Find the molar mass (molecular weight) of a chemical compound by entering the atomic masses and number of atoms for each element. Essential for stoichiometry, solution preparation, and chemical analysis.
Molar mass is the bridge between molecules (which you can't weigh) and grams (which you can). One mole of a substance contains Avogadro's number of molecules — 6.022 × 10²³ — and its mass in grams equals the sum of the atomic masses of the atoms in the molecular formula. So one mole of water (H₂O) weighs about 18.015 g. One mole of glucose (C₆H₁₂O₆) weighs about 180.16 g. The molar mass is what lets you convert "I need 0.5 moles of NaCl" into "I need 29.22 g" and pour it onto a balance.
This calculator takes up to three element atomic masses and their counts and sums them. For most chemistry problems — homework, lab prep, stoichiometry — that covers any binary or simple ternary compound. For larger molecules (proteins, polymers, organic chemistry with 5+ unique elements), repeat the calculation in pieces and add them up.
The atomic masses used are the standard IUPAC values, which are weighted averages of the natural isotopic distribution. Carbon, for example, is mostly C-12 (12.000 amu) but contains a few percent C-13 (13.003 amu) and a tiny amount of C-14, making the average atomic mass 12.011. For natural-abundance chemistry, the standard atomic masses are what you want; for isotope-labeled work (mass spec, NMR with labeled compounds), use the specific isotope mass instead.
**Scenario:** You need 100 mL of a 0.5 M NaCl solution for an experiment. **Calculation:** Molar mass NaCl: Na (22.990) + Cl (35.45) = 58.44 g/mol. Mass needed = 0.5 mol/L × 0.100 L × 58.44 g/mol = 2.922 g. **Result:** Weigh 2.922 g of NaCl on a balance with ±0.001 g precision. Dissolve in ~80 mL of water in a 100 mL volumetric flask, then fill to the mark. Final concentration: exactly 0.5 M.
**Scenario:** How many molecules of glucose are in a 5 g sample? **Calculation:** Molar mass glucose (C₆H₁₂O₆) = 180.156 g/mol. Moles = 5 / 180.156 = 0.02776 mol. Molecules = 0.02776 × 6.022 × 10²³ = 1.672 × 10²² molecules. **Result:** A 5 g pile of glucose contains about 1.7 × 10²² individual molecules — 17,000 billion-billion. The number is unwieldy, which is exactly why moles exist as a counting unit.
**Scenario:** You have 10 g of CaCO₃ and 10 g of HCl. Which is the limiting reagent for CaCO₃ + 2 HCl → CaCl₂ + H₂O + CO₂? **Calculation:** CaCO₃: Ca (40.078) + C (12.011) + 3 × O (15.999) = 100.087 g/mol → 10/100.087 = 0.0999 mol. HCl: H (1.008) + Cl (35.45) = 36.46 g/mol → 10/36.46 = 0.274 mol. Required: 2 mol HCl per mol CaCO₃ → for 0.0999 mol CaCO₃ need 0.1998 mol HCl. We have 0.274, so HCl is in excess; CaCO₃ is limiting. **Result:** CaCO₃ runs out first. Maximum yield is limited to 0.0999 mol of products. Always check both reagents in moles, not grams — the apparent "equal masses" hides very different mole counts because molar masses differ.
**Use molar mass calculation for:**
- **Solution preparation**: weighing out the right grams to dissolve in a target volume. - **Stoichiometry**: balancing chemical equations and determining limiting reagents. - **Yield calculations**: theoretical vs actual yield in synthesis problems. - **Empirical formula determination**: combustion analysis gives mass percentages that reduce to atom ratios. - **Mass spec interpretation**: peaks at specific m/z values correspond to specific molecular fragments. - **Industrial process design**: scaling reactions from milligram lab work to kilogram or ton plant production. - **Pharmaceutical dosing**: drug doses are sometimes specified in mg, sometimes in mmol — molar mass converts.
**Practical contexts:**
- **Lab homework**: nearly every quantitative chemistry problem uses molar mass somewhere. - **Brewing and chemistry kits**: when specs say "add 5 mM of X," you need molar mass to convert to grams. - **Aquarium chemistry**: ammonia, nitrite, and nitrate quantification all use molar mass conversions. - **Soil testing**: lime requirement, fertilizer N-P-K ratios — molar mass behind the scenes. - **Medication formulation**: pharmacist calculations of dose, dilution, and stability.
**Be careful about:**
- **Formula vs molecular mass**: for ionic compounds (NaCl, MgO), the "formula unit" is the smallest whole-number ratio of ions; "molecular mass" is the same thing but the term is technically inappropriate (these don't exist as molecules). - **Empirical vs molecular formula**: glucose (C₆H₁₂O₆, 180 g/mol) has empirical formula CH₂O (30 g/mol); always know which you're calculating. - **Hydrates**: copper(II) sulfate exists as CuSO₄ (anhydrous, 159.6 g/mol) and CuSO₄·5H₂O (pentahydrate, 249.7 g/mol). Make sure your formula matches your starting material. - **Average vs monoisotopic mass**: for mass spectrometry of small molecules, use monoisotopic masses (most abundant isotope of each element); for everyday chemistry, use IUPAC average atomic masses.
Calculate molarity (moles per liter) and perform dilution calculations (C1V1 = C2V2).
Calculate dilution using C1V1 = C2V2 equation. Find any missing variable.
Determine the empirical formula from mass percentages of elements.
Calculate titration endpoint including molarity, volume, and equivalence point.
Calculate boiling point elevation using ΔTb = Kb x m x i.
Calculate freezing point depression using ΔTf = Kf x m x i.
e.g. Carbon = 12.011
e.g. Hydrogen = 1.008
e.g. Oxygen = 15.999
Molar Mass
180.156 g/mol
Element 1 %
40.0%
Element 2 %
6.7%
| Parameter | Value |
|---|---|
| Element 1 Contribution | 72.0660 g/mol |
| Element 1 Mass % | 40.00% |
| Element 2 Contribution | 12.0960 g/mol |
| Element 2 Mass % | 6.71% |
| Element 3 Contribution | 95.9940 g/mol |
| Element 3 Mass % | 53.28% |
| Total Molar Mass | 180.1560 g/mol |
| Mass of 1 mmol | 0.180156 g |
| Moles in 1 g | 5.5507e-3 mol |