Calculation techniques for the higher tier to determine the empirical and molecular formulae of a substance, masses of a required substance in a reaction, and calculations involving moles.

This equation shows how relative formula mass, number of moles and mass are related:

number of moles = mass ÷ relative formula mass

Note that 1 mole of a substance contains 6.022 × 10^{23} atoms or molecules. 6.022 × 10^{23} is a constant number, known as Avogadro’s constant.

The equation can be rearranged to find the mass if the number of moles and **molar mass** (its relative formula mass in grams) are known. It can also be rearranged to find the molar mass if the mass and number of moles are known.

- Question
Calculate the number of moles of carbon dioxide molecules in 22 g of CO

_{2}.*A*_{r}(relative atomic mass) of carbon, C = 12*A*_{r}of oxygen, O = 16*M*_{r}(relative formula mass) of carbon dioxide, CO_{2}= 12 + 16 + 16 = 44number of moles = 22 ÷ 44 =

**0.5 mol**

- Question
Calculate the mass of 2 mol of carbon dioxide (CO

_{2}).mass = number of moles × relative formula mass = 2 × 44 =

**88 g**

- Question
10 mol of carbon dioxide has a mass of 440 g. What is the relative formula mass of carbon dioxide?

relative formula mass = mass ÷ number of moles = 440 ÷ 10 =

**44**

You can calculate the mass of a product or reactant using the idea of moles, a balanced equation and relevant *A*_{r} values.

Sulfuric acid and sodium hydroxide react together to make sodium sulfate and water:

H_{2}SO_{4} + 2NaOH → Na_{2}SO_{4} + 2H_{2}O

- Question
Calculate the

**mass of sodium sulfate**made when 20 g of sodium hydroxide reacts with excess sulfuric acid.*A*_{r}of hydrogen, H = 1*A*_{r}of oxygen, O = 16*A*_{r}of sodium, Na = 23*A*_{r}of sulfur, S = 32

*M*_{r}of sodium hydroxide, NaOH = 23 + 16 + 1 = 40*M*_{r}of sodium sulfate, Na_{2}SO_{4}= 23 + 23 + 32 + 16 + 16 + 16 + 16 = 142Number of moles of NaOH = mass ÷ relative formula mass = 20 ÷ 40 = 0.5 mol

From the equation, 2 mol of NaOH reacts with 1 mol of Na

_{2}SO_{4}, so 0.5 mol of NaOH will react with 0.25 mol of Na_{2}SO_{4}.mass of Na

_{2}SO_{4}= moles × relative formula mass = 0.25 × 142 =**35.5 g**The example above could also be tackled like this:

mass of Na

_{2}SO_{4}= = 35.5 g