CGP C3

Question 1

What is a relative formula mass (Mr)

Answer 1

. The relative atomic masses of all atoms in the molecular formula added together

Question 2

What is the relative atomic mass

Answer 2

. It’s the same as an elements mass number
e.g Ar of Mg = 24

Question 3

Relative formula mass worked example

Answer 3

Find the relative formula mass of MgCl2
Ar of Mg = 24 Ar of Cl = 35.5
Mg + Cl (2) = 24 x (2 x 35.5) = 95
Mr of MgCl2 = 95

Question 4

How do you calculate the % Mass of an Element in a Compound

Answer 4

Percentage mass of an element in a compound = Ar x number of atoms of that element / Mr of the compound x 100

Question 5

% Mass of an Element in a Compound worked example

Answer 5

Find the percentage mass of sodium in sodium bicarbonate : Na2CO3
Ar of Sodium = 23, Ar of Carbon = 12 Ar of Oxygen = 16
Mr of Na2CO3 = (2 x 23) + 12 + (3 x 16) = 106

(23 x 2 / 106) x 100 = 43%

Question 6

Percentage mass complicated worked example
“A mixture contains 20% iron ions by mass. What mass of iron chloride (FeCl2) would you need to provide the iron ions in 50g of the mixture?”
Ar of Fe = 56
Ar of Cl = 35.5

Answer 6

1. Find the mass of iron in the mixture
   Mixture contains 20% iron by mass, in 50g (50x0.2) there would be 10g of iron
2. Calculate percentage mass of iron in iron chloride

56 / 56 + (2 x 35.5) x 100 = 44.09%

3. Calculate the mass of iron chloride that contains 10g of iron

Iron chloride contains 44.09% iron by mass, so there will be 10g of iron in 10 / (44.09 / 100) = 23g

So you need 23g of iron chloride to provide the iron in 50g of the mixture

Question 7

Mole

Answer 7

. “The Mole” is simply the name given to an amount of a substance.
. Just like a hundred is 100 or a thousand is 1000, The Avogadro constant is 6.02 x 10^23 that’s all, it’s just a number.

Question 8

One mole of a substance

Answer 8

. One mole of any substance is just an amount of that substance that contains 6.02 x 10^23 of particles. The particles could be atoms, molecules, ions or electrons

Question 9

Why is the Avogadro constant that specific number

Answer 9

.The mass of that number of atoms or molecules of any substance os exactly the same number of grams as the Ar or Mr of the element or compound
In other words:
. One mole of atoms or molecules of any substance will have a mass in grams equal to the Ar or Mr for that substance e.g.
. Carbon’s Ar = 12, so one mole of carbon = 12g

. Nitrogen gas’ Mr = 28, so one mole of Nitrogen gas = 28g

This means that 12g of C, 28g of N2 or 44g of C02 contain one mole of atoms

Question 10

What is the formula to find the number of moles in a given mass

Answer 10

. Number of moles = Mass in g / Mr

Question 11

Finding number of moles worked example

Answer 11

How many moles are there in 66g of Carbon Dioxide (CO2)

1. Calculate Mr of Carbon dioxide = 12 + (16x2) = 44
2. Divide mass (66g) by the Mr (44) = 1.5 mol

Question 12

Conservation of mass

Answer 12

. During a chemical reaction, no atoms are destroyed and no atoms are created
. This means there are the same number and types of atoms on each side of a reaction equation

. If the mass seems to change, there’s usually a gas involved

Question 13

Moles in equations

Answer 13

. The big numbers in-front of the chemical formulas of the reactants and products tell you how many moles of each substance takes part or is formed, for example :
Mg + 2HCl = MgCl2 + H2
. In this reaction, 1 mole of magnesium and 2 moles of hydrochloric acid react together to form 1 mole of magnesium chloride and 1 mole of hydrogen gas

Question 14

Balancing equations using reacting masses

Answer 14

If you know the masses of the reactants and products that took part in a reaction, you can work out the balanced symbol equation for the reaction.

