How to Calculate Moles in a Reaction

By Kylene Arnold
Equate grams of a compound to moles using The Periodic Table of Elements.

A mole is a the quantity of a substance equal to Avogadro's Number, approximately 6.022 × 10^23. Scientists use the mole measurement because it provides a means to express large quantities with ease and the correlation between between an element's atomic weight and the number of grams in a mole of that element. You can determine the number of moles in any chemical reaction given the chemical formula and the mass of the reactants.

Determining Molar Quantities

Calculate the mass in grams of each reactant. If the reactants are not already in grams, convert the units.

For example, combine 0.050000 kg of sodium (Na) with 25.000 grams of chlorine gas (Cl2) to form NaCl or table salt. Convert 0.050000 kg of sodium to grams. Using your conversion chart, you see that 1000 g = 1 kg. Multiply 0.050000 kg by 1000 g/kg to get grams of Na.

50.000 g of Na and 25.000 g of Cl2 are used in this reaction.

Determine the atomic weight of each element using the periodic table. It will usually be written as a decimal number above or below the chemical symbol and is measured in atomic mass units (amu).

The atomic weight of Na is 22.990 amu. Cl is 35.453 amu.

Calculate the number of grams per mole (g/mol) for each reactant and product. The number of grams per mole for each single element is equal to the atomic weight of that element. Add up the masses of the elements in each compound to find the grams per mole for that compound.

For example, Na is reactant with a single element. It's atomic weight of 22.990 amu is equal to the number of grams per mole of Na -- also 22.990.

Cl2 on the other hand, is made up of two atoms of Cl. Each individually has a mass of 35.253 amu, so together the compound weighs 70.506 amu. The number of grams per mole is the same -- 70.506 g/mol

NaCl is composed of both an atom of Na and an atom of Cl. Na weighs 22.990 amu and Cl 35.253 amu, so NaCl weighs 58.243 amu and has the same number of grams per mole.

Divide the number of grams of each reactant by the number of grams per mole for that reactant.

50.000 g of Na are used in this reaction, and there are 22.990 g/mol. 50.000 / 22.990 = 2.1749. 2.1749 moles of Na are used in this reaction.

25.000 g of Cl2 are used and there are 70.506 g/mol of Cl2. 25.000 / 70.506 = 0.35458. 0.35458 moles of Cl2 are used in this reaction.

Examine your chemical formula for the reaction, noting the coefficients for each reactant and product. This ratio holds true for any quantity, whether for single atoms, dozens of atoms or more importantly, moles of atoms.

For example, in the equation 2 Na + Cl2 ---> 2 NaCl. The ratio of Na to Cl2 to NaCl is 2:1:2. Note that unlisted coefficients are assumed to be 1. Every two atoms of Na that reacts with one molecule of Cl2, yields two molecules of NaCl. The same ratio holds true for moles of atoms and molecules. Two moles of Na that reacts with one mole of Cl2, yields 2 moles of NaCl.

Calculate the limiting reactant, or the reactant which will run out first, by setting up the first of two equations. In this first equation, choose one of the reactants and multiply the moles of that reactant by the ratio of moles of reactant to moles of product.

For instance, in the example experiment, you used 2.1749 moles of Na. For every 2 moles of Na used, 2 moles of NaCl are produced. This is a 1:1 ratio, which means using 2.1749 moles of Na, also yields 2.1749 moles of NaCl.

Multiply the resulting number by the number of grams per mole of product to find the mass of product able to be produced by the given amount of reactant.

There are 2.1749 moles of NaCl and one mole equals 58.243 grams. 2.1749 * 58.243 = 126.67, so the 50.000 g of Na used in the reaction can create 126.67 g of NaCl.

Begin a second equation identical to the first, but using the other reactant.

The other reactant is Cl2, of which you have 0.35458 moles. The ratio of Cl2 to NaCl is 1:2, so for every mole of Cl2 that reacts, two moles of NaCl will be produced. 0.35458 * 2 = 0.70916 moles of NaCl.

Multiply the resulting number by the number of grams per mole of product to find the amount of product able to be produced by the second reactant.

0.70916 moles of NaCl * 58.243 g/mol = 41.304 g of NaCl.

Examine the results of both equations. Whichever equation resulted in the smaller mass of product contains the limiting reactant. Since the reaction can only proceed until this reactant is used up, however many grams of reactant is produced by this equation is the number of grams that will be produced by the entire reaction.

In the salt equation, Cl2 yielded the least number of grams of NaCl, therefore it is the limiting reactant. Only 41.304 g of NaCl will be produced by this reaction.

Determine the moles of product produced by dividing the grams of product by the grams per mole of product. You now have calculated the number of moles of every compound used in this reaction.

41.304 g of NaCl / 58.243 g/mol = 0.70917 moles of NaCl.


Do not attempt to reproduce this experiment. Sodium is a highly volatile metal and should only be handled by a professional.

About the Author

Kylene Arnold is a freelance writer who has written for a variety of print and online publications. She has acted as a copywriter and screenplay consultant for Advent Film Group and as a promotional writer for Cinnamom Bakery. She holds a Bachelor of Science in cinema and video production from Bob Jones University.