How to Solve Stoichiometry Problems: Grams to Moles to Ratio to Grams

The mole ratio is the whole method

A balanced equation counts particles, not mass. When you write N2 + 3 H2 giving 2 NH3, the 3 means three hydrogen molecules for every nitrogen molecule — it says nothing about grams, because a hydrogen molecule weighs a fourteenth of what a nitrogen molecule weighs. That is why grams cannot be compared directly and why the conversion to moles is not an optional formality: moles are the only currency in which the coefficients are meaningful.

Placing the ratio the right way up is the second half of the same step. The rule is to put the substance you want on top and the substance you have underneath, so that the unit you are leaving cancels. Going from moles of hydrogen to moles of ammonia, the factor is 2 over 3; going the other way it is 3 over 2. If your answer comes out badly wrong by a small whole-number factor, this inversion is almost always the cause.

Limiting reactant: divide by the coefficient, do not just compare moles

The worked example is built to expose the classic error. Nitrogen is present at 50.0 g against only 10.0 g of hydrogen, and it still is not the limiting reactant. In moles the gap looks even wider in hydrogen's favour: 1.785 mol of N2 against 4.960 mol of H2. Anyone who stops at that comparison concludes that nitrogen limits, and gets the wrong answer.

The fix is to compare like with like by dividing each reactant's moles by its own coefficient, which asks how many complete reaction cycles each one could sustain on its own. Nitrogen supports 1.785 cycles, hydrogen only 1.653, so hydrogen runs out first and fixes the yield at 56.3 g of ammonia. Nitrogen consumed is 1.653 mol, or 46.3 g, leaving 3.68 g sitting in the flask unreacted. That leftover is a genuine part of the answer and is often what the second half of the question asks for.

Theoretical, actual and percent yield

Theoretical yield is what the limiting reactant permits if the reaction goes to completion with nothing lost. It comes from the limiting reactant alone; the excess reagent never appears in that calculation, which is exactly why identifying the limiting one first is not a detour. Actual yield is what you weigh at the end, and percent yield is the ratio of the two multiplied by 100 — 45.0 divided by 56.3, or 79.9 percent in the example above.

A percent yield above 100 is not a violation of conservation of mass; it is a message about the sample. The usual cause is residual solvent or water in a product that was weighed before it dried fully, or an impurity carried through from the workup. Treat any figure above 100 as a signal to dry and reweigh rather than as a result to report — and note the reverse case too, since losses on filtration, transfer and recrystallisation routinely put good preparative chemistry in the 60 to 90 percent band without anything having gone wrong.

Frequently asked questions

Do I really have to balance the equation first?

Yes, without exception, because the coefficients are the ratio and there is no other source for it. An unbalanced equation still produces a number, which is what makes the omission dangerous: nothing in the arithmetic complains. Writing N2 + H2 giving NH3 and reading the ratio as one to one would put the ammonia yield out by a factor of two, and the answer would look perfectly reasonable.

What if the problem gives me litres of gas instead of grams?

Then the molar volume replaces the molar mass in the first and last steps: divide litres by 22.414 to get moles, and multiply moles by 22.414 to get litres back. Check which convention your course uses, because standard temperature and pressure has two definitions in circulation — the older 0 degrees Celsius at 1 atm gives 22.414 L/mol, while the current IUPAC standard of 0 degrees Celsius at 100 kPa gives 22.711 L/mol. The tool on this site uses 22.414. If the conditions are not standard at all, drop the shortcut and use the ideal gas law directly.

How do I know which reactant is in excess without doing the whole calculation?

You do not, and short cuts here are false economy. The one legitimate shortcut is that you only ever have to divide each reactant's moles by its coefficient — three divisions for three reactants, and the smallest wins. Everything after that concerns the limiting reactant alone, so the work you have done is not wasted: the same quotient that identified the limiter is the number you multiply by the product's coefficient in the next step.