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D3.2 · Relate Avogadro’s number, moles, and molar mass

Learn to relate avogadro’s number, moles, and molar mass through clear examples and targeted practice.

Ontario Grade 11 Chemistry

Quantities in Chemical Reactions

Ontario Grade 11 Chemistry expectation D3.2

A small sample of a substance contains an enormous number of particles. Chemists need a practical way to count those particles and connect the count to a mass that can be measured on a balance. The mole provides the particle-counting link. Molar mass provides the link between the amount in moles and the sample’s mass. This lesson develops those connections and uses them to solve a sample problem.

What you will learn

1. From tiny particles to a useful counting unit

A balance can measure the mass of a sample, but it cannot count each atom or molecule directly. In a visible sample of water, for example, the particles are far too small and numerous to count one by one. Chemists use a large counting unit instead, much as people use a dozen to mean twelve objects.
A mole is an amount of substance that contains a specified number of particles. That number is Avogadro’s number: 6.022×10236.022 \times 10^{23} particles per mole. A particle may mean an atom, molecule, or other stated chemical unit. The substance tells you which kind of particle is being counted.
The mole is not a particle itself, and it is not a mass. It is a counting unit. One mole of copper contains Avogadro’s number of copper atoms. One mole of water contains Avogadro’s number of water molecules. The particle type changes, but the number of particles in one mole does not.
The symbol nn represents amount in moles. The symbol NN represents the number of particles. Avogadro’s number is written as NAN_A. Since it is a number of particles per mole, it converts between a particle count and an amount in moles.
N=nNAN=nN_A

2. Molar mass connects amount and measurable mass

Mass is the amount of matter in a sample, measured in units such as grams. Molar mass is the mass of one mole of a substance. Its common unit is grams per mole, written as g/mol\mathrm{g/mol}. The substance matters: one mole of different substances generally has different masses.
For an element, use its atomic mass from the periodic table as the numerical value of its molar mass in grams per mole. For a compound, use its chemical formula. Add the atomic masses for all atoms shown in the formula. A subscript tells how many atoms of that element are included. If there is no subscript, the count is one.
For example, the formula H2O\mathrm{H_2O} contains two hydrogen atoms and one oxygen atom in each molecule. Its molar mass is found by adding twice the atomic mass of hydrogen to the atomic mass of oxygen. Use the periodic-table values provided in class, and keep enough digits during the calculation.
The symbol mm represents mass, usually in grams. Molar mass is represented by MM. Multiplying an amount in moles by molar mass gives mass in grams. Dividing mass by molar mass gives amount in moles. Units help show whether the calculation is set up correctly: the mole units cancel when finding mass.
m=nMm=nM

3. Use the relationships together

The two connections can be used separately or together. If you know moles, multiply by Avogadro’s number to find particles. If you know particles, divide by Avogadro’s number to find moles. If you know moles, multiply by molar mass to find mass. If you know mass, divide by molar mass to find moles.
A useful first step is to identify what is known and what is being asked. Then choose the relationship that directly connects them. If a problem asks for both mass and particles, find each from the same amount in moles. Keep the particle name attached to the answer, such as water molecules or copper atoms.
Check units as you work. Particle counts are counts, so they do not have a mass unit. Moles are written as mol\mathrm{mol}. Mass may be in grams, and molar mass may be in g/mol\mathrm{g/mol}. In a calculation, units should cancel in a way that leaves the unit requested.
n=mM=NNAn=\frac{m}{M}=\frac{N}{N_A}

4. Units, rounding, and a reasonableness check

A significant figure is a digit that communicates the precision of a measured or reported value. In multiplication and division, report the result to the same number of significant figures as the value with the fewest significant figures. Exact counting relationships, such as the defined number of particles in a mole, do not limit the reported precision in the same way as a measured value.
Do not round intermediate values too early. Carry extra digits through a calculation, then round the final answer. Include units in each calculation line. A mass should not be reported in particles, and a particle count should not be reported in grams.
Finally, ask whether the answer makes sense. A fraction of a mole contains the same fraction of Avogadro’s number of particles. A mass for a sample is greater when the amount in moles is greater, provided the substance is the same. These checks can reveal a reversed operation or a missing unit.
NA=6.022×1023 mol−1N_A=6.022 \times 10^{23}\ \mathrm{mol^{-1}}

