How to Calculate Molar Mass
Learn how to calculate the molar mass of any compound by summing atomic masses from the periodic table. Includes worked examples for simple and complex molecules.
What Is Molar Mass?
Molar mass is the mass of one mole (6.022 x 10^23 particles) of a substance, expressed in grams per mole (g/mol). It serves as the conversion factor between the microscopic world of atoms and molecules and the macroscopic world of grams that can be measured on a balance. The molar mass of an element numerically equals its atomic mass in atomic mass units (u) found on the periodic table.
How to Find Molar Mass from the Periodic Table
Locate each element in the compound on the periodic table and note its atomic mass (the decimal number, usually shown beneath the element symbol). Multiply each atomic mass by the number of times that element appears in the formula. Sum all contributions to get the molar mass of the compound. For example, water (H2O): H contributes 2 x 1.008 = 2.016 g/mol and O contributes 1 x 15.999 = 15.999 g/mol, giving a molar mass of 18.015 g/mol.
Step-by-Step Example: Glucose (C6H12O6)
Carbon: 6 x 12.011 = 72.066 g/mol. Hydrogen: 12 x 1.008 = 12.096 g/mol. Oxygen: 6 x 15.999 = 95.994 g/mol. Total molar mass of glucose = 72.066 + 12.096 + 95.994 = 180.156 g/mol. This means one mole of glucose (about 6.022 x 10^23 molecules) has a mass of 180.16 g.
Handling Parentheses in Formulas
Parentheses in a chemical formula indicate that the enclosed group is repeated by the subscript outside. For Ca3(PO4)2 (calcium phosphate), the phosphate group (PO4) appears twice. Ca: 3 x 40.078 = 120.234 g/mol. P: 2 x 30.974 = 61.948 g/mol. O: 8 x 15.999 = 127.992 g/mol. Total = 310.174 g/mol. Always distribute the outer subscript to every element inside the parentheses before summing.
Using Molar Mass in Calculations
Molar mass connects mass (g) to amount (mol) via n = m / M, where n is moles, m is mass in grams, and M is molar mass. To find the mass of 3.0 mol of NaCl (M = 58.44 g/mol): mass = 3.0 x 58.44 = 175.3 g. Conversely, 29.22 g of NaCl contains 29.22 / 58.44 = 0.500 mol. This conversion is the starting point for nearly every stoichiometry problem.
Common Errors and Tips
A common mistake is using the mass number (integer) rather than the precise atomic mass from the periodic table, which introduces rounding errors in multi-step problems. Another error is neglecting to multiply an element's atomic mass by its subscript. When working with hydrated compounds such as CuSO4 · 5H2O, remember to add the contribution of the water molecules: 5 x 18.015 = 90.075 g/mol must be added to the anhydrous salt mass.
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