1.3.2 Reacting Masses

Reacting masses calculations allow us to determine how much of a reactant is needed or how much product will be formed in a chemical reaction. These calculations are based on the mole concept and the stoichiometry of the balanced chemical equation.

Steps for Reacting Masses Calculations

1. Calculate the moles of the given substance

Use the formula:

\text{Moles} = \frac{\text{Mass}}{M_r}

Where:

Mass is the mass of the substance (in grams).
$M_r$ is the relative molecular or formula mass.

2. Use the balanced chemical equation

Use this equation to determine the mole ratio between the known and unknown species (reactant and product).

3. Calculate the moles of the unknown species

Multiply the moles of the known species by the mole ratio from the balanced equation.

4. Convert moles of the unknown species to mass

Use the same formula:

\text{Mass} = \text{Moles} \times M_r

This gives the mass of the unknown species (either reactant or product).

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Example: Calculating the Mass of Oxygen Required to React with Hydrogen Question: What mass of oxygen gas (O₂) is needed to react with hydrogen gas (H₂) to produce 9 g of water (H₂O)?

Step 1: Write the balanced chemical equation:

2H_2 + O_2 \rightarrow 2H_2O

Step 2: Calculate moles of water: The molar mass of water (H₂O) is 18 g/mol.

\text{Moles of H}_2O = \frac{9 \, \text{g}}{18 \, \text{g/mol}} = 0.5 \, \text{mol}

Step 3: Use the mole ratio: From the balanced equation, 2 moles of H₂O are produced by 1 mole of O₂.

Therefore, for 0.5 moles of water, the required moles of oxygen are:

\text{Moles of O}_2 = \frac{0.5}{2} = 0.25 \, \text{mol}

Step 4: Convert moles of O₂ to mass: The molar mass of oxygen gas (O₂) is 32 g/mol.

\text{Mass of O}_2 = 0.25 \times 32 \, \text{g/mol} = 8 \, \text{g}

Therefore, 8 g of oxygen is required.

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Example: Mass of a Reactant or Product: Question: How much calcium carbonate (CaCO₃) is required to produce 44 g of carbon dioxide (CO₂) in the reaction:

CaCO_3 \rightarrow CaO + CO_2

Solution:

$M_r$ of CO₂ = 44 g/mol
So moles of CO₂ = 44 g / 44 g/mol = 1 mol.
From the balanced equation, 1 mol of CaCO₃ produces 1 mol of CO₂
So moles of CaCO₃ = 1 mol.
$M_r$ of CaCO₃ = 100 g/mol
So the mass of CaCO₃ required = 1 mol × 100 g/mol = 100 g.

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Example: Mass of a Compound in Solution: Question: How much sodium hydroxide (NaOH) is needed to completely react with 50 cm³ of 1 mol/dm³ hydrochloric acid (HCl) in the reaction:

NaOH + HCl \rightarrow NaCl + H_2O

Solution:

Moles of HCl = 1 mol/dm³ × (50 cm³ / 1000 cm³/dm³) = 0.05 mol.
From the balanced equation, 1 mole of NaOH reacts with 1 mole of HCl.
So, moles of NaOH = 0.05 mol.
$M_r$ of NaOH = 40 g/mol.
Therefore, mass of NaOH = 0.05 mol × 40 g/mol = 2 g.

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Summary

Reacting masses calculations are essential for determining the quantities of reactants or products in a chemical reaction. The steps involve converting mass to moles, using the balanced chemical equation to find the mole ratio, and then calculating the mass or volume of the unknown species. Mastering these calculations is crucial for solving many types of chemistry problems.

Reacting Masses Simplified Revision Notes for A-Level AQA Chemistry

1.3.2 Reacting Masses

Steps for Reacting Masses Calculations

1. Calculate the moles of the given substance

2. Use the balanced chemical equation

3. Calculate the moles of the unknown species

4. Convert moles of the unknown species to mass

Summary

500K+ Students Use These Powerful Tools to Master Reacting Masses For their A-Level Exams.

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