How to Count Oxygen Atoms in a Compound: Worked Examples

A balance tells you that a sample contains 9.00 g of glucose. But how many oxygen atoms are actually sitting in that sample? There could be trillions of trillions, and you certainly aren’t going to count them individually.

The useful trick is to count in stages. First, work out how many molecules or formula units you have. Then use the chemical formula to determine how many oxygen atoms belong to each one.

Before going further, make a prediction. If you have 0.100 mol of carbon dioxide, \(\ce{CO2}\), do you have 0.100 mol or 0.200 mol of oxygen atoms?

It’s 0.200 mol. Each carbon dioxide molecule contains two oxygen atoms, so every molecule contributes two oxygen atoms to the total.

That multiplication is the key to answering the question: how to count oxygen atoms in a given amount of a compound.

01The method: from compound amount to oxygen atoms

There are two pieces of information you need:

• The amount of compound, measured in moles.
• The number of oxygen atoms in one molecule or formula unit of that compound.

The second piece comes directly from the chemical formula.

Let \(n\) represent the amount of compound in moles, \(x\) represent the number of oxygen atoms in each molecule or formula unit, and \(N_A\) represent Avogadro’s constant.

The number of oxygen atoms is:

\[ \boxed{N_{\mathrm O}=n\times x\times N_A} \]

For HSC calculations, use:

\[ N_A=6.02\times10^{23}\ \mathrm{mol^{-1}} \]

This tells you how many individual entities are contained in one mole.

If the question gives you the mass rather than the amount of compound, there is one extra step. Divide the mass by the molar mass to calculate the amount first.

\[ n=\frac{m}{M} \]

Where:

• \(m\) is the sample mass in grams (\(\mathrm g\)).
• \(M\) is the molar mass in grams per mole (\(\mathrm{g\,mol^{-1}}\)).
• \(n\) is the amount in moles (\(\mathrm{mol}\)).

Combining the two equations gives:

\[ \boxed{N_{\mathrm O}=\frac{m}{M}\times x\times N_A} \]

You now have a method that works whether the question gives you grams or moles.

What exactly are you counting?

Before using the formula, identify the entity being counted.

For a molecular compound, the entity is a molecule. For an ionic compound, such as calcium nitrate, the entity is a formula unit. A formula unit represents the simplest whole-number ratio of ions in the compound.

The distinction matters because ionic solids do not consist of separate molecules in the same way molecular substances do. However, both types of compounds can be counted using the same mole calculation.

Consider these examples:

Pay particular attention to brackets and water of crystallisation.

For example, \(\ce{Ca(NO3)2}\) contains two nitrate groups. Each nitrate group has three oxygen atoms, giving six oxygen atoms altogether.

Similarly, \(\ce{CuSO4.5H2O}\) contains four oxygen atoms in the sulfate group and another five in the water molecules. Its total is nine oxygen atoms per formula unit.

The oxygen subscript tells you how many oxygen atoms are present in the entity. If there are brackets, multiply the contents by the number outside them.

How does the mass measurement actually give an atom count?

You might reasonably wonder how a laboratory balance can possibly tell us the number of oxygen atoms in a sample.

It can’t measure individual atoms. Instead, chemists use an indirect measurement.

Imagine an experimenter weighing a pure, dry sample of a known compound.

The measurement proceeds as follows:

1. A clean, empty weighing bottle is placed on the balance and the balance is tared.
2. A known compound is transferred into the bottle, which is capped to limit moisture exchange with the air.
3. The bottle is returned to the balance, and its mass is recorded.
4. The chemical formula provides the number of oxygen atoms per entity.
5. The molar mass converts the measured sample mass into moles of compound.
6. Avogadro’s constant converts the amount of compound into a number of entities.

The balance provides an observation: the sample’s mass.

The formula and molar mass allow us to interpret that measurement and calculate an atom count.

There is an important limitation. A balance cannot tell you whether a sample is pure, whether it has absorbed water, or whether its chemical identity is correct. Those details must be established separately.

For example, leaving a moisture-sensitive sample exposed in an open weighing dish could increase its measured mass. The revised approach is to use an appropriately prepared sample and a capped weighing bottle. If the substance is a hydrate, however, its water is part of the chemical formula and must not be accidentally treated as contamination.

The entire calculation depends on knowing what substance you are actually weighing.

02Worked examples

03A common misconception: moles of compound are not moles of oxygen atoms

Suppose you have 0.100 mol of carbon monoxide, \(\ce{CO}\), and 0.100 mol of carbon dioxide, \(\ce{CO2}\).

