Convert grams to moles and particles for any chemical formula, and find the limiting reagent, theoretical yield and leftover excess of a balanced reaction.
By AdminDrafted with AI writing toolsUpdated Checked examples: 7
36.03 g of H₂O is 2 mol, or 1.204 × 10²⁴ particles.
N_A is exact in the SI since 2019.
The amount of substance links mass to particle count: n = m/M, where M is the molar mass worked out from the formula, and N = n × N_A, where the Avogadro constant N_A is 6.022 140 76 × 10²³ per mole. In limiting-reagent mode, each reactant's moles are divided by its coefficient in the balanced equation; the smallest quotient is how far the reaction can run, and it sets the theoretical yield of product.
Typical uses are weighing out reagents and predicting product mass before an experiment. The default, 36.03 g of water, is 2 mol, or 1.204 × 10²⁴ molecules. For 2 H₂ + O₂ → 2 H₂O with 10 g of hydrogen and 64 g of oxygen, oxygen runs out first and the yield is 72.06 g of water.
The yield assumes the equation is balanced and the reaction goes to completion, so measured yields come out lower.
Divide the mass by the molar mass: n = m/M. Water's molar mass is 18.015 g/mol, so 36.03 g ÷ 18.015 g/mol = 2 mol of water. The rule works for any substance with a known formula; multiplying by the Avogadro constant then gives 1.204428152 × 10²⁴ molecules.
Add the atomic weight of every atom in the formula. Water, H₂O, has two hydrogens and one oxygen: 2 × 1.0080 + 15.999 = 18.015 g/mol. Carbon dioxide, CO₂, is 12.011 + 2 × 15.999 = 44.009 g/mol.
These are the IUPAC 2021 standard atomic weights. A textbook that rounds water to 18 g/mol turns 36.03 g into 2.002 mol instead of 2.000 mol, so expect differences around the fourth significant figure.
Capital letters carry meaning. Co is cobalt, 58.933 g/mol, while CO is carbon monoxide, 28.010 g/mol. For an element-by-element breakdown of a formula, use the molar mass calculator.
Rearrange the same relationship: m = n × M gives mass and N = n × N_A gives the particle count. For example, 0.25 mol of sodium chloride at 58.44 g/mol weighs 0.25 × 58.44 = 14.61 g.
To go from a particle count to mass, divide by N_A first. 3.011 × 10²³ molecules of CO₂ is 0.499988 mol, and 0.499988 × 44.009 = 22.004 g.
The Avogadro constant, 6.02214076 × 10²³ mol⁻¹, has been exact since the 2019 SI revision fixed it to define the mole (NIST CODATA).
A particle is whatever the formula describes: atoms for C, molecules for H₂O and formula units for an ionic solid such as NaCl.
Take 2 H₂ + O₂ → 2 H₂O with 10 g of hydrogen and 64 g of oxygen:
Hydrogen is in excess even though 10 g is far less than 64 g, so comparing masses picks the wrong reactant.
Comparing moles without the coefficients can fail as well. In N₂ + 3 H₂ → 2 NH₃ with 28 g of N₂ (0.9995 mol) and 6 g of H₂ (2.976 mol), hydrogen limits, because 2.976 ÷ 3 = 0.992 is below 0.9995.
Multiply the extent by the product's coefficient for moles of product, then by its molar mass: 2 × 2.000125 × 18.015 = 72.0645 g of water. This is the theoretical yield, the most the limiting reagent can make.
The reaction consumes 2 × 2.000125 = 4.00025 mol of hydrogen, or 8.0645 g, so 1.9355 g of H₂ is left over. The water outweighs either reactant on its own because it contains atoms from both.
Percent yield compares what you actually collect with the theoretical yield. Collecting 60.0 g of water here gives 60.0 ÷ 72.0645 × 100 = 83.3%.
Transfer losses, side reactions and incomplete reaction keep real yields below 100%. A result above 100% usually points to an impure or still-wet product.
The method follows OpenStax Chemistry 2e, section 4.4.
Water conversion is 36.03 g of H₂O. Typed molar mass is 0.25 mol at 58.44 g/mol. Water reaction is the 10 g H₂ and 64 g O₂ example above, and exact ratio uses 4.032 g of H₂ with 31.998 g of O₂, which react completely with nothing left over.
| Scenario | Selected result |
|---|---|
| Water conversion | 2 mol |
| Typed molar mass | 14.61 g |
| Water reaction | 72.0645 g product |
| Exact ratio | 36.03 g product |
For concentrations in mol/L, continue with the dilution and concentration calculator. For gases, the ideal gas law calculator turns moles into a volume at a given temperature and pressure.
Checked against: 36.03 / 18.015 = 2; × 6.02214076e23 (IUPAC 2021 weights)
Checked against: SI 2019 definition of the mole (Avogadro constant exact); standard atomic weight of carbon 12.011 (IUPAC 2021)
Checked against: 0.25 × 58.44
Checked against: 3.011e23 / N_A = 0.4999883; × 44.009 g/mol
Divide the mass in grams by the molar mass in g/mol: n = m/M. Water has a molar mass of 18.015 g/mol, so 36.03 g of water is 36.03 ÷ 18.015 = 2 mol. To go back, multiply moles by the molar mass. The molar mass is the sum of the atomic weights in the formula, which the calculator adds up from IUPAC 2021 values.
It is the number of particles in one mole: exactly 6.022 140 76 × 10²³ per mole since the 2019 revision of the SI, which defines the mole by fixing this constant. One mole of carbon therefore contains 6.022 140 76 × 10²³ atoms and weighs 12.011 g. Multiply an amount in moles by this number to get molecules, atoms or formula units.
Convert each reactant's mass to moles, then divide by its coefficient in the balanced equation; the reactant with the smaller result runs out first. For N₂ + 3 H₂ → 2 NH₃ with 28 g of N₂ and 6 g of H₂, nitrogen gives 0.9995 mol ÷ 1 = 0.9995 and hydrogen gives 2.976 mol ÷ 3 = 0.992, so hydrogen limits the yield to 33.79 g of ammonia.
Theoretical yield is the product mass if the limiting reagent reacted completely, which is what this calculator reports. Percent yield compares what you actually collected: actual ÷ theoretical × 100. With a theoretical yield of 72.06 g of water and 65.0 g collected, the percent yield is 90.2 %. Transfer losses, side reactions and incomplete reaction keep real yields below 100 %.
This calculator includes 7 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.