How to Convert Moles of a Compound into Moles of Its Elements

Suppose you have \(0.250\ \mathrm{mol}\) of water molecules, \(\ce{H2O}\). How many moles of oxygen atoms does that contain?

The answer is \(0.250\ \mathrm{mol}\) of oxygen atoms. Every water molecule contains exactly one oxygen atom.

What about hydrogen? Each water molecule contains two hydrogen atoms, so the same sample contains \(0.500\ \mathrm{mol}\) of hydrogen atoms.

This is the key rule for converting between moles of a compound and moles of an element within it:

\[ \boxed{n(\text{element atoms})=k\times n(\text{compound})} \]

Here, \(k\) is the number of atoms of the element in one molecule or formula unit of the compound.

The challenge is identifying the correct value of \(k\). Brackets, polyatomic ions, and water of crystallisation can all change the atom count. Let’s work through how to get it right.

01Understand what you are counting

A mole is a counting unit. One mole contains approximately \(6.022\times10^{23}\) specified particles.

Those particles might be atoms, molecules, ions, or formula units. The chemical formula tells you which atoms are present in each compound particle.

For example:

• \(1\ \mathrm{mol}\) of \(\ce{H2O}\) contains \(1\ \mathrm{mol}\) of water molecules.
• \(1\ \mathrm{mol}\) of \(\ce{H2O}\) contains \(2\ \mathrm{mol}\) of hydrogen atoms.
• \(1\ \mathrm{mol}\) of \(\ce{H2O}\) contains \(1\ \mathrm{mol}\) of oxygen atoms.

Notice that the number of moles changes depending on which particles you are counting.

Molecules versus formula units

There is another distinction worth getting right.

Molecular compounds, such as water and carbon dioxide, consist of molecules.

Ionic compounds, such as sodium chloride and aluminium sulfate, form extended ionic lattices. We use a formula unit to represent the simplest whole-number ratio of ions in the compound.

For example, one formula unit of \(\ce{NaCl}\) represents one sodium ion for every chloride ion.

We can still use the same mole-conversion method for both types of compounds.

Think of a chemical formula as a recipe. If every batch contains two eggs and three cups of flour, four batches contain eight eggs and twelve cups of flour.

A chemical formula works similarly: it specifies the number of each type of atom per compound particle.

The analogy has a limit. Ionic compounds do not consist of separate little packets of atoms. The formula unit describes their repeating composition, rather than a discrete molecule.

02Read the chemical formula before calculating

Your most important job is to count the atoms correctly.

Consider calcium nitrate:

\[ \ce{Ca(NO3)2} \]

The small 2 outside the brackets multiplies everything inside the brackets.

Therefore, one formula unit contains:

• 1 calcium atom
• 2 nitrogen atoms
• 6 oxygen atoms

The oxygen count is particularly important:

\[ 2\times3=6 \]

A common mistake is to count only the three oxygen atoms inside \(\ce{NO3}\), forgetting that there are two nitrate groups.

What if the formula contains a polyatomic ion?

A polyatomic ion is a group of bonded atoms that carries an overall electrical charge.

You must count every atom in the group, not just the element named in the ion.

For example:

\[ \ce{Al2(SO4)3} \]

There are three sulfate groups, \(\ce{SO4}\), in each formula unit.

Therefore:

If a question asks for moles of sulfur atoms, use 3 as your multiplier.

If it asks for moles of oxygen atoms, use 12.

Do not use the same multiplier for every element in the compound.

What about hydrated compounds?

Hydrated compounds contain water molecules incorporated into their crystal structure.

For example:

\[ \ce{CuSO4.5H2O} \]

This represents one copper sulfate unit associated with five water molecules.

Count the oxygen atoms from both parts:

• The \(\ce{SO4}\) group contributes 4 oxygen atoms.
• The five water molecules contribute another 5 oxygen atoms.

Therefore, each formula unit contains:

\[ 4+5=9\text{ oxygen atoms} \]

It also contains 10 hydrogen atoms, one copper atom, and one sulfur atom.

Do not ignore the water just because it appears after the dot.

03The conversion formula

Once you know how many atoms of your target element are present in one compound particle, the calculation is straightforward.

\[ \boxed{n(E)=k\times n(C)} \]

Where:

• \(n(E)\) = amount of the element’s atoms, in mol
• \(n(C)\) = amount of the compound particles, in mol
• \(k\) = number of atoms of the target element per compound particle

For example, suppose you have \(0.250\ \mathrm{mol}\) of water.

