Mass, Moles, Formula Mass, and Particles Calculations

Good to see you again. In the previous lesson, you translated word descriptions into balanced equations and saw that coefficients represent fixed particle and mole ratios. This lesson focuses on the numerical language that sits underneath those equations: the mole.

By the end, you should be able to move confidently between a measurable mass in grams, an amount in moles, a relative formula mass, and a number of atoms, molecules, ions, or formula units. These are routine but high-mark calculations, so the emphasis is on a method that remains reliable under exam pressure.

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The mole: the bridge between grams and particles

Chemists cannot count individual atoms or molecules directly, but they can measure a substance’s mass. The mole connects these two scales.

One mole of any substance contains Avogadro’s constant:

$$N_A=6.02\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$$

So:

• $1$ mol of magnesium contains $6.02\times 10^{23}$ magnesium atoms.
• $1$ mol of water contains $6.02\times 10^{23}$ water molecules.
• $1$ mol of sodium chloride contains $6.02\times 10^{23}$ sodium chloride formula units.
• $1$ mol of chloride ions contains $6.02\times 10^{23}$ chloride ions.

The number is huge because particles are extremely small. The essential idea is simple: a mole is a counting unit, just as a dozen means 12 items — but a mole means $6.02\times 10^{23}$ items.

The diagram gives the overall route, but do not learn it as disconnected arrows. Instead, treat moles as the central checkpoint:

$$\text{mass}⟷\text{moles}⟷\text{particles}$$

If a question begins with mass and asks for particles, it must pass through moles. If it begins with particles and asks for mass, it must also pass through moles.

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Mass, molar mass, and amount in moles

The relationship between mass and amount of substance is:

$$n=\frac{m}{M}$$

where:

• $n$ is amount in moles, in $\mathrm{mol}$;
• $m$ is mass, in $\mathrm{g}$;
• $M$ is molar mass, in $\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$.

You may also see this written as:

$$\text{moles}=\frac{\text{mass}}{M_r}$$

at A-level. Strictly, $M_r$ is relative formula mass and has no unit, whereas molar mass has units of $\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. Numerically, however, they have the same value:

$$M_r({\mathrm{H}}_2\mathrm{O})=18.0$$ $$M({\mathrm{H}}_2\mathrm{O})=18.0\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$$

For an element, calculate $M_r$ from its $A_r$. For a compound, add the relative atomic masses of every atom shown in its formula.

Calculating relative formula mass accurately

For magnesium nitrate:

$$\mathrm{Mg}(\mathrm{N}{\mathrm{O}}_3{)}_2$$ $$M_r=24.3+2(14.0+3\times 16.0)$$ $$M_r=148.3$$

The brackets matter. There are two nitrate ions, meaning two nitrogen atoms and six oxygen atoms. This is one point where the formula-writing skills from the previous lesson directly protect calculation marks.

Worked example: mass to moles

Calculate the amount in $9.00\mathrm{g}$ of water.

First calculate $M_r$:

$$M_r({\mathrm{H}}_2\mathrm{O})=(2\times 1.0)+16.0=18.0$$

Then use the mole equation:

$$n=\frac{m}{M}$$ $$n=\frac{9.00}{18.0}=0.500\mathrm{mol}$$

The full answer is:

$$\boxed{0.500\mathrm{mol}{\mathrm{H}}_2\mathrm{O}}$$

Worked example: moles to mass

Calculate the mass of $0.250\mathrm{mol}$ of calcium carbonate, $\mathrm{CaC}{\mathrm{O}}_3$.

First find the molar mass:

$$M_r(\mathrm{CaC}{\mathrm{O}}_3)=40.1+12.0+(3\times 16.0)=100.1$$

Rearrange the equation:

$$m=nM$$ $$m=0.250\times 100.1=25.025$$

To three significant figures:

$$\boxed{25.0\mathrm{g}\mathrm{CaC}{\mathrm{O}}_3}$$

A useful sense check: if you have less than one mole, your mass should be less than the molar mass. Here, $0.250$ mol is one quarter of a mole, so a mass near one quarter of $100.1\mathrm{g}$ is sensible.

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Units: an easy mark to protect

The mole equation expects mass in grams. Convert before substituting values.

$$1\mathrm{kg}=1000\mathrm{g}$$ $$1\mathrm{g}=1000\mathrm{mg}$$

Therefore:

$$83.0\mathrm{mg}=0.0830\mathrm{g}$$ $$0.0347\mathrm{kg}=34.7\mathrm{g}$$

Do the unit conversion as a separate first line in your answer. It makes your method clear to the examiner and avoids a factor-of-1000 error.

