Determine the % composition of the element specified in the following: 1. % K in KMnO4 2. % P in (NH4)3PO4 3. % O in Al2(SO4)3.

Steps to Solve

1. Calculate Molar Mass of KMnO$_4$

To find the percentage of K in KMnO$_4$, we start by calculating the molar mass of the compound.

• Atomic mass of K (Potassium): 39.1 g/mol
• Atomic mass of Mn (Manganese): 54.9 g/mol
• Atomic mass of O (Oxygen): 16.0 g/mol

Using these values, the molar mass of KMnO$_4$ is calculated as follows:

$$ \text{Molar mass of KMnO}_4 = 39.1 + 54.9 + 4 \times 16.0 = 158.0 \text{ g/mol} $$

2. Calculate % K in KMnO$_4$

Next, we calculate the percentage of K in KMnO$_4$:

$$ \text{% K} = \left( \frac{\text{mass of K}}{\text{molar mass of KMnO}_4} \right) \times 100 $$

Substituting values:

$$ \text{% K} = \left( \frac{39.1}{158.0} \right) \times 100 \approx 24.77% $$

3. Calculate Molar Mass of (NH$_4$)$_3$PO$_4$

Now, we calculate the molar mass of (NH$_4$)$_3$PO$_4$:

• Atomic mass of N (Nitrogen): 14.0 g/mol
• Atomic mass of H (Hydrogen): 1.0 g/mol
• Atomic mass of P (Phosphorus): 31.0 g/mol

The molar mass of (NH$_4$)$_3$PO$_4$:

$$ \text{Molar mass of (NH}_4\text{)}_3\text{PO}_4 = 3(14.0 + 4 \times 1.0) + 31.0 + 4 \times 16.0 = 149.0 \text{ g/mol} $$

4. Calculate % P in (NH$_4$)$_3$PO$_4$

Next, for phosphorus:

$$ \text{% P} = \left( \frac{\text{mass of P}}{\text{molar mass of (NH}_4\text{)}_3\text{PO}_4} \right) \times 100 $$

Substituting values:

$$ \text{% P} = \left( \frac{31.0}{149.0} \right) \times 100 \approx 20.80% $$

5. Calculate Molar Mass of Al$_2$(SO$_4$)$_3$

Now, we calculate the molar mass of Al$_2$(SO$_4$)$_3$:

• Atomic mass of Al (Aluminum): 26.98 g/mol
• Atomic mass of S (Sulfur): 32.07 g/mol
• Atomic mass of O (Oxygen): 16.0 g/mol

Calculate the molar mass:

$$ \text{Molar mass of Al}_2\text{(SO}_4\text{)}_3 = 2(26.98) + 3(32.07 + 4 \times 16.0) = 342.15 \text{ g/mol} $$

6. Calculate % O in Al$_2$(SO$_4$)$_3$

Finally, for oxygen:

$$ \text{% O} = \left( \frac{\text{mass of O}}{\text{molar mass of Al}_2\text{(SO}_4\text{)}_3} \right) \times 100 $$

Since there are 12 O atoms in Al$_2$(SO$_4$)$_3$:

$$ \text{% O} = \left( \frac{12 \times 16.0}{342.15} \right) \times 100 \approx 56.31% $$