Mole Calculations
n = m / M
The fundamental relationship between mass, moles, and molar mass. This equation is the gateway to all stoichiometry calculations, converting between measurable mass and moles.
Variables
Amount of substance in moles
Mass of substance in grams
Molar mass in g/mol (from periodic table)
Example Calculation
Scenario
Calculate the number of moles in 25.0 g of calcium carbonate (CaCO3).
Given Data
Calculation
n = m/M = 25.0 g / 100.09 g/mol
Result
n = 0.250 mol CaCO3
Interpretation
There are 0.250 moles of calcium carbonate, which contains 0.250 mol Ca, 0.250 mol C, and 0.750 mol O atoms.
When to Use This Formula
- ✓Converting mass to moles for stoichiometry
- ✓Calculating mass from moles after stoichiometry
- ✓Finding molar mass from mass and moles
- ✓Starting point for all quantitative chemistry
Common Mistakes
- ✗Calculating molar mass incorrectly (missing subscripts)
- ✗Confusing atomic mass with molar mass
- ✗Using the wrong compound in molar mass calculation
- ✗Forgetting to include all atoms in a formula
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Common questions about this formula
A mole is 6.022 x 10^23 particles (Avogadro's number). It's the amount of substance that contains as many particles as there are atoms in exactly 12 g of carbon-12. It bridges the atomic and macroscopic scales.
Add up the atomic masses (from the periodic table) of all atoms in the formula. For CaCO3: Ca(40.08) + C(12.01) + 3×O(16.00) = 100.09 g/mol. Remember to multiply by subscripts.