, therefore moles of iron = 2.24/56 = 0.04 moles

from the equation stoichiometry it can be seen that 2 moles of iron react with 3 moles of chlorine

therefore 0.04 moles iron react with 0.04 x 3/2 moles chlorine molecules = 0.06 moles

realtive molecular mass of Cl2 = 2 x 35.5 = 71

therefore 0.06 moles of chlorine = 0.06 x 71 = 4.26g

Question 38

Q

Word equation

Answer

A

Word equations simply show the names of the reacting chemical and products. They are of limited use except for giving an overall description of the chemical reaction. They give no indication of the relative amounts of the reactants or products involved.

Sodium hydroxide + sulfuric acid sodium sulfate + water

To make equations useful, they must show the individual formulae of the reactants and products and indicate the relative quantities in which they react. These ‘formula equations’ are dealt with below.

A chemical reaction is described according to the substance or substances that are present at the beginning - the reactants, and the substances that are present having been formed in the process at the end of the reaction - the products.

Traditionally, the transition in time between the reactants and the products is shown using an arrow. The arrow is NOT an equals sign, but rather an indication that the process takes place over a period of time.

Word equations show the names of the reacting species on the left hand side, and the products of the reaction on the right hand side, both sides linked by an arrow to indicate that change has occurred.

Reactants Products
The reactants and the products are described by their names. These equations give no indication as to the relative masses or numbers of moles reacting and produced and, consequently, they are of limited, or no use in calculations.

Examples:

Sodium hydroxide + nitric acid sodium nitrate + water

sulfuric acid + sodium chloride sodium hydrogen sulfate + hydrogen chloride

Calcium hydroxide + ammonium chloride ammonia + calcium chloride + water

Question 39

Q

Reaction types - Inorganic

Answer

A

The basic reactions of inorganic chemistry can be summarised as:

 Synthesis
 Decomposition
 Neutralisation
 Precipitation
 Displacement
 Redox

Question 40

Q

Synthesis

Answer

A

Two simple substances, usually elements, combine to form a more complex compound. For example, the reaction between iron and sulfur.

Example: Synthesis of iron(II) sulfide

iron + sulfur iron(II) sulfide

The term may also be applied to more complex reactions which are designed to manufacture a specific compound, such as a pharmaceutical. In organic chemistry, for example, we may talk about aspirin synthesis:

Example: Synthesis of aspirin

ethanoic anhydride + 2-hydroxy benzene carboxylic acid aspirin + ethanoic acid

Synthesis is a rather general term that can be applied to almost any chemical change and as such, not very meaningful.

Example: Synthesis of ammonia

nitrogen + hydrogen ammonia

Question 41

Q

decomposition

Answer

A

Decomposition is when a compound is broken down, usually by heat, to produce simpler substances. An example is the thermal decomposition of calcium carbonate, producing calcium oxide and carbon dioxide.

Example: Thermal decomposition of calcium carbonate

calcium carbonate calcium oxide + carbon dioxide

Question 42

Q

neutralisation

Answer

A

Neutralisation is possibly the most common type of reaction encountered in schools. It is easy to carry out, there are many examples from everyday sources and it is easy to demonstrate using pH or litmus indicator papers. Acids occur in many foods such as fruits (lemons, oranges, apples, etc) vegetables (rhubarb), vinegar, etc.. Stomach acid is a reasonably strong solution of hydrochloric acid.

Neutralisation is the reaction of an acid and a base producing a salt and water. For example, the neutralisation of sulfuric acid by sodium hydroxide.

Example: Neutralisation of sulfuric acid by sodium hydroxide

sulfuric acid + sodium hydroxide sodium sulfate + water

Neutralisation can be extended to any compound that react with acids neutralising them. Such substances are called bases.

Question 43

Q

precipitation

Answer

A

ain and snow falls from the sky because the atmosphere cannot support the mass of particles accumulating together. Substances ‘fall out’ of solution because the water cannot dissolve them.

Precipitation (called double decomposition in old text books) is a reaction where two ionic solutions are mixed together, bringing ions that form an insoluble substance in contact with one another. Once the insoluble ionic compound is formed it falls out of the solution as a precipitate.

positive ion (aq) + negative ion (aq) insoluble salt precipitate (s)

Much use is made of this type of reaction in ‘wet’ analysis of inorganic ions, such as the use of silver nitrate solution to test for the presence of chloride ions. A white precipitate of silver chloride appears from the solution of the suspected chloride on addition of silver nitrate solution.

Question 44

Q

Displacement

Answer

A

Displacement reactions occur in several areas of chemistry.

Displacement of a weak (or volatile) acid from its salt by a stronger acid.
Displacement of a weak (or volatile) base from its salt by a stronger base.
Displacement of a metal ion from its compound by a more reactive metal (this is also a redox reaction)
Displacement of hydrogen gas from water or acids by reactive metals (this is also redox)
Example: Displacement of nitric acid from potassium nitrate

potassium nitrate + sulfuric acid potassium hydrogen sulfate + nitric acid

Example: Displacement of ammonia from ammonium chloride

ammonium chloride + calcium hydroxide calcium chloride + ammonia + water

Example: Displacement of copper from copper(II) sulfate solution

copper(II) sulfate + zinc zinc sulfate + copper

Example: Displacement of hydrogen gas from water by calcium metal

calcium + water calcium hydroxide + hydrogen

Question 45

Q

Redox

Answer

A

Redox is short for Reduction and Oxidation.

These are reactions that involve a transfer of electrons from one species (the reducing agent) to another species (the oxidising agent). The terms used are wonderfully confusing as the reducing agent gets oxidised and the oxidising agent gets reduced during the course of the reaction. Extraction of metals falls into this category, as the metal ions in the metal ore need electrons to become metal atoms. For example, the following reaction occurs in the extraction of iron from its ore, haematite.

Example: The extraction of iron (blast furnace)

iron(III) oxide + carbon iron + carbon monoxide

In this equation the iron 3+ ions get reduced by gaining three electrons from the carbon to become iron atoms.

Fe3+(aq) Fe(s)

For a full treatment of reduction and oxidation see ‘Colourful Solutions 8 - Reduction and Oxidation’.

Question 46

Q

Thermal stability

Answer

A

stability to heat is a relative issue. Enough heat will break even the strongest chemical bonds. Definitions of thermal stability usually depend on the method of heating being used. In general, a chemistry laboratory’s heat supply is provided by a Bunsen burner. If a compound is not decomposed by the hottest bunsen flame then we can say that it is thermally stable.

An example of the weakness of this definition is the case of iron(II) sulfate (green vitriol). This is generally considered to be thermally stable. However, this substance was known in the Middle Ages and used to make sulfuric acid (oil of vitriol) by heating strongly until decomposition. This produced sulfur trioxide (sulfur(VI) oxide) fumes. These fumes were lead into cold water giving the acid.

FeSO4 FeO + SO3
SO3 + H2O H2SO4

Question 47

Q

Hydrated compounds II

Answer

A

Hydrated compounds (compounds that contain water of crystallisation) also decompose when heated, losing some, or all, of their associated water molecules.

Example: Thermal decomposition of hydrated magnesium sulfate

magnesium sulfate heptahydrate Anhydrous magnesium sulfate + water

Some ionic compounds may undergo reaction between their water of crystallisation and the ions present, complicating this process. This is the case for magnesium chloride crystals, which decompose on heating to give a basic chloride salt containing chloride ions and hydroxide ions. This is not on the Syllabus.

Question 48

Q

Common bases

Answer

A

Common bases

Metal hydroxides
Metal oxides
Metal carbonates
Metal sulfites
Ammonia solution (ammonium hydroxide)
Metal hydroxides react with acids producing the corresponding salt and water.

