Quantitative Chemistry for the ESAT

Stoichiometry and Molar Quantities

A balanced chemical equation provides the exact molar quantities of the substances reacting and the products formed. This relationship is known as the stoichiometry of the reaction. For example, consider the reaction of magnesium with oxygen:

$Mg(s) + \frac{1}{2}O_2(g) \rightarrow MgO(s)$

This equation states that 1 mol of $Mg$ reacts with 0.5 mol of $O_2$ to produce 1 mol of $MgO$. These molar ratios allow us to calculate the amount of any substance in the reaction if the amount of one substance is known. To perform these calculations, we use the relationship between mass, molar mass, and moles:

$\text{Amount (mol)} = \frac{\text{mass (g)}}{\text{molar mass (g mol}^{-1}\text{)}}$

Worked Example: Calculating Product Mass

Question: What is the maximum mass of magnesium oxide that can be formed if $0.12\text{ g}$ of magnesium is completely burned in excess oxygen? ($A_r$ values: $O = 16$; $Mg = 24$)

1. Find the moles of the known substance: $n(Mg) = \frac{0.12}{24} = 0.005\text{ mol}$

2. Use the molar ratio from the equation: The ratio of $Mg$ to $MgO$ is 1:1. Therefore, $0.005\text{ mol}$ of $Mg$ will produce $0.005\text{ mol}$ of $MgO$.

3. Convert moles back to mass: $M_r(MgO) = 24 + 16 = 40$ $\text{Mass of } MgO = 0.005 \times 40 = 0.20\text{ g}$

The Limiting Reactant

In many reactions, the reactants are not mixed in the exact stoichiometric ratio. The limiting reactant is the substance that is completely used up first, thereby determining the maximum amount of product that can be formed. Other reactants are said to be in excess.

Worked Example: Identifying the Limiting Reactant

Question: In the manufacture of calcium carbide: $CaO(s) + 3C(s) \rightarrow CaC_2(s) + CO(g)$. What is the maximum mass of calcium carbide that can be obtained from $11.2\text{ kg}$ of calcium oxide and $11.2\text{ kg}$ of carbon? ($A_r$ values: $C = 12.0; O = 16.0; Ca = 40.0$)

1. Convert given masses to moles: $M_r(CaO) = 40.0 + 16.0 = 56.0$ $n(CaO) = \frac{11200\text{ g}}{56.0} = 200\text{ mol}$ $n(C) = \frac{11200\text{ g}}{12.0} = 933.3\text{ mol}$

2. Determine the limiting reactant: According to the equation, 1 mol of $CaO$ requires 3 mol of $C$. Therefore, $200\text{ mol}$ of $CaO$ would require $200 \times 3 = 600\text{ mol}$ of $C$. Since we have $933.3\text{ mol}$ of $C$, the $C$ is in excess and $CaO$ is the limiting reactant.

3. Calculate the mass of product based on the limiting reactant: The ratio of $CaO$ to $CaC_2$ is 1:1. So, $200\text{ mol}$ of $CaO$ forms $200\text{ mol}$ of $CaC_2$. $M_r(CaC_2) = 40.0 + (2 \times 12.0) = 64.0$ $\text{Mass of } CaC_2 = 200 \times 64.0 = 12800\text{ g} = 12.8\text{ kg}$

Reacting Gas Volumes

For an ideal gas, the volume of the gas molecules is considered negligible compared to the total volume occupied. Consequently, the volume of a gas depends only on temperature and pressure, not on the identity of the gas. Equal volumes of gases under the same conditions contain the same number of moles.

Molar Gas Volume

At room temperature and pressure (rtp), one mole of any gas occupies a volume of $24\text{ dm}^3$ (or $24000\text{ cm}^3$). The relationship is:

$\text{Number of moles of gas} = \frac{\text{volume}}{\text{molar volume}}$

Worked Example: Gas Volume from Mass

Question: What is the volume of hydrogen obtained when $3.25\text{ g}$ of zinc reacts with an excess of dilute sulfuric acid at rtp? ($A_r$ value: $Zn = 65$; Molar volume = $24.0\text{ dm}^3$)

1. Write the balanced equation: $Zn(s) + H_2SO_4(aq) \rightarrow ZnSO_4(aq) + H_2(g)$

2. Find moles of Zn: $n(Zn) = \frac{3.25}{65} = 0.050\text{ mol}$

3. Determine moles of gas: Ratio of $Zn$ to $H_2$ is 1:1, so $0.050\text{ mol}$ of $H_2$ is produced.

4. Calculate volume: $\text{Volume} = 0.050 \times 24.0 = 1.20\text{ dm}^3$

Constructing Equations from Data

Balanced equations can be deduced from experimental data by finding the simplest molar ratio of the reacting substances.

Example: Constructing a Combustion Equation

Question: $100\text{ cm}^3$ of a gaseous alkane required $500\text{ cm}^3$ of oxygen for complete combustion. What is the equation? (All volumes at the same conditions).

Since volumes of gases are proportional to moles, the ratio of alkane to $O_2$ is $100:500$, or $1:5$. For an alkane $C_nH_{2n+2}$, the balanced equation for oxygen atoms is: $2 \times (\text{moles of } O_2) = (2 \times n) + (n + 1)$ $10 = 2n + n + 1$ $9 = 3n$, so $n = 3$. The alkane is propane ($C_3H_8$). Equation: $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$