Mole Ratios and Their Applications in Chemical Calculations
Understanding Mole Ratios
Mole ratio
A mole ratio is the ratio of the amounts (in moles) of reactants and products in a balanced chemical equation. These ratios are determined by the stoichiometric coefficients in the equation.
- Let’s take the combustion of methane as an example: $$
\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)
$$ - From the coefficients, we can see:
- 1 mole of methane reacts with 2 moles of oxygen.
- This produces 1 mole of carbon dioxide and 2 moles of water.
- These relationships allow us to calculate the amounts of any reactant or product, provided we know the quantity of at least one substance in the reaction.
Always ensure your chemical equation is balanced before using mole ratios. An unbalanced equation gives incorrect ratios and leads to errors in calculations.
Applications of Mole Ratios
1. Calculating Masses of Reactants and Products
To find the mass of a substance in a reaction, we use the formula:
$$
\text{Mass} = \text{Moles} \times \text{Molar Mass}
$$
How much carbon dioxide is produced when 16 g of methane $\text{CH}_4 $ is completely combusted?
Solution
- Write the balanced equation:
$$
\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)
$$ - Calculate the moles of methane:
$$
n(\text{CH}_4) = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{16 \, \text{g}}{16.04 \, \text{g mol}^{-1}} = 1.00 \, \text{mol}
$$ - Use the mole ratio $ \text{CH}_4 : \text{CO}_2 = 1 : 1 $ to find the moles of $\text{CO}_2 $:
$$
n(\text{CO}_2) = 1.00 \, \text{mol}
$$ - Calculate the mass of $ \text{CO}_2 $:
$$
\text{Mass}(\text{CO}_2) = n \times \text{Molar Mass} = 1.00 \, \text{mol} \times 44.01 \, \text{g mol}^{-1} = 44.01 \, \text{g}
$$ - Answer: 44.01 g of carbon dioxide is produced.
2. Calculating Volumes of Gases
Avogadro's law
Avogadro's law states that equal volumes of all gases, measured under the same conditions of temperature and pressure, contain equal numbers of molecules.
At STP (273 K and 100 kPa), 1 mole of any gas occupies $22.7 \ \text{dm}^3$. This is invaluable for gas-related calculations.
- Many students forget to convert gas volumes to moles using the molar volume of 22.7 $\text{dm}^3 \, \text{mol}^{-1}$.
- Always check whether your problem involves gases at STP!
What volume of oxygen gas is required to completely combust 2.00 mol of propane $\text{C}_3\text{H}_8 $?
Solution
- Write the balanced equation:
$$
\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g)
$$ - Use the mole ratio $ \text{C}_3\text{H}_8 : \text{O}_2 = 1 : 5 $ to find the moles of $ \text{O}_2 $:
$$
n(\text{O}_2) = 2.00 \, \text{mol} \times 5 = 10.00 \, \text{mol}
$$ - Calculate the volume of $ \text{O}_2 $ at STP:
$$
V(\text{O}_2) = n \times 22.7 \, \text{dm}^3\ \text{mol}^{-1} = 10.00 \, \text{mol} \times 22.7 \, \text{dm}^3\ \text{mol}^{-1} = 227 \, \text{dm}^3
$$ - Answer: $227 \, \text{dm}^3$ of oxygen gas is required.
3. Reactions in Solution: Concentration and Volume
In solution-based reactions, the relationship between concentration $ C $, moles $n $, and volume $V $ is given by:
$$
C = \frac{n}{V}
$$
This formula is especially useful for titration calculations.
A 25.0 $\text{cm}^3$ solution of 0.100 $\text{mol dm}^3$ hydrochloric acid $\text{HCl} $ reacts with sodium hydroxide $\text{NaOH} $ according to:
$$
\text{HCl}(aq) + \text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)
$$
What volume of 0.200 $\text{mol dm}^3$ $ \text{NaOH} $ is required for complete neutralization?
Solution
- Calculate moles of $\text{HCl} $:
$$
n(\text{HCl}) = C \times V = 0.100 \, \text{mol dm}^{-3} \times 0.0250 \, \text{dm}^3 = 0.00250 \, \text{mol}
$$ - Use the mole ratio $ \text{HCl} : \text{NaOH} = 1 : 1 $ to find moles of $\text{NaOH} $:
$$
n(\text{NaOH}) = 0.00250 \, \text{mol}
$$ - Calculate the volume of $ \text{NaOH} $:
$$
V(\text{NaOH}) = \frac{n}{C} = \frac{0.00250 \, \text{mol}}{0.200 \, \text{mol dm}^{-3}} = 0.0125 \, \text{dm}^3 = 12.5 \, \text{cm}^3
$$ - Answer: 12.5 $\text{cm}^3$ of $\text{NaOH}$ is required.
When working with solutions, always ensure volumes are in $\text{dm}^3$ for consistency with $\text{mol dm}^3$ units.