Example: Neutralisation of hydrochloric acid by sodium hydroxide

hydrochloric acid + sodium hydroxide sodium chloride + water

Metal oxides also react with acids producing the corresponding salt and water.

Example: Neutralisation of hydrochloric acid by calcium oxide

hydrochloric acid + calcium oxide calcium chloride + water

Metal carbonates react with acids producing the corresponding salt, water and carbon dioxide.

Example: Neutralisation of hydrochloric acid by sodium carbonate

hydrochloric acid + sodium carbonate sodium chloride + water + carbon dioxide

Metal sulfites (sulfate(IV) ) react with acids producing the corresponding salt, water and sulfur dioxide (sulfur(IV) oxide).

Example: Neutralisation of hydrochloric acid by sodium sulfite (sulfate IV)

hydrochloric acid + sodium sulfite sodium chloride + water + sulfur dioxide

Question 49

Q

How do I know which ionic compounds are insoluble?

Answer

A

The short answer is - you don’t! The solubility of ionic compounds can’t be predicted, you have to learn which ones are, and aren’t, soluble. The good news is that the majority of ionic compounds are fairly soluble, so it makes sense to learn the insoluble ones.

In reality, there is no such thing as soluble or insoluble, rather that everything has different degrees of solubility. Even so-called insoluble substances dissolve to a certain degree.
Barium sulfate is generally said to be insoluble, but some does dissolve in water at room temperature, albeit only a tiny amount. The solubility is given by a concept known as the solubility product, ksp.

Similarly a soluble substance may dissolve a large mass per litre or not much at all. Solubility curves show the change in solubility with temperature and these are different for each different substance.

Question 50

Q

Formula equation and cofefficent

Answer

A

Formula equations show the formulae of the reactants and the products on either side. Balancing numbers are used - called the coefficients of the reaction - to ensure that the numbers of particles on both sides of the equation are equal.

In this equation one nitrogen molecule is needed to react with every three hydrogen molecules to produce 2 molecules of ammonia. The coefficient of nitrogen is 1, that of hydrogen is 3, and that of ammonia is 2.

To balance a chemical equation it is important to remember that the formula of the reactants and products cannot be changed and that coefficients may only be placed before the formulae, multiplying them by whole numbers.

Example: One stage in the manufacture of nitric acid is the oxidation of ammonia, as shown below:

4NH3 + _O2 _NO + _H2O
What is the coefficient for O2 when the equation is balanced?

Solution:

Balance the nitrogen by counting up the nitrogen atoms on both sides: 4NH3 + _O2 4NO + _H2O

Balance the hydrogen atoms: 4NH3 + _O2 4NO + 6H2O

Now balance the oxygen atoms: 4NH3 + 5O2 4NO + 6H2O

Correct response : Coefficient = 5

These show the actual numbers of reacting particles in a chemical reaction. The reaction must be BALANCED to give the correct number of particles on each side of the reaction arrow.

2NaOH + H2SO4 Na2SO4 + 2H2O

These equations are constructed by writing the formula of each of the compounds in the reaction, and then by counting up the number of atoms on each side to make sure they are equal. If they are not equal, balancing numbers (coefficients) are added in front of each chemical formula (where needed), so that the numbers of each type of particle on each side of the equation are the same

Step 1 - write the chemical equation

ammonia + oxygen nitrogen monoxide + water
Step 2 - write the formula of each of the reaction components

ammonia + oxygen nitrogen monoxide + water
NH3 + O2 NO + H2O
Step 3 - add coefficients IN FRONT OF the formulae to balance the equation

4NH3 + 5O2 4NO + 6H2O

Note: Whenever an exam question asks for an equation, it is the balanced formula equation that is required, unless specified otherwise.

Question 51

Q

Ionic equations

Answer

A

When ionic solutions react, the reaction only usually involves some of the ions and not others. An ionic equation shows just the ions implicated in the reaction. The other ions are often called “spectator ions”.

2NaOH + H2SO4 Na2SO4 + 2H2O

The ionic equation is written as:

H+ + OH- H2O

Although the formula equation represents the overall process, the sodium ions start off in solution as Na+(aq) and at the end of the reaction they are still Na+(aq), nothing has changed, they are merely spectator ions.

The same applies to the sulfate ions, SO42-. The only particles that actually react are the OH- ions from the sodium hydroxide and the H+ ions from the sulfuric acid.

Question 52

Q

State symbols

Answer

A

These are used to show the states of the various compounds that constitute the equation for the reaction.

(s) means that the compound is in the solid state.
(l) means that the compound is in the liquid state.
(g) means that the compound is in the gaseous state.
(aq) means that the compound is dissolved in water, i.e. it is in solution.

2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l)

The sodium hydroxide and the sulfuric acid are in solution. These make sodium sulfate in solution and water liquid.

Question 53

Q

The reaction between hydrogen and oxygen

Answer

A

When hydrogen reacts with oxygen to produce water the original reactants are molecules of hydrogen and oxygen. These consist of atoms held together by covalent bonds to form H2 and O2 molecules.

These molecules must physically collide in order to break the bonds holding the atoms together in the molecule, before new bonds can be formed. The new bonds that are formed between the oxygen and hydrogen atoms hold the atoms together in a ratio of two hydrogen atoms for each oxygen atom.

We call this arrangement of atoms a water molecule!

In terms of reacting particles, one molecule of oxygen must react with two molecules of hydrogen for the process to make two molecules of water.

In terms of moles of particles, this equates to one mole of oxygen molecules reacting with two moles of hydrogen molecules making two moles of water molecules.

The number of reacting particles is shown by the coefficients of the equation (the large numbers that appear in front of the chemical formulae in the balanced equation)

oxygen + hydrogen water
O2 + 2H2 2H2O

Question 54

Q

The law of conservation of mass

Answer

A

This idea may be stated as the “Law of Conservation of Matter”, which states that in any process the total mass of the reactants MUST equal the total mass of the products.

Considering the reaction between oxygen and hydrogen:

oxygen + hydrogen water

O2 + 2H2 2H2O

The mass of water formed MUST equal the mass of oxygen and hydrogen reacting. Notice that this does not necessarily mean the mass of oxygen and hydrogen that are mixed together. Only those molecules that actually react contribute to the mass of water formed.

O2 + 2H2 2H2O

32g + 4g 36g

So the mass of water formed is exactly equal to the mass of oxygen and hydrogen reacting.

Question 55

Q

Reactions evolving gases

Answer

A

When a gas is produced in the course of a chemical reaction it usually escapes and leaves the reaction mixture. The experimentally determined mass of the reactants seems to decrease, but this mass loss is due the mass of the gas that has left.

Gas evolution allows us to calculate reaction stoichiometry, or formulae, by mass loss or gas volume produced.

Example: When an oxide of copper was heated in a stream of hydrogen the following results were obtained.

Mass of the copper oxide before reaction = 8.0g
Mass of copper remaining after reaction = 6.4g
It seems that mass has been lost from the reaction and as we know that mass cannot be created or destroyed this means that a gas has left the original material. The oxygen has been removed and its mass can be calculated by subtraction.

Mass of oxygen leaving = Mass of copper oxide before reaction - Mass of copper remaining after reaction

Mass of oxygen leaving = 8.0 - 6.4 = 1.6g
From these results we can now work out the mole ratio of atoms in the original compound

Moles of copper = 6.4/64 = 0.1
Moles of oxygen = 1.6/16 = 0.1
Ratio of moles copper to moles oxygen = 1:1
Therefore the formula of the copper oxide = CuO

Care must be taken to ensure that all of the gas has really left, as many gases are soluble in water and will be dissolved to a certain extent in any aqueous reaction mixture. This is particularly true for gases such as ammonia or sulfur dioxide, both of which are very soluble in water.

