Quantities of chemicals are measured in grams, but the amounts that react based on a chemical reaction are expressed in moles according to the stoichiometry of the equation. The term moles refers to a collection of particles and represents a total of 6.02 x 10^23 distinct molecules. To measure directly how many particles are present, you need to convert the number of particles into a weight. Weight is measured on a balance and has units of grams. To convert the number of moles to weight requires the knowledge of the composition of the material.

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Determine the formula weight of the compound of interest. The formula weight is calculated by adding the atomic weight of the atoms in the chemical formula of the compound of interest. The atomic weights of the elements are found on the periodic table of elements. For example, you need to find the formula weight for iron rust with the chemical formula of Fe2O3. Look up the atomic weight of iron and oxygen on the periodic table. The atomic weight of iron is 55.845 and the atomic weight of oxygen is 16.000. Add the weight of each atom in the compound:

2 * 55.845 + 3 * 16.000 = 111.69 + 48.00 = 159.69.

This is the weight in grams of 1 mole of the compound.

Convert the weight of one mole to the weight of one millimole. Based in the metric system, one millimole is the same as one mole divided by 1,000. Therefore, the weight of one millimole is equal to the weight of one mole divided by 1,000. Continuing the example:

1 mole of Fe2O3 = 159.69 grams and

1 millimole Fe2O3 = 159.69 / 1000 = 0.1597 grams = 159.69 milligrams.

Calculate the number of milligrams required for the chemical reaction by multiplying the number millimoles by the weight of one millimole of the compound. Continuing the example, assume that you require 15 millimoles of iron rust to react in a chemical reaction, find the number of milligrams of iron rust you need. Based on the fact that one millimole of iron rust equals 159.69 mg, multiply the number of millimoles by the weight of one millimole:

(15 * 159.69) = 2,395.35 milligrams Fe2O3 required.