1. Determine the molar mass of the compound:
* N: 14.01 g/mol (x 2) = 28.02 g/mol
* H: 1.01 g/mol (x 9) = 9.09 g/mol
* P: 30.97 g/mol
* O: 16.00 g/mol (x 4) = 64.00 g/mol
Total molar mass of (NH₄)₂HPO₄ = 28.02 + 9.09 + 30.97 + 64.00 = 132.08 g/mol
2. Calculate the mass of nitrogen in one mole of the compound:
* Mass of nitrogen = 28.02 g/mol
3. Calculate the percent composition of nitrogen:
* Percent composition of nitrogen = (mass of nitrogen / molar mass of compound) x 100%
* Percent composition of nitrogen = (28.02 g/mol / 132.08 g/mol) x 100%
* Percent composition of nitrogen ≈ 21.22%
Therefore, the percent composition of nitrogen in (NH₄)₂HPO₄ is approximately 21.22%.