Question 56

Q

Sulfuric acid + magnesium magnesium sulfate + hydrogen

Answer

A

gas released: hydrogen

Question 57

Q

Reactions absorbing gases

Answer

A

When certain substances are heated in air they gain mass, the experimentally determined mass of the reactants increases. This is clearly impossible unless some substance has been added from the suroundings.

Oxygen from the air may react with the substance being heated causing it to gain in mass. The new substance consists of both the original particles and added oxygen particles (occasionally nitrogen or carbon dioxide may also be implicated in mass gain, although this is rather less common)

Example: When copper metal is heated in air it turns black and the product has a greater mass than the original copper. Measurement of the mass increase allows us to calculate the formula of the copper oxide produced:

Experimental data

Mass of copper before heating = 6.4g
Mass of copper after heating = 8.0g
Data processing

Mass increase due to absorbed oxygen = 1.6g

Moles of copper = 6.4/64 = 0.1 moles

Moles of oxygen absorbed = 1.6/16 = 0.1 moles

Mole ratio of copper to oxygen = 1:1

Therefore formula of copper oxide = CuO

Question 58

Q

Burning

Answer

A

reacting with oxygen (or air) producing oxides.

Question 59

Q

Combustion

Answer

A

a reaction with oxygen (or air) releasing energy

Question 60

Q

Crystallise

Answer

A

to deposit crystals from a solution

Question 61

Q

Decompose

Answer

A

to break apart a compound into simpler substances.

Question 62

Q

Dissolve

Answer

A

mix a solute (solid) into a solvent (liquid) until it disappears.

Question 63

Q

Distill

Answer

A

heat and lead the vapours produced into another container

Question 64

Q

Decant

Answer

A

pour liquid off carefully leaving any solid behind in the vessel

Answer 65

A

to come out of solution, usually as crystals, or the products laid down in the electrodes in electrolysis.

Answer 66

A

gas evolution (to effervesce = to bubble and release a gas)

Answer 67

A

pass an electric current through an ionic solution causing reactions to occur at the electrodes.

Answer 68

A

pass a liquid mixture through a filter paper to remove any undissolved solids

Answer 69

A

a solid formed by the addition of two solutions, the ions of which combine to produce an insoluble solid

Answer 70

A

heat using apparatus designed to condense and collect any vapours produced and lead them back in the reaction vessel.

Answer 71

A

to add one solution to another until a specific end point is reached (When the reaction is complete)

Answer 72

A

The concentration of a solute is measured in moles of solute per litre. If the volume of the solution is known, then the number of moles contained by the solution can be calculated.

moles = molarity x volume

For example, a 1.0 M solution contains 2 moles of solute per litre (1 litre = 1000cm3)

100 cm3 of a 1.0 M solution contains 100/1000 x 2 moles = 0.2 moles of solute

The concentration of solutions is often determined by titration

Example: 1.5g of impure calcium carbonate was added to 50cm3 of 2M HCl solution. Analysis of the resultant solution revealed that the new molarity of the HCl was 1.85M. Calculate the percentage purity of the calcium carbonate sample.

Total moles of hydrochloric acid before reaction = 0.05 x 2 = 0.1 moles

Total moles of hydrochloric acid after reaction = 0.05 x 1.8 = 0.09 moles

Therefore moles of acid reacted = 0.1 - 0.09 = 0.01 moles

Equation for the reaction:

CaCO3 + 2HCl CaCl2 + H2O

1 mole of calcium carbonate reacts with 2 moles of acid

Therefore 0.005 moles of carbonate reacted with 0.01 moles of acid

Mr of CaCO3 = 100

therefore 0.005 moles = 0.005 x 100g = 0.5 g

But the original sample had a mass of 1.5g of which only 0.5 g was calcium carbonate

Percentage purity of the sample = 0.5/1.5 x 100 = 20%

Answer 73

A

Gas particles, in common with all particles, are in constant motion. Gases are substances in which the force of attraction between the particles has been overcome by their energetic motion. The gas particles can simply no longer be held together by attractive forces.

As the particles fly around at high speed, they collide many times per second with each other and with the walls of any container. If the container has no walls, the gas particles spread out to fill all the available space.

To deal with gases in chemistry some assumptions are made:

1 The gas particles themselves occupy no volume.
2 The forces of attraction between particles are so small as to be negligible.
3 Collisions involving gas particles are perfectly elastic.

Answer 74

A

In chemistry, volumes are usually measured in litres (decimetres cubed, dm3) and centimetres cubed (cm3), rather than metres cubed (m3). The reason for this is purely practical, the metre cubed is a very large volume compared to the test tubes and flasks used in laboratories. See apparatus, section 1.62.

Gas volumes may be quoted in metres cubed (m3), litres (L), centimetres cubed (cm3) or millilitres (mL), depending on the textbook consulted.

1m3 (1 metre cubed) = 1000 dm3 (1000 decimetres cubed) = 1,000,000 cm3 (1 million centimetres cubed)

1 decimetre cubed (1 dm3) is also called 1 litre (= 1000 cm3)

1 centimetre cubed (cm3) is also called 1 millilitre (mL) as it is 1/1000 of a litre.

Answer 75

A

To convert from cm3 or ml to dm3 or litres divide by 1000

To convert from dm3 or litres to cm3 or ml multiply by 1000

Answer 76

A

The pressure of a gas is caused by the gas particles colliding with the walls of the container. Each small collision exerts a force on the wall. The sum of these forces over an area of the wall is called the gas pressure. The SI unit of force is the Newton and the unit for area is the metre squared (m2). Pressure is measured in Newtons per metre squared = Nm-2. This combined unit is called the Pascal, Pa.

1 Nm-2 = 1 Pa
The Pascal is a fairly small quantity and atmospheric pressure = 100.0 kPa (approximately) - this is the value used for gas calculations in the IB chemistry exams.

Note: In SI units atmospheric pressure = 101.3 kPa = 1.01 x 105 Pa.

Older measurements for pressure may be encountered in textbooks, for example: mmHg (Torr), where atmospheric pressure = 760 mmHg (Torr).

However, the IB is fairly consistent with the use of kPa as a pressure measurement.

Conversion of pressure units

1000 Pa = 1 kPa

To convert from Pa into kPa divide by 1000

To convert from atm (atmospheres) to kPa multiply by 101.3

To convert from kPa to atmospheres divide by 101.3

Answer 77

A

the SI unit of temperature is the kelvin (K), although problems are often set in degrees Celsius (ºC). It is important to ALWAYS carry out gas calculations using absolute (kelvin) temperature values.

0K is called absolute zero. It is the temperature at which particles have no energy. This temperature is equal to -273.16ºC, approximated to -273ºC. The magnitude of 1 kelvin is the same as that of 1º Celsius, therefore;

0K = -273ºC
273K = 0ºC
373K = 100ºC
Conversion of temperature units

To convert from degrees Celsius to Kelvin add 273

To convert from Kelvin to degrees Celsius subtract 273

Absolute temperature in Kelvin = degrees Celsius + 273

Temperature in degrees Celsius = Absolute temperature in Kelvin - 273

Answer 78

A

Ambient temperature varies according to the location and season, and atmospheric pressure depends on weather conditions and height above sea level. These variable quantities are ‘controlled’ by agreeing to a worldwide definition of standard temperature and pressure, called STP.

Standard Pressure = 1.00 x 105 Pascals (Nm-2)
Standard Temperature = 273.15 Kelvin
Chemists often measure pressure in kPa (kilo Pascals), in which standard pressure = 100 kPa (1 kPa = 1000 Pa)

However, you will occasionally see questions in atm (atmospheres) in which 1 atm = 101.3 kPa

Answer 79

A

STP is NOT standard state conditions.

