How to Convert Moles to Grams

To convert moles to grams, multiply the number of moles by the molar mass of the substance. The formula is mass = n × M, where mass is in grams, n is the amount in moles, and M is the molar mass in g mol⁻¹. This is the single calculation you need to turn a mole value from a balanced equation or a reagent list into a mass you can actually weigh out on a lab balance. The most common failure is using the wrong molar mass, treating it as a fixed number for every substance when it must be calculated from the periodic table for each compound.

The formula mass = n × M is the direct reversal of the grams-to-moles conversion n = mass / M. You use it when a problem gives you moles and asks for a mass to weigh out. The molar mass M is the sum of the standard atomic weights (from the IUPAC CIAAW table) of every atom in the chemical formula. For water, H₂O, that is (2 × 1.008) + 15.999 = 18.015 g mol⁻¹. For sodium chloride, NaCl, it is 22.990 + 35.45 = 58.44 g mol⁻¹. These values come from the OpenStax Chemistry 2e discussion of formula mass in section 3.1, and the atomic weights are published annually by the IUPAC CIAAW at ciaaw.org. The 2024 table is the most recent as of this writing.

A student who writes 18.0 g mol⁻¹ instead of 18.015 g mol⁻¹ for water introduces a rounding error of 0.015 g per mole. Across a 0.5 mol reaction, that is 0.0075 g, a small error in high school, but one that compounds in a multi-step stoichiometry problem from OpenStax Chemistry 2e section 4.3.

Example 1: Carbon dioxide (CO₂). A reaction requires 2.50 mol of CO₂. The molar mass is 44.01 g mol⁻¹. Multiply: 2.50 mol × 44.01 g mol⁻¹ = 110.025 g. Report as 110. g, because the input 2.50 has three significant figures. The failure here is forgetting to check significant figures, the problem likely expects 110. g, not 110.025 g.

Example 2: Glucose (C₆H₁₂O₆). A lab needs 0.150 mol of glucose. Molar mass is 180.156 g mol⁻¹. 0.150 mol × 180.156 g mol⁻¹ = 27.0234 g. With three significant figures in the moles, round to 27.0 g. The error to watch for is using the atomic mass of carbon (12.011) without multiplying by six atoms.

Example 3: Iron (Fe). You have 3.00 mol of iron metal. Molar mass of Fe is 55.845 g mol⁻¹. 3.00 mol × 55.845 g mol⁻¹ = 167.535 g, rounded to 168 g. This is a simple element conversion, but the failure mode is using the atomic number (26) instead of the atomic weight (55.845), a mistake that changes the mass by a factor of two.

In lab work, you rarely work in whole moles. Reagent amounts are often in millimoles (mmol) or micromoles (µmol). The prefix conversion is straightforward: 1 mmol = 10⁻³ mol, 1 µmol = 10⁻⁶ mol. To convert mmol to mg, use the same formula mass = n × M but watch the units. If the amount is in mmol and the molar mass is in g mol⁻¹, the result is in milligrams because mmol × g mol⁻¹ = mg.

For example, to weigh out 5.00 mmol of sodium hydroxide (NaOH, M = 39.997 g mol⁻¹): 5.00 mmol × 39.997 g mol⁻¹ = 199.985 mg, rounded to 200. mg. The mistake is treating the input as moles instead of millimoles, which would give 199.985 g, a 1000× error. Always confirm the unit prefix before calculating.

When a reaction procedure calls for a specific number of moles of a solid reagent, you convert that to a mass to weigh on a balance. The sequence is: read the required moles from the reaction stoichiometry (using mole ratios from the balanced equation), look up or calculate the molar mass of the reagent, apply mass = n × M, and weigh that mass. The failure mode here is using the grams-to-moles formula in reverse, dividing by the molar mass instead of multiplying. If the procedure says "add 0.250 mol of NaCl" and you calculate 0.250 g / 58.44 g mol⁻¹ = 0.00428 g, you have added a tiny fraction of what the reaction needs.

The Single Thing That Most Often Goes Wrong

The most common error in moles-to-grams conversion is using the molar mass of an element when the problem involves a diatomic molecule. Oxygen gas is O₂, not O. Its molar mass is 31.998 g mol⁻¹, not 15.999 g mol⁻¹. A student who writes 15.999 g mol⁻¹ for O₂ will calculate half the required mass, and the reaction will not proceed as expected. Check the chemical formula before you multiply.