Quantitative Aspects of Chemical Change (Grade 10 NSC Matric Physical Sciences): Revision Notes

Worked Example: Calculating Percentage Composition

Question: Work out the percentage that each element contributes to the total mass of sulphuric acid (H₂SO₄).

Solution:

Step 1: Calculate the molar masses

• Hydrogen = $2 \times 1.01 = 2.02$ g·mol⁻¹
• Sulphur = $32.1$ g·mol⁻¹
• Oxygen = $4 \times 16.0 = 64.0$ g·mol⁻¹

Step 2: Find the molecular mass of sulphuric acid

• Mass = $2.02 + 32.1 + 64.0 = 98.12$ g·mol⁻¹

Step 3: Apply the percentage formula

For hydrogen: $\frac{2.02}{98.12} \times 100\% = 2.06\%$

For sulphur: $\frac{32.1}{98.12} \times 100\% = 32.72\%$

For oxygen: $\frac{64.0}{98.12} \times 100\% = 65.23\%$

Check: These percentages should add up to 100% $(2.06\% + 32.72\% + 65.23\% = 100.01\%$ ✓)

Worked Example: Finding Empirical Formula from Percentages

Question: A compound contains 52.2% carbon, 13.0% hydrogen and 34.8% oxygen. Find its empirical formula.

Solution:

Step 1: Convert percentages to masses

• In 100g of compound: C = 52.2g, H = 13.0g, O = 34.8g

Step 2: Calculate moles of each element

• Using $n = \frac{m}{M}$:

$n(\text{Carbon}) = \frac{52.2\text{g}}{12.0 { g·mol⁻¹}} = 4.35 \text{ mol}$

$n(\text{Hydrogen}) = \frac{13.0\text{g}}{1.01 { g·mol⁻¹}} = 12.87 \text{ mol}$

$n(\text{Oxygen}) = \frac{34.8\text{g}}{16.0 { g·mol⁻¹}} = 2.18 \text{ mol}$

Step 3: Find the simplest mole ratio

• Divide each by the smallest number of moles (2.18):

• Carbon: $4.35 \div 2.18 = 2$

• Hydrogen: $12.87 \div 2.18 = 6$

• Oxygen: $2.18 \div 2.18 = 1$

Therefore the empirical formula is C₂H₆O.

Worked Example: Determining Formula from Mass Data

Question: 207g of lead combines with oxygen to form 239g of lead oxide. Work out the formula of the lead oxide. (Relative atomic masses: Pb = 207.2 u, O = 16.0 u)

Solution:

Step 1: Find the mass of oxygen

• Mass of oxygen = $239\text{g} - 207\text{g} = 32\text{g}$

Step 2: Calculate moles of each element

• Using $n = \frac{m}{M}$:

$n(\text{Lead}) = \frac{207\text{g}}{207.2 { g·mol⁻¹}} = 1 \text{ mol}$

$n(\text{Oxygen}) = \frac{32\text{g}}{16.0 { g·mol⁻¹}} = 2 \text{ mol}$

Step 3: Determine the mole ratio

• The mole ratio of Pb : O is 1 : 2, which means for every lead atom there are two oxygen atoms.

Therefore the formula is PbO₂.

Worked Example: Finding Both Empirical and Molecular Formulas

Question: Vinegar contains acetic acid with this percentage composition: 39.9% carbon, 6.7% hydrogen and 53.4% oxygen. Find the empirical and molecular formulas if the molar mass is 60.06 g·mol⁻¹.

Solution:

Step 1: Find masses in a 100g sample

• C = 39.9g, H = 6.7g, O = 53.4g

Step 2: Calculate moles

• $n(\text{C}) = \frac{39.9\text{g}}{12.0 { g·mol⁻¹}} = 3.33 \text{ mol}$
• $n(\text{H}) = \frac{6.7\text{g}}{1.01 { g·mol⁻¹}} = 6.63 \text{ mol}$
• $n(\text{O}) = \frac{53.4\text{g}}{16.0 { g·mol⁻¹}} = 3.34 \text{ mol}$

Step 3: Find empirical formula

• Dividing by the smallest (3.33):
• C : H : O = 1 : 2 : 1
• Empirical formula = CH₂O

Step 4: Find molecular formula

• Empirical formula mass = $12.0 + 2.0 + 16.0 = 30.0$ g·mol⁻¹
• Since actual molar mass = 60.06 g·mol⁻¹
• Multiple = $\frac{60.06}{30.0} = 2$

Therefore molecular formula = C₂H₄O₂ (or CH₃COOH)