1. Divide the mass of each substance by its relative formula mas to find no. of moles
2. Divide number of moles of each substance by the smallest number of moles in the reaction
3. If any of the numbers aren’t whole numbers, multiply all the numbers by the same amount so that they become whole numbers
4. Write the balanced symbol equation by putting these numbers in front of the chemical formula

Question 15

Balancing equations using reactant masses worked example

“8.1g of zinc oxide (ZnO) reacts completely with 0.6g of carbon to form 2.2g of carbon dioxde and 6.5g of zinc. Write a balanced symbol equation for this reaction.
Ar(C) =12 Ar(O)=16 Ar(Zn) =65

Answer 15

1. Work out the Mr for each substance
   ZnO = 65+16 = 81
   C = 12
   CO2 = 12 + (16x2) = 44
   Zn = 65
2. Divide the mass of each substance by its Mr to work out how many moles are produced
   ZnO = 8.1/81 = 0.1 mol
   CO2 = 2.2/44 = 0.05 mol
   C = 0.6/12 = 0.05 mol
   Zn = 6.5/65 = 0.1mol
3. Divide each number by the smallest amount of moles, in this case is 0.05
   ZnO = 0.1/0.05 = 2
   CO2 = 0.05/0.05 = 1
   C = 0.05/0.05 = 1
   Zn = 0.1/0.05 = 2
4. Now that all the numbers are whole numbers, you can substitute them into the equation
   (2)ZnO + (1)C = (1)CO2 + (2)Zn

Question 16

Limiting reactant definition

Answer 16

Reactions don’t go on forever, - you need stuff in the reaction flask that can react. If one reactant gets completely used up in a reaction before the rest, then the reaction will stop. That reactant’s called limiting

Question 17

Limiting reactants

Answer 17

When some Magnesium Chloride (MgCO3) is placed into a beaker of hydrochloric acid, you can tell a reaction is taking place because you see lots of bubbles of gas being given off.
After a while the fizzing slows down and the reaction eventually stops.

1. The reaction stops because one of the reactants have been used up completely, all the other reactants are in excess. They’re usually added in excess to make sure that the other reactant is used up
2. The reactant that i used up in a reaction is called the limiting reactant (because it limits the amount of product formed)
3. The amount of product formed is directly proportional to the amount of limiting reactant. For example, if you halve the amount of limiting reactant, the amount of product will also halve. If you double it, the amount of product formed will also double.
4. This is because if you add more reactant, there will be more reactant particles to take part in the reaction meaning more product particles

Question 18

How do we calculate the mass of product formed in a reaction

Answer 18

You can calculate the mass of a product formed in a reaction by using the mass of the limiting reactant and the balancing equation reaction:

1. Write out the balanced equation
2. Work out the Mr of the reactant and product you want
3. Find out how many moles there are of the substance you know the mass of
4. Use the balanced equation to work out how many moles there’ll be of the other substance. In this case, that’s how many moles of product will be made of this many moles of reactant
5. Use the number of moles to calculate the mass

Question 19

Calculating the mass of a product worked example
“Calculate the mass of aluminium oxide formed when 135g of aluminium is burned in the air”

Answer 19

1. Write out the balanced equation
   4Al + 3O2 = 2Al2O3
2. Calculate the Mr’s (relative formula masses)
   Al:27 Al2O3=(2x27) + (3x16) = 102 (you don’t need to find the Mr of oxygen as it’s in excess)
3. Calculate the number of moles of aluminium in 135g
   137/27=5
4. Look at the ratio of moles in the equation
   4 moles of Al react to produce 2 moles of Al2O3 - half the number of moles are produced - so 5 moles of Al will react to produce 2.5 moles of Al2O3
5. Calculate the mass of 2.5 moles of Al2O3
   2.5 x 102 = 255g

Question 20

Moles of gas

Answer 20

One mole of any gas occupies 24dm3 at 20°C
. At the same temperature and pressure, equal numbers of moles of any gas will occupy the same volume
. At room temp and pressure (20°C and 1 atm) one mole of any gas occupies 24dm3

You can use this formula to find the volume of a known mass of any gas at r.t.p
Volume of gas (dm3) = Mass of gas (g) / Mr of gas x 24

Question 21

Calculating the volume of a known mass of any gas at r.t.p worked example:
What’s the volume of 319.5g of chlorine at r.t.p?

