Calculating Mass From Chemical Equations
A chemical equation can tell us the molar ratio of the reactants and the products. Using the molar ratio we can calculate:
 the amounts of the products produced per given amount of reactants.
 the amount of the reactants that produce a certain amount of products.
Finding molar ratio and mass of products from a chemical equation
Let us look at a general representation of a chemical equation:
aW + bX → cY + dZ
whereby:
 W and X are the reactants
 Y and Z are the products
means that a moles of W react with b moles of X to produce c moles of Y and d moles of Z.
Example 1
Carbon burns in oxygen to produce carbon dioxide as follows:
C(s) + O_{2}(g) → CO_{2}(g)
This equation tells you that:
 1 mole of carbon atoms reacts with 1 mole of oxygen molecules to produce 1 mole of carbon dioxide molecules.
Example 2
Hydrogen burns in oxygen to produce water as follows:
2H_{2}(g) + O_{2}(g) → 2H_{2}O(g)
This equation tells you that:
 2 moles of hydrogen gas reacts with 1 mole of oxygen gas to produce 2 moles of water molecules.
Example 3
Calcium carbonate reacts with hydrochloric acid to produce calcium chloride, water and carbon dioxide gas as follows:
CaCO_{3}(s) + 2HCl(aq) → CaCl_{2}(aq) + H_{2}O(l) + CO_{2}(g)
Means that:
 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid to produce 1 mole of calcium chloride, 1 mole of hydrogen and 1 mole of carbon dioxide.
Now let us look at a more complex example:
Question 1
Iron is obtained by reducing iron(III) oxide using the gas carbon monoxide to produce molten iron and carbon dioxide gas in the blast furnace.
 Write down the word equation for the reaction
 Write down the balanced chemical equation for the reaction
 Calculate the relative formula mass of iron(III) oxide.
 How many moles of iron(III) oxide are there in 1600 kg of iron(III) oxide?
 How many moles of iron are obtained from 1 mole of iron(III) oxide?
 Calculate how many moles of iron atoms are obtained from 1600 kg of iron(III) oxide.
 Calculate the mass of iron (in kg) obtained from 1600 kg of iron(III) oxide?
Solution 1

The word equation is straight forward here since the reactants and the products are all mentioned in the question. So the word equation is:
Iron(III) oxide + carbon monoxide → iron + carbon dioxide
 Fe_{2}O_{3}(s) + 3CO(g) → 2Fe(s) + 3CO_{2}(g)

Relative formula mass (M_{r}) is calculated the same way as relative molecular mass (M_{r}).
So M_{r} of Fe_{2}O_{3}(s) = (56 × 2) + (16 × 3) = 160.

When finding number of moles you first convert mass to grams
1 600 kg = 1 600 000 g
Number of moles = $latex \frac{mass}{molar\ mass}$
= $latex \frac{1\ 600\ 000\ g}{160\ g}$
= 10 000 moles
 From the equation, 1 mole of Fe_{2}O_{3} contains 2 moles of Fe.

From the equation, 1 mole of Fe_{2}O_{3} produces 2 moles of Fe.
From the calculations above, 1 600 kg of iron(III) oxide = 10 000 moles. Therefore 1 600 kg of iron(III) oxide produces 20 000 of iron.
 Mass = number of moles × molar mass
= 20 000 × 56
= 1 120 000 g
= 1 120 kg of iron.
Finding mass of reactants given the amount of products
Given the mass of products, we can find the mass of the reactants as shown in the example below:
Question 2
Calcium carbonate is an important raw material in the industry.
 Name a rock made up of calcium carbonate.
 Calculate the relative formula mass of CaCO_{3}.
 Calculate the relative formula mass of CaO.
 If 7.00 kg of calcium oxide was formed, what mass of calcium carbonate was heated?
When calcium carbonate is heated strongly, it decomposes as follows:
CaCO_{3} → CaO + CO_{2}
Solution 2
 Limestone
 Relative formula mass of CaCO_{3} is the sum of the atomic masses of the elements involved
So M_{r} of CaCO_{3} = 40 + 12 + (16 × 3)
= 40 + 12 + 48
= 100
 M_{r} of CaO = 40 + 16
= 56
 First we find the number of moles in 7.00 kg of calcium oxide was formed.
N = $latex \frac{m}{M_r}$
= $latex \frac{7\ 000}{56}$
= 125
The equation, CaCO_{3} → CaO + CO_{2}, shows us that 1 mole of CaCO_{3} produces 1 mole of CaO. So, 125 moles of CaCO_{3} decomposed to produce 125 moles of CaO.
Whats left is to find the mass of 125 moles CaCO_{3}.
Mass = number of moles × molar mass
= 125 × 100
= 12 500 g
= 12,5 kg