Determine the percent composition for ammonium phosphate, (NH4)3PO4, a compound used as a fertilizer.

Step 1:Determine the molar mass of (NH4)3PO4
Step 2:Determine the percentages of each element present.

Step 1:

3 moles of N atoms= 3(14.007 g)= 42.021 g
12 moles of H atoms= 12(1.008 g)= 12.096 g
1 mole of P atoms= 1(30.9738 g)= 30.9738 g
4 moles of O atoms= 4(15.999 g)= 63.996 g
1 mole of (NH4)3PO4149.0868 g

Step 2:

Percent composition
N: [42.021 g N / 149.0868 g (NH4)3PO4] x 100%= 28.19 % N
H: [12.096 g H / 149.0868 g (NH4)3PO4] x 100%= 8.11 % H
P: [30.9738 g P / 149.0868 g (NH4)3PO4] x 100%= 20.78 % P
O: [63.996 g O / 149.0868 g (NH4)3PO4] x 100%= 42.93 % O

Note that the sum of the percentages is 100%