This causes confusion amongst students, as the definitions of standard state conditions are (usually) 298 K and atmospheric pressure. Standard state is used in thermodynamics calculations and is symbolised using the standard state subway sign. (more correctly called a superscript plimsole line)

For example

The definition of standard enthalpy of combustion ΔHc, is the energy change due to combustion when 1 mole of a substance burns in excess oxygen measured at 25ºC and 1 atmosphere pressure.

Table showing difference between STP and standard state

Temperature Pressure

STP 273 K 100 kPa
Standard state 298 K 100 kPa

Answer 80

A

In 1808 the Freanch chemist Gay Lussac investigating the reactions of gases came to the conclusion that when gases combine chemically they do so in volumes that have a simple ratio to one another, and to any gaseous product, provided that all gases are measured at the same temperature and pressure. This became known as Gay Lussac’s law.

For example, in the reaction between hydrogen and oxygen making water:

hydrogen + oxygen water
2H2 + O2 2H2O
The volume of hydrogen needed for complete reaction is always double the volume of oxygen, provided the temperature and pressure of the two gases are the same.

Answer 81

A

Extending Gay Lussac’s work in 1811, Avogadro suggested that “equal volumes of all gases contain equal number of molecules (the gases being measured at the same temperature and pressure). This became known as Avogadro’s hypothesis or Avogadro’s law.

This means that in the reaction:

N2 + 3H2 2NH3

For a given volume of nitrogen, three times the volume of hydrogen is needed for complete reaction. The volume of nitrogen contains a certain number of molecules and there are three times as many molecules in the volume of hydrogen.

N2
+
3H2

2NH3
1 molecule
+
3 molecules

2 molecules
1 volume
+
3 volumes

2 volumes
Avogadro’s hypothesis allows us to substitute the coefficients of any balanced gaseous equation for volumes of gas, for example:

Example: Find the volume of hydrogen required to react completely with 200 cm3 of nitrogen according to the equation:

N2 + 3H2 2NH3

The equation tells us that 1 volume of nitrogen reactrs completely with 3 volumes of hydrogen:

Therefore volume of hydrogen = 3 x volume of nitrogen

Therefore volume of hydrogen = 3 x 200 cm3 = 600 cm3

Example:

Using the equation:

N2 + 3H2 2NH3

if 100 cm3 of nitrogen was used (this is now equivalent to 1 ‘volume’) it would need 3 times as much hydrogen for complete reaction i.e. 3 volumes = 300 cm3.

The reaction would produce 2 volumes of ammonia i.e. 2 x 100 cm3 = 200 cm3

Answer 82

A

Avogadro realised that if gas volumes contain the same number of particles as one another (under the same temperature and pressure conditions) then an Avogadro number of molecules of each gas will occupy the same volume. This volume is called the gas molar volume

When the temperature is 273K and the pressure is standard (100 kPa) the gas molar volume is equal to 22.7 dm3 ( to 3 significant figures)

This means that the number of moles of a gas can be calculated from the volume, providing the temperature and the pressure are known.

It is important to be consistent with the volume units used. In the equation triangle above, the units are litres for all volume measurements..

Example: Calculate the number of moles of present in 10 dm3 of gas at STP

1 mole of gas occupies 22.7 dm3 at STP

therefore moles of gas = volume (in litres) /22.7

Moles of gas = 10/22.7

Moles of gas = 0.441 moles

If the conditions are not STP (standard temperature and pressure) then the ideal gas equation PV = nRT mus

Answer 83

A

This was one of the first means of establishing relative masses. The number of moles can be ascertained using the volume data at STP and the mass of the gas can be worked out from density or other, more direct, weighing methods.

Example: Calculate the relative molecular mass of a gas if 0.254g of it occupies 80cm3 at STP

1 mole of gas occupies 22.7 dm3 at STP

therefore 80 cm3 of gas = 3.52 x 10-3 moles

3.52 x 10-3 moles of gas has a mass = 0.254g

Therefore relative molecular mass = 0.254/(3.52 x 10-3) = 72.1

Answer 84

A

It gives a measure of the amount of matter there is per unit volume, how ‘dense’ something is. The density of a gas is the mass of a given volume of gas divided by its volume.

Gases have a much lower density than liquids or solids. Hydrogen is the least dense gas.

Density = mass/volume

units: g/cm3 or g cm-3

Example: A block of wood was found to have a mass = 36 g. When fully immersed in water it displaced a volume of 126 cm3. Calculate its density.

Density = mass/volume

Density = 36/126 = 0.286 gcm-3

To find the density of a gas, the volume and the mass must be known. At STP both of these quantities are known, as 1 mole of any gas occupies 22700 cm3. It is, therefore, a simple matter of dividing the molar mass by 22700 to obtain the density. The values produced, however, are very small and gas densities are instead often quoted in grams per litre for this very reason.

For example, hydrogen has a density at STP of 2/22700 = 8.8 x 10-5 g/cm3

Gas densities at STP (g/dm3)
Hydrogen	0.088
Helium	0.176
Oxygen	1.410
Carbon dioxide	1.938
Xenon	5.784
Radon	9.780

Answer 85

A

Water has a density of about 1g per cm3 which can make it useful in calculations. Relative density (see relative measurements) is the density relative to the density of water, so if a substance has a relative density of 13.8 this means that it is 13.8 times as dense as water.

Water density = 1g per cm3, so something with a relative density of 13.8 has a density = 13.8 g per cm3.

Example: An evacuated sealed flask is weighed = m1

The flask is filled with an unknown gas and reweighed = m2

The flask is filled with water and the water poured into a measuring cylinder to find the volume = V

The mass of the unknown gas = mass of the gas filled flask - mass of the empty flask = m2 - m1

The density of the gas = mass of the gas/volume of the flask = (m2 - m1)/V

Once the density of a gas is known at a specific temperature, its mass can be obtained from the volume occupied.

Note: We are still referring to STP conditions; Under other temperature and pressure conditions the gas laws must be used.

Example: An unknown gas has a density of 1.79 x 10-4 g/cm3 at STP, calculate its relative molecular mass.

At STP 1 mole of gas occupies 22.7 dm3 = 22700 cm3

Therefore mass of 1 mol of the unknown gas at STP = 22700 x 1.79 x 10-4 g = 4.06 g

The relative molecular mass of unknown gas = 4.06

Note: According to the gas laws, the volume of a gas is dependent on the temperature and so then is the density. As a gas increases in temperature it becomes less dense as the volume occupied by the same mass of gas increases. This is why hot air rises.

The hot air has a lower density than the surrounding colder air and is ‘pushed up’ by the colder air, in the same way as a piece of wood tends to be pushed up to the surface of water. The wood is less dense than water and therefore floats.

Answer 86

A

Robert Boyle found that for a given number of moles of gas the volume of the gas is proportional to its pressure provided the temperature is kept constant.

This can be expressed as:

PV = constant
The pressure of a gas derived from collisions of the gas particles with the walls of the container. If the volume is decreased then the gas particles are resticted to less space and there are more collisions per second at the walls and consequently a greater pressure is exerted.

Boyles law allows calculation of new volume and/or pressure conditions (at constant temperature) in, for example, a piston or gas syringe.

Example: Calculate the pressure when 100cm3 of gas at 1 atmosphere pressure is compressed to 10cm3 at constant temperature.

Boyle’s law states that PV is constant therefore the initial product of the pressure times the volume must equal the final product of the pressure times the volume, i.e.:

P1V1 = P2V2

P1 = 1.01 x 105 Pa, V1 = 0.1dm3, V2 = 0.01dm3, P2 = ?

