Convert grams to moles and particles for any chemical formula, and find the limiting reagent, theoretical yield and leftover excess of a balanced reaction.
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.
Checked against: Python 3.8 decimal: 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: Python 3.8 decimal: 0.25 × 58.44
Checked against: Python 3.8 decimal: 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 %.
Accuracy depends on your inputs and the method's assumptions. Decimal arithmetic uses 50 significant digits, but estimates, numerical methods and source data can be less precise; the displayed rounding does not remove those limits. It is checked against 7 worked examples whose answers come from independent sources; for example, “36.03 g of water is 2 mol” is checked against Python 3.8 decimal: 36.03 / 18.015 = 2; × 6.02214076e23 (IUPAC 2021 weights).
CODATA 2022 / SI 2019 — Avogadro constant N_A = 6.022 140 76 × 10²³ mol⁻¹ (exact); OpenStax Chemistry 2e, §4.4 Reaction yields (limiting reactant, theoretical yield).
7 worked examples with independently sourced answers ship with this calculator. They run in the test suite; you can run them here too.
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