Answer 21

Volume = mass of gas/mr of gas x 24 =
319.5/71 x 24 =108dm3

Question 22

Using the volume of one gas to to find the volume of another:
“How much carbon dioxide is formed when 30 dm3 of oxygen reacts with carbon monoxide 2CO + O2 = 2CO2

Answer 22

1 mol of O2 = 2 moles of CO2
1 volume of O2 = 2 volumes of CO2
30 dm3 of O2 = 60 dm3 of CO2

Question 23

What is concentration

Answer 23

Concentration is a measure of how crowded things are

Question 24

Concentration explained

Answer 24

. Lots of reactions in chemistry take place between substances that are dissolved in a solution. The amount of a substance (e.g the mass or the number of moles) in a certain volume of solution is called it’s concentration

. The more solute (the substance that’s dissolved) there is in a given volume, the more concentrated the solution

Question 25

How do we measure concentration

Answer 25

. One way to measure the concentration of a solution is by calculating the mass of a substance in a given volume of solution. (The units will be units of mass/units of volume)

Question 26

How do we measure the concentration of a solution

Answer 26

concentration (g/dm3) or (mol/dm3) = mass of solute(g) or (mol)/volume of solvent(dm3)

Question 27

Finding concentration of a solution worked example: "What's the concentration in g/dm3 of a solution of sodium chloride where 30g of sodium chloride is dissolved in 0.2dm3 of water"

Answer 27

30 / 0.2 = 150

Question 28

What is 1dm3 in cm3

Answer 28

1000cm3

Question 29

How do we convert cm3 into dm3

Answer 29

dividing by 1000

Question 30

What's the concentration in mol/dm3 of a solution with 2 moles of salt in 500cm3

Answer 30

500cm3/1000 = 0.5dm3 2 / 0.5 = 4 mol/dm3

Question 31

What is a titration

Answer 31

Titrations are experiments that allow you to find the volume needed for two solutions to react together completely. If you know the concentration of one of the solutions, you can use the volumes from the titration experiment, as well as the reaction equation, to find the concentration of the solution

Question 32

A student started with 30.0cm3 of sulfuric acid H2SO4 of unknown concentration in a flash. She found that it took an average of 25.0cm3 of 0.100 mold/dm3 sodium hydroxide (NaOH) to neutralise the sulfuric acid. Find the concentration of the acid in mol/dm3. The balanced symbol equation for the reaction is: 2NaOH + H2SO4 = Na2SO4 + 2H2O

Answer 32

1. Work out how many moles of the known substance you have using the formula (no. moles = conc. x volume) 0.100 mol/dm3 x (25.0 / 1000) dm3 = 0.00250 moles of NaOH 2. Use the reaction equation to work out how many moles of the "unknown" stuff you must have had Using the equation, you can see that two moles of sodium hydroxide reacts with one mole of sulfuric acid. So 0.00250 moles of NaOH must have reacted with 0.00250 / 2 = 0.00125 moles of H2SO4 3. Work out the concentration of the unknown stuff Concentration = number of moles / volume = 0.00125 mol / (30.0 / 1000) dm3 = 0.41666... mol/dm3 = 0.0417 mol/dm3

Question 33

Converting mol/dm3 to g/dm3

Answer 33

1. To find the concentration in g/dm3, start by finding the concentration in mol/dm3 using the steps above 2. Then, convert the concentration in mol/dm3 to g/dm3 using the equation mass = moles x Mr

Question 34

Converting mol/dm3 to g/dm3 worked example: "What's the concentration in g/dm3, of the sulfuric acid solution in the example above?"

