The per cent of `N "in" 66%` pure `(NH_(4))_(2)SO_(4)` sample is:

Options

A. `32`
B. `28`
C. `14`
D. none of these

Explanation

To find the percentage of nitrogen in a 66% pure sample of ammonium sulfate \((NH_4)_2SO_4\), we can follow these steps:

### Step 1: Calculate the molar mass of \((NH_4)_2SO_4\)
The molar mass can be calculated as follows:
- Nitrogen (N): 14 g/mol (2 nitrogen atoms in the formula)
- Hydrogen (H): 1 g/mol (8 hydrogen atoms in the formula)
- Sulfur (S): 32 g/mol (1 sulfur atom in the formula)
- Oxygen (O): 16 g/mol (4 oxygen atoms in the formula)

Calculating the total:
\[ \text{Molar mass of } (NH_4)_2SO_4 = (2 \times 14) + (8 \times 1) + (32) + (4 \times 16) = 28 + 8 + 32 + 64 = 132 \text{ g/mol} \]

### Step 2: Determine the mass of nitrogen in 132 g of \((NH_4)_2SO_4\)
From the formula, there are 2 nitrogen atoms, contributing a total of:
\[ \text{Mass of nitrogen} = 2 \times 14 = 28 \text{ g} \]

### Step 3: Calculate the mass of \((NH_4)_2SO_4\) in the 66% pure sample
If we have 132 g of \((NH_4)_2SO_4\) and it is 66% pure, the actual mass of the pure compound in the sample is:
\[ \text{Mass of pure } (NH_4)_2SO_4 = 132 \text{ g} \times 0.66 = 87.12 \text{ g} \]

### Step 4: Calculate the mass of nitrogen in the pure sample
Now, we need to find out how much nitrogen is present in the 66% pure sample:
\[ \text{Mass of nitrogen in the sample} = \left(\frac{28 \text{ g of N}}{132 \text{ g of } (NH_4)_2SO_4}\right) \times 87.12 \text{ g} = \frac{28}{132} \times 87.12 \approx 18.00 \text{ g} \]

### Step 5: Calculate the percentage of nitrogen in the sample
Finally, to find the percentage of nitrogen in the 66% pure sample:
\[ \text{Percentage of nitrogen} = \left(\frac{\text{Mass of nitrogen}}{\text{Mass of pure } (NH_4)_2SO_4}\right) \times 100 = \left(\frac{18.00}{87.12}\right) \times 100 \approx 20.65\% \]

Thus, the percentage of nitrogen in the 66% pure \((NH_4)_2SO_4\) sample is approximately **20.65%**.