Chemistry » Quantitative Aspects of Chemical Change » Stoichiometric Calculations

# Stoichiometric Calculations

## Stoichiometric calculations

Stoichiometry is the calculation of the quantities of reactants and products in chemical reactions. It is important to know how much product will be formed in a chemical reaction, or how much of a reactant is needed to make a specific product.

The following diagram shows how the concepts that we have learnt in this section relate to each other and to the balanced chemical equation:

## Example: Stoichiometric Calculation

### Question

What volume of oxygen at S.T.P. is needed for the complete combustion of $$\text{2}$$ $$\text{dm^{3}}$$ of propane ($$\text{C}_{3}\text{H}_{8}$$)? (Hint: $$\text{CO}_{2}$$ and $$\text{H}_{2}\text{O}$$ are the products in this reaction (and in all combustion reactions))

### Step 1: Write the balanced equation

$\text{C}_{3}\text{H}_{8}\text{(g)} + 5\text{O}_{2}\text{(g)} \rightarrow 3\text{CO}_{2}\text{(g)} + 4\text{H}_{2}\text{O (g)}$

### Step 2: Find the ratio

Because all the reactants are gases, we can use the mole ratios to do a comparison. From the balanced equation, the ratio of oxygen to propane in the reactants is $$5:1$$.

### Step 3: Find the answer

One volume of propane needs five volumes of oxygen, therefore $$\text{2}$$ $$\text{dm^{3}}$$ of propane will need $$\text{10}$$ $$\text{dm^{3}}$$ of oxygen for the reaction to proceed to completion.

## Example: Stoichiometric Calculation

### Question

What mass of iron (II) sulphide is formed when $$\text{5.6}$$ $$\text{g}$$ of iron is completely reacted with sulphur?

### Step 1: Write the balanced equation

$\text{Fe (s)} + \text{S (s)} \rightarrow \text{FeS (s)}$

### Step 2: Calculate the number of moles

We find the number of moles of the given substance:

\begin{align*} n & = \frac{m}{M} \\ & = \frac{\text{5.6}\text{ g}}{\text{55.8}\text{ g·mol$^{-1}$}} \\ & = \text{0.1}\text{ mol} \end{align*}

### Step 3: Find the mole ratio

We find the mole ratio between what was given and what you are looking for. From the equation $$\text{1}$$ $$\text{mol}$$ of Fe gives $$\text{1}$$ $$\text{mol}$$ of $$\text{FeS}$$. Therefore, $$\text{0.1}$$ $$\text{mol}$$ of iron in the reactants will give $$\text{0.1}$$ $$\text{mol}$$ of iron sulphide in the product.

### Step 4: Find the mass of iron sulphide

\begin{align*} m & = nM \\ & = (\text{0.1}\text{ mol})(\text{87.9}\text{ g·mol$^{-1}$}) \\ & = \text{8.79}\text{ g} \end{align*}

The mass of iron (II) sulphide that is produced during this reaction is $$\text{8.79}$$ $$\text{g}$$.