Question

calculate the percent nitrogen in ammonium hexacyanoiron

Answer 1

To calculate the percent nitrogen in ammonium hexacyanoiron, we first need to write its chemical formula. Ammonium hexacyanoiron can be represented as \([ \text{NH}_4 ] [ \text{Fe}(\text{CN})_6 ]\).

1. **Determine the formula mass** of ammonium hexacyanoiron:

- Ammonium ion (NH₄⁺):
- N: 1 × 14.01 g/mol = 14.01 g/mol
- H: 4 × 1.01 g/mol = 4.04 g/mol
- Total for NH₄⁺ = 14.01 + 4.04 = 18.05 g/mol

- Hexacyanoiron:
- Fe: 1 × 55.85 g/mol = 55.85 g/mol
- C: 6 × 12.01 g/mol = 72.06 g/mol
- N (from cyanides): 6 × 14.01 g/mol = 84.06 g/mol
- Total for \(\text{Fe(CN)}_6\) = 55.85 + 72.06 + 84.06 = 211.97 g/mol

2. **Combine the masses** to get the total molar mass of ammonium hexacyanoiron:

\[
\text{Total molar mass} = 18.05 \, \text{g/mol (NH₄⁺)} + 211.97 \, \text{g/mol (Fe(CN)₆)} = 230.02 \, \text{g/mol}
\]

3. **Determine the total mass of nitrogen** in the entire compound:

- There are 1 nitrogen from the ammonium ion and 6 nitrogen atoms from the cyanide groups:
\[
\text{Total N} = 1 + 6 = 7
\]
- Total mass of nitrogen:
\[
7 \, \text{N atoms} \times 14.01 \, \text{g/mol} = 98.07 \, \text{g/mol}
\]

4. **Calculate the percent nitrogen** in ammonium hexacyanoiron:

\[
\text{Percent nitrogen} = \left(\frac{\text{Total mass of nitrogen}}{\text{Total molar mass}} \right) \times 100
\]
\[
\text{Percent nitrogen} = \left(\frac{98.07}{230.02}\right) \times 100 \approx 42.7\%
\]

Thus, the percent nitrogen in ammonium hexacyanoiron is approximately **42.7%**.