How this instrument works
A molar ratio (or mole ratio) is the proportion, taken straight from a balanced chemical equation's coefficients, between the amounts of any two substances involved in a reaction. If a balanced equation reads 2 A + 3 B -> products, the molar ratio of A to B is 2:3, meaning every 2 moles of A that react consume exactly 3 moles of B. This instrument applies that ratio directly: molB = molA x (coeffB / coeffA).
The coefficients work as a ratio because a balanced equation is fundamentally a statement about relative particle counts, not absolute amounts — 2 A + 3 B -> products is equally true whether you're reacting 2 molecules with 3, or 2 moles (roughly 1.2 x 10^24 molecules) with 3 moles. Balancing the equation is what guarantees the atom count matches on both sides; once balanced, the coefficients become a fixed conversion factor between any two species in the reaction, usable at any scale.
This is the core arithmetic step in stoichiometry, the branch of chemistry concerned with quantitative relationships between reactants and products. It sits between two other common conversions: turning a measured mass into moles (using molar mass) on one side, and turning the resulting moles of a different substance back into a mass or volume on the other. The molar ratio itself is the bridge in the middle — it only works between substances that appear in the same balanced equation.
- Enter the balanced equation's coefficient for the substance you already know the amount of into Stoichiometric coefficient of A.
- Enter the coefficient for the substance you want to find into Stoichiometric coefficient of B.
- Enter the known amount, in moles, into Moles of A (mol).
- Read Moles of B (mol) beneath the inputs — the instrument applies molA x (coeffB / coeffA) automatically.
- Make sure both coefficients come from the same balanced equation — a molar ratio only holds between substances that actually appear together in one balanced reaction.
Worked example — a 2:3 stoichiometric ratio
Enter 2 into Stoichiometric coefficient of A, 3 into Stoichiometric coefficient of B, and 4 into Moles of A (mol) — describing a balanced equation of the form 2 A + ... -> 3 B + ..., with 4 moles of A available. Moles of B (mol) reads 6.0: 4 x (3/2) = 6.0.
The ratio 3/2 = 1.5 means every mole of A produces one and a half moles of B, so 4 moles of A scales up to 4 x 1.5 = 6 moles of B. Because the coefficient of B (3) is larger than the coefficient of A (2), the reaction always yields more moles of B than the moles of A consumed, regardless of how much A you start with.
Questions
Where do the coefficients come from?
From a correctly balanced chemical equation for the specific reaction you're working with — they are not something you choose, but something you determine (or look up) by balancing the equation so the number of atoms of each element matches on both sides. An unbalanced equation gives coefficients that don't represent a real, physically valid molar ratio, so balancing must come before this calculation.
Does the molar ratio work the same for products as for reactants?
Yes — the coefficient ratio applies to any two species in the balanced equation, whether both are reactants, both are products, or one is a reactant and the other a product. The equation 2 A + 3 B -> 4 C, for instance, lets you convert between A and B, between A and C, or between B and C, always using the two relevant coefficients as the ratio.
Why is the ratio coeffB/coeffA and not the reverse?
Because you're solving for moles of B starting from moles of A, so the coefficient of the substance you're solving for (B) belongs on top and the coefficient of the substance you already have (A) belongs on the bottom — this cancels the 'per mole of A' unit correctly. Flipping the ratio by mistake is the most common error in mole-ratio stoichiometry, and it produces a reciprocal (and usually implausible) answer.
Can I use a molar ratio to convert directly between masses?
Not directly — the molar ratio only converts between moles. To go from a mass of A to a mass of B, first convert the mass of A to moles using A's molar mass, apply the molar ratio to get moles of B, then convert that back to a mass using B's molar mass. Skipping the moles step and applying the coefficient ratio to masses directly gives a wrong answer, since the ratio is defined in terms of particle counts, not mass.
What if the reaction has more than two substances I care about?
Apply the same two-coefficient ratio repeatedly, once for each pair. If a balanced equation involves A, B, and C, you'd run this calculation once using A and B's coefficients to get moles of B, and again using A and C's (or B and C's) coefficients to get moles of C — each pairing uses only its own two coefficients from the same balanced equation.