Calculate the percentage of each element in calcium carbonate (CaCO3) and ethanol (C2H5OH).

Here's the solution to question 24:

To calculate the percentage composition, we use the formula:

$$\text{Percentage of element} = \frac{\text{Total mass of element in compound}}{\text{Molar mass of compound}} \times 100\%$$

We will use the following atomic masses: $\text{H} = 1.0$, $\text{C} = 12.0$, $\text{N} = 14.0$, $\text{O} = 16.0$, $\text{Na} = 23.0$, $\text{Mg} = 24.3$, $\text{S} = 32.1$, $\text{Cl} = 35.5$, $\text{Ca} = 40.1$, $\text{Co} = 58.9$, $\text{Cu} = 63.5$.

a) Calculate the percentage of each element in the following compounds:

i) Calcium carbonate, $\text{CaCO}_3$

Step 1: Calculate the molar mass of $\text{CaCO}_3$.

$$\text{Molar mass of } \text{CaCO}_3 = 40.1 (\text{Ca}) + 12.0 (\text{C}) + (3 \times 16.0 (\text{O})) = 40.1 + 12.0 + 48.0 = 100.1 \text{ g/mol}$$

Step 2: Calculate the percentage of each element.

$$\text{Percentage of Ca} = \frac{40.1}{100.1} \times 100\% = 40.06\%$$ $$\text{Percentage of C} = \frac{12.0}{100.1} \times 100\% = 11.99\%$$ $$\text{Percentage of O} = \frac{48.0}{100.1} \times 100\% = 47.95\%$$

The percentages are: $\boxed{\text{Ca: 40.06\%}, \text{C: 11.99\%}, \text{O: 47.95\%}}$.

ii) Ethanol, $\text{C}_2\text{H}_5\text{OH}$ (or $\text{C}_2\text{H}_6\text{O}$)

Step 1: Calculate the molar mass of $\text{C}_2\text{H}_6\text{O}$.

$$\text{Molar mass of } \text{C}_2\text{H}_6\text{O} = (2 \times 12.0 (\text{C})) + (6 \times 1.0 (\text{H})) + 16.0 (\text{O}) = 24.0 + 6.0 + 16.0 = 46.0 \text{ g/mol}$$

Step 2: Calculate the percentage of each element.

$$\text{Percentage of C} = \frac{24.0}{46.0} \times 100\% = 52.17\%$$ $$\text{Percentage of H} = \frac{6.0}{46.0} \times 100\% = 13.04\%$$ $$\text{Percentage of O} = \frac{16.0}{46.0} \times 100\% = 34.78\%$$

The percentages are: $\boxed{\text{C: 52.17\%}, \text{H: 13.04\%}, \text{O: 34.78\%}}$.

iii) Hydrated copper sulphate, $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$

Step 1: Calculate the molar mass of $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$.

$$\text{Molar mass of } \text{CuSO}_4 = 63.5 (\text{Cu}) + 32.1 (\text{S}) + (4 \times 16.0 (\text{O})) = 63.5 + 32.1 + 64.0 = 159.6 \text{ g/mol}$$ $$\text{Molar mass of } 5\text{H}_2\text{O} = 5 \times ((2 \times 1.0 (\text{H})) + 16.0 (\text{O})) = 5 \times (2.0 + 16.0) = 5 \times 18.0 = 90.0 \text{ g/mol}$$ $$$$