Both samples contain the same amount of compound particles. Does that mean they contain the same amount of oxygen atoms?

No.

The carbon dioxide sample contains twice as many oxygen atoms because each molecule contains two oxygen atoms instead of one.

This is why you should never automatically treat the amount of compound as the amount of oxygen atoms.

The same principle applies to ionic compounds. One mole of \(\ce{Al2(SO4)3}\) contains one mole of formula units, but twelve moles of oxygen atoms.

A useful decision table

When you see an oxygen-counting question, first identify what information you have.

One final distinction is worth remembering: if a question asks for the amount of oxygen atoms in moles, stop after calculating \(n_{\mathrm O}\). If it asks for the number of oxygen atoms, multiply by Avogadro’s constant.

04Questions and solutions

Try each question before reading its solution. The questions start with a straightforward application and gradually introduce purity, ionic compounds, hydrates, and experimental measurements.

Question 1

A sample contains 0.250 mol of carbon dioxide, \(\ce{CO2}\).

Calculate the total number of oxygen atoms in the sample.

Solution 1

The sample contains \(3.01\times10^{23}\) oxygen atoms.

Each carbon dioxide molecule contains two oxygen atoms. Therefore, the amount of oxygen atoms is twice the amount of carbon dioxide molecules:

\[ n_{\mathrm O}=2\times0.250 =0.500\ \mathrm{mol} \]

Now convert into the number of oxygen atoms:

\[ \begin{aligned} N_{\mathrm O}&=0.500\times6.02\times10^{23}\\ &=3.01\times10^{23} \end{aligned} \]

\[ \boxed{3.01\times10^{23}\text{ oxygen atoms}} \]

The common mistake is to multiply 0.250 mol directly by Avogadro’s constant. That would count carbon dioxide molecules, not oxygen atoms. You must account for the two oxygen atoms in each molecule first.

Question 2

An impure sample has a total mass of 15.0 g. It contains 80.0% sodium carbonate, \(\ce{Na2CO3}\), by mass.

Assume the remaining impurity is inert and contains no oxygen.

Calculate the total number of oxygen atoms in the sample.

Use the relative atomic masses \(Na=23.0\), \(C=12.0\), and \(O=16.0\).

Solution 2

The sample contains \(2.04\times10^{23}\) oxygen atoms.

The key is to calculate the mass of pure sodium carbonate before calculating its amount. Using the entire 15.0 g would incorrectly assume that the impurity is also sodium carbonate.

Step 1: Calculate the mass of pure sodium carbonate.

\[ \begin{aligned} m_{\mathrm{pure}}&=15.0\times\frac{80.0}{100}\\ &=12.0\ \mathrm g \end{aligned} \]

Only 12.0 g contributes oxygen atoms.

Step 2: Calculate the molar mass of sodium carbonate.

\[ \begin{aligned} M&=2(23.0)+12.0+3(16.0)\\ &=106.0\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Step 3: Calculate the amount of sodium carbonate.

\[ \begin{aligned} n&=\frac{12.0}{106.0}\\ &=0.1132\ \mathrm{mol} \end{aligned} \]

Step 4: Calculate the amount of oxygen atoms.

Each formula unit of \(\ce{Na2CO3}\) contains three oxygen atoms.

\[ \begin{aligned} n_{\mathrm O}&=3\times0.1132\\ &=0.3396\ \mathrm{mol} \end{aligned} \]

Step 5: Calculate the total number of oxygen atoms.

\[ \begin{aligned} N_{\mathrm O}&=0.3396\times6.02\times10^{23}\\ &=2.04\times10^{23} \end{aligned} \]

\[ \boxed{N_{\mathrm O}=2.04\times10^{23}\text{ oxygen atoms}} \]

The important reasoning is that purity changes the amount of compound present, not the number of oxygen atoms in each formula unit. Sodium carbonate still contains three oxygen atoms per formula unit, but there is less sodium carbonate available to count.

Question 3

A student dissolves 0.0800 mol of aluminium sulfate, \(\ce{Al2(SO4)3}\), in water.

The student claims that the solution contains:

• 0.0800 mol of sulfate ions;
• 0.320 mol of oxygen atoms.

Assuming the sulfate ions remain intact, identify both errors and calculate the correct number of oxygen atoms in the solution.

