Atoms are far too small to count one by one, yet a chemist who needs 0.250 mol of calcium carbonate simply weighs out 25.0 g of it. The mole is the unit that makes this possible: it links a mass you can measure on a balance to a number of particles you could never count.

Every calculation in this lesson moves along one path: grams to moles with the molar mass, and moles to particles with Avogadro’s number (Figure 1). Later lessons on formulas and stoichiometry use the same steps.

1. startMassgrams (g)
2. $\times \dfrac{1}{M}$molar mass (g/mol)
3. Amountmoles (mol)
4. $\times N_A$Avogadro’s number
5. answerParticlesatoms, molecules or formula units

Going left: multiply by M, or divide by N_A.Grams and particles always meet at moles.

Learning Objectives

After this lesson, you should be able to:

1. Calculate a molar mass from atomic weights.
2. Convert between mass, amount (mol) and number of particles with the molar mass and Avogadro’s number.

Key Ideas

1. The Mole Is a Counting Unit

A dozen means 12 of anything. A mole (mol) means $6.022 \times 10^{23}$ of anything: atoms, molecules or formula units. This number is Avogadro’s number, $N_A$.

One mole of any substance contains the same number of particles, but different substances have different molar masses, because their particles have different masses (Figure 2). One mole of carbon atoms has a mass of about 12.01 g; one mole of water molecules, 18.02 g.

Key idea: one mole is always 6.022 × 10²³ particles. Only the mass of one mole changes from substance to substance.

2. Molar Mass from Atomic Weights

The molar mass $M$ is the mass of one mole, in g/mol. For an element it equals the atomic weight from the periodic table, in g. For a compound, add the atomic weights of every atom in the formula:

Carbon dioxide, $\mathrm{CO_2}$, has one C atom and two O atoms: $12.011 + 2 \times 15.999 = 44.009$, so $M_{\mathrm{CO_2}} = 44.01\,\text{g/mol}$ (Figure 3).

A subscript after a closing parenthesis multiplies everything inside it. In $\mathrm{Ca(NO_3)_2}$ the group $\mathrm{NO_3}$ appears twice, so there are 2 N atoms and 6 O atoms. In these lessons the sum is rounded to 0.01 g/mol at the end.

Key idea: count every atom first (subscripts, and parentheses times their subscript), then multiply each count by its atomic weight and add.

3. Converting Mass, Moles and Particles

Two conversion factors connect the three quantities (Figure 4). The molar mass links grams and moles; Avogadro’s number links moles and particles:

• $m$ = mass (g), $M$ = molar mass (g/mol)
• $n$ = amount (mol)
• $N$ = number of particles (atoms, molecules or formula units)

Grams and particles are never connected directly: always go through moles.

Worked Example 1: Molar Mass of a Compound

Problem: Calcium nitrate ($\mathrm{Ca(NO_3)_2}$) is used in fertilizers. Calculate its molar mass. Use the atomic weights Ca = 40.078, N = 14.007 and O = 15.999 g/mol.

Step 1. Count the atoms of each element in one formula unit. The subscript 2 after the parenthesis multiplies the N and the three O atoms.

• $\mathrm{Ca}:\ 1\ \text{atom}$
• $\mathrm{N}:\ 1 \times 2 = 2\ \text{atoms}$
• $\mathrm{O}:\ 3 \times 2 = 6\ \text{atoms}$

Step 2. Multiply each count by the atomic weight.

• $1 \times 40.078\,\text{g/mol} = 40.078\,\text{g/mol}$
• $2 \times 14.007\,\text{g/mol} = 28.014\,\text{g/mol}$
• $6 \times 15.999\,\text{g/mol} = 95.994\,\text{g/mol}$

Step 3. Add the contributions and round to 0.01 g/mol.

Final Answer: 164.09 g/mol

Oxygen supplies more than half of the molar mass: 95.994 of 164.086 g/mol (Figure 5).

Your turn: molar mass

Same three steps as Worked Example 1: count the atoms, multiply by the atomic weights, add and round to 0.01 g/mol.

Worked Example 2: Moles from a Mass

Problem: How many moles of calcium carbonate ($\mathrm{CaCO_3}$) are in a 25.0 g sample of chalk? $M_{\mathrm{CaCO_3}} = 100.09\,\text{g/mol}$.

Step 1. List the given information.

• $m = 25.0\,\text{g}$
• $M = 100.09\,\text{g/mol}$
• $n = ?$

Step 2. Divide the mass by the molar mass.

