Chapter 4 - Stoichiometry

Question 1

What is the structural formula

Answer 1

The structural formula tells you the way in which the atoms in a particular molecule are bonded. This can be done by either a diagram (displayed formula) or written (simplified structural formula)
E.g CH3CH2CH2CH3 simplified

Question 2

What is the empirical formula

Answer 2

The empirical formula tells you the simplest whole-number ratio of atoms in a compound
E.g simplified from C4H10 to C2H5

Question 3

What is the molecular formula

Answer 3

The molecular formula tells you the actual number of atoms of each element in one molecule of the compound or element e.g. H2 has 2 hydrogen atoms, HCl has 1 hydrogen atom and 1 chlorine atom
E.g C4H10

Question 4

What does the forward arrow imply in chemical reactions

Answer 4

The arrow implies the conversion of reactants into products

Question 5

In naming compounds with metal and non metal what do you do

Answer 5

If one is a metal and the other a nonmetal, then the name of the metal atom comes first and the ending of the second atom is replaced by adding –ide. Though this still works with hydrogen being a non metal. E.g hydrogen chloride

Question 6

What happens to ionic compounds in aqueous solutions

Answer 6

In aqueous solutions ionic compounds dissociate into their ions, meaning they separate into the component atoms or ions that formed them
E.g. HCL —> H + CL

Question 7

Write the ionic equation for the reaction of aqueous chlorine and aqueous potassium iodide.

Answer 7

It goes from 2KI + CL2 —> 2KCL + I2
to 2K + 2I + 2CL —> 2K + 2CL + 2I then it goes
To 2I + CL2 —> 2CL + I2

Question 8

How do you calculate relative atomic mass

Answer 8

This is calculated from the mass number and relative abundances of all the isotopes of a particular element
Basically finding the average weight that a element has with all their isotopes

Question 9

How do you calculate relative formula (molecular mass)

Answer 9

To calculate the Mr of a substance, you have to add up the Relative Atomic Masses of all the atoms present in the formula.
E.g H20 = H( 2 x H) + O( O x 1) = 18

Question 10

What is the mole

Answer 10

This is the mass of a substance containing the same number of fundamental units as there are atoms in exactly 12.000 g of Carbon

Question 11

What is the Avogadros constant

Answer 11

6.02 x 10^23 is known as the Avogadro Number
One mole is the amount of a substance that contains 6.02 x 10^23 particles (Atoms, Molecules or Formulae) of a substance.
Examples: 1 mole of Sodium (Na) contains 6.02 x 10^23 Atoms of Sodium
1 mole of Hydrogen (H2) contains 6.02 x 10^23 Molecules of Hydrogen.
1 mole of Sodium Chloride (NaCl) contains 6.02 x 10^23 Formula units of Sodium Chloride

Question 12

What is the mole and relative atomic mass close

Answer 12

One mole of any element is equal to the relative atomic mass of that element in grams
For example one mole of carbon, that is if you had 6.02 x 10^23 atoms of carbon in your hand, it would have a mass of 12g
For compounds we add up the relative atomic mass to find how many grams in one mole.

Question 13

If hydrogen has one atomic mass how much is the mass of carbon

Answer 13

Hydrogen which has an atomic mass of 1 is therefore equal to 1/12 the mass of a 12C atom
So one carbon atom has the same mass as 12 hydrogen atoms

Question 14

What’s the molar volume

Answer 14

This is the volume that one mole of any gas (be it molecular such as CO2 or monoatomic such as helium) will occupy
It’s value is 24dm^3 or 24,000 cm^3 at room temperature and pressure (r.t.p.)

Question 15

How do you calculate amount of gas(mol)

Answer 15

Amount of gas (mol) = Volume of gas (dm^3) ÷ 24
or

Amount of gas (mol) = Volume of gas (cm^3) ÷ 24000

Question 16

What is dm^3

Answer 16

It’s a volume unit equivalent to a litre

Question 17

How do you calculate the volume of gas that a particular amount of moles occupies

Answer 17

Equation:

Volume of gas (dm^3)= Amount of gas (mol) x 24
or

Volume of gas (cm^3)= Amount of gas (mol) x 24000

Question 18

How to Calculate the moles in a particular volume of gas

Answer 18

Equation:

