Stoichiometry: The Mole, Empirical Formula and the Calculations That Follow

A mole is a number. That is all it is.

It is 6,02 × 1023, in the same way that a dozen is 12 and a ream is 500. Chemists use such an enormous one because atoms are so small that any sensible amount of a substance contains an absurd number of them.

Get that straight and the whole of this chapter is one calculation done from different starting points. Get it muddled and every question looks like a new problem.

The mole and Avogadro's number

One mole of anything contains 6,02 × 1023 particles. That figure is Avogadro's number, NA, and it is on your data sheet, so it does not need memorising.

n = N ÷ NA

where n is the number of moles and N is the number of particles.

Say what the particles are. A mole of water is 6,02 × 1023 water molecules, which is 1,81 × 1024 atoms, because each molecule holds three. Questions exploit that difference constantly.

Molar mass, and why it is the same number as the periodic table

Molar mass, M, is the mass of one mole of a substance, in g·mol-1.

It is numerically the same as the relative atomic or molecular mass, so you read it straight off the periodic table and add up. Carbon is 12, so one mole of carbon is 12 g. Water is 2(1) + 16 = 18, so one mole of water is 18 g.

n = m ÷ M

This is the most used equation in Grade 11 chemistry. Mass in grams, never kilograms. Putting 0,020 kg into it instead of 20 g gives an answer a thousand times too small and it happens in every test.

Molar volume, for gases only

One mole of any gas occupies 22,4 dm3 at STP.

n = V ÷ Vm, with Vm = 22,4 dm3·mol-1

The remarkable part is the word any. A mole of hydrogen and a mole of carbon dioxide occupy the same volume at the same temperature and pressure, despite one weighing 2 g and the other 44 g. At these separations the size of the molecules barely matters, only how many there are.

STP means 0 °C and 101,3 kPa, which gives 22,4 dm3·mol-1. You will find 24,5 dm3·mol-1 quoted elsewhere online, for 25 °C. That is the wrong value for a South African exam. Use 22,4.

And it only works for gases. There is no molar volume of a solid or a liquid worth using.

Concentration

Concentration is moles per cubic decimetre of solution.

c = n ÷ V, with V in dm3

Combine it with n = m/M and you get the version that saves a step:

c = m ÷ (M × V)

1 dm3 = 1 litre = 1 000 cm3. Volumes in chemistry arrive in millilitres and cubic centimetres and the formula wants cubic decimetres, so divide by 1 000 before you start. This is the second most common arithmetic slip in the chapter after grams and kilograms.

It is the volume of the solution, not the volume of water you added. Dissolving 5,85 g of salt and topping up to 1 dm3 gives a 0,1 mol·dm-3 solution. Dissolving it in 1 dm3 of water gives something slightly more concentrated, because the salt takes up room.

Where this fits in the curriculum

Subject Physical Sciences
Grade 11
Term 2
Topic Chemical change, quantitative aspects
Status Theory. No practical and no formal assessment

Percentage composition

What fraction of a compound's mass is a given element.

% element = mass of the element ÷ mass of the compound × 100

For water, M = 18 g·mol-1, of which 16 is oxygen:

%O = 16 ÷ 18 × 100 = 88,9 %

Water is nearly nine tenths oxygen by mass and one tenth hydrogen, which surprises people who are thinking about the two hydrogens in the formula. Formula counts atoms. Percentage composition weighs them.

Empirical and molecular formula

The empirical formula is the smallest whole number ratio of the elements in a substance. The molecular formula is what the molecule actually is.

They are often different, and several compounds can share one empirical formula.

Compound Molecular formula Empirical formula
Formaldehyde CH2O CH2O
Acetic acid C2H4O2 CH2O
Glucose C6H12O6 CH2O

All three have the same empirical formula and nothing else in common. That is why an empirical formula on its own is not enough to identify a substance, and why questions give you the molar mass as well.

Finding the empirical formula, in five steps

Take 0,4026 g of a substance made of 0,1610 g carbon, 0,0268 g hydrogen and 0,2148 g oxygen.

