Stoichiometry Calculator | Mole, Mass & Yield Solver

Stoichiometry starts with a balanced equation

Stoichiometry connects measured quantities to the proportions in a chemical reaction. For the equation 2H₂ + O₂ → 2H₂O, two moles of hydrogen react with one mole of oxygen to make two moles of water. The same relationship describes two hydrogen molecules and one oxygen molecule forming two water molecules. It does not say that two grams of hydrogen react with one gram of oxygen.

A coefficient is the number before a chemical formula; a subscript is part of the substance's identity. You may change coefficients to balance an equation. Changing H₂O to H₂O₂ changes water into a different substance. Count each element on both sides before using the calculator. The chemical equation balancer provides a separate place to check that step.

The eight modes above keep the units visible through each calculation. They cover molar mass, mole ratios, mass-to-mass conversion, two-reactant limiting problems, three forms of the yield equation, empirical formulas, solution concentrations and ideal gases. They are educational models, not a determination that a proposed reaction actually occurs.

How to use the calculator without losing the chemistry

1. Write and balance the reaction. Check its chemical plausibility using your course material. The tool applies the coefficients you enter; it cannot verify a complete reaction from separate fields.
2. Choose the information you have. Mass-to-Mass takes grams. Solution Stoich takes molarity in mol/L and volume in mL. Gas Stoich takes litres, absolute pressure in atm and temperature in kelvin.
3. Enter coefficients separately from formulas. For 2H₂ + O₂ → 2H₂O, enter H2 and H2O as formulas and 2 and 2 as coefficients. The coefficient of O2 is 1 even when the equation leaves it unwritten.
4. Read the step table. Verify formula atom counts, molar masses, mole conversion and ratio before accepting the final answer.
5. Round at the end. The decimal selector changes display precision. It does not infer experimental significant figures. Keep guard digits during intermediate work and apply the rounding required by the question.

Formula syntax: H2O, Ca(OH)2, Al2(SO4)3 and CuSO4·5H2O are accepted, as are Unicode subscripts such as H₂O. Matched parentheses, square brackets and braces work for nested groups. A dot or middle dot separates hydrate components; in this parser a period always means a hydrate separator, never a decimal subscript. Element capitalization matters: Co is cobalt; CO contains carbon and oxygen.

Use positive whole-number atom counts. Do not enter ionic charge symbols, isotope notation, state labels such as (aq), reaction arrows or a leading reaction coefficient in a formula field. For an ordinary ion, its element composition can be entered without the charge for an approximate classroom molar mass; the tool neither checks charge balance nor models electron-mass corrections. Isotope-specific and non-stoichiometric materials need a different mass model.

The mass display selector changes calculated mass outputs to g, kg or mg. Inputs explicitly labelled “in g” remain grams. In Percent Yield, the selected unit applies to both actual and theoretical yield; convert them to the same unit before entering them. Editing an input or switching a mode clears the old answer so it cannot be mistaken for a fresh calculation.

The four-step stoichiometry method

For known substance A and target substance B, use the route mass A → moles A → moles B → mass B. With mass m in grams, molar mass M in g/mol, amount n in mol and positive balanced-equation coefficients ν:

Watch the units cancel: g A × mol A/g A × mol B/mol A × g B/mol B leaves g B. A ratio with the target substance on top and the known substance below is much easier to check than an unexplained multiplier. OpenStax's reaction-stoichiometry chapter uses this balanced-equation approach; the worked examples below show each stage explicitly.

Molar mass: count every atom once

For water, M(H₂O) = 2(1.008) + 15.999 = 18.015 g/mol. For calcium hydroxide, the final 2 applies to the entire OH group: Ca(OH)₂ contains one Ca, two O and two H, giving 40.078 + 2(15.999 + 1.008) = 74.092 g/mol.

Al₂(SO₄)₃ contains two aluminium atoms, three sulfur atoms and twelve oxygen atoms. A hydrate dot adds the following component: CuSO₄·5H₂O includes five water units for each copper(II) sulfate unit. It is not an instruction to multiply the molar mass of CuSO₄ by the molar mass of water.

