1. Calculations in Chemistry
IGCSE Chemistry

2. Mole concept

3. Relative atomic mass and Relative molecular mass
Why atomic mass or molecular mass has no unit?
Atoms and molecules are too small to weigh.
Relative atomic mass and Relative molecular mass are comparative values.
This shows how many times an atom or a molecule heavier than th of a Carbon atom.

4. The mole….
Atomic mass or molecular mass with g after the number is called a mole.
Example:
Atomic mass of Oxygen is 16.
16g of oxygen(O) is 1 mole of oxygen
atoms.
Molecular mass of Oxygen O2 is 32.
32g of Oxygen (O2) is 1 mole of oxygen molecules.
Molecular mass of water is 18.
18g of water is called 1 mole of water.

5. Number of moles =
Mass (g) =Number of moles X Molar mass

6. Questions 1. How many moles are there in 4g of Sodium hydroxide?
Number of moles =
= 0.10moles
2. Find the number of moles present in 1.28g of oxygen molecules.
= 0.04 moles

7. 3. What is the mass of 0.01 moles of sodium hydroxide?
Mass = X 40g
= 0.40g
4. How many grams 1.5 moles of sulphuric acid weighs?
Mass = 1.5 X 98g
= 147g

8. Try the following questions:
1. What is the mass of 0.02 moles of carbon dioxide gas?
0.88g
2. How many moles are present in 1g of calcium carbonate,CaCO3?
0.01moles
3. Find the number of moles present in 3g of potassium hydroxide, KOH.
0.054moles
4. Calculate the mass of 100 moles of copper oxide, CuO
8000g
g of sodium hydroxide,NaOH or 3g of potassium hydroxide, KOH contains more number of moles? Show your working clearly.
NaOH = 0.13 moles KOH = moles. So NaOH

9. Avogadro constant One mole of any substance contains
6.02 X 10 atoms or molecules.
6.02 X is a constant and is called Avogadro constant or Avogadro Number.
In other words, Avogadro Number is the number of atoms or number of molecules present in ONE mole of a substance.
Avogadro constant is used to find number of particles or number of moles.

10. Number of moles X Avogadro constant
Number of moles = Number of molecules
Avogadro constant
Number of molecules =
Number of moles X Avogadro constant

11. Calculations
Remember always change mass or number of molecules into moles.

12. Examples:
How many molecules of Carbon dioxide are there in 0.44g of the gas?
Answer:
Number of moles = 0.44g/44g
= 0.01moles
Number of molecules = X 6.02 X 10
= 6.02 X 10

13. Calculate how many oxygen atoms are present in 1.96g of sulphuric acid?
Answer:
Moles of sulphuric acid = 1.96g/98g
= 0.02 moles
Number of molecules of sulphuric acid
= X 6.02 X 10
=1.20 X 10
Each sulphuric acid contains 4 oxygen atoms.
So number of oxygen atoms 0.02 moles of H2SO4 is
= 4 X 1.20 X = 4.80 X 10

14. Mole ratio
When chemical reactions take place, the reactants and products are in a simple ratio (mole ratio)
For example, Zinc and hydrochloric acid react according to the following reaction:
Zinc + Hydrochloric acid  Zinc chloride + Hydrogen gas
Zn + 2HCl  ZnCl2 + H2
It means, zinc, Hydrochloric acid, Zinc chloride and Hydrogen moles are in the ratio 1:2:1:1
One mole of Zinc reacts with 2 moles of Hydrochloric acid to produce, One mole of Zinc chloride and 1 mole of Hydrogen gas.
The reactants and products are always in a ratio by moles and not by mass.

15. Find how many grams of copper oxide can be made from 3.2g of copper.
Answer:
2Cu + O2  2CuO
The balanced equation shows that two moles of copper reacts with 1 mole of oxygen to form 2 moles of copper(II) oxide.
So change 3.2g to moles
Moles of copper = 3.2g/64g =0.05moles
Copper :Copper oxide is 1:1 ratio.
So moles of copper(II) oxide = 0.05moles
Mass (g) = X 80g
= 4g

16. Moles of phosphorus(V)oxide= 8g/142g = 0.056moles
When Posphorus burns in oxygen, Phosporus(V)oxide is produced. In an experiment, 8g of phsphorus(V)oxide is produced. Find the mass of Phosphorus reacted with oxygen.
Answer:
Equation: 4P + 5O2  2P2O5
Mole ratio is 4:5:2
Moles of phosphorus(V)oxide= 8g/142g
= 0.056moles
Mole ratio Phosphorus oxide:phosphorus
2:4
Moles of phosphorus burnt = X 2
= 0.112
Mass of phosphorus = X 31g
= 3.472g

17. Limiting reagent and excess
When there is a chemical reaction, the reactant which is used up completely is the limiting reagent.
The chemical which is left some unreacted at the end is called excess.
When the limiting reagent is finished, the reaction stops.
All mole ratios should be done using the limiting reagent.

