The mole and stoichiometry: 4 steps

In Year 3 of the liceo scientifico (Italian science-focused high school), the mole and stoichiometric calculations are taught between late September and October, just after (or together with) the periodic table and chemical bonds. The test typically arrives between late November and early December, nine to eleven weeks from now. It is no accident that this is the topic behind most failing grades in science in November and December: it is the first point of the year where knowing the theory is no longer enough. You also have to do the sums correctly, in the right order.

Why the mole sends half the class into a panic

Every year I see the same thing in my lessons. The student has understood what a mole is as an idea (“it is a number, like a dozen, only it is worth $6.022 \times 1 0^{23}$ ”), but gets stuck as soon as they have to use it in an exercise with grams. The problem is almost never conceptual: it is procedural. There is no fixed method to repeat the same way every time, and without one every exercise becomes a new problem to invent from scratch.

One thing worth saying straight away: I do not explain here how to balance an equation. I have already done that, step by step, in this post on balancing chemical equations, and in Year 3 that method is taken for granted. Nor do I explain the naming of compounds again. You will find it in this post on traditional and IUPAC nomenclature, which is useful here only for reading correctly the formulas that appear in the reactions. The heart of this article is the calculations: mass-to-mole conversions, stoichiometric ratios, limiting reagent, percentage yield. If your school track has fewer hours of chemistry a week, the same content at the level of the liceo delle scienze umane (human sciences high school) is in this other post.

Molar mass: the bridge between grams and moles

Before any calculation you need one piece of data: the molar mass of each species, that is, how many grams one mole of that substance weighs. You read it directly from the periodic table, element by element, and then add it up over the formula. Here are the values I use most often (rounded, as you almost always do in a test):

Element	Molar mass (g/mol)
H	$1$
C	$12$
N	$14$
O	$16$
Na	$23$
S	$32$
Fe	$56$

For a compound you add up the molar mass of every atom present, counted as many times as it appears: the molar mass of ammonia $N H_{3}$ is $14 + 3 \times 1 = 17$ g/mol; that of carbon dioxide $C O_{2}$ is $12 + 2 \times 16 = 44$ g/mol.

The four-step method

This is the scheme I give to all my Year 3 students, and it never changes, however complicated the reaction looks:

Balance the equation. If it is not already balanced, go no further. Every wrong coefficient spoils the rest of the calculation (full method in the post linked above).
Go from grams to moles. Use the molar mass of the reagent whose mass you know, with $n = m / M$ : $n$ is the number of moles, $m$ the mass in grams, $M$ the molar mass.
Use the mole ratio given by the balanced coefficients to go from the moles of the species you know to the moles of the species you want.
Go back from moles to grams, if the question asks for it, by multiplying by the molar mass of the target species: $m = n \times M$ .

The conversion in step 2 is the one you use on every line of every exercise:

n = \frac{m}{M}

Four steps, always in the same order: balance, grams→moles, mole ratio, moles→grams. There is no fifth way, and no need to invent a different one each time.

Limiting reagent: who runs out first

When an exercise gives you the quantities of both reagents, the hidden question is always the same: which of the two runs out first and stops the reaction? That one is the limiting reagent, and it alone determines how much product actually forms. The other reagent is partly left over, in excess.

The safest way to find it: calculate the moles of each available reagent, divide each by its own coefficient in the balanced equation, and compare the two ratios. The reagent with the smaller ratio is the limiting one. All the later calculations (product formed, excess reagent left) are always done starting from the moles of the limiting reagent, never from the other one.

Percentage yield: what you really get

The stoichiometric calculation with the four-step method gives you the theoretical yield: how much product you would get if all the limiting reagent were transformed exactly as the equation says. In the laboratory (and in many test problems) the amount actually obtained is lower, because of incomplete reactions, losses of product or side reactions. The ratio between the two is called the percentage yield:

η = \frac{actual mass}{theoretical mass} \times 100

It is always a number less than or equal to 100.