Amount of gas (mol) = Volume of gas (dm3) ÷ 24
or

Amount of gas (mol) = Volume of gas (cm3) ÷ 24000

Question 19

How to calculate moles

Answer 19

Moles = mass of substance in grams / molecular mass
N= m/M

Question 20

How to calculate mass of substance

Answer 20

Mass of substance = moles x molecular mass

Or m = N x M

Question 21

How to calculate relative formula mass

Answer 21

Molecular mass = mass of substance in grams / moles of substance

Question 22

How to calculate percentage composition

Answer 22

calculating the percentage by mass of each particular element in a compound

Question 23

How do you calculate percentage of oxygen in CO2

Answer 23

Step 1 – Calculate the molar mass of the compound

Molar mass CO2 = (2 x 16) + 12 = 44

Step 2 – Add the atomic masses of the element required as in the question (oxygen)

16 + 16 = 32

Step 3 – Calculate the percentage

% of oxygen in CO2 = 32/44 x 100 = 72.7%

Question 24

How to calculate concentration of solutions

Answer 24

Concentration (mol / dm^3) = Amount of substance (mol) ÷ Volume of solution (dm^3)
Or C = N/V

Question 25

How to calculate moles using concentration and volume of substance

Answer 25

Equation: Amount of Substance (mol) = Concentration x Volume of Solution (dm3)

Question 26

How to calculate concentration

Answer 26

Equation: Concentration (mol / dm3) = Amount of substance (mol) ÷ Volume of solution (dm3) Or C = n/V

Question 27

What to always remember with cm^3 and dm^3

Answer 27

You need to remember to always when it says out it in the form of /dm^3 you convert the cm^3 into dm^3 by dividing by 1000

Question 28

What is the limiting reactant

Answer 28

The limiting reactant is the reactant which is not present in excess in a reaction It is always the first reactant to be used up which then causes the reaction to stop

Question 29

How much is a tonne in grams

Answer 29

10^6g or 1000000g

Question 30

An example: 9.2g of sodium is reacted with 8.0g of sulfur to produce sodium sulfide, NaS. Which reactant is in excess and which is the limiting reactant?

Answer 30

Step 1 – Calculate the moles of each reactant Moles = Mass ÷ Molecular mass Moles Na = 9.2/23 = 0.40 Moles S = 8.0/32 = 0.25 Step 2 – Write the balanced equation and determine the molar ratio 2Na + S → Na2S so the molar ratios is 2 : 1 Step 3 – Compare the moles. So to react completely 0.40 moles of Na require 0.20 moles of S and since there are 0.25 moles of S, then S is in excess. Na is therefore the limiting reactant.

Question 31

What is the molar ratio

Answer 31

A mole ratio is ​the ratio between the amounts in moles of any two compounds involved in a chemical reaction. Mole ratios are used as conversion factors between products and reactants in many chemistry problems. The mole ratio is also called the mole-to-mole ratio.

