VDOE SOL (tech-enhanced, fill in the blank)-style practice: A sample of water, H_2O, has a mass of 36.0 g. (a) Calculate the number of moles. (b) Calculate the number of molecules.

A 2-point conversion item. (a) Moles (1 point): the molar mass of water is 2(1.0) + 16.0 = 18.0 g/mol, so moles = 36.0\ g18.0\ g/mol = 2.00 mol. (b) Molecules (1 point): multiply moles by Avogadro's number, 2.00 × 6.02 × 10^23 = 1.20 × 10^24 molecules. Markers reward dividing mass by molar mass for the moles, then multiplying by Avogadro's number for the particle count, with the units cancelling correctly.

VirginiaChemistrySyllabus dot point

How does the mole connect the mass of a sample to the number of particles it contains?

The mole and molar mass: use the mole, molar mass and Avogadro's number to convert between mass, moles and number of particles.

A focused Virginia SOL Chemistry answer on the mole under CH.3: Avogadro's number, finding the molar mass from the periodic table, and converting between mass, moles and number of particles, the master skill behind all chemical calculations.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-13

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
The mole and Avogadro's number
Molar mass
The mass-mole-particle chain
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What this topic is asking

Standard CH.3 treats the mole as the central unit of chemistry. Virginia expects you to use the mole, molar mass and Avogadro's number to convert between the mass of a sample (grams you can weigh), the number of moles, and the number of particles it contains. This conversion underpins percent composition, stoichiometry, molarity and every quantitative item on the test, so fluency here pays off across the whole exam.

The mole and Avogadro's number

Atoms are far too small and numerous to count individually, so chemists weigh a sample and use the mole to know how many particles it holds. The representative particle depends on the substance: atoms for an element such as neon, molecules for a molecular compound such as $\text{CO}_2$ , and formula units for an ionic compound such as $\text{NaCl}$ .

Molar mass

The periodic-table atomic mass is already the natural-abundance weighted average of the element's isotopes, so you do not average isotopes again to find a molar mass. Remember to multiply each atomic mass by its subscript before summing.

The mass-mole-particle chain

The central relationships are:

\text{moles} = \frac{\text{mass}}{\text{molar mass}}, \qquad \text{particles} = \text{moles} \times (6.02 \times 10^{23})

Rearranged, mass $=$ moles $\times$ molar mass. Lay the calculation out as a chain so the units cancel; if your units do not cancel to what the question asks for, a step is the wrong way round. The molar mass acts as the conversion factor between grams and moles, and Avogadro's number acts as the conversion factor between moles and particles, so the chain is really two factor-label steps stacked together.

A formula is itself a statement about moles: one mole of $\text{C}_6\text{H}_{12}\text{O}_6$ contains six moles of carbon atoms, twelve of hydrogen and six of oxygen, which is the basis of percent composition and stoichiometry. Reading a formula this way also lets you count moles of a particular atom within a compound: one mole of $\text{CaCl}_2$ contains one mole of calcium ions and two moles of chloride ions, so it holds three moles of ions in total. Keeping clear about whether a question wants moles of compound, moles of a specific atom, or a particle count is the difference between a right and a wrong answer.

Mass to particles

A sample of carbon dioxide has a mass of $22.0$ g. How many molecules of $\text{CO}_2$ are present?

step 1 Find the molar mass

$M(\text{CO}_2) = 12.0 + 2(16.0) = 44.0$ g/mol.

step 2 Convert mass to moles

$\text{moles} = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{22.0\ \text{g}}{44.0\ \text{g/mol}} = 0.500$ mol.

step 3 Convert moles to molecules

$\text{molecules} = 0.500\ \text{mol} \times 6.02 \times 10^{23}\ \text{mol}^{-1} = 3.01 \times 10^{23}$ molecules.

step 4 Check

Half a mole should give about half of Avogadro's number, and $3.01 \times 10^{23}$ is half of $6.02 \times 10^{23}$ , so the result is consistent.

Try this

Q1. Calculate the number of moles in $9.0$ g of water (molar mass $18.0$ g/mol). [1 point]

Cue. $\text{moles} = \dfrac{9.0}{18.0} = 0.50$ mol.

Q2. State the number of atoms in $1.0$ mole of helium gas. [1 point]

Cue. $6.02 \times 10^{23}$ atoms, because helium is monatomic (one atom per particle).

Exam-style practice questions

Practice questions written in the style of VDOE exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SOL (multiple choice)1 marksWhat is the molar mass of calcium carbonate,

\text{CaCO}_3

? (A)

57.1

g/mol (B)

68.1

g/mol (C)

100.1

g/mol (D)

116.1

g/mol

Show worked answer →

The answer is (C) $100.1$ g/mol.

Add the molar masses from the periodic table: calcium $40.1$ , carbon $12.0$ , and three oxygen at $16.0$ each ( $48.0$ ). The total is $40.1 + 12.0 + 48.0 = 100.1$ g/mol.

The trap is using only one oxygen; the subscript $3$ means three oxygen atoms, so multiply $16.0$ by $3$ before adding.

SOL (tech-enhanced, fill in the blank)2 marksA sample of water,

\text{H}_2\text{O}

, has a mass of

36.0

g. (a) Calculate the number of moles. (b) Calculate the number of molecules.

Show worked answer →

A 2-point conversion item.

(a) Moles (1 point): the molar mass of water is $2(1.0) + 16.0 = 18.0$ g/mol, so moles $= \dfrac{36.0\ \text{g}}{18.0\ \text{g/mol}} = 2.00$ mol.
(b) Molecules (1 point): multiply moles by Avogadro's number, $2.00 \times 6.02 \times 10^{23} = 1.20 \times 10^{24}$ molecules.

Markers reward dividing mass by molar mass for the moles, then multiplying by Avogadro's number for the particle count, with the units cancelling correctly.

Related dot points

Sources & how we know this

2018 Science Standards of Learning - Chemistry — Virginia Department of Education (2018)
Chemistry Curriculum Framework — Virginia Department of Education (2018)