Stoichiometry Calculator
A stoichiometry calculator converts a mass of one substance in a chemical reaction into the mass of another substance, using the balanced equation's mole ratio as the conversion factor. It takes the given mass to moles, scales by the coefficient ratio between the two species, then converts that mole amount back to mass.
The Mole Concept: Bridging Mass and Particles
Stoichiometry works because the mole is a fixed count of particles tied to a fixed mass, the molar mass. Every calculation on this page rests on one conversion: mass divided by molar mass gives moles, and moles multiplied by molar mass gives mass back. For the full breakdown of that single conversion, including how to read atomic masses off a formula, see grams to moles and molar mass. On this page that conversion is only the entry and exit point of a larger calculation. What happens in between, scaling an amount of one substance into an amount of another, is what a balanced equation makes possible.
Why a Balanced Equation Gives Mole Ratios, Not Mass Ratios
A balanced equation reports how many moles of each species react and form, never how many grams. The coefficients are a mole ratio, and reading them as a mass ratio is one of the most common errors in an introductory course. Take 2 H2 + O2 -> 2 H2O. Read carelessly, the coefficients look like they might scale straight to grams, so a student guesses that 2 g of H2 reacts with 1 g of O2. It does not. The equation says 2 mol of H2 reacts with 1 mol of O2. Since H2 has a molar mass near 2.02 g/mol and O2 near 32.00 g/mol, 2 mol of H2 is about 4.04 g while 1 mol of O2 is about 32.00 g, a mass ratio near 1:8, not 2:1. The mole ratio from the equation only ever holds in moles. Turning it into a usable mass comparison requires the molar mass of each species, which is why the method below always routes through moles instead of jumping from one mass straight to another. For a closer look at isolating and applying that ratio on its own, see mole ratio.
The General Mass-to-Mass Method
Every mass-to-mass stoichiometry problem follows the same moves no matter what the reaction is. Convert the mass you were given to moles using its molar mass, multiply by the mole ratio, the product coefficient over the reagent coefficient, taken from the balanced equation, then convert the resulting moles of the target substance back to mass using its own molar mass. The calculator above runs these three conversions live and shows each intermediate value, so the arithmetic can be checked one step at a time rather than trusting only a final number.
Worked Example: Combustion of Methane
Methane burns according to CH4 + 2 O2 -> CO2 + 2 H2O. Given 4.00 g of CH4 (molar mass 16.04 g/mol), find the mass of CO2 produced (molar mass 44.01 g/mol).
- Moles of CH4: 4.00 g / 16.04 g/mol = 0.2494 mol.
- The mole ratio of CO2 to CH4 in the equation is 1:1, so moles of CO2 = 0.2494 mol.
- Mass of CO2: 0.2494 mol x 44.01 g/mol = 10.98 g.
Entering the same values above, with both coefficients set to 1, reproduces this result and shows the running moles at each stage.
Worked Example: Iron and Sulfur Forming Iron(II) Sulfide
Iron and sulfur combine directly: Fe + S -> FeS. Given 5.00 g of Fe (molar mass 55.85 g/mol), find the mass of FeS produced (molar mass 87.91 g/mol).
- Moles of Fe: 5.00 g / 55.85 g/mol = 0.0895 mol.
- Both coefficients are 1, so moles of FeS = 0.0895 mol.
- Mass of FeS: 0.0895 mol x 87.91 g/mol = 7.87 g.
Worked Example: Neutralizing NaOH with HCl
Sodium hydroxide neutralizes hydrochloric acid in a 1:1 reaction: NaOH + HCl -> NaCl + H2O. Given 2.00 g of NaOH (molar mass 40.00 g/mol), find the mass of NaCl produced (molar mass 58.44 g/mol).
- Moles of NaOH: 2.00 g / 40.00 g/mol = 0.0500 mol.
- The mole ratio of NaCl to NaOH is 1:1, so moles of NaCl = 0.0500 mol.
- Mass of NaCl: 0.0500 mol x 58.44 g/mol = 2.92 g.
None of these three reactions involves a limiting reagent, since only one reactant's mass is given in each case. If the mass of the second reactant were also given, checking which one runs out first would come before any product mass is calculated, and that check is covered on limiting reactant.
When Two Reactant Masses Are Given
The method above assumes one known mass and one target mass. Real lab problems often supply two reactant masses instead, for example both the NaOH and the HCl, or both the Fe and the S above. In that case the mole-ratio calculation has to be run once for each reactant to see which one produces less product; that reactant is the limiting one, and it alone sets the actual mass of product formed while the other is left in excess. The full method for identifying which reagent runs out first, and for finishing the calculation once it is identified, is covered on limiting reactant rather than repeated here. Once the limiting reagent sets a theoretical yield, comparing it against a measured actual yield is handled by theoretical yield and the percent yield calculator on the home page.
