What does stoichiometry predict?
Type a reaction, the substance you know the amount of, and the one you want. The stoichiometry calculator balances the equation, uses the mole ratio, and gives the moles and grams of every substance.
- Mass to find
- 131.737
44 g (0.9978 mol) of C3H8 gives 131.737 g (2.9934 mol) of CO2.
- Moles to findmol
- 2.9934
- Known molesmol
- 0.9978
- Known massg
- 44
- Balanced equation
- C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
- Mole ratio
- 3 mol CO₂ per 1 mol C₃H₈
- Every substance
- C₃H₈: 0.9978 mol, 44 g; O₂: 4.989 mol, 159.638 g; CO₂: 2.9934 mol, 131.737 g; H₂O: 3.9912 mol, 71.9015 g
Mass to find: 131.737. 44 g (0.9978 mol) of C3H8 gives 131.737 g (2.9934 mol) of CO2.
How much of each substance?
How to calculate
Balances a chemical equation and turns a known mass or number of moles of one substance into the moles and grams of every other substance in the reaction.
Example with the default inputs (Chemical equation C3H8 + O2 = CO2 + H2O, Known substance C3H8, Known amount 44, Amount in Grams, Find CO2): 44 g (0.9978 mol) of C3H8 gives 131.737 g (2.9934 mol) of CO2.
Method: Balance the equation; known moles n = grams ÷ molar mass (or as typed); moles of X = n × (coefficient of X ÷ coefficient of the known); grams of X = moles of X × molar mass of X.
- The reaction goes to completion as written, and the known substance sets the amount (no limiting reactant check).
- Molar masses use the CIAAW 2024 abridged standard atomic weights; electrons count as having no mass.
Worked examples
Each example is checked against the calculator on every build.
- Chemical equation Al + I2 = AlI3, Known substance Al, Known amount 0.429, Amount in Moles, Find I2 gives Moles to find 0.6435, Balanced equation 2Al + 3I₂ → 2AlI₃.Source: OpenStax, Chemistry 2e, §4.3 Reaction Stoichiometry, https://openstax.org/books/chemistry-2e/pages/4-3-reaction-stoichiometry (Example 4.8: 0.429 mol Al × 3 mol I₂ ÷ 2 mol Al = 0.644 mol I₂)
- Chemical equation MgCl2 + NaOH = Mg(OH)2 + NaCl, Known substance Mg(OH)2, Known amount 16, Amount in Grams, Find NaOH gives Mass to find 21.946604, Known moles 0.274353.Source: OpenStax, Chemistry 2e, §4.3 Reaction Stoichiometry, https://openstax.org/books/chemistry-2e/pages/4-3-reaction-stoichiometry (Example 4.10: about 22 g NaOH); CIAAW, Abridged Standard Atomic Weights (2024), https://ciaaw.org/abridged-atomic-weights.htm
- Chemical equation C8H18 + O2 = CO2 + H2O, Known substance C8H18, Known amount 702, Amount in Grams, Find O2 gives Mass to find 2,458.001698, Balanced equation 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O.Source: OpenStax, Chemistry 2e, §4.3 Reaction Stoichiometry, https://openstax.org/books/chemistry-2e/pages/4-3-reaction-stoichiometry (Example 4.11: 2.46 × 10³ g O₂); CIAAW, Abridged Standard Atomic Weights (2024), https://ciaaw.org/abridged-atomic-weights.htm
- Chemical equation C3H8 + O2 = CO2 + H2O, Known substance C3H8, Known amount 44, Amount in Grams, Find CO2 gives Mass to find 131.736581, Moles to find 2.993401.Source: OpenStax, Chemistry 2e, §4.3 Reaction Stoichiometry, https://openstax.org/books/chemistry-2e/pages/4-3-reaction-stoichiometry (Example 4.9: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O); CIAAW, Abridged Standard Atomic Weights (2024), https://ciaaw.org/abridged-atomic-weights.htm
How it works
- Balance the equation. Each element, and the charge for ions, must match on both sides. The page finds the smallest whole-number coefficients exactly, as the chemical equation balancer does; coefficients you type in front of formulas are ignored.
- Find the substances. The known substance and the one to find are matched to the equation by their atoms and charge, so
H2OmatchesH2O(l). - Molar mass M = the sum of each element’s abridged standard atomic weight (CIAAW 2024) × its count in the formula.
- Known moles n = grams ÷ M of the known substance, or the moles you typed.