Worked example: milligrams to moles

Calculate the amount in $83.0\mathrm{mg}$ of magnesium hydroxide, $\mathrm{Mg}(\mathrm{OH}{)}_2$.

Convert the mass first:

$$83.0\mathrm{mg}=0.0830\mathrm{g}$$

Calculate the molar mass:

$$M_r(\mathrm{Mg}(\mathrm{OH}{)}_2)=24.3+2(16.0+1.0)=58.3$$

Now calculate moles:

$$n=\frac{0.0830}{58.3}$$ $$n=1.42\times 10^{-3}\mathrm{mol}$$ $$\boxed{1.42\times 10^{-3}\mathrm{mol}\mathrm{Mg}(\mathrm{OH}{)}_2}$$

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Avogadro’s constant: moles to particles

The equation connecting amount and particle number is:

$$N=n{N}_A$$

where:

• $N$ is the number of particles;
• $n$ is the amount in moles;
• $N_A$ is Avogadro’s constant, $6.02\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$.

To go in the opposite direction:

$$n=\frac{N}{N_A}$$

Watch this section of Mole Conversions Tutorial by Crash Chemistry Academy. It reinforces the central idea that conversion factors should be arranged so unwanted units cancel, then combines mass–mole and mole–particle calculations.

Worked example: moles to molecules

Calculate the number of water molecules in $0.200\mathrm{mol}$ of water.

$$N=n{N}_A$$ $$N=0.200\times 6.02\times 10^{23}$$ $$N=1.204\times 10^{23}$$

To three significant figures:

$$\boxed{1.20\times 10^{23}\text{water molecules}}$$

The wording matters. The calculation has found the number of water molecules, not the total number of atoms.

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Identify the particle before calculating

Many questions are designed to check whether you distinguish between molecules, atoms, ions, and formula units. Avogadro’s constant gives the number of particles specified by the question — but you may need an additional multiplication based on the chemical formula.

For water:

$${\mathrm{H}}_2\mathrm{O}$$

Every molecule contains:

• $2$ hydrogen atoms;
• $1$ oxygen atom;
• $3$ atoms in total.

If $0.200$ mol of water contains:

$$1.20\times 10^{23}\text{water molecules}$$

then it contains:

$$2\times 1.20\times 10^{23}=2.40\times 10^{23}\text{hydrogen atoms}$$

and:

$$3\times 1.20\times 10^{23}=3.60\times 10^{23}\text{atoms in total}$$

These are different answers to different questions.

Particle-language checklist

If the question asks for…	First calculate…	Then…
molecules of $\mathrm{C}{\mathrm{O}}_2$	number of $\mathrm{C}{\mathrm{O}}_2$ molecules	stop
oxygen atoms in $\mathrm{C}{\mathrm{O}}_2$	number of $\mathrm{C}{\mathrm{O}}_2$ molecules	multiply by $2$
total atoms in $\mathrm{C}{\mathrm{O}}_2$	number of $\mathrm{C}{\mathrm{O}}_2$ molecules	multiply by $3$
formula units of $\mathrm{CaC}{\mathrm{l}}_2$	number of formula units	stop
chloride ions in $\mathrm{CaC}{\mathrm{l}}_2$	number of formula units	multiply by $2$

For ionic compounds, use formula units, not molecules. Sodium chloride is an extended ionic lattice, not a collection of individual NaCl molecules. In most calculation questions, the numerical method is identical; the correct scientific label earns and protects marks.

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Two-step calculations: mass to particles

When mass is given but the question asks for particles, use two equations in sequence:

$$\text{mass}\to \text{moles}\to \text{particles}$$

Worked example: number of molecules from mass

Calculate the number of carbon dioxide molecules in $4.40\mathrm{g}$ of carbon dioxide.

First find $M_r$:

$$M_r(\mathrm{C}{\mathrm{O}}_2)=12.0+(2\times 16.0)=44.0$$

Calculate moles:

$$n=\frac{4.40}{44.0}=0.100\mathrm{mol}$$

Convert moles to molecules:

$$N=n{N}_A$$ $$N=0.100\times 6.02\times 10^{23}$$ $$\boxed{N=6.02\times 10^{22}{\text{CO}}_2\text{ molecules}}$$

Write the calculation in stages, even if you could combine it on a calculator. In a multi-mark question, the intermediate mole value is a clear method mark and makes errors much easier to spot.

