Molar Ratio Calculator

Scale a balanced equation to the moles and masses you actually need.

Covers up to three reactants and three products, with the molecular weight built from any two elements.

Updated October 2, 2026

CreatorFrank Zhao

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Introduction

A balanced equation already tells you how the amounts of its reactants and products are related. What it does not tell you is what those amounts become when you actually run the reaction with the quantity sitting on your bench. This calculator bridges that gap: enter the coefficients from a balanced equation, enter one real quantity for any species, and every other mole count and mass falls into place.

It handles up to three reactants and three products at once, which covers the great majority of equations met in introductory chemistry and in routine laboratory work. You choose how much detail you want to supply: just the coefficients for the ratios, add mole counts to scale the equation, or add molecular weights to get masses in grams, milligrams, micrograms, nanograms or atomic mass units.

The one idea worth remembering

Every mole count in a balanced equation is the coefficient multiplied by a single shared factor. That is the whole mechanism behind this page:

n_i = c_i \times k

Type a mole count into any one species and $k$ is fixed for the whole equation. Everything else follows from that one number, in either direction.

Before you start, the coefficients have to come from an equation you have already balanced. If yours is not balanced yet, the chemical equation balancer will do that first.

How to Use

Set the calculation type to molar ratio, moles and mass, choose how many reactants and products your equation has, then fill the species boxes in this order.

1Set Reactants and Products to the number of species on each side of your balanced equation.

2Type each coefficient into its Coefficient in balanced reaction box.

3Give the Molecular Weight of each species. Use the element menus to build it from atoms, or type the figure directly.

4Enter a real mole count for any one species. The remaining mole counts and every mass are then calculated.

Worked example: burning 3 mol of methane

The balanced equation for complete combustion of methane is

\mathrm{CH_4} + 2\,\mathrm{O_2} \rightarrow \mathrm{CO_2} + 2\,\mathrm{H_2O}

Coefficients 1, 2, 1, 2 go into the four Coefficient boxes. Starting from 3 mol of $\mathrm{CH_4}$ gives the shared factor

k = \frac{n_{\mathrm{CH_4}}}{c_{\mathrm{CH_4}}} = \frac{3}{1}

=

3\ \mathrm{mol\ per\ coefficient}

so every other species follows from its own coefficient:

n_{\mathrm{O_2}} = 2 \times 3 = 6\ \mathrm{mol}

\quad

n_{\mathrm{CO_2}} = 1 \times 3 = 3\ \mathrm{mol}

\quad

n_{\mathrm{H_2O}} = 2 \times 3 = 6\ \mathrm{mol}

Each mass is that mole count times its molecular weight:

m_{\mathrm{H_2O}} = 6\ \mathrm{mol} \times 18.0148\ \mathrm{g/mol}

=

108.0888\ \mathrm{g}

The same run gives 48.1248 g of $\mathrm{CH_4}$ , 191.988 g of $\mathrm{O_2}$ and 132.024 g of $\mathrm{CO_2}$ . Your equation, scaled:

1\,\mathrm{CH_4} + 2\,\mathrm{O_2} \rightarrow 1\,\mathrm{CO_2} + 2\,\mathrm{H_2O}

The reduced ratios it reports are worth a second look, because they are not always just the coefficients: $\mathrm{CH_4} : \mathrm{H_2O} = 1 : 1$ even though the coefficients are 1 and 2. That is covered in the method section.

Reading the results

Your equation, scaled lists each species with its coefficient, mole count, molecular weight and mass in one table. Values the calculator worked out for you are shown in blue; anything you typed stays in the normal text colour. Below it, the molar ratio table gives every pair of species reduced to the smallest whole numbers that describe the same proportion.

Calculation Method

Three relationships do all the work. Each of them reads in either direction, which is why you can start from a mass, a mole count or a molecular weight and still get a complete answer.