The calculator uses rounded educational atomic weights based on the CIAAW abridged table. Natural isotopic composition and the table supplied in an examination can change the last digits. Results shown to six decimal places do not make the underlying atomic weights that precise. Elements without a standard atomic weight require isotope-specific information and are not assigned a silent mass-number substitute.

Coefficients compare amounts, not masses

Multiplying every coefficient by the same positive factor does not change a reaction ratio. The equations 2H₂ + O₂ → 2H₂O and 4H₂ + 2O₂ → 4H₂O therefore give the same answer. For a target of water from oxygen, the ratio is 2 mol H₂O per 1 mol O₂. For hydrogen from oxygen, it is 2 mol H₂ per 1 mol O₂.

If only one reactant amount is given, the usual classroom assumption is that every other required reactant is available in excess. State that assumption. Otherwise a single-reactant calculation may predict more product than the real mixture can produce.

Mass-to-mass conversions and conservation checks

Grams must pass through moles because substances have different molar masses. For 4.00 g H₂ and excess oxygen, the theoretical water mass is (4.00/2.016) × (2/2) × 18.015 = 35.744… g, or 35.7 g to three significant figures. A product mass larger than the hydrogen mass is reasonable because oxygen supplies the rest of the product's mass.

Conservation of mass applies to the complete system: all consumed reactants, all products and any remaining reactants. Do not compare one product only with one reactant and infer that mass was created. Nor should a yield calculation assume every product atom originated in the substance whose mass was supplied.

Find the limiting reactant by comparing reaction extent

For each reactant, divide its available moles by its coefficient: ξ_A = n_A/ν_A and ξ_B = n_B/ν_B. The smaller value gives the maximum reaction extent. Product moles equal ξ_min × ν_product. Remaining reactant A equals n_A − ξ_minν_A, and similarly for B.

The lighter sample is not necessarily limiting. Compare equivalent reaction extents, not grams or unadjusted moles. If the extents match, the inputs are stoichiometric: neither reactant is in excess in this idealized model. The tool identifies a numerical tie within a relative tolerance of 10⁻¹², which is not a claim about experimental uncertainty.

This mode compares two entered reactants. It assumes any additional reactants are in sufficient supply, the stated reaction goes to completion and competing reactions are absent. It does not predict equilibrium, reaction rate or a safe mixing procedure.

Theoretical yield, actual yield and percent yield

Theoretical yield is the amount predicted from the balanced reaction and available limiting reactant under the stated ideal assumptions. Actual yield is what is measured or recovered. Their ratio is:

Both amounts must refer to the same product and use compatible units. The calculator disables the field being solved so an old value in it cannot affect the result. A zero actual yield with a positive theoretical yield is 0%; trying to infer an unknown theoretical yield from 0% cannot determine a unique answer.

A result above 100% is a reason to investigate. Retained water, solvent, impurities, an incorrect theoretical basis, unit mismatch or measurement error can inflate a measured mass. Report and explain the discrepancy rather than clipping the result to 100%. Product loss, incomplete reaction and side reactions can reduce actual yield. The calculator performs the comparison; it does not diagnose the experiment.

Empirical formulas need evidence, not forced rounding

An empirical formula expresses the simplest whole-number atom ratio. Divide each measured element mass by that element's atomic weight, then divide every resulting mole value by the smallest. A C:H:O ratio close to 1:2:1 suggests CH₂O. It does not establish a molecular formula: C₂H₄O₂ and C₆H₁₂O₆ have the same simplest ratio.

For percentages, use a hypothetical 100 g sample: 40.0% means 40.0 g. Use all relevant elements and one consistent basis; percentages should account for the whole analyzed material within the stated uncertainty. The tool also accepts raw masses, so it cannot automatically require that your entries add to 100.

