Calculating Mass from Moles

Calculating mass from moles is the reverse of the more common moles-from-mass step. It is typically the final step in a multi-step calculation: you have already found the moles of your target substance using the mole ratio, and now you need to convert back to a mass the question can mark.

The method

1. Identify the number of moles of the substance (given or calculated from a previous step).
2. Calculate or look up the Mr of the substance.
3. Multiply the moles by Mr to get the mass in grams.

The formula

mass = n x Mr

Where:

• mass = mass of the substance in grams (g)
• n = number of moles (mol)
• Mr = relative formula mass

Worked examples

Example 1: Mass of water from moles

Question: Calculate the mass of 3.0 mol of water, H2O. (Ar: H = 1, O = 16)

Working:

• Mr of H2O = (2 x 1) + 16 = 18
• mass = n x Mr = 3.0 x 18 = 54 g

Answer: 54 g

Examiner note: Straightforward application. One mark for Mr, one for the final answer.

Example 2: Mass of carbon dioxide produced

Question: A reaction produces 0.15 mol of carbon dioxide, CO2. Calculate the mass of carbon dioxide produced. (Ar: C = 12, O = 16)

Working:

• Mr of CO2 = 12 + (2 x 16) = 44
• mass = n x Mr = 0.15 x 44 = 6.6 g

Answer: 6.6 g

Examiner note: This often appears as the final step of a reacting masses question. Make sure you are using the Mr of the product asked about, not the reactant.

Example 3: Mass of copper deposited in electrolysis

Question: During electrolysis, 0.025 mol of copper is deposited at the cathode. Calculate the mass of copper deposited. (Ar: Cu = 64)

Working:

• Ar of Cu = 64
• mass = n x Ar = 0.025 x 64 = 1.6 g

Answer: 1.6 g

Examiner note: Copper is an element, so use Ar directly. This is a common final step in electrolysis calculations.

Example 4: Mass of sodium hydroxide

Question: Calculate the mass of 0.40 mol of sodium hydroxide, NaOH. (Ar: Na = 23, O = 16, H = 1)

Working:

• Mr of NaOH = 23 + 16 + 1 = 40
• mass = n x Mr = 0.40 x 40 = 16 g

Answer: 16 g

Examiner note: When the question gives you the moles directly, the calculation is two steps: Mr then multiply. Do not overcomplicate it.

Example 5: Mass of hydrated salt

Question: Calculate the mass of 0.10 mol of hydrated copper sulfate, CuSO4.5H2O. (Ar: Cu = 64, S = 32, O = 16, H = 1)

Working:

• Mr of CuSO4.5H2O = 64 + 32 + (4 x 16) + 5[(2 x 1) + 16] = 64 + 32 + 64 + 5(18) = 64 + 32 + 64 + 90 = 250
• mass = n x Mr = 0.10 x 250 = 25 g

Answer: 25 g

Examiner note: You must include the water of crystallisation in the Mr. Anhydrous CuSO4 has Mr = 160, which would give a wrong answer of 16 g. Read the formula carefully.

Example 6: Answer required in kilograms

Question: An industrial process produces 500 mol of ammonia, NH3. Calculate the mass of ammonia produced. Give your answer in kilograms. (Ar: N = 14, H = 1)

Working:

• Mr of NH3 = 14 + (3 x 1) = 17
• mass = n x Mr = 500 x 17 = 8500 g
• Convert to kg: 8500 / 1000 = 8.5 kg

Answer: 8.5 kg

Examiner note: The question specifies kilograms. Giving 8500 g when the question asks for kg may lose the final answer mark. Always check the required unit.

Common mistakes

1. Dividing instead of multiplying. mass = n x Mr, not n / Mr. If your answer seems far too small, check the operation.
2. Using the wrong Mr. In multi-step problems, make sure you are using the Mr of the substance the question asks about, not the substance from a previous step.
3. Ignoring water of crystallisation. If the formula shows .xH2O, include those water molecules in Mr.
4. Not converting units when asked. If the question requests kilograms, divide grams by 1000.
5. Dropping significant figures. Keep at least 3 significant figures to match exam expectations.

When this appears

This calculation appears as the final step in reacting mass calculations and titration calculations. It is the reverse of moles from mass. You also use it when converting moles found from concentration calculations or gas volume calculations back into a mass. The full theory is in the stoichiometry topic.