Moles follow the coefficients

n_i = c_i \times k

\qquad

k = \frac{n_i}{c_i}

The shared factor $k$ is the same for every species in the equation. Type a mole count into any one species and it is fixed for the whole equation.

This is also where the molar ratio comes from: because the factor cancels, the mole ratio of two species equals the ratio of their coefficients.

\frac{n_A}{n_B} = \frac{c_A}{c_B}

Mass is moles times molecular weight

m_i = n_i \times M_i

The molecular weight is entered in either grams per mole or atomic mass units. These are the same number, and this calculator treats them as interchangeable: one dalton is one twelfth of a carbon-12 atom and one mole contains exactly Avogadro's number of them, so a formula mass in amu and a molar mass in g/mol carry identical digits. This is the convention every stoichiometry table uses.

Molecular weight from a formula

M = \sum_{j=1}^{2} a_j \times A_j

Each element row multiplies its atom count by its atomic weight and the results are added. Water, built from two hydrogens and one oxygen:

M_{\mathrm{H_2O}} = 2 \times 1.0079 + 1 \times 15.999 = 18.0148\ \mathrm{g/mol}

Both element rows have to be filled before this relationship can produce a result, so a single-element substance such as $\mathrm{O_3}$ needs its molecular weight typed in rather than assembled.

Reduced molar ratios

The ratio of any two coefficients is divided by their greatest common divisor, leaving the smallest pair of whole numbers that still describes the same proportion. For the methane equation:

\mathrm{CH_4} : \mathrm{O_2} = 1 : 2

\quad

\mathrm{CH_4} : \mathrm{CO_2} = 1 : 1

\quad

\mathrm{CH_4} : \mathrm{H_2O} = 1 : 2

Note the middle one. The coefficients say 1 and 2, but 3 mol of $\mathrm{CH_4}$ and 3 mol of $\mathrm{CO_2}$ stand in a one-to-one ratio on that run, because $\mathrm{CO_2}$ was formed in the same amount that was consumed.

Reduction only applies when both coefficients are whole numbers. A fractional coefficient is shown exactly as entered rather than forced into a whole-number form.

Units

Mole counts work in $\mathrm{mol}$ or in individual molecules. The conversion uses the Avogadro constant, whose exact SI value is $6.022\,140\,76 \times 10^{23}\ \mathrm{mol^{-1}}$ as fixed by the International Bureau of Weights and Measures, so no rounding creeps in. Six moles is 3,613,284,456,000,000,000,000,000 particles exactly.
Molecular weights work in $\mathrm{g/mol}$ or $\mathrm{amu}$ , as described above.
Masses work in $\mathrm{g}$ , $\mathrm{mg}$ , $\mathrm{\mu g}$ , $\mathrm{ng}$ or $\mathrm{u}$ . Switching unit never changes the underlying quantity, only how it is written.

Atomic weights used

The element menus fill conventional atomic weights rounded for classroom work, such as 1.0079 for hydrogen and 12.01 for carbon. These are close to, but not identical with, the current published standard atomic weights, which the International Commission on Isotopic Abundances and Atomic Weights gives as intervals reflecting natural variation, for example [1.00784, 1.00811] for hydrogen and [12.0096, 12.0116] for carbon.

The difference is far below the precision of almost any coursework answer, but if your course specifies particular figures, type them into the Atomic mass box and they are used instead.

To look up the molar mass of a whole formula at once rather than atom by atom, the molar mass calculator works from a formula string.

Worked Examples

Starting backwards from a mass you already weighed out

Same methane equation, but you have weighed 100 g of oxygen and need to know what it will consume. Enter 100 into the second reactant's Mass box instead of entering a mole count anywhere.

n_{\mathrm{O_2}} = \frac{100\ \mathrm{g}}{31.998\ \mathrm{g/mol}}

=

3.1251953\ \mathrm{mol}

That fixes the shared factor, so:

n_{\mathrm{CH_4}} = 1.5625977\ \mathrm{mol} \rightarrow 25.066567\ \mathrm{g}

\quad

n_{\mathrm{H_2O}} = 3.1251953\ \mathrm{mol} \rightarrow 56.299769\ \mathrm{g}

About 25.07 g of methane and, in theory, 56.30 g of water. Compare that theoretical figure with whatever you actually isolate and the gap is your yield.