P2 = (1.01 x 105 x 0.1)/0.01

P2 = 1.01 x 106 Pa

Answer 87

A

e found that for a given number of moles of gas the volume of the gas is inversely proportional to its temperature, provided the pressure is kept constant.

This may be expressed as:

V/T = constant
This relationship can be explained by considering the effect that temperature has on the average kinetic energy of the gas particles. Increase the temperature and the particles increase in energy and speed. Collisions between particles force the gas apart and if the pressure is kept constant then the gas expands.

Charles’ Law is useful for calculating the change in a gas volume under different temperature conditions. Gases are not usually dealt with under standard conditions and so this law allows correction for the current temperature.

Note: the temperature referred to is ALWAYS the absolute temperature in Kelvin, where 0 Kelvin = -273ºC.

Forgetting to change the temperature data into kelvin is a common mistake in calculations involving the gas laws.

Example: Calculate the volume occupied by 1 mole of gas at 35ºC at atmospheric pressure.

At STP (273K, 1 atmosphere pressure) 1 mole of any gas occupies 22.7 dm3

According to Charles’ Law:

V1/T1=V2/T2
T1
T2

V1T2/T1=V2

(22.7 x 308/273)=V2
New volume = 25.6 dm3 (3 sig. figs)

Answer 88

A

he equations of Boyle’s law (PV = constant) and Charles’ law (V/T = constant) may be combined to give the expression: PV/T = constant. This is called the equation of state. This new combined gas equation can be used to find the new volume, pressure and temperature of a gas if the conditions are changed.

The equation of state: P1xV1/T1=P2xV2/T2

Answer 89

A

The equation of state refers to a fixed mass of gas. From Avogadro’s law we know that the same volume of all gases contain the same number of moles and from this, it follows that the volume is proportional to the number of moles.

Volume ∝ number of moles (n)

P1xV1/T1= constant

These two equations can be combined to obtain an expression involving all the quantities:

P1xV1/T1=k’ x n

fter rearrangement, for ‘n’ moles of gas the proportionality constant is called the Universal Gas Constant and is given the symbol ‘R’

This gives the ideal gas equation:

Ideal Gas Equation: PV = nRT

where:

P = pressure in Pa
V = volume in m3
n = number of moles of gas
R = Universal Gas constant = 8.314 JK-1mol-1
T = the absolute temperature in Kelvin
It is often more convenient to express the pressure in kPa and the volume in litres (dm3). This leaves the value of R the same (see below).

Example: Calculate the number of moles of gas present in 2.6 dm3 at a pressure of 1.01 x 105 Pa and 300 K.

PV = nRT

6 dm3 = 0.0026 m3
0026 x 1.01 x 105 = n x 8.314 x 300

n = 0.0026 x 1.01 x 105 / 8.314 x 300

n = 0.105 moles

There are several units used for gas volume, gas pressure and temperature. It is important to be consistent with the use of units when carrying out gas law calculations. The Syllabus states that SI units will be used wherever possible.

Answer 90

A

The SI units of P, V and T give rise to the previously used value for the universal gas constant, R = 8.314 J K-1 mol-1.

How does this happen when chemists do not use these SI units?

Remember:

1 litre = 1 dm3 = 1000 cm3

Consequently, if litres are used in the Ideal Gas equation then the pressure units must also be divided by 1000 (as PV = constant). Pressure is measured in Pa or Nm-1, and so the unit of the kPa correct for the difference in volume units.

Atmospheric pressure in Pa = 1.00 x 105 Pa

Atmospheric pressure in kPa = 1.00 x 102 kPa

Provided that you are consistent with the application of units there will be no problem. It is always a good idea when carrying out calculations to look at the value of your answer and ask yourself, “does it seem reasonable?”

The IBO is consistent with the use of litres (dm3) and kPa in gas law questions.

Answer 91

A

The SI units of P, V and T give rise to the previously used value for the universal gas constant, R = 8.314 J K-1 mol-1.

How does this happen when chemists do not use these SI units?

Remember:

1 litre = 1 dm3 = 1000 cm3

Consequently, if litres are used in the Ideal Gas equation then the pressure units must also be divided by 1000 (as PV = constant). Pressure is measured in Pa or Nm-1, and so the unit of the kPa correct for the difference in volume units.

Atmospheric pressure in Pa = 1.00 x 105 Pa

Atmospheric pressure in kPa = 1.00 x 102 kPa

Provided that you are consistent with the application of units there will be no problem. It is always a good idea when carrying out calculations to look at the value of your answer and ask yourself, “does it seem reasonable?”

The IBO is consistent with the use of litres (dm3) and kPa in gas law questions.

Answer 92

A

Real gases deviate from ideal behaviour but only by small amounts under low pressures.

Answer 93

A

At high pressures, distances between gas particles are fairly small, and attractive forces between them becomes significant. Neighboring molecules exert an attractive force, which reduces the interaction of molecules with the container walls. The apparent pressure will be less than ideal (PV/RT will consequently be less than ideal).

As pressures increase, the volume of the gas molecules themselves becomes significant in relationship to the container volume and PV/RT will be higher than ideal (V is higher).

Answer 94

A

At high temperatures, the kinetic energy of the molecules is much more important than the forces of attraction and the gases behave more ideally
The relationship between pressure and ideality may be shown by a graph of PV against Pressure (at constant temperature).

The graph shows that an ideal gas maintains the value of PV=constant over a very large range of different pressures. A non-ideal gas (real gas) shows a deviation from the relationship PV= constant as the pressure increases to high levels.

The curve obtained from real gases differs according to the gas used. Fortunately, experiments at pressures normally used in the laboratory give good agreement between the gas laws and the observed behaviour.

Answer 95

A

Experimentation on the pressure and volume characteristics of real gases was carried out by Van der Waal, but will not be required of IB students.

Following the results of his experiments, Van der Waal introduced some modifications to the ideal gas law to compensate for the actual volume occupied by the particles themselves and the forces that exist between the particles. As stated above, the pressure of real gases is reduced by attractions between particles, and the available volume for movement is reduced due to the volume occupied by the particles themselves.

The ideal gas equation, PV = nRT becomes:

Where a and b are constants that are unique to each type of gas.

Answer 96

A

The rate of diffusion is dependent on the density of the gas and hence its relative molecular mass. The smaller the relative molecular mass the faster the rate of diffusion.

Logically this makes sense. It would be easier for a small ch¡ld to find his way through a crowd of adults than for a very large man to do the same.

The actual dependency follows the relationship:

Rate of diffusion = k √1/Mr
For example hydrogen gas, Mr = 2, diffuses fastest, as the term 1/Mr is the largest.

A comparison of ammonia (Mr = 17 and hydrogen chloride (Mr = 36.5) shows that the relative rates of diffusion are √1/17 for ammonia and √1/36.5 for hydrogen chloride, respectively.

This approximates to a relative diffusion rate of 1/4 for ammonia to 1/6 for hydrogen chloride, i.e. ammonia diffuses 3/2 times faster than hydrogen chloride (obtained by multiplying both fractions by 6). You would expect ammonia to diffuse 3/2 times the distance diffused by HCl over the same time period.

Demonstration Experiment

Place wads of cotton wool in both ends of a horizontally clamped 1 m long glass tube. Soak one wad with concentrated HCl and the other with concentrated ammonia solution and observe.

After a while a white ring of ammonium chloride forms in the tube. The ring appears 60cm from the ammonia soaked wad and 40cm from the HCl soaked wad.

Calculations involving Graham’s law are not required for standard or higher level, but are required for option C (our thanks to Jeffrey Frankel for pointing this out).

