Questions: What is the % composition of (NH4)3P?

Solution Steps

First, we need to find the molar mass of each element in the compound \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{P}\).

• Nitrogen (N): \(14.007 \, \text{g/mol}\)
• Hydrogen (H): \(1.008 \, \text{g/mol}\)
• Phosphorus (P): \(30.974 \, \text{g/mol}\)

Next, we calculate the total molar mass of \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{P}\).

• There are 3 ammonium ions \((\mathrm{NH}_{4})\) in the compound.
• Each ammonium ion contains 1 nitrogen and 4 hydrogens.

So, the molar mass of \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{P}\) is: \[ 3 \times (14.007 \, \text{g/mol} + 4 \times 1.008 \, \text{g/mol}) + 30.974 \, \text{g/mol} \] \[ = 3 \times (14.007 + 4.032) + 30.974 \] \[ = 3 \times 18.039 + 30.974 \] \[ = 54.117 + 30.974 \] \[ = 85.091 \, \text{g/mol} \]

Now, we calculate the mass contribution of each element in the compound.

• Nitrogen: \(3 \times 14.007 \, \text{g/mol} = 42.021 \, \text{g/mol}\)
• Hydrogen: \(3 \times 4 \times 1.008 \, \text{g/mol} = 12.096 \, \text{g/mol}\)
• Phosphorus: \(30.974 \, \text{g/mol}\)

Finally, we calculate the percent composition of each element by dividing the mass contribution of each element by the total molar mass and multiplying by 100.

• Percent composition of Nitrogen: \[ \frac{42.021}{85.091} \times 100 \approx 49.37\% \]

• Percent composition of Hydrogen: \[ \frac{12.096}{85.091} \times 100 \approx 14.21\% \]

• Percent composition of Phosphorus: \[ \frac{30.974}{85.091} \times 100 \approx 36.42\% \]

Final Answer