Answer 34

1. Work out the relative formula mass for the acid Mr (H2SO4) = (2x1) + 32 + (4x16) = 98 2. Convert the concentration in moles (that you've already worked out) into concentration in grams. So, in 1dm3 Mass in grams = moles x relative formula mass = 0.041666 x 98 = 4.08333...g So the concentration in g/dm3 = 4.08 g/dm3

Question 35

Atom economy explained

Answer 35

1. A lot of reactions make more than one product. Some of them will be useful, but others will be a waste. 2. The atom economy (atom utilisation) of a reaction tells you how much mass of the mass of the reactants is wasted when manufacturing a chemical and how much ends up as useful products 3. 100% atom economy means that all the atoms in the reactants have been turned into useful (desired products). The higher the atom economy, the 'greener' the process.

Question 36

Atom economy equation

Answer 36

Atom economy = (relative formula mass of desired products) / (relative formula mass of all reactants) x 100

Question 37

Atom economy worked example: "Calculate the atom economy of the following reaction to produce hydrogen gas: CH4(g) + H2O(g) = CO(g) + 3H2(g)

Answer 37

1. Identify the desired product - hydrogen gas 2. Work out the Mr of all the reactants: Mr (CH4) = 12 + (4x1) = 16 Mr (H2O) = (2x1) + 16 = 18 16 + 18 = 34 3. Work out the Mr of the desired product 3 x Mr(H2) = 3 x (2x1) = 6 4. Use the formula to calculate atom economy: 6/34 x 100 = 17.6% 17.6% of starting materials are useful Over 80% of the starting materials are wasted

Question 38

High atom economy

Answer 38

1. Reactions with low atom economy use up resources quickly and make lots of waste material that have to be disposed of somehow. That tends to make these reactions unsustainable - the raw materials will run out and the wate has to go somewhere 2. Low atom economy reactions aren't usually profitable. Raw materials are often expensive to buy, and waste products can be expensive to remove and dispose of responsibly 3. The best way around the problem is to find a use for the waste products rather than throwing them away. There's often more than one way to make the product you want, so the trick is to come up with a reaction that gives useful "by-products" rather than useless ones 4. Reactions with the highest atom economy are the ones that only have one product. Those reactions have an atom economy of 100%. The more products there are, the lower the atom economy is likely to be

Question 39

What does percentage yield tell you

Answer 39

Percentage yield tells you about the overall success of an experiment. It compares what you think you should get (theoretical yield) with what you get in practise (actual yield).

Question 40

Percentage yield explanation

Answer 40

. The amount of product you get is known as the yield. The more reactants you start with, the higher the actual yield will be. But the percentage yield doesn't depend on the amount of reactants you started with - it's a percentage. . Percentage yield is always somewhere between 0 and 100%. 100% yield means that you got all the product you expected to get. 0% yield means that no reactants were converted into product i.e no product at all was made . Industrial processes should have as high a percentage yield as possible to reduce waste and reduce costs.

Question 41

Yield in real life

Answer 41

Yields are always less than 100% . In real life, you never get a 100% yield. Some product reactant always gets lost along the way and that goes for big industrial processes as well as school lab experiments. How this happens depends on what sort of reaction it is and what apparatus is being used. Lots of things can go wrong, but three common problems are: . Not all reactants react to make a product. . There might be side reactions . You lose some product when you separate it from reaction mixture

Question 42

1. Not all reacts react to make a product

Answer 42

. In reversible reactions, the products can turn back into reactants, so the yield will never be 100% For example in the Haber process, at the same time N2 + 3H2 = 2NH3 is taking place, the reverse reaction 2NH3 = N2 + 3H2 is also happening This means the reaction N2 + 3H2 = 2NH3 never goes to completion (the reactants don't all get used up)

Question 43

2. There might be side reactions

Answer 43

The reactants sometimes react differently to how you expect. They might react with gases in the air, or impurities in the reaction mixture, so they end up forming extra products other than the ones you want

Question 44

3. You lose some product when you separate it from the reaction mixture

Answer 44

When you filter a liquid to remove solid particles, you nearly always lose a bit of liquid or a bit of solid. . If you want to keep the liquid, you'll lose the bit that remains with the solid and filter paper (as they always stay a bit wet) . If you want to keep the solid, some of it'll get left behind when you scrape it off the filter paper. You'll also lose a bit of material when you transfer it from one container to another - even if you manage not to spill it. Some of it always gets left behind on the inside surface of the old container.