Checking a molar mass against its own formula

A supplier lists a drying agent at 18.015 g/mol and you want to confirm that against the formula. Pick oxygen for the first element row and hydrogen for the second, then set the atom counts.

1 \times 15.999 + 2 \times 1.0079 = 18.0148\ \mathrm{g/mol}

That matches. This route is the one to use when a figure on a label or in a paper needs a sanity check rather than a blind substitution.

Plausible-magnitude masses for very large molecules

Molecular weights span an enormous range, so units matter. A 6 mol portion of water is about 108 g, but expressed in nanograms that same portion is 108,088,800,000 ng, and counting individual molecules gives 3,613,284,456,000,000,000,000,000 of them. Switching units never changes the arithmetic, so it is safe to enter the value in whichever unit your protocol uses.

Tips & Best Practices

Enter one quantity, not all of them

The whole equation is driven by a single factor. If you type mole counts for two species that do not agree with the coefficients, the calculator has to choose which value gives way, and it will silently hand the answer to a different box. Enter one reliable quantity and let the rest be calculated.

Watch for the box that loses its value

When you edit a figure the calculator had worked out, that result moves elsewhere to keep the equation self-consistent. If a number you expected to stay put changes instead, look for the box that turned blue: that is the one being recomputed, and it usually means the two values you entered disagreed.

Fill in every element row you use

A molecular weight can only be assembled once both element rows are filled. For a two-element compound that is natural. For a single-element substance, or for anything with three or more elements, type the molecular weight straight into its box instead.

Use the reduced ratios to check your reading

If the reduced ratio between two species is not what your textbook states, work out which run produces that relationship. Most often it is a genuine result for the quantity you entered, as with $\mathrm{CH_4} : \mathrm{CO_2} = 1 : 1$ above.

Frequently Asked Questions

Why is the ratio between two species not always the same as their coefficients?

The coefficients describe the equation as written, which is always reduced to the smallest whole numbers that describe it. A ratio between two specific coefficients can still reduce further on its own. With 1 and 2 the pair reduces to 1 and 2, but with 1 and 1 the pair is already 1 and 1. Both are correct; the ratio table reports the simplest form for the pair you ask about.

Can I enter masses and get mole counts?

Yes. Mass is mole count multiplied by molecular weight, so entering a mass with a known molecular weight solves back to moles and then on to every other species. Each species needs its molecular weight filled in before its mass carries any meaning.

Why does the molecular weight stay blank after I fill in one element?

Both element rows have to be complete before the molecular weight is determined, because an unfinished formula has no single value to report. Fill the second row, or type the molecular weight directly, and it will appear.

Do I have to use the atomic weights the element menus fill in?

No. Each atomic mass box accepts any value you type, which is the right approach if your course, your supplier or a paper specifies different figures. The element menus simply save you looking them up.

Limitations

•This calculator does not balance equations. Every coefficient you enter is used exactly as typed, so an unbalanced equation produces confident but meaningless results.
•It models a single reaction to completion. It knows nothing about equilibrium, reversible steps, side reactions, excess reagents or temperature effects on yield.
•Only two element rows are available per species, so longer formulas need their molecular weight typed in. Purity, hydration and isotope differences are not modelled.
•Coefficients must be greater than zero and mole counts, molecular weights and masses cannot be negative. A molecular weight above 2000 g/mol is flagged, since nothing in the periodic table comes close.
•These are theoretical figures. Real bench results depend on losses, incomplete reaction and measurement uncertainty, so treat them as the target rather than the outcome.

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