Calculate the number of moles in 22.0 g of carbon dioxide, CO_2 (gram-formula mass 44.0 g/mol). [2 points]

New YorkChemistrySyllabus 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 and gram-formula mass to convert between the mass of a substance, the number of moles, and the number of particles.

A focused Regents Chemistry answer on the mole and gram-formula mass: Avogadro's number, how to find the molar mass from the periodic table, and the mass-mole-particle conversions, using the mole formulas on Table T of the Reference Tables.

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

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What this topic is asking
The mole and Avogadro's number
Gram-formula mass
The mass-mole-particle chain
Reading a formula in moles
Try this

What this topic is asking

The Core Curriculum asks you to use the mole as the bridge between the mass of a sample (grams you can weigh) and the number of particles it contains. You must find the gram-formula mass (molar mass) from the periodic table and convert between mass, moles and particles. The mole formula is on Table T of the Reference Tables, and this skill underpins every quantitative question in the course.

The mole and Avogadro's number

The mole exists because atoms are far too small and numerous to count one at a time. By fixing an enormous count, a chemist can weigh a sample and 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 $\text{NaCl}$ .

Gram-formula mass

The periodic-table atomic mass is already the natural-abundance weighted average of an element's isotopes, so you do not need to average isotopes again to find a gram-formula mass. The Regents often calls this quantity "gram-formula mass" rather than "molar mass", but they mean the same thing.

The mass-mole-particle chain

Table T provides the central relationship:

\text{number of moles} = \frac{\text{given mass}}{\text{gram-formula mass}}

Rearranged, mass $=$ moles $\times$ gram-formula mass. To move between moles and particles, use Avogadro's number:

\text{number of particles} = \text{moles} \times (6.02 \times 10^{23})

Lay the calculation out as a chain so that the units cancel; if your units do not cancel to what the question asks for, you have a step the wrong way round.

Mass to particles

A sample of pure water has a mass of $9.0$ g. How many molecules of water are present?

step 1 Find the gram-formula mass

$M(\text{H}_2\text{O}) = 2(1.0) + 16.0 = 18.0$ g/mol.

step 2 Convert mass to moles (Table T)

$\text{moles} = \dfrac{\text{given mass}}{\text{gram-formula mass}} = \dfrac{9.0\ \text{g}}{18.0\ \text{g/mol}} = 0.50$ mol.

step 3 Convert moles to molecules

$\text{molecules} = 0.50\ \text{mol} \times 6.02 \times 10^{23}\ \text{mol}^{-1} = 3.0 \times 10^{23}$ molecules.

step 4 Check the answer

Half a mole should give about half of Avogadro's number, and $3.0 \times 10^{23}$ is indeed half of $6.02 \times 10^{23}$ , so the result is consistent.

Reading a formula in moles

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 moles of hydrogen atoms and six moles of oxygen atoms. This mole-ratio reading of a formula is the basis of percent composition and of all reaction stoichiometry, so being fluent with grams-to-moles now pays off across the whole exam.

Try this

Q1. Calculate the number of moles in $22.0$ g of carbon dioxide, $\text{CO}_2$ (gram-formula mass $44.0$ g/mol). [2 points]

Cue. $\text{moles} = \dfrac{22.0}{44.0} = 0.500$ mol.

Q2. State the number of molecules in $2.0$ mol of any gas. [1 point]

Cue. $2.0 \times 6.02 \times 10^{23} = 1.2 \times 10^{24}$ molecules.

Exam-style practice questions

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

Regents (Part B-2 style)3 marksA sample of glucose,

\text{C}_6\text{H}_{12}\text{O}_6

, has a mass of

36.0

g. (a) Show the numerical setup for the gram-formula mass of glucose. (b) Calculate the gram-formula mass. (c) Determine the number of moles in the sample.

Show worked answer →

A 3-point constructed-response item testing gram-formula mass and the mole formula from Table T.

(a) Setup (1 point): $6(12.0) + 12(1.0) + 6(16.0)$ .
(b) Gram-formula mass (1 point): $72.0 + 12.0 + 96.0 = 180.\,\text{g/mol}$ .
(c) Moles (1 point): using Table T, $\text{moles} = \dfrac{\text{given mass}}{\text{gram-formula mass}} = \dfrac{36.0\ \text{g}}{180.\ \text{g/mol}} = 0.200$ mol.

Markers reward a correct mass setup using periodic-table values, the correct sum, and dividing mass by gram-formula mass for the moles.

Regents (Part A style)1 marksWhat is the total number of atoms in

1.0

mole of

\text{CO}_2

? (1)

6.0 \times 10^{23}

(2)

1.2 \times 10^{24}

(3)

1.8 \times 10^{24}

(4)

3.0 \times 10^{23}

Show worked answer →

A 1-point Part A item on Avogadro's number. The answer is (3) $1.8 \times 10^{24}$ .

One mole of $\text{CO}_2$ contains $6.0 \times 10^{23}$ molecules. Each molecule has three atoms (one carbon and two oxygen), so the number of atoms is $3 \times 6.0 \times 10^{23} = 1.8 \times 10^{24}$ atoms.

The trap is stopping at the number of molecules; the question asks for atoms, so multiply by the atoms per molecule.

Related dot points

Sources & how we know this

Physical Setting/Chemistry Core Curriculum — New York State Education Department (2002)
Reference Tables for Physical Setting/Chemistry, 2011 Edition — New York State Education Department (2011)