Do not round 1:1.5 to 1:2. Multiply all ratios by 2 to obtain 2:3. The calculator searches common multipliers from 1 through 12, accepts a maximum absolute discrepancy of 0.02 in each normalized ratio, then reduces common factors. It labels a successful fit as a candidate empirical formula. The tolerance is a transparent numerical heuristic, not an uncertainty analysis or proof of compound identity. No acceptable small ratio produces a request to recheck the data, not a fabricated rounded formula.

Enter each element once with a positive amount. Combine duplicate rows and include both fields in every row you use. An isotopically enriched sample, omitted element or impure sample can make even a neat ratio misleading.

Solutions and gases: keep the conventions explicit

For concentration c in mol/L and solution volume V in litres, n = cV. Convert millilitres to litres by dividing by 1000. A 0.500 mol/L solution with volume 100.0 mL contains 0.0500 mol solute. The balanced-equation ratio then connects that amount to the target. This calculation assumes that the other reactant is sufficient and that the relevant stoichiometric reaction is complete; it does not automatically identify a titration endpoint.

For an ideal gas, n = PV/(RT). This tool uses P in atm, V in L, T in K and R = 0.082057366080960 L·atm·mol⁻¹·K⁻¹, derived from the SI gas constant listed by NIST. Temperature must be absolute: T(K) = t(°C) + 273.15. Pressure must be absolute, not gauge pressure. The model ignores non-ideal-gas corrections and water-vapour corrections for wet gases.

STP is not a unique classroom convention. At 273.15 K and 1 atm (101.325 kPa), ideal molar volume is about 22.41397 L/mol. At 273.15 K and 100 kPa, the IUPAC standard-gas convention, it is about 22.71095 L/mol. Both choices are explicit in the mode selector. Follow the conditions stated in your question rather than treating “22.4 L” as universal. At room temperature, use the actual temperature and pressure or the molar volume specified by the course.

For a separate exploration of pressure, volume and temperature relationships, use the gas laws calculator. Gas stoichiometry still requires the mole ratio after the gas amount is found.

Eight worked examples

Ten practice problems with explained answers

Use the calculator's stated atomic weights. Assume each balanced reaction proceeds completely and any unlisted co-reactant is sufficient. Try each question before opening its answer. Extra digits below show the calculation; round the final measurement as your course requires.

1. Grouped formula: Find the molar mass of Ca₃(PO₄)₂.

   Show answer and reasoning

   The atom counts are Ca₃P₂O₈. Add 3(40.078) + 2(30.974) + 8(15.999) = 310.174 g/mol. The outer 2 multiplies the whole phosphate group.

2. Hydrate: Find the molar mass of MgSO₄·7H₂O.

   Show answer and reasoning

   Count Mg₁S₁O₁₁H₁₄. Then 24.305 + 32.06 + 11(15.999) + 14(1.008) = 246.466 g/mol. Seven water units contribute seven oxygen atoms in addition to the four in sulfate.

3. Mole ratio: In N₂ + 3H₂ → 2NH₃, how much ammonia forms from 4.50 mol hydrogen?

   Show answer and reasoning

   Use the target-to-known coefficient ratio: 4.50 × 2/3 = 3.00 mol NH₃. Its calculated mass is 3.00 × 17.031 = 51.093 g, or 51.1 g to three significant figures.

4. Mass conversion: For CaCO₃ → CaO + CO₂, what mass of carbon dioxide follows from 25.0 g calcium carbonate?

   Show answer and reasoning

   Moles CaCO₃ = 25.0/100.086 = 0.2497851847 mol. The ratio is 1:1, so multiply by 44.009 g/mol: 10.99279620 g. The final answer is 11.0 g CO₂ to three significant figures.

5. Limiting and excess: In 4Al + 3O₂ → 2Al₂O₃, combine 5.40 g aluminium and 9.00 g oxygen. Identify the limiting reagent, the theoretical product mass and the oxygen left.

