3.6 - Reacting Mass Calculations
Reacting masses by proportion
- A balanced equation shows the proportions substances react in, and relative formula masses turn those proportions into masses. In:
- C + O2 → CO2
- Carbon has a relative atomic mass of 12, O2 a relative formula mass of 32, and CO2 one of 44.
- So 12 g of carbon reacts with 32 g of oxygen to form 44 g of carbon dioxide, and any reacting quantities keep to that ratio.
Numbers in front of the formulae
- Where a number is written in front of a formula, multiply that substance's relative formula mass by it. In:
- CH4 + 2O2 → CO2 + 2H2O
- Water has a relative formula mass of 18, but the 2 in front means the mass of water here is 2 × 18 = 36.
- You only need to calculate the masses for the two substances named in the question.
The method
- Read the equation for the two substances named, noting the number in front of each.
- Work out the mass of each of those two in the equation by multiplying its relative formula mass by the number in front of it.
- Find the ratio between those two masses by dividing the wanted substance's mass by the known substance's.
- Multiply the mass in the question by that ratio.
Worked example - Product mass by proportion
Methane burns completely in oxygen:
CH4 + 2O2 → CO2 + 2H2O
Calculate the mass of water formed when 7.00 g of methane is burned in a plentiful supply of oxygen. Relative atomic masses: H = 1, C = 12, O = 16.
Step 1 - Read the equation for the two named substances
No number in front of CH4, so one methane takes part. A 2 in front of H2O, so two waters form.
Step 2 - Work out the mass of each in the equation
- CH4: 12 + (4 × 1) = 16, and there is 1 of it, so 16
- H2O: (2 × 1) + 16 = 18, and there are 2 of them, so 2 × 18 = 36
So 16 g of methane forms 36 g of water.
Step 3 - Find the ratio between the two masses
36 ÷ 16 = 2.25
The mass of water is always 2.25 times the mass of methane burnt.
Step 4 - Multiply the question's mass by that ratio
2.25 × 7.00 = 15.75 g
The mass in the question is given to three significant figures, so the answer is too:
Mass of water = 15.8 g
Reacting masses through the mole
- The numbers in front of the formulae in a balanced equation show amounts in moles. In:
- Mg + 2HCl → MgCl2 + H2
- One mole of magnesium reacts with two moles of hydrochloric acid to give one mole of magnesium chloride and one mole of hydrogen.
Why you go via moles
- Two masses can't be compared directly, because a gram of one substance contains a different number of particles from a gram of another.
- The route runs: mass → amount in moles → across the equation → mass.
The method
- Write out the balanced equation.
- Work out the relative formula mass () of the known substance and the wanted substance.
- Turn the known mass into an amount in moles - .
- Use the balancing numbers in the equation to find the moles of the wanted substance.
- Turn that amount back into a mass - .
A relative formula mass is always for one unit of the formula, so the number in front plays no part in step 2 - it only applies at step 4 when comparing moles.
Worked example - Product mass through the mole
Lithium burns in air to form lithium oxide:
4Li + O2 → 2Li2O
Calculate the mass of lithium oxide formed when 105 g of lithium is burned in air. Relative atomic masses: O = 16, Li = 7.
Step 1 - Write out the balanced equation
4Li + O2 → 2Li2O
Step 2 - Work out the relative formula masses
- Li: 7
- Li2O: (2 × 7) + 16 = 30
Step 3 - Turn the known mass into an amount
Moles = 105 ÷ 7 = 15.0 mol of lithium
Step 4 - Use the numbers in front to get the other amount
The equation shows 4 units of Li giving 2 units of Li2O, so half as many moles of oxide are made as of metal:
15.0 ÷ 2 = 7.50 mol of Li2O
Step 5 - Turn that amount back into a mass
Mass = 7.50 × 30 = 225 g
Mass of lithium oxide = 225 g