Answer 97

A

The pressure of a gas is caused by the collisions of the gas particles with the walls of the container. Such collisions are totally elastic and as each collision occurs a resultant force is produced. Gas pressure is the sum of all of these tiny forces per unit area. It is measured in Pa, which is a combined dimension from Nm-2, i.e. the force in Newtons per metre squared.

A particle collides with the wall of the container causing a force acting perpendicular to the wall.

The pressure is the sum of these forces per unit area. This pressure is measured in Newtons per metre squared = Pascals

Gas particles in a container Nm-2 = Pa

Answer 98

A

This seems to be contrary to what we already know about colliding objects. It can be explained by considering that the walls of the container also have vibrating particles that can, in turn, transfer energy to the gas particles. Overall there is no energy exchange if the gas and the walls are at the same temperature (as they must be when the system is at equilibrium).

Answer 99

A

The definition of mole fraction proves to be very useful in several areas of stoichiometry. It is defined as:

The number of moles of a substance divided by the total number of moles.

Example: A mixture contains three gases A, B and C, in proportions of 1 mole to 1 mole to 8 moles respectively. Calculate the mole fraction of gas A.

The mole fraction of A is 1 mole divided by the total number of moles present = 1 + 1 + 8 = 10.

Therefore the mole fraction of A = 1/10 = 0.1

Answer 100

A

Two facts emerge from what is known about the causes of gas pressure:

In gas mixtures of two or more component gases:

1 Collisions by the molecules of each gas cause a pressure on the walls of the container, called the partial pressure.

2 The total pressure of a mixture is the sum of all of the individual partial pressures.

The ideal gas equation:

PV = nRT

when T and V are constant,

P n

i.e. the pressure of any gas is directly proportional to the number of moles of gas.

For a two component system, A & B

total moles = moles gas A + moles gas B

Hence, the total pressure is the sum of the partial pressures

P(total) = P(gas A) + P (gas B)

Finally, the partial pressure of an individual gas in a mixture is given by the mole fraction of the gas multiplied by the total pressure

P(gas A) moles(gas A)

From the above equations it can be derived that:

P(gas A) = P (total) x moles A/(total moles)

John Dalton put these ideas together into what is now known as Dalton’s law:

The partial pressure of a gas is equal to the mole fraction of that gas multiplied by the total pressure.
The total pressure of a gas mixture is equal to the sum of the partial pressures of the individual gases.
Example: 2 moles of oxygen and 3 moles of hydrogen are contained in a vessel at 100 kPa pressure. Calculate the partial pressures of each gas.

The total number of moles of gas = 2 + 3 = 5

The mole fraction of this total that is oxygen = 2/5

Therefore oxygen’s contribution to the total pressure is 2/5 x total pressure = 2/5 x 100 = 40 kPa

Similarly the partial pressure of hydrogen = 3/5 x 100 = 60 kPa

Answer 101

A

The concentration of a solution is the quantity of solute that it contains per unit volume.

This may be given in grams per 100cm3 or grams per litre, but it is usually given in terms of molarity as this gives a direct measure of the number of solute particles contained by the solution.

Answer 102

A

The concept of molarity arises from the need to know the amount of solute present in a solution in moles. 1 mole of any substance contains an Avogadro number of particles of that substance = 6.02 x 1023. A 1 molar solution contains 1 mole of solute, dissolved in 1 litre of solution.

Note: the definition is not per 1 litre of solvent, but per 1 litre of solution. This allows us to measure a volume of solution and work out the number of moles, and hence the number of particles, that it contains.

Molarity = number of moles of solute per litre of solution (1 litre = 1000cm3)

The molarity is denoted by the capital letter M, and given the units mol dm-3

Example: Calculate the molarity of a solution containing 0.15 moles of potassium nitrate in 100cm3 of solution.

Molarity = moles/litres

100cm3 = 0.1dm3

Molarity = 0.15/0.1 = 1.5 M

The molarity of a specific ion within an ionic solution may also be considered separately. In a 1M solution of copper sulfate (CuSO4) the copper ions are separate from the sulfate ions. The solution may be said to be both 1 molar in terms of copper 2+ ions and 1 molar in terms of sulfate 2- ions.

A 1 molar (1M) solution of copper nitrate (Cu(NO3)2), however, is 1 molar with respect to copper 2+ ions but 2M with respect to nitrate ions.

Example: Calculate the molarity of hydrogen ions in a 0.15 molar solution of sulfuric acid.

The formula of sulfuric acid is H2SO4. It dissociates in solution according to the following equation:

H2SO4 2H+ + SO42-

Hence, if a solution is 1 molar in sulfuric acid, it must be double that in hydrogen ions.

Molarity of the solution = 0.15M in sulfuric acid,

Therefore the molarity in hydrogen ions = 0.15 x 2 = 0.3 mol dm-3

Using:

In conjunction with

We can calculate the mass needed to prepare solutions or the mass contained in solutions of known concentration.

Example: Calculate the mass of iron(II) sulfate in 100 cm3 of 0.1 mol dm-3 solution.

The formula of sulfuric acid is FeSO4. It has a relative formula mass = 56 + 32 + 64 = 152

Number of moles in 100cm3 of 0.1 mol dm-3 solution = 0.1 x 0.1 = 0.01 moles

Therefore mass of iron /(II) sulfate = moles x relative formula mass = 0.01 x 152

Therefore mass of iron(II) sulfate = 1.52 g

The first of the above mathematical formulae can be manipulated by rearrangement to obtain any of the three factors, moles, molarity or volume of solution.

Answer 103

A

If we wish to make up a solution of known concentration, then the starting material must also be ‘pure’. In reality, it is very difficult to obtain chemical compounds in a very pure state and, even if pure when bought, reaction with moisture or carbon dioxide in the air, light sensitivity or decomposion on standing can all reduce the level of purity.

Reactions in solution are often carried out for analytical and determination/estimation purposes. If the compound being used cannot be guaranteed to be pure, then it first must be standardised against a solution whose concentration can be determined to a high degree of accuracy. Such substances are called primary standards.

A suitable primary standards must have the following characteristics:

obtainable in a pure state
stable in the air
readily soluble
react with the reagents of choice
high relative molecular mass
The high relative molecular mass is simpy to reduce percentage errors in the weighing stage.

primary standards
formula
relative mass

sodium carbonate
Na2CO3
105.99

potassium hydrogen phthalate
KHC8H4O4
204.23

potassium hydrogen iodate
KH(IO3)2
389.92

potassium dichromate
K2Cr2O7
294.19

Answer 104

A

In the laboratory, specially designed pieces of glassware are used to ensure accuracy, when dealing with solutions. These are:

Pipette
Burette
Volumetric (graduated) flask
A pipette is used in conjunction with a pipette filler. The filler draws the solution into the glassware by means of suction, until the liquid reaches the correct level, indicated by a small circular line etched (drawn) around the upper thin part of the pipette. The pipette is emptied by allowing the liquid to run freely into its final receptacle and then, when it has all drained out, touching the tip of the pipette on the surface of the liquid to allow the surface tension of the aqueous liquid surface to remove the final part.

The volumetric flask also has a circle around the upper narrow part of the flask and when the lower part of the liquid meniscus is level with the line, the volumetric flask holds the correct amount.

Answer 105

A

The desired quantity of solute is weighed out on an accurate balance.
This is then dissolved in a small quantity of water and the solution transferred to the volumetric flask.
Any remaining drops are rinsed into the volumetric flask using distilled water taking care to avoid splashing or spillage.
The volumetric flask is then filled up to the mark using distilled water.

The solution is then shaken well by repeatedly inverting the flask.

Example: To prepare 250cm3 of 0.1M sodium hydroxide solution.

0.1 M means that 1 litre of solution contains 0.1 moles of solute. The solute in this case is sodium hydroxide. Sodium hydroxide has a relative formula mass of 40 (Na=23, O=16, H=1).

