Reacting Masses (WJEC GCSE Chemistry: Combined Science)

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:

	hydrogen	oxygen
Write the mass of each element	10 g	80 g
Calculate the number of moles (Divide each mass by the Ar)	= 10	= 5
Find the simplest whole number molar ratio (Divide by the smallest number)	= 2	= 1

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%

	carbon	hydrogen	oxygen
Convert % to g (Assume 100 g of substance is present)	31.58 g	5.26 g	63.16 g
Calculate the number of moles (Divide each mass by the Ar)	= 2.63	= 5.26	= 3.95
Find the simplest molar ratio (Divide by the smallest number)	= 1	= 2	= 1.5
Obtain a whole number ratio (Multiply all by 2)	1 x 2 = 2	2 x 2 = 4	1.5 x 2 = 3

So, the simplest formula = C₂H₄O₃

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
  • = 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 = =  = 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 = =  = 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 =  = 27 tonnes

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 = 
  
  • C = = 0.1 moles 
  • ZnO =  = 0.2 moles 
  • CO₂ = = 0.1 moles 
  • Zn =  = 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