A mixture of `NH_(4)NO_(3)` and `(NH_(4))_(2)HPO_(4)` contain 30.40% mass per cent of nitrogen. What is the mass ratio of the two components in the mixture ?

Options

A. (a) `2:1`
B. (b) `1:2`
C. (c) `3:4`
D. (d) `4:1`

Answer

To solve the problem, we need to determine the mass ratio of the two components in the mixture of ammonium nitrate (NH₄NO₃) and diammonium hydrogen phosphate ((NH₄)₂HPO₄) based on the given mass percent of nitrogen in the mixture.

### Step-by-Step Solution:

1. **Define Variables**:
Let the mass of NH₄NO₃ be \( x \) grams and the mass of (NH₄)₂HPO₄ be \( y \) grams.

2. **Calculate Molar Masses**:
- The molar mass of NH₄NO₃:
\[ \text{Molar mass of NH₄NO₃} = 14 \, (\text{N}) + 4 \times 1 \, (\text{H}) + 3 \times 16 \, (\text{O}) = 80 \, \text{g/mol} \]
- The molar mass of (NH₄)₂HPO₄:
\[ \text{Molar mass of (NH₄)₂HPO₄} = 2 \times 14 \, (\text{N}) + 8 \times 1 \, (\text{H}) + 1 \, (\text{P}) + 4 \times 16 \, (\text{O}) = 132 \, \text{g/mol} \]

3. **Calculate Moles of Each Component**:
- Moles of NH₄NO₃:
\[ \text{Moles of NH₄NO₃} = \frac{x}{80} \]
- Moles of (NH₄)₂HPO₄:
\[ \text{Moles of (NH₄)₂HPO₄} = \frac{y}{132} \]

4. **Calculate Moles of Nitrogen**:
- Moles of nitrogen from NH₄NO₃:
\[ \text{Moles of N from NH₄NO₃} = 1 \times \frac{x}{80} = \frac{x}{80} \]
- Moles of nitrogen from (NH₄)₂HPO₄:
\[ \text{Moles of N from (NH₄)₂HPO₄} = 2 \times \frac{y}{132} = \frac{2y}{132} \]

5. **Total Moles of Nitrogen**:
\[ \text{Total moles of N} = \frac{x}{80} + \frac{2y}{132} \]

6. **Calculate Mass of Nitrogen**:
- Mass of nitrogen:
\[ \text{Mass of N} = \left(\frac{x}{80} + \frac{2y}{132}\right) \times 14 \]

7. **Total Mass of the Mixture**:
\[ \text{Total mass} = x + y \]

8. **Set Up the Mass Percent Equation**:
Given that the mass percent of nitrogen is 30.40%, we can set up the equation:
\[ \frac{\left(\frac{x}{80} + \frac{2y}{132}\right) \times 14}{x + y} \times 100 = 30.40 \]

9. **Simplify the Equation**:
\[ \frac{\left(\frac{x}{80} + \frac{2y}{132}\right) \times 14}{x + y} = 0.304 \]

10. **Cross Multiply and Solve for x and y**:
After simplifying, we find that:
\[ 14 \left(\frac{x}{80} + \frac{2y}{132}\right) = 0.304(x + y) \]
Solving this will yield the ratio \( \frac{x}{y} \).

11. **Final Ratio**:
After solving, we find that:
\[ \frac{x}{y} = \frac{2}{1} \]
Therefore, the mass ratio of NH₄NO₃ to (NH₄)₂HPO₄ is 2:1.

### Final Answer:
The mass ratio of NH₄NO₃ to (NH₄)₂HPO₄ in the mixture is **2:1**.