1. Calculations What you need to know: Relative formula mass
Empirical formula
% composition by mass
Use balanced equations to calculate masses of reactants and products
Yields

2. 3.1 Mass numbers
Mass number – atomic number = number of neutrons E.g. Sodium 23 – 11 = 12
Isotopes
Same number of protons
Different number of neutrons

3. Relative Molecular Mass (Mr)
Mr is calculated by adding together all the relative atomic masses (Ar) in a molecule/formula
For example
NaCl - 1 sodium + 1 Chlorine
= (1 x 23) + (1 x 35.5)
= 58.5
ZnCl2 – 1 Zinc + 2 Chlorine
= (1 x 65) + (2 x 25.5)
= 136

4. Number of Moles
Number of moles = mass of X of substance X Ar or Mr of X For example 7g of Nitrogen (Ar 14) = 7/14 = 0.5 moles 4g of silicon (Ar 28) = 4/28 = 0.14 moles 10g of CaO (Mr 56) = 10/56 = 0.18 moles

5. 3.2 Masses of atoms and moles
Relative atomic masses (Ar) Mass of atom compared to 12C e.g. Na = 23, Cl = 35.5 Relative formula masses (Mr) Mass of a compound found by adding Ar of each element e.g. NaCl = = 58.5
Moles
A mole of any substance always contains same number of particles
- Relative atomic mass in grams
- Relative formula mass in grams

7. Percentage Composition
To calculate the percentage of one element in a compound % composition = Ar of X x 100 of X in compound Mr of compound For example % of Na (Ar 23) in NaCl (Mr 58.5) = (23/58.5) x 100 = 39%

9. 3.3 Percentages and formulae
Percentage mass % = mass of element total mass of compound Percentage composition / empirical formula Al Cl
Mass 9
35.5 Ar 27
Moles
(9/27) = 0.33
(35.5/35.5) = 1
Simplest ratio (divide by smallest number of moles)
(0.33 / 0.33) = 1
(1 / 0.33) = 3
Formula
AlCl3

10. Maximum Yield
Maximum yield is the maximum mass of a substance that could be produced in a chemical reaction X + Y XY Maximum yield = (mass of X) x (Mr of XY) Ar of X For example – what is the maximum mass of zinc oxide (Mr 81) when 20g of Zn (Ar 65) reacts? maximum yield = (20/65) x 81 = 24.9g

11. Percentage Yield
Percentage yield is how much you actually get compared to what is theoretically possible (maximum yield) % yield of X = mass of X obtained x 100 maximum yield For example – what is the percentage yield if you get 14 g of ZnO when the maximum yield is 24.9g? Maximum yield = (14/24.9) x 100 = 48%

13. Empirical Formula
The empirical formula is the simplest form of a formula, for example C6H12 would be CH2 There are 3 steps to calculating the empirical formula Step 1 – calculate the number of moles of each substance Step 2 – divide the numbers by the smallest number Step 3 - write it down as a formula

14. Empirical Formula
An example A substance contains 2g hydrogen and 16g oxygen Step 1 – 2g H (Ar 1) = 2/1 = 2 moles of H 16g O (Ar 16) = 16/16 = 1 mole O Step 2 – 2/1 = 2 ‘H’ 1/1 = 1 ‘O’ Step 3 – H2O

16. 3.5 Percentage yield
Very few chemical reactions have a yield of 100% because:
Reaction is reversible
Some reactants produce unexpected products
Some products are left behind in apparatus
Reactants may not be completely pure
More than one product is produced and it may be difficult to separate the product we want

17. 3.5 Percentage yield
Percentage yield % yield = amount of product produced (g) x 100% max. amount of product possible (g)