To calculate the amount of oxygen atoms:

\[ \begin{aligned} n(\mathrm O) &=0.250\times1\\ &=0.250\ \mathrm{mol} \end{aligned} \]

To calculate the amount of hydrogen atoms:

\[ \begin{aligned} n(\mathrm H) &=0.250\times2\\ &=0.500\ \mathrm{mol} \end{aligned} \]

Why does multiplying work?

Let’s connect the formula to what a mole actually counts.

Every water molecule contains one oxygen atom. Therefore, the number of oxygen atoms must equal the number of water molecules.

More generally, if each compound particle contains \(k\) atoms of an element:

\[ N(E)=kN(C) \]

Here, \(N\) represents the actual number of particles.

Since one mole of either type of particle contains Avogadro’s number of particles, dividing both sides by Avogadro’s constant gives:

\[ n(E)=kn(C) \]

This is why the multiplier comes directly from the chemical formula.

Check the units using dimensional analysis

You can write the conversion as a unit-cancelling calculation:

\[ \begin{aligned} &0.250\ \mathrm{mol\ H_2O}\\ &\times \frac{1\ \mathrm{mol\ O\ atoms}} {1\ \mathrm{mol\ H_2O}}\\ &=0.250\ \mathrm{mol\ O\ atoms} \end{aligned} \]

The water units cancel, leaving the required unit: moles of oxygen atoms.

This is a useful exam technique. If your final units still say moles of the compound when the question asks for moles of an element, you have not completed the conversion.

04Experimental connection: determining the composition of a hydrate

The relationship between compound particles and their constituent atoms also helps chemists investigate the composition of unknown substances.

Consider a sample of blue copper(II) sulfate crystals. The crystals contain water of crystallisation, but how many water molecules are associated with each copper sulfate unit?

An experiment can answer this by measuring mass before and after heating.

How the experiment works

The apparatus consists of:

• A balance to measure mass.
• A porcelain crucible and lid to hold the crystals.
• A Bunsen burner and tripod to heat the sample.

The process follows a particular sequence.

First, place the empty, dry crucible and lid on the balance. Add a known mass of hydrated copper sulfate crystals and record the sample’s initial mass.

Next, place the crucible on the tripod and heat the crystals gently. Keep the lid slightly open so that water vapour can escape.

As water leaves the crystal structure, the blue crystals become white or pale. The remaining solid is anhydrous copper sulfate, assuming the heating conditions have not caused further decomposition.

Finally, cool the crucible and weigh the remaining solid.

What should the experiment show?

If the sample is a pure hydrate, and heating removes only water, its mass should decrease.

The mass lost represents the water that has escaped.

However, a single heating cycle may not remove all the water. If you weigh the sample too early, the remaining mass will be too high.

A better design is to repeat the heating, cooling, and weighing process until the mass becomes constant.

This reduces the chance of incorrectly calculating the amount of water lost.

There are still limitations. A constant mass does not independently prove that only water was lost. Spattering, decomposition, or loss of solid material could also reduce the measured mass.

For the calculations below, we assume that no copper sulfate is lost and that water is the only substance removed.

The key idea is that the experiment determines the ratio of water molecules to copper sulfate units. Once that ratio is known, the chemical formula tells us how many oxygen atoms are present.

05Worked examples

When the question gives you mass instead of moles

Sometimes a question gives you the mass of a compound rather than its amount in moles.

In that case, you need one additional step.

First, calculate the amount of the compound using:

\[ n(C)=\frac{m(C)}{M(C)} \]

Where:

• \(m(C)\) is the compound’s mass in grams.
• \(M(C)\) is its molar mass in \(\mathrm{g\,mol^{-1}}\).
• \(n(C)\) is its amount in mol.

Then multiply by the number of atoms of your target element per compound particle.

Your complete method is:

\[ \boxed{ n(E)=\frac{m(C)}{M(C)}\times k } \]

This connects mass, amount of compound, and amount of element atoms.

If you need help calculating the molar mass first, see Molar Mass of Elements and Compounds: HSC Chemistry Guide.

06Common mistakes to avoid

Before attempting exam questions, check these four common errors.