Worked example: total atoms from mass

Calculate the total number of atoms in $3.60\mathrm{g}$ of water.

First calculate moles of water:

$$n=\frac{3.60}{18.0}=0.200\mathrm{mol}$$

Find the number of water molecules:

$$N=0.200\times 6.02\times 10^{23}$$ $$N=1.20\times 10^{23}\text{molecules}$$

Each water molecule contains three atoms, so:

$$\text{total atoms}=3\times 1.20\times 10^{23}$$ $$\boxed{3.61\times 10^{23}\text{atoms}}$$

The final multiplication by 3 is the step commonly omitted in exams. Before calculating, underline the exact particle named in the question.

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The reverse route: particles to mass

For a particles-to-mass question, reverse the pathway:

$$\text{particles}\to \text{moles}\to \text{mass}$$

Worked example: number of molecules to mass

Calculate the mass of $3.01\times 10^{22}$ carbon dioxide molecules.

First convert molecules to moles:

$$n=\frac{N}{N_A}$$ $$n=\frac{3.01\times 10^{22}}{6.02\times 10^{23}}$$ $$n=5.00\times 10^{-2}\mathrm{mol}$$

Then convert moles to mass. Since $M_r(\mathrm{C}{\mathrm{O}}_2)=44.0$:

$$m=nM$$ $$m=5.00\times 10^{-2}\times 44.0$$ $$\boxed{m=2.20\mathrm{g}}$$

Notice the powers of ten:

$$\frac{10^{22}}{10^{23}}=10^{-1}$$

Your answer must be less than one mole because the given number of molecules is less than Avogadro’s constant. That provides a quick plausibility check.

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A dependable exam routine

For every mole-conversion calculation, follow this sequence:

1. Write what you have and what you need.
   For example: “$4.40\mathrm{g}\mathrm{C}{\mathrm{O}}_2$ to molecules.”

2. Identify the route.

   • mass to moles: use $n=m/M$;
   • moles to mass: use $m=nM$;
   • moles to particles: use $N=n{N}_A$;
   • particles to moles: use $n=N/N_A$.

3. Convert mass units to grams first, if necessary.

4. Calculate $M_r$ explicitly when it is not given.

5. Use moles as the middle step for mass-to-particles or particles-to-mass questions.

6. Check the particle wording.
   Does the question want molecules, formula units, a particular atom, a particular ion, or total atoms?

7. Round only at the end.
   Keep extra calculator digits during intermediate steps, then give your final answer to an appropriate number of significant figures.

A compact answer layout might look like this:

$$M_r(\mathrm{C}{\mathrm{O}}_2)=44.0$$ $$n=\frac{4.40}{44.0}=0.100\mathrm{mol}$$ $$N=0.100\times 6.02\times 10^{23}$$ $$N=6.02\times 10^{22}\text{molecules}$$

This is concise, logical, and easy for a marker to award full credit.

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Errors to eliminate from now on

Error	Why it is wrong	Fix
Using mass in mg directly	The mole equation uses grams	Convert units before substitution
Forgetting brackets in $M_r$	This gives the wrong number of atoms	Count atoms from the full formula, including brackets
Multiplying by $N_A$ when finding moles	This creates an impossibly large answer	Particles to moles means divide by $N_A$
Giving “atoms” when you calculated molecules	The particle type is scientifically incorrect	Copy the particle label from the question
Forgetting the formula multiplier	Molecules and atoms are not the same count	Inspect subscripts before your final answer
Rounding too early	It can change a final multi-step answer	Keep calculator values until the final line

For your calculation-error log, use four headings this time: formula mass, unit conversion, route, and particle count. When you check marked work, record which category caused any lost mark rather than simply copying the correct answer.

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Key takeaways

The mole is the central link between measurable mass and invisible particles.

$$n=\frac{m}{M}$$ $$m=nM$$ $$N=n{N}_A$$ $$n=\frac{N}{N_A}$$

Remember:

• calculate $M_r$ carefully from the full formula;
• use grams in mass–mole calculations;
• use $N_A=6.02\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$;
• go through moles for every mass-to-particle or particle-to-mass problem;
• state whether your answer is atoms, molecules, ions, or formula units.

Next, you will use the same mole method to determine empirical formulae from composition data and then derive molecular formulae from molar mass.