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MasterMath

Stoichiometry Calculator

From the grams of a reactant to the grams of a product, going through moles and the ratio in the balanced equation. It works the molar masses out from the formulas, so you only have to write those.

Mass of product

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Mass of product—
Moles of reactant—
Moles of product—
Molar mass of the reactant—
Molar mass of the product—
Molecules of product—

How this was worked out

    The formula

    grams → moles → equation ratio → moles → grams

    What it means

    A balanced equation says nothing about grams: it says how many molecules react with how many, and that means moles. So the route has three legs. First grams become moles by dividing by the molar mass; then the coefficient ratio is applied, which is where the chemistry lives; and finally moles become grams again by multiplying by the product's molar mass. Skipping the moles and doing a straight proportion with grams is the single most repeated mistake in the topic.

    How to work it out by hand

    1. Work out the molar mass of the reactant and of the product from their formulas
    2. Divide the grams of reactant by its molar mass to get moles
    3. Multiply by the product's coefficient and divide by the reactant's
    4. Multiply the moles of product by its molar mass to get back to grams

    What is worth knowing

    This assumes the reactant you entered is the limiting one and that the reaction goes to completion. With two reactants in given amounts you first have to work out which runs out first, and if the reaction has an eighty per cent yield you multiply the answer by 0.8: the reaction yield calculator does that. The coefficients come from the balanced equation; if you do not have it yet, the equation balancer will give you the four numbers.

    Frequently asked questions

    Where do the coefficients come from?

    From the balanced equation. If you do not have it, the chemical equation balancer gives you all four numbers.

    Why can I not just scale the grams?

    Because the equation relates molecules, not masses. Two grams of hydrogen and two grams of oxygen are nowhere near the same amount of matter.

    Does it handle the limiting reactant?

    No: it assumes the one you enter is the one that runs out. With two given amounts you have to check which limits first.

    What about the actual yield?

    This is the theoretical yield, the maximum. What you really get is always less, and that ratio is worked out separately.

    What are the molecules of product?

    The moles multiplied by Avogadro's number, 6.022 × 10²³. It is the same quantity counted one by one.