Q9. A compound contains 46.7% of silicon (Si) and 53.3% oxygen (O) by mass. Find the empirical formula of the compound, given the relative atomic mass of silicon (Si) = 28 and oxygen (O) = 16. Step 1: Assume a 100 g sample. This means we have 46.7 g of Si and 53.3 g of O. Step 2: Convert mass to moles for each element. $$ \text{Moles of Si} = \frac{\text{Mass of Si}}{\text{Relative atomic mass of Si}} = \frac{46.7 \text{ g}}{28 \text{ g/mol}} \approx 1.6679 \text{ mol} \\ \text{Moles of O} = \frac{\text{Mass of O}}{\text{Relative atomic mass of O}} = \frac{53.3 \text{ g}}{16 \text{ g/mol}} \approx 3.3313 \text{ mol} $$ Step 3: Find the simplest whole number ratio of moles. Divide each number of moles by the smallest number of moles (1.6679 mol). $$ \text{Ratio of Si} = \frac{1.6679 \text{ mol}}{1.6679 \text{ mol}} = 1 \\ \text{Ratio of O} = \frac{3.3313 \text{ mol}}{1.6679 \text{ mol}} \approx 1.997 \approx 2 $$ The simplest whole number ratio of Si:O is 1:2. Step 4: Write the empirical formula. The empirical formula is $\boxed{\text{SiO}_2}$. --- Q10. Caffeine is a compound found in coffee and tea. The percentage composition of caffeine is 49.1% carbon, 5.1% hydrogen, 16.5% oxygen and 28.9% nitrogen. The relative molecular mass of caffeine is 195. Find the molecular formula of caffeine, given the atomic mass of carbon (C) = 12, hydrogen (H) = 1, oxygen (O) = 16, and nitrogen (N) = 14. Step 1: Assume a 100 g sample and convert percentages to masses. Mass of C = 49.1 g Mass of H = 5.1 g Mass of O = 16.5 g Mass of N = 28.9 g Step 2: Convert mass to moles for each element. $$ \text{Moles of C} = \frac{49.1 \text{ g}}{12 \text{ g/mol}} \approx 4.0917 \text{ mol} \\ \text{Moles of H} = \frac{5.1 \text{ g}}{1 \text{ g/mol}} = 5.1 \text{ mol} \\ \text{Moles of O} = \frac{16.5 \text{ g}}{16 \text{ g/mol}} \approx 1.0313 \text{ mol} \\ \text{Moles of N} = \frac{28.9 \text{ g}}{14 \text{ g/mol}} \approx 2.0643 \text{ mol} $$ Step 3: Find the simplest whole number ratio of moles. Divide each number of moles by the smallest number of moles (1.0313 mol for O). $$ \text{Ratio of C} = \frac{4.0917 \text{ mol}}{1.0313 \text{ mol}} \approx 3.967 \approx 4 \\ \text{Ratio of H} = \frac{5.1 \text{ mol}}{1.0313 \text{ mol}} \approx 4.945 \approx 5 \\ \text{Ratio of O} = \frac{1.0313 \text{ mol}}{1.0313 \text{ mol}} = 1 \\ \text{Ratio of N} = \frac{2.0643 \text{ mol}}{1.0313 \text{ mol}} \approx 2.001 \approx 2 $$ The simplest whole number ratio of C:H:O:N is 4:5:1:2. Step 4: Write the empirical formula. The empirical formula is $\text{C}_4\text{H}_5\text{ON}_2$. Step 5: Calculate the empirical formula mass. $$ \text{Empirical formula mass} = (4 \times 12) + (5 \times 1) + (1 \times 16) + (2 \times 14) \\ = 48 + 5 + 16 + 28 \\ = 97 \text{ g/mol} $$ Step 6: Determine the factor 'n'. Given relative molecular mass = 195. $$ n = \frac{\text{Relative molecular mass}}{\text{Empirical formula mass}} = \frac{195}{97} \approx 2.01 \approx 2 $$ Step 7: Write the molecular formula. Multiply the subscripts in the empirical formula by n=2. $$ \text{Molecular formula} = (\text{C}_4\text{H}_5\text{ON}_2)_2 = \boxed{\text{C}_8\text{H}_{10}\text{O}_2\text{N}_4} $$