Mistake 1: Multiplying by the total number of atoms

Water contains three atoms per molecule, but if the question asks for oxygen atoms, your multiplier is 1, not 3.

Always count the element being requested.

Mistake 2: Forgetting brackets

For \(\ce{Al2(SO4)3}\), there are 12 oxygen atoms, not 4.

The outside subscript multiplies every atom inside the brackets.

Mistake 3: Ignoring water of crystallisation

For \(\ce{CuSO4.5H2O}\), there are nine oxygen atoms, not four.

Include the atoms in the water molecules when counting the complete hydrated compound.

Mistake 4: Confusing atoms with molecules

This is particularly important for elements that exist as diatomic molecules.

Consider oxygen gas, \(\ce{O2}\).

One oxygen molecule contains two oxygen atoms. Therefore:

\[ 1\ \mathrm{mol\ O_2} = 2\ \mathrm{mol\ O\ atoms} \]

Compare this with water:

\[ 1\ \mathrm{mol\ H_2O} = 1\ \mathrm{mol\ O\ atoms} \]

Both examples contain oxygen, but the number of oxygen atoms per particle is different.

Always identify whether the question asks for oxygen atoms or oxygen molecules.

07Questions and solutions

Try each question before reading its solution. The questions become progressively more demanding, so pay attention to the assumptions as well as the calculations.

Question 1

A sample contains \(0.180\ \mathrm{mol}\) of ammonia, \(\ce{NH3}\).

Calculate:

a. The moles of nitrogen atoms.

b. The moles of hydrogen atoms.

Solution 1

The sample contains \(0.180\ \mathrm{mol}\) of nitrogen atoms and \(0.540\ \mathrm{mol}\) of hydrogen atoms.

Each ammonia molecule contains one nitrogen atom and three hydrogen atoms.

For nitrogen:

\[ \begin{aligned} n(\mathrm N) &=0.180\times1\\ &=0.180\ \mathrm{mol} \end{aligned} \]

For hydrogen:

\[ \begin{aligned} n(\mathrm H) &=0.180\times3\\ &=0.540\ \mathrm{mol} \end{aligned} \]

The amount of hydrogen atoms is three times the amount of ammonia because there are three hydrogen atoms in every ammonia molecule.

Question 2

A sample contains \(0.0750\ \mathrm{mol}\) of aluminium sulfate, \(\ce{Al2(SO4)3}\).

Calculate:

a. The moles of aluminium atoms.

b. The moles of sulfur atoms.

c. The moles of sulfate ions.

d. The moles of oxygen atoms.

e. The total number of oxygen atoms.

Use \(N_A=6.022\times10^{23}\ \mathrm{mol^{-1}}\).

Solution 2

The sample contains \(0.150\ \mathrm{mol}\) of aluminium atoms, \(0.225\ \mathrm{mol}\) of sulfur atoms, \(0.225\ \mathrm{mol}\) of sulfate ions, and \(0.900\ \mathrm{mol}\) of oxygen atoms. This corresponds to approximately \(5.42\times10^{23}\) oxygen atoms.

Start by reading the formula:

\[ \ce{Al2(SO4)3} \]

One formula unit contains:

• 2 aluminium atoms.
• 3 sulfur atoms.
• 3 sulfate ions.
• 12 oxygen atoms.

Part a: Aluminium atoms

\[ \begin{aligned} n(\mathrm{Al}) &=0.0750\times2\\ &=0.150\ \mathrm{mol} \end{aligned} \]

Part b: Sulfur atoms

\[ \begin{aligned} n(\mathrm S) &=0.0750\times3\\ &=0.225\ \mathrm{mol} \end{aligned} \]

Part c: Sulfate ions

There are three sulfate ions per formula unit.

\[ \begin{aligned} n(\mathrm{SO_{4}}^{2-}) &=0.0750\times3\\ &=0.225\ \mathrm{mol} \end{aligned} \]

Notice that the amount of sulfur atoms and sulfate ions is numerically identical. However, these are different counted entities. Each sulfate ion contains one sulfur atom.