   Show answer and reasoning

   Compare (5.40/26.982)/4 = 0.05003335557 with (9.00/31.998)/3 = 0.09375585974. Aluminium limits. Product moles = 2(0.05003335557); multiplying by 101.961 g/mol gives 10.20290193 g, or 10.2 g Al₂O₃. Oxygen left = 9.00 − 3(0.05003335557)(31.998) = 4.197098065 g, or 4.20 g O₂.

6. Exact proportions: In 2H₂ + O₂ → 2H₂O, use 3.00 mol hydrogen and 1.50 mol oxygen. Which is in excess, and how much water forms?

   Show answer and reasoning

   The two reaction extents are 3.00/2 = 1.50 and 1.50/1 = 1.50. Neither reactant is in excess in this ideal model. Product = 1.50 × 2 = 3.00 mol H₂O. Its mass is 54.045 g, or 54.0 g to three significant figures.

7. Percent yield: A reaction gives 12.6 g product against a theoretical yield of 15.0 g. Find the percent yield.

   Show answer and reasoning

   Percent yield = 12.6/15.0 × 100 = 84.0%. Use actual yield on top and theoretical yield below, with the same product and units.

8. Empirical formula: A compound contains 28.014 g nitrogen and 79.995 g oxygen. Find its simplest atom ratio.

   Show answer and reasoning

   The amounts are 28.014/14.007 = 2.0000 mol N and 79.995/15.999 = 5.0000 mol O. Dividing by the smaller gives 1:2.5. Multiply both by 2 to get 2:5, giving N₂O₅.

9. Solution ratio: For H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, what amount and mass of NaOH are required by 35.0 mL of 0.200 mol/L sulfuric acid?

   Show answer and reasoning

   Convert 35.0 mL to 0.0350 L. Acid amount = 0.200 × 0.0350 = 0.00700 mol. Multiply by 2/1: 0.0140 mol NaOH. Its mass is 0.01400 × 39.997 = 0.559958 g, or 0.560 g.

10. Name the gas convention: In 2H₂ + O₂ → 2H₂O, 11.207 L oxygen is measured at 273.15 K and 1 atm. How much water is theoretically possible? What changes if the same volume is instead measured at 100 kPa?

    Show answer and reasoning

    At 1 atm, n(O₂) = 11.207/22.413969545 = 0.5000006794 mol. The 2/1 ratio gives 1.000001359 mol H₂O and 18.01502448 g, reported as 18.015 g to five significant figures. The rounded 22.414 L/mol shortcut gives exactly 1.0000 mol water within that approximation.

    At 100 kPa, use 22.710954641 L/mol instead. Water amount becomes 0.9869246077 mol and mass 17.77944681 g, or 17.779 g. Pressure changes the gas amount even when the volume and coefficient ratio stay the same.

Common mistakes and a quick self-check

• Using grams as coefficient units: convert to moles first
• Changing subscripts to balance: preserve chemical formulas and adjust coefficients
• Choosing the smallest mass as limiting: compare moles divided by coefficients
• Ignoring a hydrate multiplier: count every water atom as well as the anhydrous compound
• Using mL directly in n = cV: convert to litres when concentration is mol/L
• Using Celsius or gauge pressure in PV = nRT: use kelvin and absolute pressure
• Equating display decimals with significant figures: decide precision from the given data
• Forcing an empirical formula: examine residuals, missing elements and uncertainty

Before submitting an answer, name the substance, attach the unit, state any excess-reactant or ideal-gas assumption, and check its scale. A scientific calculator is useful for independently checking a single step without copying an entire unexamined result.

Sources & References

Methods and conventions checked against the following references on 5 October 2026. Existing worked examples are retained and expanded; additional examples, practice questions and teaching figures were prepared for this guide.

• OpenStax Chemistry 2e, Reaction Stoichiometry
• OpenStax Chemistry 2e, Reaction Yields
• CIAAW, Abridged Standard Atomic Weights
• IUPAC Gold Book, Standard Conditions for Gases
• NIST, CODATA fundamental physical constants