1. Stoichiometric Calculations
Introduction
Dr Seemal Jelani

2. PROTON NUMBER = number of protons (obviously)
Atomic mass
RELATIVE ATOMIC MASS, Ar (“Mass number”) = number of protons + number of neutrons He 2 4
SYMBOL
PROTON NUMBER = number of protons (obviously)
Dr Seemal Jelani

3. Relative formula mass, Mr
The relative formula mass of a compound is the relative atomic masses of all the elements in the compound added together.
Relative atomic mass of O = 16
E.g. water H2O:
Relative atomic mass of H = 1
Therefore Mr for water = 16 + (2x1) = 18
Work out Mr for the following compounds:
HCl
NaOH
MgCl2
H2SO4
K2CO3
H=1, Cl=35 so Mr = 36
Na=23, O=16, H=1 so Mr = 40
Mg=24, Cl=35 so Mr = 24+(2x35) = 94
H=1, S=32, O=16 so Mr = (2x1)+32+(4x16) = 98
K=39, C=12, O=16 so Mr = (2x39)+12+(3x16) = 138
Dr Seemal Jelani

4. More examples
CaCO3
x16
100
HNO3
x16
2MgO
2 x ( ) 80
3H2O
3 x ((2x1) + 16)
4NH3
2KMnO4
3C2H5OH
4Ca(OH)2
Moles – The relative formula mass of a substance, in grams, is known as 1 mole of that substance E.g. 18g of H2O = 1 mole of H2O
Dr Seemal Jelani

5. Calculating percentage mass
If you can work out Mr then this bit is easy…
Percentage mass (%) =
Mass of element Ar
Relative formula mass Mr
x100%
Calculate the percentage mass of magnesium in magnesium oxide, MgO:
Ar for magnesium = 24 Ar for oxygen = 16
Mr for magnesium oxide = = 40
Therefore percentage mass = 24/40 x 100% = 60%
Calculate the percentage mass of the following:
Hydrogen in hydrochloric acid, HCl
Potassium in potassium chloride, KCl
Calcium in calcium chloride, CaCl2
Oxygen in water, H2O
Dr Seemal Jelani

6. A classic exam question:
Empirical formulae
27/02/2019
Empirical formulae is simply a way of showing how many atoms are in a molecule (like a chemical formula). For example, CaO, CaCO3, H20 and KMnO4 are all empirical formulae. Here’s how to work them out:
A classic exam question:
Find the simplest formula of 2.24g of iron reacting with 0.96g of oxygen.
Step 1: Divide both masses by the relative atomic mass:
For iron 2.24/56 = For oxygen 0.96/16 = 0.06
Step 2: Write this as a ratio and simplify:
0.04:0.06 is equivalent to 2:3
Step 3: Write the formula:
2 iron atoms for 3 oxygen atoms means the formula is Fe2O3
Dr Seemal Jelani

7. Example questions
27/02/2019
Find the empirical formula of magnesium oxide which contains 48g of magnesium and 32g of oxygen.
Find the empirical formula of a compound that contains 42g of nitrogen and 9g of hydrogen.
Find the empirical formula of a compound containing 20g of calcium, 6g of carbon and 24g of oxygen.
Dr Seemal Jelani

8. Applications of balancing
Dr Seemal Jelani

9. Calculating the mass of a product
27/02/2019
Calculating the mass of a product
E.g. what mass of magnesium oxide is produced when 60g of magnesium is burned in air?
IGNORE the oxygen in step 2 – the question doesn’t ask for it
Step 1: READ the equation:
2Mg + O MgO
Step 2: WORK OUT the relative formula masses (Mr):
2Mg = 2 x 24 = MgO = 2 x (24+16) = 80
Step 3: LEARN and APPLY the following 3 points:
48g of Mg makes 80g of MgO
1g of Mg makes 80/48 = 1.66g of MgO
60g of Mg makes 1.66 x 60 = 100g of MgO
Dr Seemal Jelani