18. Moles of Al2O3 (excess) left after reaction is 1.00-0.67 = 0.33moles.
Example 1
3Mg + Al2O3  3MgO + 2Al
If 2 moles of Magnesium is used with 1 moles of aluminium oxide, which is the limiting agent? Which chemical is in excess? How many moles in excess?
Mole ratio of Mg and Al2O3 is 3:1
It means for 2 moles of Mg, 0.67 moles of Al2O3 is enough. But we have 1 mole of Al2O3. So Al2O3 is excess and Mg is the limiting reagent.
Moles of Al2O3 (excess) left after reaction is
= 0.33moles.

19. If mass is given, change all masses into moles first.
Example 2
2Cu + O2  2CuO
In a reaction, 1.92g of copper is oxidised with 0.64g of oxygen.
Find which chemical is the limiting agent
How many g of the other chemical (excess) remaining at the end?
What mass of Copper (II)oxide produced?
Answer
Change mass in to moles
Number of moles of copper = 1.92g/64g = moles
Number of moles of oxygen = 0.64g/32g = moles
Cu : O2 mole ratio is 2:1
It means, for 0.03 moles of copper only moles of oxygen is enough. But we have 0.02 moles.
Copper is the limiting reagent and oxygen is excess.

20. Excess oxygen = 0.020 – 0.015 = 0.005 moles
Mass of excess oxygen = X 32g = 0.16g
Moles of copper oxide (make mole ratio with limiting reagent) 2:2 That is 0.03moles
Mass of copper oxide = 0.03 X 80g = 2.40g
Now try out the following:
2Na + S  Na2S
If 0.46g of sodium and 0.30g of sulphur reacted,
(a) find the limiting reagent and excess reagent.
Sulphur
(b) find the mass of excess reagent left.
moles Na which is 0.028g
(c) find what mass of sodium sulfide formed.
Moles of Na2S is means 0.73g

21. Molar volume
Gases have volume. Volume of a gas depends on two factors :
Temperature and Pressure
Volume decreases when pressure is increased.
Volume increases when the temperature is increased.

22. Molar Volume
One mole of any gas at Room Temperature and Pressure has a volume of 24dm^3 (24000 cm^3)
Room temperature is considered as
20 degree C and atmospheric pressure is
1 atmosphere (760 mm of mercury)
It means 1 mole of oxygen gas is cm3
3 moles of oxygen gas is 3X24000 cm3 at RTP.
24dm3(1mole) is called molar volume of gases

23. Moles X molar volume = volume of a gas
Moles = volume of a gas/molar volume
Remember to convert volume or mass to moles first.
Example 1
What is the volume of 1.1g of carbon dioxide gas at RTP?
First convert mass in to moles
Moles of CO2 = 1.1g/44g = 0.025moles
Volume = moles X molar volume
= X cm3
= 600cm3

24. Example 2
Find out the mass of 8000 cm3 of sulphur dioxide gas, SO2 at RTP.
First change the volume into moles.
Moles = Volume/molar volume
8000 cm3/24000 cm3
= 0.33 moles
Now change the moles in to mass
Mass = moles X molar mass
0.33 X 64g = 21.12g

25. % Yield
We can calculate and predict how much of a product is formed during a reaction using the mole ratio.
How're in a real experiment, the predicted (calculated)amount may be different that what we get in actual experiment.
This can be due to:
1. Some reactants may be left unreacted.
2. Change of substances in to other forms such as gases
3.Some reactants may be lost to surrounding by overflowing, splashing etc
4. Experimental errors and low quality measuring devices.

26. %yield = Actual yield X 100
Theoretical yield
Example 1:
In an experiment, 20g of Calcium carbonate when heated gave 10g of calcium oxide is formed. Find the percentage yield of calcium oxide.
CaCO3  CaO + CO2
To work out the above problem, we need the actual yield and theoretical yield(Calculated yield).
Actual yield is given as 10g of Calcium oxide.
Theoretical yield should be calculated by using mole ratio method.

27. CaCO3  CaO + CO2
We can find the mass of calcium oxide formed from the mass of calcium carbonate given.
But first convert mass in to moles.
Moles of CaCO3 = 20g/100g
= 0.20moles
Mole ratio CaCO3:CaO is 1:1
So, moles of CaO is also 0.20
Mass of CaO = X 56g
= g
Now insert the values
% yield = Actual yield X 100%
Theoretical yield
That is g X 100% = %
11.20g

28. Example 2
2Na + 2H2O  2NaOH + H2
115g of sodium was reacted with water which produced 4.2g of hydrogen gas. What is the % yield of hydrogen gas?
Answer
Moles of sodium = moles
Moles of H = moles (2:1)
Mass of H = 5.00g
% yield = 84%

29. End of part 1