Reacting Masses (WJEC GCSE Science (Double Award): Chemistry): Revision Note

• Although, there are questions that ask you to 'calculate the empirical formula of', the exam does not usually use the term 'empirical formula'

• Typically, the exam will use phrases like:

  • Find the simplest formula of...

  • Calculate the simplest formula of...

A sample of a compound was found to contain 10 g of hydrogen and 80 g of oxygen.

Calculate the simplest formula of this compound.

A_r(H) = 1   A_r(O) = 16

Answer:

So, the simplest formula = H₂O

Carbohydrate X was analysed and found to contain 31.58% carbon and 5.26% hydrogen by mass.

Find the simplest formula of carbohydrate X.

A_r(H) = 1   A_r(C) = 162   A_r(O) = 16

Answer:

A carbohydrate contains carbon, hydrogen and oxygen

The percentages do not add up to 100%, which means that you need to calculate the percentage of oxygen needs to be calculated

Percentage of oxygen = 100 - 31.58 - 5.26 = 63.16%

So, the simplest formula = C₂H₄O₃

• The molar ratio must be a whole number

  • If you don't get a whole number when calculating the ratio of atoms in an empirical formula, such as 1.5, multiply that and the other ratios to achieve whole numbers

• Make sure that you show your working

  • Your answer may look like an incorrect formula for a chemical that you know

  • For example, data from the incomplete combustion of magnesium in oxygen could result in a formula of Mg₂O for magnesium oxide instead of MgO

Substance X with the simplest formula C₄H₁₀S is found to have a relative molecular mass of 180.

Find its molecular formula.

A_r(C) = 12   A_r(H) = 1   A_r(S) = 32

Answer

• Step 1 - Calculate the relative formula mass of the simplest formula

  • M_r = (12 x 4) + (1 x 10) + (32 x 1) = 90

• Step 2 - Divide the relative formula mass of X by the relative formula mass of the simplest formula

  • $\left(\frac{180}{90}\right)$ = 2

• Step 3 - Multiply each number of elements by 2

  • C_{(4 x 2)}

  • H_{(10 x 2)}

  • S_{(1 x 2)}     

• So, the molecular formula of X = C₈H₂₀S₂

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 = $\left(\frac{\mathrm{mass}}{M_r}\right)$ = $\left(\frac{6}{24}\right)$ = 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

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 = $\left(\frac{\mathrm{mass}}{M_r}\right)$ = $\left(\frac{51000000}{102}\right)$ = 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 = $\left(\frac{27000000}{{10}^6}\right)$ = 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

Example 3:

Zinc is extracted from zinc oxide, ZnO.

When carrying out the reaction 1.2 g of carbon reacted with 16.2 g of zinc oxide to produce 4.4 g of carbon dioxide and 13.0 g of zinc.

Determine the balanced equation for the reaction.

A_r(C) = 12   A_r(O) = 16   A_r(Zn) = 65

Answer:

• Step 1 - calculate the masses

  • C = 12

  • ZnO = 65 + 16 = 81

  • CO₂ = 12 + (2 x 16) = 44

  • Zn = 65

• Step 2 - calculate the moles of each substance using moles = $\left(\frac{\mathrm{mass}}{M_r}\right)$

  • C = $\left(\frac{1.2}{12}\right)$ = 0.1 moles 

  • ZnO = $\left(\frac{16.2}{81}\right)$ = 0.2 moles 

  • CO₂ = $\left(\frac{4.4}{44}\right)$ = 0.1 moles 

  • Zn = $\left(\frac{13.0}{65}\right)$ = 0.2 moles

• Step 3 - convert to the simplest whole number ratio

  • C = 1

  • ZnO = 2

  • CO₂ = 1

  • Zn = 2

• Step 4 - apply the values in front of the chemicals in the equation

  • C + 2ZnO → CO₂ + 2Zn

• These questions look hard but they are very methodical, so 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