Part d: Oxygen atoms

There are 12 oxygen atoms per formula unit.

\[ \begin{aligned} n(\mathrm O) &=0.0750\times12\\ &=0.900\ \mathrm{mol} \end{aligned} \]

Part e: Number of oxygen atoms

Convert the amount in moles into an actual number of atoms:

\[ N=nN_A \]

\[ \begin{aligned} N(\mathrm O) &=0.900\times6.022\times10^{23}\\ &=5.4198\times10^{23} \end{aligned} \]

To three significant figures:

\[ \boxed{N(\mathrm O)=5.42\times10^{23}\text{ atoms}} \]

This question tests two different counting skills: interpreting brackets correctly and converting between moles and the actual number of particles.

Question 3

A laboratory has a \(2.65\ \mathrm g\) sample of pure, anhydrous sodium carbonate, \(\ce{Na2CO3}\).

Calculate:

a. The moles of carbon atoms.

b. The moles of oxygen atoms.

c. The total number of oxygen atoms.

Use the following relative atomic masses:

• Na = 22.99
• C = 12.01
• O = 16.00

Solution 3

The sample contains \(0.0250\ \mathrm{mol}\) of carbon atoms and \(0.0750\ \mathrm{mol}\) of oxygen atoms, corresponding to approximately \(4.52\times10^{22}\) oxygen atoms.

This question requires you to convert mass into moles before counting atoms.

Step 1: Calculate the molar mass of sodium carbonate.

Each formula unit contains two sodium atoms, one carbon atom, and three oxygen atoms.

\[ \begin{aligned} M(\ce{Na2CO3}) &=2(22.99)+12.01+3(16.00)\\ &=45.98+12.01+48.00\\ &=105.99\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Step 2: Convert the sample mass into moles of sodium carbonate.

\[ \begin{aligned} n(\ce{Na2CO3}) &=\frac{m}{M}\\ &=\frac{2.65}{105.99}\\ &=0.0250\ \mathrm{mol} \end{aligned} \]

Step 3: Calculate the moles of carbon atoms.

There is one carbon atom per formula unit.

\[ \begin{aligned} n(\mathrm C) &=0.0250\times1\\ &=0.0250\ \mathrm{mol} \end{aligned} \]

Step 4: Calculate the moles of oxygen atoms.

There are three oxygen atoms per formula unit.

\[ \begin{aligned} n(\mathrm O) &=0.0250\times3\\ &=0.0750\ \mathrm{mol} \end{aligned} \]

Step 5: Calculate the number of oxygen atoms.

\[ \begin{aligned} N(\mathrm O) &=n(\mathrm O)N_A\\ &=0.0750\times6.022\times10^{23}\\ &=4.5165\times10^{22} \end{aligned} \]

Therefore:

\[ \boxed{N(\mathrm O)=4.52\times10^{22}\text{ atoms}} \]

The important detail is that you cannot multiply the original mass directly by the number of oxygen atoms. You must first convert the mass into moles of the compound using its molar mass.

Question 4

A sealed reaction vessel initially contains:

• \(0.0400\ \mathrm{mol}\) of carbon monoxide, \(\ce{CO}\).
• \(0.0200\ \mathrm{mol}\) of carbon dioxide, \(\ce{CO2}\).

A chemist adds \(0.0300\ \mathrm{mol}\) of oxygen gas, \(\ce{O2}\), and ignites the mixture.

The carbon monoxide reacts completely according to:

\[ \ce{2CO + O2 -> 2CO2} \]

Assume no other reactions occur.

Calculate:

a. The initial moles of oxygen atoms in the original mixture, before oxygen gas is added.

b. The final moles of carbon dioxide molecules.

c. The final moles of unreacted oxygen molecules.

d. The total moles of oxygen atoms in the final vessel.

Explain why the final oxygen atom count is different from the original oxygen atom count.

Solution 4

The original mixture contains \(0.0800\ \mathrm{mol}\) of oxygen atoms. After the reaction, the vessel contains \(0.0600\ \mathrm{mol}\) of carbon dioxide, \(0.0100\ \mathrm{mol}\) of unreacted oxygen gas, and \(0.140\ \mathrm{mol}\) of oxygen atoms in total.

The key is to count the atoms in each compound separately, then account for the oxygen added during the reaction.