Final Answer: 0.250 mol CaCO₃

1. CaCO₃ given$25.0\,\text{g}$
2. Step 2$\times \dfrac{1\,\text{mol}}{100.09\,\text{g}}$grams to moles
3. Amount$0.2498\,\text{mol} \approx 0.250\,\text{mol}$Final answer to 3 significant figures, set by 25.0 g

Exam tip: check the size. 25.0 g is about a quarter of 100.09 g, the mass of one mole, so the answer must be about a quarter of a mole.

Your turn: moles from a mass

Same method as Worked Example 2: divide the mass in g by the molar mass.

Worked Example 3: Mass from Moles

Problem: What is the mass of 0.750 mol of carbon dioxide ($\mathrm{CO_2}$)? $M_{\mathrm{CO_2}} = 44.01\,\text{g/mol}$.

Step 1. List the given information.

• $n = 0.750\,\text{mol}$
• $M = 44.01\,\text{g/mol}$
• $m = ?$

Step 2. Multiply the moles by the molar mass.

Final Answer: 33.0 g CO₂

1. CO₂ given$0.750\,\text{mol}$
2. Step 2$\times 44.01\,\text{g/mol}$moles to grams
3. Mass$33.01\,\text{g} \approx 33.0\,\text{g}$Final answer to 3 significant figures, set by 0.750 mol

Exam tip: the units guide you: mol × g/mol leaves g. If the units do not cancel to the unit you want, the operation is upside down.

Your turn: mass from moles

Same method as Worked Example 3: multiply the moles by the molar mass.

Worked Example 4: Counting Molecules

Problem: How many water molecules are in 10.0 g of water ($\mathrm{H_2\!O}$)? $M_{\mathrm{H_2\!O}} = 18.02\,\text{g/mol}$.

Step 1. List the given information.

• $m = 10.0\,\text{g}$
• $M = 18.02\,\text{g/mol}$
• $N_A = 6.022 \times 10^{23}\,\text{mol}^{-1}$

Step 2. Convert the mass to moles.

Step 3. Multiply the moles by Avogadro’s number.

Final Answer: 3.34 × 10²³ water molecules

1. Water given$10.0\,\text{g}$
2. Step 2$\times \dfrac{1\,\text{mol}}{18.02\,\text{g}}$grams to moles
3. Amount$0.5549\,\text{mol}$
4. Step 3$\times 6.022 \times 10^{23}\,\text{mol}^{-1}$moles to molecules
5. Molecules$3.342 \times 10^{23} \approx 3.34 \times 10^{23}$Final answer to 3 significant figures

Exam tip: to type the answer in the practice problems, write 3.34e23 or 3.34 x 10^23.

Your turn: counting molecules

Same two steps as Worked Example 4: grams to moles, then moles to molecules.

Common Mistakes

Common mistake: Ignoring subscripts. Adding one atomic weight per element for $\mathrm{Ca(NO_3)_2}$ gives 70.084 g/mol. Multiply each atomic weight by the number of atoms of that element.

Common mistake: Forgetting the parentheses. The 2 after $\mathrm{(NO_3)}$ doubles both N and O. Counting only 1 N and 3 O gives 102.082 g/mol instead of 164.09 g/mol.

Common mistake: Multiplying by the molar mass to get moles. In Worked Example 2, 25.0 g × 100.09 g/mol is about 2.50 × 10³, with units of $\frac{\text{g}^{2}}{\text{mol}}$. To get moles, divide: $n = \frac{m}{M}$.

Common mistake: Dividing the wrong way round. $\frac{M}{m} = \frac{100.09}{25.0}$ = 4.00 is not a number of moles. The mass goes on top. Going from moles to grams, multiply: dividing 0.750 mol by 44.01 g/mol gives 0.0170, not a mass.

Common mistake: Stopping at moles when particles are asked. 0.5549 mol is the amount of water, not the number of molecules. Multiply by $N_A$: dividing by it instead gives 9.21 × 10^-25, far less than one molecule.

Common mistake: Rounding too early. Keep 4 significant figures in intermediate values and round only the final answer to 3. Round molar masses to 0.01 g/mol only after adding all the atomic weights.

Key Takeaway

The mole counts particles by weighing them. Add atomic weights to get the molar mass, divide grams by the molar mass to get moles, and multiply moles by Avogadro’s number to count particles.

Practice: randomized problems

Each problem below uses fresh numbers every time: work it out, check your answer, then open the worked solution to compare your steps.

Your results

This summary collects every practice problem you try on this page and suggests what to review next.

Figures: original diagrams by Dr. Ray’s Learning Center, licensed CC BY 4.0. Atomic weights: IUPAC 2021 abridged standard atomic weights. The practice problems are generated and checked by an audited engine; see the lesson audit log.