Question 32

``` For the reaction: 2 H2(g) + O2(g) → 2 H2O(g) ``` What is the mole ratio

Answer 32

The mole ratio between O2 and H2O is 1:2. For every 1 mole of O2 used, 2 moles of H2O are formed. The mole ratio between H2 and H2O is 1:1. For every 2 moles of H2 used, 2 moles of H2O are formed. If 4 moles of hydrogen were used, then 4 moles of water would be produced.

Question 33

Calculate the Mass of Magnesium Oxide that can be made by completely burning 6 g of Magnesium in Oxygen Magnesium (s) + Oxygen (g) → Magnesium Oxide (s)

Answer 33

Step 1 – Calculate the moles of Magnesium Used in reaction Moles = Mass ÷ Mr Moles = 6 ÷ 24 = 0.25 Step 2 – Find the Ratio of Magnesium to Magnesium Oxide using the balanced Chemical Equation which is 1:1 Step 3 – Find the Mass of Magnesium Oxide Moles of Magnesium Oxide = 0.25 Mass = Moles x Mr Mass = 0.25 x 40 = 10 g Mass of Magnesium Oxide Produced = 10 g

Question 34

Calculate the Mass, in Tonnes, of Aluminium that can be Produced from 51 Tonnes of Aluminium Oxide ``` Aluminium Oxide (s) → Aluminium (s) + Oxygen (g) Or 2Al2O3 → 4Al + 3O2 ```

Answer 34

Step 1 – Calculate the moles of aluminium oxide used Mass of Aluminium Oxide in Grams = 51 x 106 = 51,000,000 g Moles = Mass ÷ Ar Moles = 51,000,000 ÷ 102 = 500,000 Step 2 – Find the ratio of aluminium oxide to aluminium using the balanced chemical equation which is aluminium oxide to aluminium 1:2 Step 3 – Find the mass of aluminium Moles of aluminium = 1,000,000 Mass in grams = Moles x Ar Mass = 1,000,000 x 27 = 27,000,000 Mass in Tonnes = 27,000,000 ÷ 106 = 27 Tonnes Mass of Aluminium Produced = 27 Tonnes

Question 35

What is the empirical formula

Answer 35

gives the simplest whole number ratio of atoms of each element in the compound Calculated from knowledge of the ratio of masses of each element in the compound

Question 36

What is the empirical formula for 10 grams of Hydrogen and 80 grams of Oxygen

Answer 36

Empirical Formula of H2O. This can be shown by the following calculations: you have to first find out the moles Amount of Hydrogen Atoms = Mass in grams ÷ Ar of Hydrogen = (10 ÷ 1) = 10 moles Amount of Oxygen Atoms = Mass in grams ÷ Ar of Oxygen = (80 ÷ 16) = 5 moles The ratio is hydrogen to oxygen 10:5 or 2:1

Question 37

What’s the molecular formula

Answer 37

gives the exact numbers of atoms of each element present in the formula of the compound Divide the relative formula mass of the molecular formula by the relative formula mass of the Empirical Formula Multiply the number of each element present in the Empirical Formula by this number to find the Molecular Formula

Question 38

Find out the question? | The Empirical Formula of X is C4H10S1 and the Relative Formula Mass of X is 180. What is the Molecular Formula of X?

Answer 38

Step 1 – Calculate Relative Formula Mass of Empirical Formula (C x 4) + (H x 10) + (S x 1) = (12 x 4) + (1 x 10) + (32 x 1) = 90 Step 2 – Divide Relative Formula Mass of X by Relative Formula Mass of Empirical Formula 180 / 90 = 2 Step 3 – Multiply Each Number of Elements by 2 (C4 x 2) + (H10 x 2) + (S1 x 2) = (C8) + (H20) + (S2) Molecular Formula of X = C8H20S2

Question 39

What’s percentage yield

Answer 39

Percent yield is the percent ratio of actual yield to the theoretical yield. It is calculated to be the experimental yield / theoretical yield x 100%. If the actual and theoretical yield ​are the same, the percent yield is 100%. Usually, percent yield is lower than 100% because the actual yield is often less than the theoretical value

Question 40

Formula for percentage yield

Answer 40

Equation: Percentage Yield = (Yield Obtained / Theoretical Yield) x 100

Question 41

In an experiment to displace copper from copper sulfate, 6.5 g of Zinc was added to an excess of copper (II) sulfate solution. The copper was filtered off, washed and dried. The mass of copper obtained was 4.8 g. Calculate the percentage yield of copper?

Answer 41

Step 1: Calculate the Amount, in Moles of Zinc Reacted Moles of Zinc = 6.5 ÷ 65 = 0.10 moles Step 2: Calculate the Maximum Amount of Copper that could be formed from the Molar ratio Maximum Moles of Copper = 0.10 moles (Molar ratio is 1:1) Step 3: Calculate the Maximum Mass of Copper that could be Formed Maximum Mass of Copper = ( 0.10 x 64 ) = 6.4 g Step 4: Calculate the Percentage of Yield of Copper Percentage Yield = ( 4.8 ÷ 6.4 ) x 100 = 75% Percentage Yield of Copper = 75%

Question 42

What’s percentage purity

Answer 42

Often the product you are trying to fabricate may become contaminated with unwanted substances such as unreacted reactants, catalysts etc. Equation: Percentage Purity = (Mass of pure substance / Mass of impure substance) x 100

Question 43

Calculate the question in an experiment 7.0g of impure calcium carbonate were heated to a very high temperature and 2.5g of carbon dioxide were formed. Calculate the percentage purity of the calcium carbonate.

Answer 43

Ask teacher Equation Of Reaction: CaCO3 (s) → CaO(s) + CO2 (g) Step 1: Calculate the relative formula masses 1 mole CaCO3 → 1 mole CO2 40+12+(3×16) 12+(2×16) 100 → 44 Step 2: Calculate the theoretical mass of calcium carbonate used if pure From 2.5g CO2 we would expect 2.5/44 x 100 = 5.68g Step 3: Calculate the percentage purity (Mass of pure substance / mass of impure substance) x 100 = 5.68/7.0 x 100 = 81.1% Ask teacher