In basic stoichiometry, what is the gram formula mass (molar mass) of ammonium phosphate, (NH4)3PO4, expressed in grams per mole?

Introduction / Context:
Calculating gram formula mass, also called molar mass, is a fundamental skill in chemistry that underlies stoichiometric calculations. The gram formula mass of a compound tells you the mass of one mole of its formula units. Ammonium phosphate, written as (NH4)3PO4, is a common example used in problems related to fertilisers and inorganic salts. This question tests whether you can correctly interpret the formula and sum the atomic masses to obtain the correct gram formula mass in grams per mole.

Given Data / Assumptions:
- The chemical formula is (NH4)3PO4, ammonium phosphate. - Approximate atomic masses used are: N about 14 g per mol, H about 1 g per mol, P about 31 g per mol, and O about 16 g per mol. - There are three ammonium ions NH4 plus and one phosphate group PO4 3 minus in the formula unit. - We assume standard rounding of atomic masses to two decimal places where needed.

Concept / Approach:
To find the gram formula mass of a compound, multiply the atomic mass of each element by the number of atoms of that element in the formula and then add all contributions. For (NH4)3PO4, we first count how many nitrogen, hydrogen, phosphorus, and oxygen atoms are present. Then we use approximate atomic masses for each element. The total gives the molar mass of one mole of ammonium phosphate formula units, which we can compare with the options given to select the closest value.

Step-by-Step Solution:
Step 1: Expand the formula (NH4)3PO4. The subscript 3 applies to NH4, so there are 3 nitrogen atoms and 12 hydrogen atoms from the ammonium ions. Step 2: Count atoms of each element. There are 3 N atoms, 12 H atoms, 1 P atom, and 4 O atoms in one formula unit of (NH4)3PO4. Step 3: Use approximate atomic masses: nitrogen about 14.01 g per mol, hydrogen about 1.008 g per mol, phosphorus about 30.97 g per mol, and oxygen about 16.00 g per mol. Step 4: Calculate the total mass contribution of each element: N: 3 * 14.01 ≈ 42.03 g per mol; H: 12 * 1.008 ≈ 12.10 g per mol; P: 1 * 30.97 ≈ 30.97 g per mol; O: 4 * 16.00 = 64.00 g per mol. Step 5: Add all contributions. Total mass ≈ 42.03 + 12.10 + 30.97 + 64.00 = 149.10 g per mol (rounded, about 149.08 g per mol). Step 6: Compare with options and choose 149.08 g per mol as the closest and correct gram formula mass.

Verification / Alternative check:
As a quick check, you can round atomic masses more roughly: N ≈ 14, H ≈ 1, P ≈ 31, O ≈ 16. Then the mass is 3 * 14 + 12 * 1 + 31 + 4 * 16 = 42 + 12 + 31 + 64 = 149 g per mol. This back of the envelope calculation agrees closely with the more precise value and confirms that a result near 149 g per mol is correct. Since 149.08 g per mol is the only option near 149, it must be the correct answer.

Why Other Options Are Wrong:
Option A (139.54 g per mol) is too low and would imply missing several oxygen or nitrogen atoms in the calculation. Option B (140.70 g per mol) is also significantly below the correct value and does not match the sum of the individual atomic contributions. Option D (153.24 g per mol) is higher than the calculated value and would correspond to counting extra atoms or using incorrect atomic masses.

Common Pitfalls:
Students sometimes forget to apply the outer subscript 3 to both N and H in NH4, leading to incorrect counts of nitrogen and hydrogen atoms. Another common mistake is to misread the formula as NH4 3PO4 without recognising the grouping, which changes the stoichiometry. Using inaccurate or rounded atomic masses without checking the result can also introduce large errors. Always carefully count each atom in the formula, multiply by the proper atomic mass, and then sum the contributions to get the correct gram formula mass.

Final Answer:
The correct answer is: 149.08 g per mol.