Relative mass formula, atomic mass, and empirical formula

Relative formula mass, Mr

The relative formula mass of a compound is blatantly the relative atomic masses of all the elements in the compound added together.E.g. water H2O:

Therefore Mr for water = 16 + (2x1) = 18

Work out Mr for the following compounds:1) HCl

2) NaOH

3) MgCl2

4) H2SO4

5) 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

Relative atomic mass of O = 16

Relative atomic mass of H = 1

More examples

CaCO3 40 + 12 + 3x16 100

HNO3 1 + 14 + 3x16

2MgO 2 x (24 + 16) 80

3H2O 3 x ((2x1) + 16)

4NH3

2KMnO4

3C2H5OH

4Ca(OH)2

Relative atomic mass

• The mass of an isotopic element relative to Carbon-12. • Example: chlorine occurs in isotope forms Cl-35 (75.5%)

and Cl-37 (24.5%)• Relative atomic mass = • ((75.5x35)+(24.5x37))/(75.5+24.5)=35.5• Try this: neon-20 (90.9%), neon-21 (0.3%), and neon-22

(8.8%)

Calculating percentage mass

If you can work out Mr then this bit is easy…

Calculate the percentage mass of magnesium in magnesium oxide, MgO:

Ar for magnesium = 24 Ar for oxygen = 16

Mr for magnesium oxide = 24 + 16 = 40

Therefore percentage mass = 24/40 x 100% = 60%

Percentage mass (%) =

Mass of element Ar

Relative formula mass Mr

x100%

Calculate the percentage mass of the following:

1) Hydrogen in hydrochloric acid, HCl

2) Potassium in potassium chloride, KCl

3) Calcium in calcium chloride, CaCl2

4) Oxygen in water, H2O

Calculating the mass of a product

E.g. what mass of magnesium oxide is produced when 60g of magnesium is burned in air?

Step 1: READ the equation:

2Mg + O2 2MgO

IGNORE the oxygen in step 2 – the question

doesn’t ask for it

Step 3: LEARN and APPLY the following 3 points:

1) 48g of Mg makes 80g of MgO

2) 1g of Mg makes 80/48 = 1.66g of MgO

3) 60g of Mg makes 1.66 x 60 = 100g of MgO

Step 2: WORK OUT the relative formula masses (Mr):

2Mg = 2 x 24 = 48 2MgO = 2 x (24+16) = 80

Work out Mr: 2H2O = 2 x ((2x1)+16) = 36 2H2 = 2x2 = 4

1. 36g of water produces 4g of hydrogen

2. So 1g of water produces 4/36 = 0.11g of hydrogen

3. 6g of water will produce (4/36) x 6 = 0.66g of hydrogen

Mr: 2Ca = 2x40 = 80 2CaO = 2 x (40+16) = 112

80g produces 112g so 10g produces (112/80) x 10 = 14g of CaO

Mr: 2Al2O3 = 2x((2x27)+(3x16)) = 204 4Al = 4x27 = 108

204g produces 108g so 100g produces (108/204) x 100 = 52.9g of Al2O3

1) When water is electrolysed it breaks down into hydrogen and oxygen:

2H2O 2H2 + O2

What mass of hydrogen is produced by the electrolysis of 6g of water?

3) What mass of aluminium is produced from 100g of aluminium oxide?

2Al2O3 4Al + 3O2

2) What mass of calcium oxide is produced when 10g of calcium burns?

2Ca + O2 2CaO

Another method

Try using this equation:

Mass of product IN GRAMMES

6g

4

36

So mass of product = (4/36) x 6g = 0.66g of hydrogen

Calculating the volume of a product

At normal temperature and pressure the Relative Formula Mass (Mr) of a gas will occupy a volume of 24 litres

e.g. 2g of H2 has a volume of 24 litres

32g of O2 has a volume of 24 litres

44g of CO2 has a volume of 24 litres etcQ. When water is electrolysed it breaks down into hydrogen and

oxygen:

2H2O 2H2 + O2

What VOLUME of hydrogen is produced by the electrolysis of 6g of water?• On the previous page we said that the MASS of hydrogen produced was 0.66g

• 2g of hydrogen (H2) will occupy 24 litres (from the red box above),

• So 0.66g will occupy 0.66/2 x 24 = 8 litres

Example questions

1) What volume of hydrogen is produced when 18g of water is electrolysed?

2H20 2H2 + O2

2) Marble chips are made of calcium carbonate (CaCO3). What volume of carbon dioxide will be released when 500g of CaCO3 is reacted with dilute hydrochloric acid?

CaCO3 + 2HCl CaCl2 + H2O + CO2

3) In your coursework you reacted magnesium with hydrochloric acid. What volume of hydrogen would be produced if you reacted 1g of magnesium with excess acid?

Mg + 2HCl MgCl2 + H2

Empirical formulae

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 = 0.04 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

Example questions

1) Find the empirical formula of magnesium oxide which contains 48g of magnesium and 32g of oxygen.

2) Find the empirical formula of a compound that contains 42g of nitrogen and 9g of hydrogen.

3) Find the empirical formula of a compound containing 20g of calcium, 6g of carbon and 24g of oxygen.