Mass-to-Mass Stoichiometry

Three conversions collapsed into one line: grams of A to moles of A, moles of A to moles of B through the balanced coefficients, then moles of B back to grams. Written out it is m_B = (m_A/M_A)(b/a)M_B, and it is the single most-used calculation in a teaching lab, because a balance is the only instrument on the bench that measures anything directly.

Worked: burning hydrogen, 2H₂ + O₂ → 2H₂O. Take 4.032 g of hydrogen, molar mass 2.016 g/mol, so 2.000 mol. The coefficients on H₂ and H₂O are both 2, so 2.000 mol of water. Times 18.015 g/mol gives 36.03 g. Note that the mass grew ninefold while the moles did not change at all — the mole ratio was 1:1 and every gram of the difference came from the oxygen.

This answer is the theoretical yield, and it is only true if A is the limiting reagent. The equation as written assumes there is enough of everything else. In a real flask there rarely is. The test is to run this calculation from each reactant in turn and take the smallest answer: that reactant is the limiter, and its figure is the real ceiling. Doing it from whichever reactant the question mentions first is how a calculation comes out generously and confidently wrong.

And limiting is decided by moles, not by mass. Given 100 g of hydrogen and 100 g of oxygen, the masses match and the amounts do not — 49.6 mol against 3.13 mol, with the reaction needing two hydrogens per oxygen. Oxygen limits by a factor of eight. Hydrogen simply looks generous because it is light.

What comes out of the flask is always less than this. Percent yield compares the two, and a figure above 100% means the product is still carrying solvent it has not finished drying out of — stoichiometry itself never returns more than it predicts.