10. Mr: 2Al2O3 = 2x((2x27)+(3x16)) = 204 4Al = 4x27 = 108
When water is electrolysed it breaks down into hydrogen and oxygen:
2H2O H2 + O2
What mass of hydrogen is produced by the electrolysis of 6g of water?
27/02/2019
Work out Mr: 2H2O = 2 x ((2x1)+16) = H2 = 2x2 = 4
36g of water produces 4g of hydrogen
So 1g of water produces 4/36 = 0.11g of hydrogen
6g of water will produce (4/36) x 6 = 0.66g of hydrogen
2) What mass of calcium oxide is produced when 10g of calcium burns?
2Ca + O CaO
Mr: 2Ca = 2x40 = CaO = 2 x (40+16) = 112
80g produces 112g so 10g produces (112/80) x 10 = 14g of CaO
3) What mass of aluminium is produced from 100g of aluminium oxide?
2Al2O Al + 3O2
Mr: 2Al2O3 = 2x((2x27)+(3x16)) = Al = 4x27 = 108
204g produces 108g so 100g produces (108/204) x 100 = 52.9g of Al2O3
Dr Seemal Jelani

11. Actual Yield
27/02/2019
Even though no atoms are ever gained or lost in a chemical reaction, it is not always possible to obtain the calculated amount of product. Because:
The reaction may not totally finish – it may be reversible
Some of the product may be lost when it is separated from the reaction mixture – filtered
Some of the reactants may react in different ways to the expected reaction
Dr Seemal Jelani

12. Mr: 2Al2O3 = 2x((2x27)+(3x16)) = 204 4Al = 4x27 = 108
27/02/2019
Percentage yield
The amount of product obtained is known as the yield. When compared to the maximum theoretical (calculated) amount as a percentage, it is called percentage yield.
Percentage yield (%) =
Actual yield made
Maximum yield possible
x100%
E.g. What mass of aluminium is produced from 100g of aluminium oxide?
2Al2O Al + 3O2
Mr: 2Al2O3 = 2x((2x27)+(3x16)) = Al = 4x27 = 108
204g produces 108g so 100g produces (108/204) x 100 = 52.9g of Al
However, only 35.6g of Al was actually obtained during the experiment. What is the percentage yield.
Percentage yield (%) =
35.6
52.9
x100%
Dr Seemal Jelani
= 67.3%

13. Reversible Reactions
27/02/2019
Some chemical reactions are reversible. In other words, they can go in either direction: A + B C D
e.g. Ammonium chloride
Ammonia + hydrogen chloride
NH4Cl
NH3 + HCl
When a reversible reaction occurs in a closed system (Where nothing can escape), equilibrium is reached when both reactions occur at exactly the same rate in each direction.
The relative amounts of all the reacting substances at equilibrium depend on the conditions of the reaction.
Dr Seemal Jelani

14. Making Ammonia Nitrogen + hydrogen Ammonia N2 + 3H2 2NH3
27/02/2019
Fritz Haber,
Guten Tag. My name is Fritz Haber and I won the Nobel Prize for chemistry. I am going to tell you how to use a reversible reaction to produce ammonia, a very important chemical. This is called the Haber Process.
Nitrogen + hydrogen Ammonia
N2 + 3H NH3
To produce ammonia from nitrogen and hydrogen you have to use three conditions:
High pressure
450O C
Iron catalyst
Mixture of NH3, H2 and N2. This is cooled causing NH3 to liquefy.
Nitrogen
Hydrogen
Recycled H2 and N2
Dr Seemal Jelani

15. Ammonia + nitric acid Ammonium nitrate
Uses of Ammonia
27/02/2019
Ammonia is a very important chemical as it can be used to make plant fertilisers and nitric acid:
Water and oxygen
Nitrogen monoxide
Ammonia gas
Oxygen
Hot platinum catalyst
Nitrogen monoxide
Nitric acid
Cooled
More ammonia can then be used to neutralise the nitric acid to produce AMMONIUM NITRATE (a fertiliser rich in nitrogen).
Ammonia + nitric acid Ammonium nitrate
NH3 + HNO NH4NO3
The trouble with nitrogen based fertilisers is that they can also create problems – they could contaminate our drinking water.
Dr Seemal Jelani

16. Haber Process Summary
27/02/2019
A low temperature increases the yield of ammonia but is too slow
A high temperature improves the rate of reaction but decreases the yield too much
A high pressure increases the yield of ammonia but costs a lot of money
To compromise all of these factors, these conditions are used to make a reasonable Yield of ammonia, quickly:
200 atm pressure
450O C
Iron catalyst
Recycled H2 and N2
Nitrogen
Hydrogen
Mixture of NH3, H2 and N2. This is cooled causing NH3 to liquefy.
Dr Seemal Jelani