Part a: Initial oxygen atoms

Carbon monoxide contains one oxygen atom per molecule:

\[ n(\mathrm O\text{ in CO})=0.0400\times1 \]

\[ =0.0400\ \mathrm{mol} \]

Carbon dioxide contains two oxygen atoms per molecule:

\[ n(\mathrm O\text{ in CO}_2)=0.0200\times2 \]

\[ =0.0400\ \mathrm{mol} \]

Adding these gives:

\[ \begin{aligned} n(\mathrm O)_{\mathrm{initial}} &=0.0400+0.0400\\ &=0.0800\ \mathrm{mol} \end{aligned} \]

Part b: Final carbon dioxide

The balanced equation is:

\[ \ce{2CO + O2 -> 2CO2} \]

Two moles of carbon monoxide produce two moles of carbon dioxide.

Therefore, the mole ratio between CO and CO2 is 1:1.

The \(0.0400\ \mathrm{mol}\) of carbon monoxide produces \(0.0400\ \mathrm{mol}\) of carbon dioxide.

Including the carbon dioxide already present:

\[ \begin{aligned} n(\ce{CO2})_{\mathrm{final}} &=0.0400+0.0200\\ &=0.0600\ \mathrm{mol} \end{aligned} \]

Part c: Unreacted oxygen gas

The reaction requires one mole of oxygen gas for every two moles of carbon monoxide.

Therefore:

\[ \begin{aligned} n(\ce{O2})_{\mathrm{used}} &=\frac{0.0400}{2}\\ &=0.0200\ \mathrm{mol} \end{aligned} \]

The vessel initially received \(0.0300\ \mathrm{mol}\) of oxygen gas.

The remaining amount is:

\[ \begin{aligned} n(\ce{O2})_{\mathrm{remaining}} &=0.0300-0.0200\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

Part d: Final oxygen atoms

First, count the oxygen atoms in the final carbon dioxide:

\[ \begin{aligned} n(\mathrm O\text{ in CO}_2) &=0.0600\times2\\ &=0.120\ \mathrm{mol} \end{aligned} \]

Now count the oxygen atoms in the remaining oxygen gas.

Each \(\ce{O2}\) molecule contains two oxygen atoms:

\[ \begin{aligned} n(\mathrm O\text{ in O}_2) &=0.0100\times2\\ &=0.0200\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ \begin{aligned} n(\mathrm O)_{\mathrm{final}} &=0.120+0.0200\\ &=0.140\ \mathrm{mol} \end{aligned} \]

Why did the oxygen amount change?

The original mixture contained \(0.0800\ \mathrm{mol}\) of oxygen atoms.

However, the chemist added another \(0.0300\ \mathrm{mol}\) of oxygen molecules. Each molecule contains two oxygen atoms, so this added:

\[ 0.0300\times2=0.0600\ \mathrm{mol} \]

of oxygen atoms.

The total available oxygen atoms therefore became:

\[ 0.0800+0.0600=0.140\ \mathrm{mol} \]

This matches the final calculation.

The important distinction is that chemical reactions rearrange atoms. They do not create or destroy them. But the total amount of an element in your vessel can change when that element is added or removed from the system.

Question 5

A student has a \(5.00\ \mathrm g\) sample of pure copper(II) sulfate pentahydrate:

\[ \ce{CuSO4.5H2O} \]

The sample is heated gently, causing some of its water of crystallisation to escape. The final solid has a mass of \(3.92\ \mathrm g\).

Assume:

• No copper sulfate is lost.
• Only water leaves the solid.
• The final solid is a single stoichiometric hydrate, \(\ce{CuSO4.xH2O}\), where \(x\) is a whole number.
• The measured masses are accurate to the stated precision.

Use:

\[ \begin{aligned} M(\ce{CuSO4.5H2O})&=249.68\ \mathrm{g\,mol^{-1}}\\ M(\ce{CuSO4})&=159.61\ \mathrm{g\,mol^{-1}}\\ M(\ce{H2O})&=18.02\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Calculate:

a. The value of \(x\) in the final hydrate.

b. The moles of oxygen atoms remaining in the final solid.

c. Explain why multiplying the original amount of hydrate by nine gives an incorrect answer for part b.

Solution 5

The final solid is approximately \(\ce{CuSO4.2H2O}\), and it contains \(0.120\ \mathrm{mol}\) of oxygen atoms.

The important insight is that the number of copper sulfate units stays constant, but the number of water molecules associated with each unit decreases.

This means you cannot use the original formula unchanged when calculating the oxygen content of the final solid.

Step 1: Calculate the original moles of hydrate.

