Masses of reactants and products from balanced equations

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Relative masses

Relative atomic mass: The average mass of one atom of an element, taking the natural abundances of the isotopes of that element into account. Relative atomic mass is measured on a scale whereby one carbon-12 atom has a mass of exactly 12.

Relative atomic mass = \(\frac{\text{Total mass of the atoms}}{\text{Total number of atoms}} \)

Most periodic tables show the relative atomic mass instead of the atomic mass.

Relative formula mass: The average mass of one molecule of a compound, where one carbon-12 atom has a mass of one unit.

The relative formula mass of a compound can be found by adding up all the relative atomic masses of all the atoms in that compound.

Example: Find the relative formula mass of calcium carbonate.

Answer: Calcium carbonate has the formula \(CaCO_3\). This means one molecule of calcium carbonate contains one calcium atom, one carbon atom and three oxygen atoms. Its relative formula mass is the sum of the relative atomic masses of all these atoms.

Relative atomic mass of carbon: 12.01

Relative atomic mass of oxygen: 15.99

Relative atomic mass of calcium: 40.08.

Therefore, the relative formula mass of calcium carbonate can be given by: \[(Ca \times 1)+(C \times 1) + (O \times 3) = (40.08 \times 1)+(12.01 \times 1) + (15.99 \times 3) = 100g\]

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Calculating product mass from balanced equation

A balanced equation can be used to calculate the mass of products or reactants involved in a reaction. This is best explained by example.

Consider the equation: \(CaCO_3 \rightarrow CaO +CO_2\)

Mole: One mole of a substance is the amount that contains \(6 \times 10^{23} \) atoms or molecules of that substance. The mass of one mole of a element/compound is the same as the relative atomic/formula mass in grams.

For example, one mole of calcium carbonate will weigh 100g.

The relative formula mass of calcium oxide is 56 so \(6 \times 10^{23} \) molecules of calcium oxide will weigh 56g.

The relative formula mass of carbon dioxide is 44 so one mole of carbon dioxide will weigh 44g.

The balanced equation is expressed in terms of moles - One mole of calcium carbonate will decompose to give one mole of calcium oxide and one mole of carbon dioxide.

This means that 100g of calcium carbonate should decompose to give 56g of calcium oxide and 44g of carbon dioxide.

These figures can be scaled to fit whatever mass of reactant we have.

Imagine we have 10g of calcium carbonate. This means we have 1/10 of a mole of calcium carbonate. This should decompose to produce 1/10 mole of calcium oxide and 1/10 of a mole of carbon dioxide. Therefore, with 10g of calcium carbonate we would expect to produce 5.6g of calcium oxide and 4.4g of carbon dioxide.

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Calculating reactant mass from balanced equation

Now consider this equation: \(2Fe_2O_3 + 3C \rightarrow 4Fe +3CO_2 \)

The compounds have the following relative formula masses:

Iron(III) oxide: 160g

Carbon: 12

Iron: 56

Carbon dioxide: 44

Suppose I need 112g of iron. How many grams of iron(III) oxide will I need?

112g is 2 moles of iron. The balanced equation tells us that 2 moles of iron(III) oxide are needed to produce 4 moles of iron. Therefore, we know that one mole of iron(III) oxide is needed to produce 2 moles of iron.

One mole of iron(III) oxide weighs 160g - so 160g of iron(III) oxide is required.

3 moles of carbon are required to produce 4 moles of iron. I need half of this amount, so I only require 1.5 moles of carbon. This will weigh 66g.