  1. Percentage composition.
    %C = 0,1610 ÷ 0,4026 × 100 = 40,0 %
    %H = 0,0268 ÷ 0,4026 × 100 = 6,7 %
    %O = 0,2148 ÷ 0,4026 × 100 = 53,3 %
  2. Moles of each element in 100 g. The percentages become grams.
    n(C) = 40 ÷ 12 = 3,3 mol
    n(H) = 6,7 ÷ 1 = 6,7 mol
    n(O) = 53,3 ÷ 16 = 3,3 mol
  3. Write the mole ratio. C : H : O = 3,3 : 6,7 : 3,3
  4. Divide by the smallest. 3,3/3,3 : 6,7/3,3 : 3,3/3,3 = 1 : 2 : 1
  5. Write the formula. CH2O

Step 2 is the one that gets skipped. Learners divide the percentages by each other, or take the ratio of the percentages directly. You cannot compare masses, only moles, because the atoms have different masses. Converting to moles is the entire point of the step.

And then the molecular formula

You need the molar mass, which the question will give you.

Empirical formula CH2O has a mass of 12 + 2 + 16 = 30 g·mol-1. If the substance has M = 180 g·mol-1, then 180 ÷ 30 = 6, so the molecular formula is C6H12O6. Glucose.

The route through every calculation in this chapter

Nine different-looking questions, one method.

given mass or volume → moles → the balanced ratio → moles asked → mass or volume asked

Every stoichiometry question is that road. What changes is only which door you come in by:

You are given Get to moles with
A mass n = m / M
A volume of gas at STP n = V / 22,4
A volume and a concentration n = cV
A number of particles n = N / NA

The middle of the road is always the same: convert to moles, use the coefficients in the balanced equation to cross to the substance you were asked about, then convert back out.

Balance the equation first, every time. The ratio step reads the coefficients, so an unbalanced equation gives a confidently wrong answer with no warning.

The limiting reagent, and the mistake almost everyone makes

When you are given amounts of two reactants, one of them runs out first. That one is the limiting reagent, and it decides how much product you can get. The other is in excess and some of it is left over.

The mistake: picking the reactant with the smaller mass.

That is wrong roughly half the time. It is about moles and the balanced ratio, not grams. Two grams of hydrogen is a whole mole; two grams of iodine is barely a hundredth of one.

How to actually do it

  1. Convert both given amounts to moles
  2. Divide each by its coefficient in the balanced equation
  3. The smaller answer is the limiting reagent

For 2H2 + CO → CH3OH, starting with 6,0 g H2 and 37,25 g CO:

  • n(H2) = 6,0 ÷ 2 = 3,0 mol, divided by its coefficient 2 gives 1,5
  • n(CO) = 37,25 ÷ 28 = 1,33 mol, divided by its coefficient 1 gives 1,33

1,33 is smaller, so carbon monoxide is limiting, even though there is over six times as much of it by mass.

Everything after that point is calculated from the limiting reagent and nothing else. The excess reactant takes no further part in the arithmetic except when you are asked how much of it is left.

Percentage yield

The theoretical yield is what the stoichiometry says you should get. The actual yield is what comes out of the flask, and it is almost always less.

percentage yield = actual yield ÷ theoretical yield × 100 %

This one has a page of its own, because there is a practical attached to it and because yields over 100 % need explaining properly. See percentage yield, Grade 11.

The five mistakes that cost the most marks

Mistake What it does
Mass in kilograms in n = m/M Answer 1 000 times too small
Volume in cm3 in c = n/V Answer 1 000 times too big
Not balancing the equation first The ratio step is wrong, and nothing warns you
Picking the limiting reagent by mass Wrong about half the time
Taking a ratio of percentages instead of moles Empirical formula comes out wrong

Four of those five are unit errors, not chemistry errors. The chemistry in this chapter is one road walked four ways. The marks are lost on grams and cubic decimetres.

A worked problem, all the way through

What mass of magnesium oxide forms when 4,86 g of magnesium burns completely in oxygen?

  1. Balance. 2Mg + O2 → 2MgO
  2. Given to moles. M(Mg) = 24,3 g·mol-1, so n(Mg) = 4,86 ÷ 24,3 = 0,200 mol
  3. Ratio. 2 Mg makes 2 MgO, so 1 : 1. n(MgO) = 0,200 mol
  4. Moles to mass. M(MgO) = 24,3 + 16 = 40,3 g·mol-1. m = nM = 0,200 × 40,3 = 8,06 g

The product is heavier than the metal you started with, which learners often assume must be a mistake. It is not. The oxygen came from the air and it has mass.

Related pages

The one piece of apparatus this chapter needs

A balance. Every mass-based calculation on this page assumes a learner can weigh something to a sensible precision, and the percentage yield practical is impossible without one.

They are in the balances and scales range. For Grade 11 stoichiometry a 0,01 g readability is ample; the 0,001 g precision balances are for Grade 12 titration work.