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

3.1 Mass numbers

Mass number – atomic number = number of neutrons

E.g. Sodium23 – 11 = 12

Isotopes• Same number of

protons• Different number of

neutrons

Relative Molecular Mass (Mr)

Mr is calculated by adding together all the relative atomic masses (Ar) in a molecule/formula

For exampleNaCl - 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

Number of Moles

Number of moles = mass of Xof 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

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 = 23 + 35.5 = 58.5

Moles• A mole of any substance always contains same number of particles

- Relative atomic mass in grams

- Relative formula mass in grams

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%

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 35.5

Moles (9/27) = 0.33 (35.5/35.5) = 1

Simplest ratio (divide by smallestnumber of moles)

(0.33 / 0.33) = 1 (1 / 0.33) = 3

Formula AlCl3

Maximum Yield

Maximum yield is the maximum mass of a substance that could be produced in a chemical reactionX + 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

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 100maximum 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%

Empirical FormulaThe empirical formula is the simplest form of a formula, for example C6H12 would be CH2

There are 3 steps to calculating the empirical formulaStep 1 – calculate the number of moles of each substanceStep 2 – divide the numbers by the smallest numberStep 3 - write it down as a formula

Empirical Formula

An exampleA 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

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

3.5 Percentage yield

Percentage yield

% yield = amount of product produced (g) x 100%

max. amount of product possible (g)