- Moles of any substance X = n × (coefficient of X ÷ coefficient of the known substance).
- Grams of X = moles of X × M of X.
“Every substance” lists the moles and grams of each substance in the equation for this amount of reaction.
Rules. The known amount is more than 0 and at most 10¹². The page gives no answer, with a message, when: the equation has no single = or arrow; a formula cannot be read; an element is on one side only; no whole numbers balance it, or it balances in more than one way; a coefficient would be 0 or negative; the known or wanted substance is not in the equation; the substance is an electron; or a substance contains an element with no standard atomic weight (such as Tc or Pu).
Exact arithmetic. Atomic weights are exact decimals and the typed amount is read as its exact decimal, so every step is an exact fraction, rounded once at the end.
Output format. Decimals and significant figures below are the most shown; trailing zeros are dropped (23.0 shows as 23, money keeps its cents). Masses and moles show 6 significant figures, rounded half up from the decimal value. “Every substance” lists each substance in the order of the balanced equation, separated by “; ”, as formula: moles mol, grams g (for example CO₂: 3 mol, 132.027 g); an electron is listed as e⁻: moles mol with no grams. Each number there is rounded to 6 significant figures half up from its exact value (33.00675 shows 33.0068), with trailing zeros dropped. The mole ratio reads a mol X per b mol Y, a and b the balanced coefficients of the substance to find and the known one. The balanced equation is written with subscripts and arrows (2Al + 3I₂ → 2AlI₃).
Assumptions
- The reaction goes to completion as written, and the known substance is the one that sets the amount: no limiting reactant or percent yield.
- Electrons in ionic half-equations count as having no mass.
Worked examples by hand
Aluminium and iodine (OpenStax Example 4.8). 2Al + 3I₂ → 2AlI₃. 0.429 mol Al × 3 ÷ 2 = 0.6435 mol I₂ (the book rounds to 0.644).
Magnesium hydroxide (Example 4.10). MgCl₂ + 2NaOH → Mg(OH)₂ + 2NaCl. M(Mg(OH)₂) = 24.305 + 2 × (15.999 + 1.0080) = 58.319 g/mol; M(NaOH) = 22.990 + 15.999 + 1.0080 = 39.997 g/mol. 16 ÷ 58.319 = 0.274353 mol; × 2 = 0.548706 mol NaOH; × 39.997 = 21.9466 g (the book: 22 g).
Octane (Example 4.11). 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O. M(C₈H₁₈) = 8 × 12.011 + 18 × 1.0080 = 114.232. 702 ÷ 114.232 = 6.14539 mol; × 25 ÷ 2 = 76.8174 mol O₂; × 31.998 = 2,458.0 g (the book: 2.46 × 10³ g).
Propane, the default. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. M(C₃H₈) = 3 × 12.011 + 8 × 1.0080 = 44.097; 44 ÷ 44.097 = 0.997800 mol; × 3 = 2.99340 mol CO₂; × 44.009 = 131.737 g.
Other questions people ask
How do I solve a stoichiometry problem?
Balance the equation, change the known mass to moles with its molar mass, multiply by the mole ratio from the coefficients, and change back to grams. For 16 g of Mg(OH)₂ in MgCl₂ + 2NaOH → Mg(OH)₂ + 2NaCl: 16 ÷ 58.319 = 0.2744 mol, × 2 = 0.5487 mol NaOH, × 39.997 = 21.95 g.
What is a mole ratio?
The ratio of two coefficients in the balanced equation. In 2Al + 3I₂ → 2AlI₃, 3 mol of I₂ react for every 2 mol of Al, so 0.429 mol of Al needs 0.429 × 3 ÷ 2 = 0.6435 mol of I₂.
Does the equation need to be balanced first?
No. Type it balanced or not; the calculator works out the smallest whole-number coefficients itself and shows the balanced equation.
How do I find the limiting reactant?
Run the calculator once for each reactant you have, finding the same product. The reactant that gives the least product is limiting, and that smaller amount is the theoretical yield. This page assumes the known substance sets the amount.
Which atomic weights does the calculator use?
The abridged standard atomic weights of CIAAW (2024), the same table as the molar mass calculator: H 1.0080, C 12.011, O 15.999 and so on. Textbooks that round to H = 1.01 or O = 16.00 get slightly different answers.
How many grams of CO₂ come from burning propane?
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. 44 g of propane is 44 ÷ 44.097 = 0.9978 mol, which makes 3 × 0.9978 = 2.9934 mol of CO₂, or 131.74 g.