The Mole: Counting by Weighing

A single drop of water holds about 1.7 × 10²¹ molecules. Nobody counts those. What a chemist does instead is weigh the drop, and then convert — and the conversion factor is the whole content of this lesson.

The mole gets taught as a definition to memorize, which is why so many students can recite 6.022 × 10²³ and still stall on a two-step problem. Treat it instead as a unit of counting, exactly like dozen or ream. A dozen eggs is 12 eggs whatever the eggs weigh. A mole of anything is 6.022 × 10²³ of it, whatever it weighs.

One number, three jobs

Avogadro’s number, N_A = 6.022 × 10²³ mol⁻¹, is the bridge between the scale you can measure (grams, on a balance) and the scale reactions actually happen at (individual particles). It is the only bridge, and every quantitative problem in this course crosses it at least once.

Molar mass is the mass of one mole of a substance in grams. For an element it is the number printed on the periodic table, read as g/mol. For a compound it is the sum of the molar masses of every atom in the formula. That is the entire rule — there is nothing else to it.

[Diagram — see the figure in the print workbook.]

Going toward particles: divide by molar mass, then multiply by N_A. Going away: multiply by molar mass, divide by N_A. If you can state which direction you are walking before you touch the calculator, you will not invert a conversion factor.

Molar mass of a compound is just addition

Work through Ca(NO₃)₂ once, slowly, because the parentheses are where marks are lost. The subscript 2 outside the bracket multiplies everything inside it: there are one calcium, two nitrogens and six oxygens, not three.

Molar mass of calcium nitrate. The oxygen row is the one students get wrong: the outer subscript multiplies through the bracket.

“Particles” means whatever the formula names

A mole of H₂O is 6.022 × 10²³ water molecules. But it contains twice that many hydrogen atoms, and three times that many atoms in total. A question that asks for ‘atoms of oxygen’ and one that asks for ‘molecules’ want different numbers from the same sample.

For an ionic compound the unit is the formula unit, not a molecule. One mole of MgCl₂ is 6.022 × 10²³ formula units, which releases one mole of Mg²⁺ and two moles of Cl⁻ when it dissolves. Unit 4 leans on that constantly.

[Diagram — see the figure in the print workbook.]

Worked example 1

Calculate the molar mass of aluminum sulfate, Al₂(SO₄)₃.

1. 1.Count atoms, taking the outer subscript through the bracket: 2 Al, 3 S, and 3 × 4 = 12 O.
2. 2.Al: 2 × 26.98 = 53.96 g/mol.
3. 3.S: 3 × 32.06 = 96.18 g/mol.
4. 4.O: 12 × 16.00 = 192.00 g/mol.
5. 5.Add: 53.96 + 96.18 + 192.00 = 342.14 g/mol.

The oxygen count is the whole question. Twelve oxygens contribute more than half the mass of this compound, so miscounting them as four does not give a slightly wrong answer — it gives one that is out by a factor of nearly two.

Answer: 342.14 g/mol

Worked example 2

How many moles are in 25.0 g of CO₂?

1. 1.Molar mass: 12.01 + 2(16.00) = 44.01 g/mol.
2. 2.Grams to moles, so divide by molar mass.
3. 3.25.0 g × (1 mol / 44.01 g) = 0.568 mol.
4. 4.Sanity check: 25.0 g is a bit over half of 44.01 g, so a bit over half a mole is exactly what to expect.

That last line is worth building as a habit. Most mole answers can be checked against ‘is this more or less than one mole?’ in two seconds, and it catches an inverted conversion factor immediately — the wrong version here gives 1100, which fails the check on sight.

Answer: 0.568 mol CO₂

Worked example 3

How many oxygen atoms are in 12.0 g of glucose, C₆H₁₂O₆?

1. 1.Molar mass: 6(12.01) + 12(1.008) + 6(16.00) = 72.06 + 12.10 + 96.00 = 180.16 g/mol.
2. 2.Moles of glucose: 12.0 g / 180.16 g/mol = 0.0666 mol.
3. 3.Each molecule holds 6 oxygen atoms, so moles of O atoms = 6 × 0.0666 = 0.3996 mol.
4. 4.Atoms: 0.3996 mol × 6.022 × 10²³ mol⁻¹ = 2.41 × 10²³.

Notice where the 6 entered: after converting to moles of glucose, not before. Multiplying the mass by 6 first is a real and common error, and it does not even have the right units.

Answer: 2.41 × 10²³ oxygen atoms

Worked example 4

A sample contains 3.011 × 10²³ formula units of NaCl. What is its mass, and how many chloride ions does it contain?

1. 1.Moles: 3.011 × 10²³ / 6.022 × 10²³ mol⁻¹ = 0.500 mol.
2. 2.Molar mass of NaCl: 22.99 + 35.45 = 58.44 g/mol.
3. 3.Mass: 0.500 mol × 58.44 g/mol = 29.2 g.
4. 4.Each formula unit gives one Cl⁻, so there are 0.500 mol of chloride, i.e. 3.011 × 10²³ ions.

The 1:1 ratio makes the last part look trivial, and that is the point of asking it — the same question about MgCl₂ would have doubled the chloride count. Read the formula, do not assume.

Answer: 29.2 g, containing 3.011 × 10²³ chloride ions

Worked example 5

Which contains more atoms: 10.0 g of helium or 10.0 g of neon? Justify with a calculation.

1. 1.He: 10.0 g / 4.003 g/mol = 2.498 mol.
2. 2.Ne: 10.0 g / 20.18 g/mol = 0.4956 mol.
3. 3.Both are monatomic gases, so moles of atoms equals moles of substance.
4. 4.2.498 mol > 0.4956 mol, by a factor of about 5.

Equal masses never mean equal counts. The lighter the particle, the more of them fit into a fixed mass — which is the same reasoning that makes hydrogen the extreme case at both ends of this course.

Answer: Helium, by roughly a factor of 5 (2.50 mol against 0.496 mol).

Common trap: the balance never reads moles

Every mole problem starts with a quantity you can actually measure, and that is almost always a mass or a volume of solution. The single most common structural error is starting the arithmetic before converting to moles — multiplying grams by a mole ratio, for instance, which is dimensionally meaningless and earns nothing.

Write the units into every line. If they do not cancel to the unit the question asked for, the number is wrong no matter how tidy the algebra looked. AP readers give credit for a correctly set-up conversion with an arithmetic slip; they give nothing for a right-looking number that arrived by an invalid route.

Hack: say the sentence before you press a key

Before any conversion, say out loud: “I have grams, I want particles, so I go grams → moles → particles.” Then write that arrow chain down the page and hang one fraction under each arrow.

It sounds childish and it removes almost every inverted-conversion error, because the direction is now on the paper instead of in your head. Under time pressure in May, that is what fails first.