The mistakes I see most often

Skipping the conversion to moles and working directly on grams. The stoichiometric ratio given by the balanced coefficients holds for moles, not for grams: two moles of one compound and two moles of another do not weigh the same at all. Anyone who tries to “balance” the masses directly always gets it wrong, even when the working looked smooth.
Confusing the mole ratio with the ratio of the masses given in the problem. The mole ratio comes only from the coefficients of the balanced equation, not from the quantities you happen to have in the lab that day.
Not checking the limiting reagent when the exercise gives both masses. If you are given the quantities of two reagents, it is almost always a limiting-reagent problem in disguise. Ignoring this step and using “the first reagent you find” gives a wrong result even if all the later calculations are flawless.
Forgetting the units in the intermediate steps. Always writing “mol” or “g” next to every number, step by step, lets you notice at once if you have skipped a step.
Rounding too early. Rounding the moles to the second decimal place and then multiplying by the molar mass amplifies the error in the final result. Keep as many decimal places as possible until the last step and round only the result.

Three worked examples

1. From the mass of a reagent to the mass of the product

The synthesis of ammonia follows the already balanced equation:

N_{2} + 3 H_{2} \to 2 N H_{3}

You react $14$ g of $N_{2}$ with hydrogen in excess. How many grams of ammonia form?

Step 1, balancing: already given, coefficients $1$ , $3$ , $2$ .

Step 2, grams → moles of $N_{2}$ (molar mass $= 28$ g/mol, that is $2 \times 14$ ):

n_{N_{2}} = \frac{14}{28} = 0.5 mol

Step 3, mole ratio $N_{2} : N H_{3} = 1 : 2$ , so:

n_{N H_{3}} = 0.5 \times 2 = 1.0 mol

Step 4, moles → grams of $N H_{3}$ (molar mass $= 17$ g/mol):

m_{N H_{3}} = 1.0 \times 17 = 17 g

From $14$ g of nitrogen, with enough hydrogen, $17$ g of ammonia form.

For this type of exercise, and for stoichiometry in general, we do not yet have a dedicated quiz in our catalogue of practice tests. You will, however, find all the science tests (and those for the other subjects) already available in the practice tests for high school, with the correct answer free and the guided solution that unlocks with your first lesson.

2. Limiting reagent

$5.6$ g of iron react with $4.8$ g of sulfur according to the balanced reaction:

Fe + S \to FeS

How many grams of iron sulfide form, and how much excess reagent is left?

First the starting moles (molar mass Fe $= 56$ g/mol, S $= 32$ g/mol):

n_{Fe} = \frac{5.6}{56} = 0.1 mol, n_{S} = \frac{4.8}{32} = 0.15 mol

The ratio is $1 : 1$ , so I divide each quantity by its own coefficient (which is $1$ for both) and compare:

\frac{0.1}{1} = 0.1 \frac{0.15}{1} = 0.15

The smaller value is the iron’s: iron is the limiting reagent. All the following calculations start from its moles.

Moles of $FeS$ formed ( $1 : 1$ ratio with the iron): $0.1$ mol. Molar mass of $FeS = 56 + 32 = 88$ g/mol:

m_{FeS} = 0.1 \times 88 = 8.8 g

Sulfur used up: $0.1$ mol ( $1 : 1$ ratio with the iron), that is $0.1 \times 32 = 3.2$ g. Sulfur left over in excess:

4.8 - 3.2 = 1.6 g

3. Percentage yield

Propane is reacted with oxygen in excess according to the combustion:

C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O

From $22$ g of propane, $55$ g of carbon dioxide are obtained in the laboratory. What is the percentage yield of the reaction?

Molar mass $C_{3} H_{8} = 3 \times 12 + 8 \times 1 = 44$ g/mol:

n_{C_{3} H_{8}} = \frac{22}{44} = 0.5 mol

Mole ratio $C_{3} H_{8} : C O_{2} = 1 : 3$ :

n_{C O_{2}, theoretical} = 0.5 \times 3 = 1.5 mol

Molar mass $C O_{2} = 44$ g/mol, so the theoretical yield is:

m_{C O_{2}, theoretical} = 1.5 \times 44 = 66 g

Comparing with the actual mass obtained:

η = \frac{55}{66} \times 100 \approx 83.3%

About 11 g of theoretical product did not form, because the reaction was not 100% complete. This happens almost every time, and it is exactly the point of percentage yield.

Before the test

If this topic is giving you trouble in class, or if you are already looking ahead to the admission tests, where stoichiometry turns up regularly, you will find the overview of the subject on science and ready-made tests in the high school practice tests. When you move on to gases, the mole becomes a volume through $P V = n R T$ : the exercises are in Gas laws in general chemistry: worked examples.

If you would like a hand with this topic, the first chat is free → contact.

The mole and stoichiometry: a four-step method