1 mole of NaOH = 40g

therefore 0.1 moles of NaOH = 4.0g

However, we don’t require 1 litre (1000cm3) of solution, only 250cm3

therefore the mass of sodium hydroxide needed = 250/1000 x 4.0g = 1.0g

Answer 106

A

A calculated quantity of solution is transferred into a graduated (volumetric) flask using a pipette.

The flask is filled up to the mark using distilled water.

Example: To prepare 250cm3 of 0.1 mol dm-3 hydrochloric acid using 2.0 mol dm-3 standardised acid.

In this case it is better to proceed by calculating the number of moles required in the final solution. Then, the volume of standardised solution that contains this number of moles can be calculated.

Final solution required: 250cm3 of 0.1 M hydrochloric acid

Moles = molarity x volume (litres)

This contains 0.1 x 0.25 moles of HCl = 0.025 moles

We wish to obtain 0.025 moles from the 2.0 M solution

volume (litres) = moles/molarity

Volume = 0.025/2 = 0.0125 dm3

Volume required = 12.5 cm3

Procedure:

Measure 12.5cm3 of the standardised 2.0 M HCl into a volumetric flask using a burette.

Add distilled water up to the mark mixing carefully.

Answer 107

A

Substances dissolve because they form bonds with the solvent molecules. This bond formation (an exothermic process), coupled with the entropy increase on forming a solution, makes the process thermodynamically favourable.

For insoluble substances, the opposite is true; either the bonding within the solid is too strong and requires too much energy to break it, or the bonds formed with the solute molecules are too weak to make the process favourable.

Dissolution, then, is a balance between several factors, making prediction of solubility difficult. Unfortunately, solubility is something that has to be learned, but fortunately, there are a few facts that make the task a little easier.

Reactions occur easily in solution, as the particles are in constant motion and can come into contact with one another. The majority of solutions dealt with in chemistry are ionic. The ions in the structure of an ionic solid are broken apart by interaction with the water molecules. Energy is required to overcome the strong electrostatic forces within the ionic lattice.

Six water molecules approach a
positive sodium Na+ ion
The six water molecules surround the positive sodium Na+ ion
This happens because water is a polar solvent with partial positive charges on the hydrogen atoms and a partial negative charge on the oxygen atom. The partially positive hydrogen ‘end’ of the molecule is attracted to the negative ions (anions), and the partially negative charged oxygen atom in the water molecule is attracted to the positive ions (cations).

The bonds formed between the ions and the solvent water molecules are strong enough to make the process energetically favourable. The energy release on association of water molecules with ions is called the hydration enthalpy of the ions. The smaller the ion the larger its hydration enthalpy. This is dealt with below

Answer 108

A

Water molecules are polar (they have a permanent dipole) and bond to the two types of ion in solution formed when an ionic substance dissolves, surrounding them in an octahedral arrangement. The energy released acts as one of the driving forces behind the dissolution process. However, energy is required to break the crystal lattice of the ionic solid and so any overall energy change (the enthalpy of solution) is the sum of the following three processes.

The Lattice enthalpy

Breaking the lattice requires energy, it is an endothermic process. It is represented by the equation:

NaCl(s) –> Na+(g) + Cl-(g)
The stronger the electrostatic forces of attraction between oppositely charged ions, the higher the lattice energy and the more energy required to break the lattice.

The Hydration enthalpy of the positive ion (cation)

This is an exothermic (energy releasing) process as the ions are forming bonds with the water molecules. It can be represented by the equation:

Na+(g) + nH2O(l) –> [Na(H2O)6]+(aq)
3. The Hydration enthalpy of the negative ion (anion)

This is also an exothermic process as the ions are forming bonds (attractions) with the water molecules. It may be represented by the equation:

Cl-(g) + nH2O(l) –> [Cl(H2O)6]- (aq)
NOTE: Although written as such above, the hydrated ions are not considered to be complexes in the same way as the transition metal complex ions.

Finally, the process of dissolution involves an increase in entropy of the system as the ions go from their highly ordered crystal lattice to a more disordered arrangement in solution. The Gibb’s free energy change of solution determines whether a substance is soluble and, if so, the extent to which it is soluble.

ΔG = ΔH -TΔS
ΔG = Gibb's Free Energy, ΔH= enthalpy of solution, T= Absolute temperature, ΔS= entropy change on dissolution

From the equation you can see that the term -TΔS gets larger as the temperature increases. This means that ΔG gets more negative as the temperature gets higher. The process becomes more favourable (more spontaneous) and solubility usually increases as the temperature rises for this reason.

Answer 109

A

If two ions collide that produce an insoluble solid, then a precipitate appears as the new solid is formed. This forms the basis of most simple analytical tests for the presence of ions in solution.

sodium chloride(aq) + silver nitrate(aq)  silver chloride(s) + sodium nitrate(aq)
NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
Although it seems as though two new substances have been formed, in fact only the silver chloride can truly claim to have undergone any chemical change. To appreciate this, consider the substances present in the mixture both before and after reaction:

Species present in solution before reaction
Na+(aq), Cl-(aq), Ag+(aq), NO3-(aq)

Species present in solution after reaction
Na+(aq), NO3-(aq), AgCl(s)

It may be seen that the sodium ions and the nitrate ions do not undergo any change or transformation. They start off in solution and they end up in solution. We call ions such as these spectator ions. The ions simply ‘watch’ the processes going on around them.

The ionic equation shows only those ions actually involved in reaction forming the white precipitate of silver chloride:

Cl-(aq) + Ag+(aq) AgCl(s)
Calculations involving ionic solutions must take into account the stoichiometric relationships between the ions involved in reaction.

Example: Calculate the mass of silver chloride precipitated when 25cm3 0.1M silver nitrate reacts with 50cm3 0.1M sodium chloride solution

Moles of silver nitrate = 0.025 x 0.1 = 0.0025 moles

Moles of sodium chloride = 0.05 x 0.1 = 0.0050 moles

From the equation for the reaction:

Cl-(aq) + Ag+(aq) AgCl(s)
We can see that moles of Cl- = moles of Ag+

Therefore silver nitrate is the limiting reagent and 0.0025 moles of silver chloride is formed

Relative formula mass (silver chloride) = 108 + 35.5 = 143.5

Mass of silver chloride formed = 143.5 x 0.0025 = 0.359g

Answer 110

A

Acids are substances that release hydrogen ions in solution. These hydrogen ions can react in solution with certain other ions, forming water. This removes the hydrogen ions from the solution. We call this type of ionic reaction, neutralisation. Bases contain ions that can react with the hydrogen ions from acids - they neutralise the acids.

Ions that can neutralise acids include:

hydroxide ions OH-
oxide ions O2-
carbonate ions CO32-
sulfite ions SO32-
Hydroxides

These contain the hydroxide ion usually with a metal ion such as sodium or potassium. Most other metal hydroxides are insoluble and so cannot react with acids from solution. They can however be added as solids to a solution of an acid and neutralise it.

