What is the relative formula mass of #(NH_4)_3PO_4#?

We're asked to calculate the formula mass of ammonium phosphate, #"(NH"_4)_3"PO"_4#.

To calculate the formula mass of any compound, we can multiply the relative atomic masses (or atomic weights or atomic masses) of each of the elements by by how many elements are in the compound, do it for each element, and sum the total.

You may need to refer to a periodic table to find the relative atomic masses of each element; nevertheless, here's how to do it (it's pretty straightforward):

First, let's multiply the atomic mass of nitrogen (#"N"#) by however many #"N"# atoms are in the compound, which is #3# (the subscript outside the ammonium species):

#3(14.01"amu") = 42.03"amu"#

("#"amu"#" stands for atomic mass units. The mass of a substance in #"amu"# is equivalent to the molar mass of that substance in #"g"/"mol"#. Thus, nitrogen has both an atomic mass of #14.01 "amu"# and a molar mass of #14.01"g"/"mol"#.)

Let's repeat this process for each of the elements in the compound, and sum the total:

#overbrace(42.03 "amu")^"N" + overbrace(12(1.01"amu"))^"H" + overbrace(1(30.97"amu"))^"P" + overbrace(4(16.00"amu"))^"O" = color(red)(149.12 "amu"#

Thus, the formula mass of #"(NH"_4)_3"PO"_4# is #149.12"amu"#.