Stoichiometric Calculations (Grade 11 NSC Matric Physical Sciences): Revision Notes

Worked Example: Determining Limiting Reagents

Question: Sulfuric acid (H₂SO₄) reacts with ammonia (NH₃) to produce ammonium sulfate ((NH₄)₂SO₄):

H₂SO₄(aq) + 2NH₃(g) → (NH₄)₂SO₄(aq)

What is the maximum mass of ammonium sulfate that can be obtained from 2,0 kg of sulfuric acid and 1,0 kg of ammonia?

Solution:

Step 1: Convert masses to moles

For sulfuric acid: $n = \frac{m}{M} = \frac{2000}{98} = 20,4 \text{ mol}$

For ammonia: $n = \frac{m}{M} = \frac{1000}{17} = 58,8 \text{ mol}$

Step 2: Determine the limiting reagent

Look at how many moles of product each reactant can produce:

From H₂SO₄: The mole ratio is 1:1, so 20,4 mol H₂SO₄ can produce 20,4 mol (NH₄)₂SO₄

From NH₃: The mole ratio is 2:1, so 58,8 mol NH₃ can produce $58,8 ÷ 2 = 29,4$ mol (NH₄)₂SO₄

Since H₂SO₄ produces the smaller amount (20,4 mol), sulfuric acid is the limiting reagent.

Step 3: Calculate the maximum mass of product

Using the limiting reagent result: $m = nM = (20,4)(132) = 2692,8 \text{ g} = 2,69 \text{ kg}$

The maximum mass of ammonium sulfate that can be produced is 2,69 kg.

Worked Example: Calculating Percent Yield

Question: Using the same reaction as before, a factory worker gets 2,5 kg of ammonium sulfate from 2,0 kg of sulfuric acid and 1,0 kg of ammonia. What is the percentage yield?

Solution:

Step 1: Identify the limiting reagent and theoretical yield

From the previous example, we know sulfuric acid is limiting and the theoretical yield is 2,69 kg.

Step 2: Apply the percent yield formula

$\text{\%yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$

$\text{\%yield} = \frac{2,5}{2,694} \times 100 = 92,8\%$

This reaction has a high percent yield (92,8%), making it useful for industrial purposes.

Worked Example: Determining Empirical and Molecular Formulae

Question: Acetic acid has the following percentage composition: 39,9% carbon, 6,7% hydrogen, and 53,4% oxygen. The molar mass is 60,06 g·mol⁻¹. Find the empirical and molecular formulae.

Solution:

Step 1: Find the mass of each element in 100 g In 100 g of acetic acid: 39,9 g C, 6,7 g H, 53,4 g O

Step 2: Convert masses to moles $n = \frac{m}{M}$

$n_C = \frac{39,9}{12} = 3,325 \text{ mol}$

$n_H = \frac{6,7}{1,01} = 6,6337 \text{ mol}$

$n_O = \frac{53,4}{16} = 3,3375 \text{ mol}$

Step 3: Find the simplest ratio

Divide each by the smallest number (3,325):

• C: $3,325 ÷ 3,325 = 1$
• H: $6,6337 ÷ 3,325 = 2$
• O: $3,3375 ÷ 3,325 = 1$

Empirical formula: CH₂O

Step 4: Find the molecular formula

Empirical formula mass = $12 + 2(1) + 16 = 30,02 { g·mol}^{-1}$

Molecular formula factor = $60,06 ÷ 30,02 = 2$

Molecular formula: C₂H₄O₂ (or CH₃COOH)

Worked Example: Basic Percent Purity Calculation

Question: A shell weighs 5 g. After experiments, the mass of calcium carbonate and crucible is 3,2 g, and the crucible alone is 0,5 g. How much calcium carbonate is in the shell?

Solution:

Step 1: Find the mass of the product Mass of CaCO₃ = $3,2 - 0,5 = 2,7 \text{ g}$

Step 2: Calculate percent purity

$\text{\%purity} = \frac{\text{mass of compound}}{\text{mass of sample}} \times 100$

$\text{\%purity} = \frac{2,7}{5} \times 100 = \text{54\%}$

Worked Example: Percent Purity Through Chemical Reaction

Question: Jake adds HCl to a 3,5 g limestone sample. The reaction produces 3,6 g of CaCl₂. What is the percent purity?

CaCO₃(s) + 2HCl(aq) → CO₂(g) + CaCl₂(aq) + H₂O(l)

Solution:

Step 1: Calculate moles of CaCl₂ produced

$n = \frac{m}{M} = \frac{3,6}{111} = 0,032 \text{ mol}$

Step 2: Use stoichiometry to find moles of CaCO₃

From the equation, the ratio is 1:1, so moles of CaCO₃ = 0,032 mol

Step 3: Calculate mass of CaCO₃

$m = nM = (0,032)(100) = 3,24 \text{ g}$

Step 4: Calculate percent purity

$\text{\%purity} = \frac{3,3}{3,5} \times 100 = \text{94,3\%}$