The following reaction shows a solution of sodium hydroxide neutralising hydrochloric acid.

sodium hydroxide(aq) + hydrochloric acid(aq) –>sodium chloride(aq) + water(l)

NaOH(aq) + HCl(aq) –> NaCl(aq) + H2O(l)
Once again, consideration of the ions present before and after the reaction shows:

Species present before reaction
Na+(aq), OH-(aq), H+(aq), Cl-(aq)

Species present after reaction	 Na+(aq), Cl-(aq), H2O(l)

The sodium ions and the chloride ions are spectator ions, they do not take part in the reaction. The only ions that participate in this reaction are the hydrogen and hydroxide ions:

H+(aq) + OH-(aq) H2O(l)

Example: Calculate the volume of 2M hydrochloric acid needed to neutralise 20cm3 1M sodium hydroxide solution

Equation for the reaction:

HCl + NaOH NaCl + H2O
Moles of sodium hydroxide = molarity x volume (litres) = 1 x 0.02 = 0.02 moles

From the equation moles sodium hydroxide = moles hydrochloric acid

Therefore moles hydrochloric acid = 0.02 moles

Volume = moles/Molarity = 0.02/2 = 0.01 dm3

Volume of hydrochloric acid needed = 10cm3

Answer 111

A

ther species that can act as a base by neutralising (removing) the H+ ions in solution are:

1 Carbonate ions: CO32-

2H+(aq) + CO32-(aq) H2O(l) + CO2(g)

2 Oxide ions: O2-

2H+(aq) + O2-(s) H2O(l)

3 Sulfite (sulfate(IV)) ions: SO32-

2H+(aq) + SO32-(aq) H2O(l)+ SO2(g)

4 Ammonia (NH3):

NH3 + H+(aq) NH4+

Answer 112

A

The procedure involves adding one solution to another solution, in the presence of an indicator that shows when the two solutions have completely reacted. If the concentration and volume of one of the solutions is known and the volume of the other solution is known, then the unknown concentration can be calculated providing the stoichiometry of the reaction is known.

The experimental procedure is very accurate and liquids are carefully measured, using specially designed glass apparatus with low margins of error. See errors and inaccuracies.

Titrations are intended for analytical and determination/estimation purposes. If the compound being used for the titration cannot be obtained in a very pure state then it first must be standardised against a solution whose concentration can be determined to a high degree of accuracy. Such substances are called primary standards.

In summary:

Measure 25cm3 of one solution into a conical flask
Add a few drops of indicator solution
Add a second solution very slowly from a burette until the indicator just changes colour
Repeat the process until concordant results are obtained (results within 0.1 cm3 of one another)

Click on the ‘titrate’ button to see the general experimental procedure.

Titration is carried out several times until results within 0.1cm3 of one another are obtained. These are then said to be ‘concordant results’

Readings on the burette should be taken at eye level; the bottom of the meniscus at the surface of the solution is usually read.

Once concordant results have been obtained an average of these is used for calculations.

Measure 25cm3 of one solution into a conical flask.

Add a few drops of indicator solution.

The reading on the burette is recorded taking into account the tolerance or error limits (e.g. ±0.05 cm3).

Close to the ‘end-point’, i.e. the point at which the solute in the flask has been completely reacted, the solution is added from the burette 1 drop at a time, swirling the solution in the flask gently between each addition, until the colour just reaches the ‘end-point’.

The solution is added slowly from the burette into the reagent solution in the flask until the colour of the indicator starts to change.

The final reading on the burette is recorded, once again taking into account the error limits.

NOTE The solution in the conical flask does not have to be the unknown, the choice is usually dependent on the ease of observation of the indicator colour change. For example, it is easier to see the first hint of pink appearing from a colourless solution in a flask on a white tile, than to carry out the titration the other way round.

Answer 113

A

The calculation is carried out as follows:

The number of moles of solute in the known solution is calculated using the relationship:

moles = molarity x volume
Then the stoichiometric relationship between this substance and the other reacting solution is used to determine the number of moles of solute in the unknown solution.

Example of equation stoichoimetry:

Sodium hydroxide	+	nitric acid		sodium nitrate	+ water
NaOH	+	HNO3		
NaNO3
\+ H2O
1 mole	≡	1 mole			
Here the stoichiometric relationship is:

moles of base = moles of acid

The molarity of the unknown solution can now be calculated from the number of moles present and the volume used:

molarity = moles/volume(litres)

Answer 114

A

The basic concept is used in many walks of life. If you go into a shop with 8.00 €uros to buy, for example, a rubber duck, you can find out the cost of the article by looking at the change the shop assistant gives back.

You walk into the shop with coins worth ...	And you walk out of the shop with coins worth...	The rubber duck must have cost the difference between the coins.
8.00 €uros
4.30 €uros
8 - 4.30 = 3.70 €uros
Total money = 8.00 €uros

Change received = 4.30 €uros

Cost of item = 8.00 - 4.30 = 3.70 €uros

In the case of chemistry this would be:

Total acid = 2.0 moles

Acid remaining = 1.6 moles

Acid used up in initial reaction = 2.0 - 1.6 = 0.4 moles

The experimental procedure, then, must focus on finding out the amount of acid remaining after the initial reaction. All of the other factors can be calculated from the amount of acid remaining and the other directly recorded data (mass of solid, initial molarity and volume of the acid before reaction).

NOTE Although all of the examples discussed here involve acids, back titration is not their exclusive domain - the principles involved here can also be applied to other reaction systems.

Answer 115

A

React a known mass of the solid to be analysed with an excess (but known) amount of acid.
Make up the excess acid to a specific volume and titrate against a standard base.
Calculate the amount of acid remaining (the excess).
Calculate the amount of acid used up in the original reaction by subtraction from the initial number of moles.
Calculate the number of moles present in the original solid by consideration of the stoichiometry of the reaction.

Answer 116

A

Finding the relative formula mass of an unknown carbonate
Identifying the metal in an unknown metal oxide
Finding the purity of an known carbonate mixture
Finding the percentage metal in an alloy

Answer 117

A

procedure

Weigh out about 2.5 g of the unknown carbonate
Dissolve the unknown carbonate in 50 cm3 of 2M HCl (an excess of acid - it will react with all of the carbonate and there will be some acid left over)
Make up the reaction mixture to 250 cm3 in a volumetric flask
Titrate 25 cm3 aliquots against a standardised solution of sodium hydroxide (0.1M) using phenolphthalein indicator
Typical results

Mass of unknown carbonate = 2.64g

Treatment of results

Volume of 0.1M sodium hydroxide used in titration = 37.15 cm3

Moles of sodium hydroxide = 0.1 x 0.03715 = 0.003715 moles

NaOH + HCl NaCl + H2O
Moles of sodium hydroxide = moles of hydrochloric acid = 0.003715 moles

But only 25 cm3 samples taken from a 250cm3 volumetric flask were titrated, therefore the total moles of hydrochloric acid in the volumetric flask was 0.003715 moles x 250/25 = 0.03715 moles

Original moles of hydrochloric acid = molarity x volume = 2 x 0.05 = 0.1

Therefore, moles of hydrochloric acid neutralised in the original reaction with the unknown carbonate = 0.1 - 0.03715 = 0.06285 moles

Reaction of an acid with a carbonate is:

2H+ + CO32- CO2 + H2O
From the stoichiometry 2 moles of acid is required to react with 1 mole of carbonate

Moles of hydrochloric acid = 0.06285 moles therefore moles of carbonate = 0.06285/2 moles = 0031425

The mass of the unknown carbonate = 2.64g

Therefore the relative formula mass of the unknown carbonate = mass/moles = 2.64/0.031425 = 84.01

The carbonate group CO32- has a relative mass of 12 + 48 = 60

Therefore the metal in the unknown carbonate has a relative mass of 84 - 60 = 24

The unknown carbonate is magnesium carbonate

Answer 118

A

Redox titrations

Redox titrations involve reduction - oxidation reactions, i.e. there is a transfer of electrons from one species to another. Some transition metals change from one oxidation state to another with an accompanying colour change. This means that they are self-indicating and no third substance needs to be added.

In the majority of cases however, some means of indicating the end-point of the titration is needed.

Potassium manganate(VII)
Sodium thiosulfate
Potassium iodide
Potassium dichromate(VI)
Hydrogen peroxide

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Stoichiometric relationships 1 Flashcards

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