You balanced equations in Types of Chemical Reactions and treated the coefficients as a bookkeeping device — whatever numbers make the atoms come out even. Then The Mole gave those numbers a second job. This page is what happens when you take that second job seriously, and it is the reason the whole unit exists.
The question stoichiometry answers is the one every chemist actually asks: I have this much of that. How much of the other thing will I get?
The coefficients are a ratio of particles, and therefore of moles
Look at
That equation says two hydrogen molecules react with one oxygen molecule. It does not say two grams of hydrogen react with one gram of oxygen, and if you use it that way every answer after it is wrong.
Here is the step that makes it usable. If two molecules of hydrogen react with one molecule of oxygen, then two million react with one million, and two moles react with one mole — the ratio survives being scaled by any number, including . So the coefficients are a mole ratio, and moles are something you can reach from a balance reading.
The mass ratio, for comparison, is 4.03 g of hydrogen to 32.00 g of oxygen. There is no way to see 2 : 1 in those numbers. That is why the conversion to moles is not an extra step you could skip if you were clever — it is the only place where the equation’s information can get into your calculation.
One road, and it is always the same road
Every stoichiometry problem is this journey, or part of it.
graph LR A["mass of A<br/>grams"] -->|"divide by molar mass of A"| B["moles of A"] B -->|"multiply by the mole ratio<br/>from the coefficients"| C["moles of B"] C -->|"multiply by molar mass of B"| D["mass of B<br/>grams"]
Three steps, always in that order. The middle one is the only step that uses the chemistry; the two on the outside are arithmetic with the periodic table. Later pages hang extra approaches onto the two ends — for a solution in Concentration, molar volume for a gas in The Gas Laws — but the road through the middle never changes.
Two things follow that are worth saying plainly:
- The equation must be balanced before you start. An unbalanced equation has no mole ratio to give you. This is not a tidiness rule.
- You never convert mass directly to mass. If you find yourself multiplying a mass by a coefficient, stop — that is the single most common error in this unit, and it produces answers that look reasonable.
Worked: how much carbon dioxide from 10.0 g of propane
Propane burns completely as
The molar mass of propane is g/mol, and of carbon dioxide g/mol.
Ten grams of fuel, thirty grams of carbon dioxide. The extra mass is the oxygen from the air, which is exactly what the equation said would happen — five molecules of it per molecule of propane. Notice also that the answer carries three significant figures because 10.0 g did; the intermediate values were kept to four and rounded only at the end, which is the habit set out in Significant Figures and Units.
This number is the honest version of a claim people make loosely. A barbecue cylinder does not “release its weight” in carbon dioxide. It releases roughly three times its weight, and you can prove it with a periodic table.
What goes wrong, in order of frequency
- The equation was not balanced. Every subsequent step is then arithmetic on a false premise. Balance first, count the atoms, and only then start converting.
- The ratio was used upside down. Write it as a fraction with units in it — — and the wrong way up becomes visible immediately, because the units will not cancel.
- A mass was treated as a mole count. Especially when the numbers are convenient. 44.11 g of propane is one mole; 44 g of anything else is not.
- Rounding partway through. Rounding at each step and then again at the end can move the final digit. Carry one or two extra digits through and round once.
- The answer has no units. A number without units is not an answer to a chemistry question, and “29.9” could be grams, moles, or molecules.
Practise the road in both directions in Stoichiometry Practice.
So far every problem has quietly assumed you have as much of the other reactant as you need. Real reactions rarely oblige, and what happens when one reactant runs out first is Limiting Reagent and Yield — which is also where the number you calculated meets the number you actually recovered.
Curriculum connection
D3.4
explain the quantitative relationships expressed in a balanced chemical equation, using appropriate units of measure (e.g., moles, grams, atoms, ions, molecules)
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D2.5
calculate the corresponding mass, or quantity in moles or molecules, for any given reactant or product in a balanced chemical equation as well as for any other reactant or product in the chemical reaction [AI]
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