## 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.

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### 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 involvedSo 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`, shows us that 1 mole of CaCO_{3}→ CaO + CO_{2}_{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