Initially, the sample contains \(5.00\ \mathrm g\) of copper sulfate pentahydrate.

\[ \begin{aligned} n_{\mathrm{initial}} &=\frac{5.00}{249.68}\\ &=0.02003\ \mathrm{mol} \end{aligned} \]

Each original formula unit contains one \(\ce{CuSO4}\) unit.

Since no copper sulfate is lost, the amount of copper sulfate units remains:

\[ n(\ce{CuSO4})=0.02003\ \mathrm{mol} \]

This amount is conserved during dehydration.

Step 2: Calculate the effective molar mass of the final hydrate per mole of copper sulfate units.

The final solid has a mass of \(3.92\ \mathrm g\).

Dividing this by the unchanged amount of copper sulfate units gives:

\[ \begin{aligned} M_{\mathrm{final}} &=\frac{3.92}{0.02003}\\ &=195.8\ \mathrm{g\,mol^{-1}} \end{aligned} \]

This is the mass of one mole of the final hydrate formula units.

Step 3: Calculate the number of water molecules per copper sulfate unit.

The final hydrate consists of one \(\ce{CuSO4}\) unit plus \(x\) water molecules.

Therefore:

\[ M_{\mathrm{final}} =M(\ce{CuSO4})+xM(\ce{H2O}) \]

Rearranging:

\[ x=\frac{M_{\mathrm{final}}-M(\ce{CuSO4})}{M(\ce{H2O})} \]

Substitute:

\[ \begin{aligned} x&=\frac{195.8-159.61}{18.02}\\ &=2.01 \end{aligned} \]

The small difference from a whole number comes from rounding the measured masses.

Therefore:

\[ \boxed{x=2} \]

The final solid is approximately:

\[ \ce{CuSO4.2H2O} \]

Step 4: Count the oxygen atoms in each final formula unit.

The final formula contains:

• 4 oxygen atoms from \(\ce{SO4}\).
• 2 oxygen atoms from the two water molecules.

Therefore:

\[ k=4+2=6 \]

Step 5: Calculate the moles of oxygen atoms remaining.

\[ \begin{aligned} n(\mathrm O) &=0.02003\times6\\ &=0.1202\ \mathrm{mol} \end{aligned} \]

To three significant figures:

\[ \boxed{n(\mathrm O)=0.120\ \mathrm{mol}} \]

Step 6: Understand why the tempting alternatives are wrong.

There are two plausible mistakes here.

The first is to use the original amount of hydrate and multiply by nine:

\[ 0.02003\times9=0.180\ \mathrm{mol} \]

This is the amount of oxygen atoms in the original pentahydrate, not the final partially dehydrated solid.

Three water molecules per formula unit have escaped, taking their oxygen atoms with them.

The second mistake is to divide the final mass by the molar mass of the original pentahydrate:

\[ \frac{3.92}{249.68}=0.0157\ \mathrm{mol} \]

This assumes the final solid is still pentahydrate. It isn’t.

The correct approach is to recognise what has remained constant: the number of copper sulfate units. The number of associated water molecules has changed.

As a final check, three water molecules were lost per copper sulfate unit:

\[ n(\ce{H2O\ lost})=3\times0.02003 \]

\[ =0.0601\ \mathrm{mol} \]

That means \(0.0601\ \mathrm{mol}\) of oxygen atoms were lost as well.

The original oxygen amount was \(0.180\ \mathrm{mol}\), so:

\[ 0.180-0.0601\approx0.120\ \mathrm{mol} \]

This agrees with our answer.

The key lesson is that a formula tells you the number of atoms per compound particle, but you must use the formula that describes the substance at the stage you are investigating.

07The method to remember

When converting between moles of a compound and moles of an element, follow this sequence:

1. Identify the compound’s molecules or formula units.
2. Read the formula carefully, including brackets and waters of crystallisation.
3. Count the number of atoms of the specific element requested.
4. If given mass, calculate moles of compound using \(n=m/M\).
5. Multiply the amount of compound by the number of target atoms per particle.
6. Include the correct units in your final answer.

The most important distinction is between the particles you start with and the atoms you are trying to count.

Once you can identify those two quantities clearly, the conversion becomes a simple mole ratio. From there, you can move confidently into more involved stoichiometry problems, including reactions, limiting reagents, and quantitative composition analysis.