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Generated on: July 20, 2026 at 09:33 PM

Mass-to-Mass Stoichiometry Calculator

Inputs

Mass of known substance10 g
Molar mass of known substance18.015 g/mol
Coefficient of known substance2
Coefficient of target substance2
Molar mass of target substance2.016 g/mol

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Chemistry

Mass-to-Mass Stoichiometry Calculator

Convert the mass of a known substance into the mass of another substance in a reaction. Uses the mole-ratio bridge — grams to moles, mole ratio from the balanced equation, then moles back to grams — and reports both the moles and mass of the target.

Inputs

g/mol
g/mol

Results

Enter a value to see results.

\begin{aligned} m_B &= \dfrac{m_A}{M_A} \cdot \dfrac{b}{a} \cdot M_B \\ &= \dfrac{10\,\text{g}}{18.015\,\text{g/mol}} \cdot \dfrac{2}{2} \cdot 2.016\,\text{g/mol} \\ &= ?\,\text{g} \end{aligned}

Details

\begin{aligned} n_A &= \dfrac{m_A}{M_A} \\ &= \dfrac{10\,\text{g}}{18.015\,\text{g/mol}} \\ &= ?\,\text{mol} \end{aligned}
\begin{aligned} n_B &= n_A \cdot \dfrac{b}{a} \\ &= \left(?\,\text{mol}\right) \cdot \dfrac{2}{2} \\ &= ?\,\text{mol} \end{aligned}

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What mass-to-mass stoichiometry answers

A balanced chemical equation says how the amounts of substances in a reaction relate to one another — but it speaks in moles, while a laboratory balance reads in grams. Mass-to-mass stoichiometry closes that gap: given the mass of one substance, it finds the mass of another that reacts with or forms from it. Typical questions are "how many grams of product does this much reactant make?" and "how many grams of one reactant does another require?"

The route from one mass to another never goes straight across. It passes through moles, because only moles obey the whole-number ratios of the equation.

The mole-ratio bridge

The calculation is three linked conversions, often drawn as a bridge from grams to grams:

Step 1 — grams to moles

Divide the known mass by its molar mass to get moles of the known substance:

nA=mAMAn_A = \frac{m_A}{M_A}

Step 2 — moles to moles

Multiply by the mole ratio, read straight off the coefficients of the known substance (aa) and the target substance (bb) in the balanced equation:

nB=nA⋅ban_B = n_A \cdot \frac{b}{a}

Step 3 — moles to grams

Multiply the target moles by its molar mass to return to a mass:

mB=nB⋅MBm_B = n_B \cdot M_B

Chained into a single expression, the three steps read:

mB=mAMA⋅ba⋅MBm_B = \frac{m_A}{M_A} \cdot \frac{b}{a} \cdot M_B
SymbolQuantityUnit
m_AMass of known substanceg
M_AMolar mass of known substanceg/mol
aCoefficient of known substance—
bCoefficient of target substance—
M_BMolar mass of target substanceg/mol
n_AMoles of known substancemol
n_BMoles of target substancemol
m_BMass of target substanceg

Worked example

Water decomposes by electrolysis according to

2 H2O→2 H2+O22\,\text{H}_2\text{O} \rightarrow 2\,\text{H}_2 + \text{O}_2

How many grams of hydrogen form from 10 g of water?

  1. Grams to moles: n(H₂O) = 10 ÷ 18.015 = 0.5551 mol.
  2. Mole ratio: H₂O has coefficient 2 and H₂ has coefficient 2, so n(H₂) = 0.5551 × 2/2 = 0.5551 mol.
  3. Moles to grams: m(H₂) = 0.5551 × 2.016 = 1.119 g.

So 10 g of water yields about 1.119 g of hydrogen gas. The oxygen produced could be found the same way, using the O₂ coefficient of 1 and its molar mass of 32.00 g/mol.

When the ratio is not 1:1

The mole ratio does the real work, and it is only 1 when the two coefficients happen to match. Consider the synthesis of ammonia:

N2+3 H2→2 NH3\text{N}_2 + 3\,\text{H}_2 \rightarrow 2\,\text{NH}_3

Starting from 28.02 g of nitrogen (1 mol), the ratio to ammonia is 2/1, so 2 mol of NH₃ form — 34.06 g. Had the masses been scaled directly by the coefficients instead of through moles, the answer would be wrong, because N₂ and NH₃ have different molar masses.

Where the molar masses come from

Each molar mass is the sum of the atomic masses in the substance's formula, taken from the periodic table. Water is 2 × 1.008 + 16.00 = 18.02 g/mol; ammonia is 14.01 + 3 × 1.008 = 17.03 g/mol. The known and target substances are usually different compounds, so their molar masses are entered separately.

The mole ratio at the centre of this bridge is the same one every stoichiometry problem turns on. When the target is the maximum product a limiting reactant can make, a theoretical-yield calculation frames the same arithmetic around a single product; when two reactants compete, a limiting-reagent analysis decides which one runs out first. The grams-to-moles and moles-to-grams steps on either side are the mole–mass conversion used throughout chemistry.

Frequently Asked Questions (FAQ)

What is mass-to-mass stoichiometry?

Mass-to-mass stoichiometry finds the mass of one substance in a reaction from the mass of another. Balances weigh substances in grams, but a balanced equation relates them in moles, so the calculation routes through moles: convert the known mass to moles, apply the mole ratio from the equation, then convert the target moles back to a mass. The result tells you how many grams of product a given mass of reactant makes, or how many grams of one reactant another requires.

How does the mole-ratio bridge work?

The bridge is three linked steps. First, grams to moles: divide the known mass by its molar mass, n = m ÷ M. Second, moles to moles: multiply by the mole ratio b/a, read straight off the coefficients of the known (a) and target (b) in the balanced equation. Third, moles to grams: multiply the target moles by its molar mass. Written together, mass of target = (mass of known ÷ molar mass of known) × (b/a) × molar mass of target.

Where do the molar masses come from?

A molar mass is the sum of the atomic masses of every atom in a substance’s chemical formula, taken from the periodic table and expressed in grams per mole. Water (H₂O) is 2 × 1.008 + 16.00 = 18.02 g/mol; carbon dioxide (CO₂) is 12.01 + 2 × 16.00 = 44.01 g/mol. Enter the molar mass of the known substance and of the target substance separately, since they are usually different compounds.

Why convert through moles instead of relating masses directly?

A balanced equation conserves atoms, not mass of individual substances, so the coefficients count particles — moles — rather than grams. Two substances with the same coefficient rarely have the same mass because their molar masses differ. Converting to moles lets the coefficient ratio apply cleanly, and converting back to grams gives an answer you can weigh out. Skipping the mole step and scaling masses by the coefficients directly would give the wrong result whenever the two molar masses are not equal.

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