Question

What is the percent composition by mass of sulfur in ammonium sulfate (NH4)2SO4 ?
Responses

32%
32%

6.7%
6.7%

28%
28%

24%

Answer 1

To find the percent composition by mass of sulfur in ammonium sulfate \((NH_4)_2SO_4\), we first need to determine the molar mass of the compound.

1. **Calculate the molar mass:**
- Nitrogen (N): 14.01 g/mol (2 N in \((NH_4)_2\) = 2 × 14.01 = 28.02 g/mol)
- Hydrogen (H): 1.01 g/mol (8 H in \((NH_4)_2\) = 8 × 1.01 = 8.08 g/mol)
- Sulfur (S): 32.07 g/mol (1 S)
- Oxygen (O): 16.00 g/mol (4 O in \((SO_4)\) = 4 × 16.00 = 64.00 g/mol)

Now, we will sum these values to get the total molar mass of \((NH_4)_2SO_4\):

\[
\text{Molar mass} = 28.02 + 8.08 + 32.07 + 64.00 = 132.17 \, \text{g/mol}
\]

2. **Find the mass contribution of sulfur:**
- The mass of sulfur in one mole of ammonium sulfate is 32.07 g.

3. **Calculate the percent composition by mass of sulfur:**
\[
\text{Percent composition of S} = \left(\frac{\text{mass of S}}{\text{molar mass of } (NH_4)_2SO_4}\right) \times 100
\]
\[
\text{Percent composition of S} = \left(\frac{32.07}{132.17}\right) \times 100 \approx 24.24\%
\]

Thus, the closest response is **24%**.