Worked example

Find the mass and number of molecules in a water sample

A water sample contains 0.250 mol0.250\ \mathrm{mol} of water. Find its mass and the number of water molecules. Use atomic masses H=1.01\mathrm{H}=1.01 and O=16.00\mathrm{O}=16.00.
  1. Find the molar mass
    Each water molecule has two hydrogen atoms and one oxygen atom. Add their atomic masses using the formula’s subscripts. The result is the mass of one mole of water.
    M(H2O)=2(1.01)+16.00=18.02 g/molM(\mathrm{H_2O})=2(1.01)+16.00=18.02\ \mathrm{g/mol}
  2. Calculate the mass
    Multiply the amount by the molar mass. The mole units cancel, leaving grams. Keep the unrounded product until the final step.
    m=(0.250 mol)(18.02 g/mol)=4.505 gm=(0.250\ \mathrm{mol})(18.02\ \mathrm{g/mol})=4.505\ \mathrm{g}
  3. Calculate the molecule count
    Multiply the amount in moles by Avogadro’s number. The result counts water molecules because water is made of molecules.
    N=(0.250 mol)(6.022×1023 mol−1)=1.5055×1023N=(0.250\ \mathrm{mol})(6.022\times10^{23}\ \mathrm{mol^{-1}})=1.5055\times10^{23}
  4. Round the results
    The given amount has three significant figures. Round each final result to three significant figures, and name the particle type.
    m=4.51 g,N=1.51×1023 water moleculesm=4.51\ \mathrm{g},\qquad N=1.51\times10^{23}\ \text{water molecules}
Answer: The sample has a mass of 4.51 g4.51\ \mathrm{g} and contains 1.51×10231.51\times10^{23} water molecules.
Check: The calculated mass is close to the mass of one quarter of a mole of water, since the sample contains 0.250 mol0.250\ \mathrm{mol}. The particle count is one quarter of Avogadro’s number, as expected.

Common mistakes and how to avoid them

Treating a mole as a mass or as a single particle.
Correction: A mole is an amount that counts 6.022×10236.022\times10^{23} specified particles.
Using molar mass as though it were the mass of any sample.
Correction: Molar mass is for one mole. Multiply it by the sample’s amount in moles to find that sample’s mass.
Forgetting subscripts while calculating a compound’s molar mass.
Correction: Count each element according to its subscript before adding the atomic masses.
Writing a particle count without saying what the particles are.
Correction: State whether the count refers to atoms, molecules, or another chemical unit.

Lesson summary

Check your understanding

Question 1

How many molecules are in 0.500 mol0.500\ \mathrm{mol} of a substance?
  1. 3.011×10233.011\times10^{23} molecules
  2. 1.204×10241.204\times10^{24} molecules
  3. 0.5000.500 molecules
  4. correctIndex
Show answer and explanation
3.011×10233.011\times10^{23} molecules
Multiply moles by Avogadro’s number: 0.500×6.022×1023=3.011×10230.500\times6.022\times10^{23}=3.011\times10^{23} molecules.

Question 2

What is the molar mass of CO2\mathrm{CO_2} using C=12.01\mathrm{C}=12.01 and O=16.00\mathrm{O}=16.00?
  1. 28.01 g/mol28.01\ \mathrm{g/mol}
  2. 44.01 g/mol44.01\ \mathrm{g/mol}
  3. 44.01 g44.01\ \mathrm{g}
  4. correctIndex
Show answer and explanation
44.01 g/mol44.01\ \mathrm{g/mol}
There is one carbon atom and two oxygen atoms: 12.01+2(16.00)=44.01 g/mol12.01+2(16.00)=44.01\ \mathrm{g/mol}.

Question 3

A sample has a mass of 9.00 g9.00\ \mathrm{g} and a molar mass of 18.0 g/mol18.0\ \mathrm{g/mol}. How much substance is present?
  1. 0.500 mol0.500\ \mathrm{mol}
  2. 162 mol162\ \mathrm{mol}
  3. 2.00 mol2.00\ \mathrm{mol}
  4. correctIndex
Show answer and explanation
0.500 mol0.500\ \mathrm{mol}
Divide mass by molar mass: 9.00 g/(18.0 g/mol)=0.500 mol9.00\ \mathrm{g}/(18.0\ \mathrm{g/mol})=0.500\ \mathrm{mol}.

Key terms

Mole
An amount of substance containing 6.022×10236.022\times10^{23} specified particles.
Avogadro’s number
The number of particles in one mole, 6.022×10236.022\times10^{23}.
Molar mass
The mass of one mole of a substance, commonly expressed in g/mol\mathrm{g/mol}.
Significant figures
Digits that communicate the precision of a measured or reported value.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Chemistry (SCH3U), expectation D3.2. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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