Amounts of Substances in Equations | AQA GCSE Chemistry Revision Notes 2018

Calculating masses from balanced equations

Higher tier only

Chemical / symbol equations can be used to calculate:
- The moles of reactants and products
- The mass of reactants and products
To do this:
- Information from the question is used to find the amount in moles of the substances being considered
- Then, the ratio between the substances is identified using the balanced chemical equation
- Once the moles have been determined they can then be converted into grams using the relative atomic or relative formula masses

Quantitative chemistry example 1:

Magnesium undergoes combustion to produce magnesium oxide.

The overall reaction that is taking place is shown in the equation below.

2Mg (s) + O₂ (g) ⟶ 2 MgO (s)

Calculate the mass of magnesium oxide that can be made by completely burning 6.0 g of magnesium in oxygen in the following reaction:

A_r(O) = 16 A_r(Mg) = 24

Answer:

Step 1 - calculate the moles of magnesium
- Moles = $(\frac{mass}{M_{r}})$ = $(\frac{6}{24})$ = 0.25
Step 2 - use the molar ratio from the balanced symbol equation
- 2 moles of magnesium produce 2 moles of magnesium oxide
- The ratio is 1 : 1
- Therefore, 0.25 moles of magnesium oxide is produced
Step 3 - calculate the mass of magnesium oxide
- Mass = moles x M_r = 0.25 moles x (24 + 16) = 10 g

Quantitative chemistry example 2:

In theory, aluminium could decompose as shown in the equation below.

2Al₂O₃ ⟶ 4Al + 3O₂

Calculate the maximum possible mass of aluminium, in tonnes, that can be produced from 51 tonnes of aluminium oxide.

A_r(O) = 16 A_r(Al) = 27

Answer:

Step 1 - calculate the moles of aluminium oxide
- Mass = 51 tonnes x 10⁶ = 51 000 000 g
- Moles = $(\frac{mass}{M_{r}})$ = $(\frac{51000000}{102})$ = 500 000
Step 2 - use the molar ratio from the balanced symbol equation
- 2 moles of aluminium oxide produces 4 moles of aluminium
- The ratio is 1 : 2
- Therefore, 2 x 500 000 = 1 000 000 moles of aluminium is produced
Step 3 - calculate the mass of aluminium
- Mass = moles x M_r = 1 000 000 moles x 27 = 27 000 000 g
- Mass in tonnes = $(\frac{27000000}{10^{6}})$ = 27 tonnes

When you see any of the following phrases in a question, make sure you work methodically through the process of calculating moles, using molar ratios and calculating mass
- Calculate the mass...
- Calculate the minimum mass...
- Calculate the maximum mass...
As long as you are consistent, it doesn't matter whether you work in grams or tonnes or any other mass unit as the reacting masses will always be in proportion to the balanced equation

If the masses of reactants and products of a reaction are known then we can use them to write a balanced equation for that reaction
This is done by converting the masses to moles and simplifying to find the molar ratios

A student reacts 1.2 g of carbon with 16.2 g of zinc oxide. The resulting products are 4.4 g of carbon dioxide and 13 g of zinc.

Determine the balanced equation for the reaction.

Answer:

Step 1: Write the unbalanced chemical equation:
- _C + _ZnO → _CO₂ + _Zn
Step 2: Write down the masses of each substance:
- C = 12
- ZnO = 65 + 16 = 81
- CO₂ = 12 + (2 x 16) = 44
- Zn = 65
Step 3: Calculate the moles of each substance:
- C = 1.2 / 12 = 0.1
- ZnO = 16.2 / 81 = 0.2
- CO₂ = 4.4 / 44 = 0.1
- Zn = 13 / 65 = 0.2
Step 4: Deduce the whole number ratio of substances:
- C : ZnO : CO₂ : Zn
- 0.1 : 0.2 : 0.1 : 0.2
- 1 : 2 : 1 : 2
Step 5: Apply the whole number ratio to the unbalanced equation:
- 1C + 2ZnO → 1CO₂ + 2Zn
- C + 2ZnO → CO₂ + 2Zn

These questions look hard but they are actually quite easy to do, as long as you follow the steps and organise your work neatly.

Remember the molar ratio of a balanced equation gives you the ratio of the amounts of each substance in the reaction.