4. (NH_(4))_(3)PO_(4)=underline ( )g % N=underline ( ) % H=underline ( ) % P=underline ( ) % O=underline ( )

To find the percentage composition of each element in $(NH_{4})_{3}PO_{4}$, we need to calculate the molar mass of the compound and then determine the mass of each element as a percentage of the total molar mass.<br /><br />The molar mass of $(NH_{4})_{3}PO_{4}$ is calculated as follows:<br />- Nitrogen (N): 14.01 g/mol × 3 = 42.03 g/mol<br />- Hydrogen (H): 1.01 g/mol × 12 = 12.12 g/mol<br />- Phosphorus (P): 30.97 g/mol × 1 = 30.97 g/mol<br />- Oxygen (O): 16.00 g/mol × 4 = 64.00 g/mol<br /><br />Total molar mass = 42.03 + 12.12 + 30.97 + 64.00 = 149.12 g/mol<br /><br />Now, we can calculate the percentage composition of each element:<br /><br />- Nitrogen (N): (42.03 g/mol / 149.12 g/mol) × 100% = 28.21%<br />- Hydrogen (H): (12.12 g/mol / 149.12 g/mol) × 100% = 8.15%<br />- Phosphorus (P): (30.97 g/mol / 149.12 g/mol) × 100% = 20.88%<br />- Oxygen (O): (64.00 g/mol / 149.12 g/mol) × 100% = 42.86%<br /><br />Therefore, the percentage composition of each element in $(NH_{4})_{3}PO_{4}$ is as follows:<br />- Nitrogen (N): 28.21%<br />- Hydrogen (H): 8.15%<br />- Phosphorus (P): 20.88%<br />- Oxygen (O): 42.86%