The Four Stoichiometry Conversion Types
| Conversion | What it involves | Tool |
|---|---|---|
| Mass to mass | Given mass of one substance, find mass of another, through moles and the mole ratio | Stoichiometry calculator (this page) |
| Mass to mole | Given mass of a substance, find how many moles it represents | grams to moles |
| Mole to mole | Given moles of one substance, find moles of another using only the equation's coefficients | mole ratio |
| Mole to mass | Given moles of a substance, find its mass using molar mass | molar mass |
Frequently asked questions
What is stoichiometry in simple terms?
Stoichiometry is mole-based arithmetic that converts an amount of one substance in a chemical reaction into an amount of another substance, using the coefficients of a balanced equation as the conversion factor. It answers questions like how many grams of product form from a given mass of reactant. The mole, not the gram, is the unit the equation actually speaks in, which is why every calculation on this page passes through moles first.
How do you do a mass-to-mass stoichiometry calculation?
Convert the given mass to moles using its molar mass, multiply by the mole ratio between the two substances taken from the balanced equation's coefficients, then convert that mole amount to a mass using the target substance's molar mass. This three-step path, mass to moles to moles to mass, is exactly what the calculator on this page performs and displays at each stage.
Why can't you use mass ratios directly from a balanced equation?
Because the coefficients in a balanced equation count moles of each species, not grams, and different substances have different molar masses. In 2 H2 + O2 -> 2 H2O, the mole ratio is 2:1, but the mass ratio works out closer to 1:8 once each side is converted using its own molar mass. Skipping the mole step and reading coefficients as grams gives a wrong answer every time.
What is the mole ratio and where does it come from?
The mole ratio is the ratio between the coefficients of two substances in a balanced chemical equation, and it comes directly from balancing that equation, nothing else. It tells you how many moles of one species correspond to how many moles of another. See <a href="/mole-ratio/">mole ratio</a> for how to isolate and apply this ratio on its own without a full mass-to-mass conversion.
Do you need a balanced equation before doing stoichiometry?
Yes, every mole ratio used in a stoichiometry calculation comes from the coefficients of a correctly balanced equation, so balancing has to happen first. An unbalanced equation gives coefficients that don't reflect real atom counts, which produces a wrong mole ratio and a wrong mass result no matter how careful the arithmetic is afterward. See <a href="/balanced-equation/">balanced equation</a> for the balancing method itself.
What if I'm given two reactant masses instead of one?
Given two reactant masses, run the mass-to-mole-to-mass calculation for each reactant separately and compare which one yields less product. That reactant is the limiting one, and it alone sets the actual product mass while the other remains partly unused. The full method for making that comparison, rather than just the single-mass case covered here, is on <a href="/limiting-reactant/">limiting reactant</a>.
Can stoichiometry give you percent yield directly?
No, stoichiometry alone only gives a theoretical mass of product, the maximum a reaction can form on paper. Percent yield also requires an actual yield measured in the lab, then divides that actual mass by the theoretical mass and multiplies by 100. See <a href="/theoretical/">theoretical yield</a> for the theoretical side and <a href="/formula/">the percent yield formula</a> for the final ratio.
What's the difference between stoichiometry and mole ratio calculations?
A mole ratio calculation converts moles of one substance to moles of another using only the equation's coefficients, with no mass involved at either end. A full stoichiometry calculation, as done on this page, adds a mass-to-mole conversion at the start and a mole-to-mass conversion at the end, so it begins and ends in grams. See <a href="/mole-ratio/">mole ratio</a> for the isolated, moles-only version.
How many significant figures should a stoichiometry answer have?
A stoichiometry answer should carry the same number of significant figures as the least precise measured value in the problem, which is almost always the given mass. Molar masses read from a periodic table are usually known to four or five significant figures, so they rarely limit precision. A mass given as 4.00 g supports three significant figures in the final answer, for example 10.98 g rather than 10.9758 g.
What happens if the equation isn't balanced first?
An unbalanced equation gives coefficients that don't match the atoms actually reacting, so any mole ratio pulled from it is wrong and every mass calculated afterward is wrong too, even with correct arithmetic. Balancing has to happen before any mole ratio is used. See <a href="/balanced-equation/">balanced equation</a> for the method of finding correct coefficients before running a stoichiometry calculation.