Solution 3

The correct number of oxygen atoms is \(5.78\times10^{23}\). The solution contains 0.240 mol of sulfate ions and 0.960 mol of oxygen atoms.

The mistake is overlooking the subscript 3 outside the sulfate brackets.

Step 1: Determine the number of sulfate ions per formula unit.

The formula is:

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

Each formula unit contains three sulfate ions.

Therefore, the amount of sulfate ions is:

\[ \begin{aligned} n_{\mathrm{SO4^{2-}}}&=3\times0.0800\\ &=0.240\ \mathrm{mol} \end{aligned} \]

The student incorrectly assumed there was only one sulfate ion per formula unit.

Step 2: Calculate the amount of oxygen atoms.

Each sulfate ion, \(\mathrm{SO_{4}}^{2-}\), contains four oxygen atoms.

Therefore:

\[ \begin{aligned} n_{\mathrm O}&=4\times0.240\\ &=0.960\ \mathrm{mol} \end{aligned} \]

You can also check this directly from the original formula. Each formula unit contains:

\[ 3\times4=12\text{ oxygen atoms} \]

And:

\[ 12\times0.0800=0.960\ \mathrm{mol} \]

Step 3: Convert into the number of oxygen atoms.

\[ \begin{aligned} N_{\mathrm O}&=0.960\times6.02\times10^{23}\\ &=5.7792\times10^{23} \end{aligned} \]

\[ \boxed{N_{\mathrm O}=5.78\times10^{23}\text{ oxygen atoms}} \]

Dissolving the compound separates its ions, but it does not automatically break the sulfate ions into individual sulfur and oxygen atoms. The four oxygen atoms remain part of each sulfate ion.

The useful habit here is to count the complete formula unit before considering what happens when the compound dissolves.

Question 4

A 24.96 g sample of copper(II) sulfate pentahydrate has the formula \(\ce{CuSO4.5H2O}\).

A student calculates the oxygen atom count using only the four oxygen atoms in the sulfate group.

Explain why this is incorrect and calculate the actual number of oxygen atoms in the sample.

Use the relative atomic masses \(Cu=63.5\), \(S=32.1\), \(O=16.0\), and \(H=1.0\).

Solution 4

The sample contains \(5.42\times10^{23}\) oxygen atoms. The student has forgotten to count the five oxygen atoms contributed by the water of crystallisation.

Step 1: Count the oxygen atoms in one formula unit.

The formula is:

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

There are four oxygen atoms in \(\ce{SO4}\), and five water molecules, each containing one oxygen atom.

Therefore:

\[ x=4+5=9 \]

Each formula unit contains nine oxygen atoms.

Step 2: Calculate the molar mass.

\[ \begin{aligned} M={}&63.5+32.1+4(16.0)\\ &+5(2(1.0)+16.0)\\ ={}&249.6\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Step 3: Calculate the amount of copper(II) sulfate pentahydrate.

\[ \begin{aligned} n&=\frac{24.96}{249.6}\\ &=0.1000\ \mathrm{mol} \end{aligned} \]

Step 4: Calculate the amount of oxygen atoms.

\[ \begin{aligned} n_{\mathrm O}&=9\times0.1000\\ &=0.9000\ \mathrm{mol} \end{aligned} \]

Step 5: Calculate the number of oxygen atoms.

\[ \begin{aligned} N_{\mathrm O}&=0.9000\times6.02\times10^{23}\\ &=5.418\times10^{23} \end{aligned} \]

\[ \boxed{N_{\mathrm O}=5.42\times10^{23}\text{ oxygen atoms}} \]

The important distinction is that water of crystallisation is part of the compound’s formula. Its oxygen atoms belong in the original atom count.

If you had counted only the sulfate oxygen atoms, you would have obtained an answer that was too small by a factor of \(9/4\).

Question 5

A chemist is given a 10.80 g sample containing a mixture of anhydrous magnesium sulfate, \(\ce{MgSO4}\), and hydrated magnesium sulfate, \(\ce{MgSO4.7H2O}\).

The sample is heated under controlled conditions until all water of crystallisation has been removed. Assume that:

• All water is removed from the sample.
• No magnesium sulfate decomposes or escapes.
• The only material lost from the sample is water.
• The final solid is pure, anhydrous \(\ce{MgSO4}\).

After heating and cooling, the remaining solid has a mass of 6.00 g.

Calculate the total number of oxygen atoms that were present in the original 10.80 g sample.

Also determine whether the measured mass loss is consistent with the sample being a mixture of the two stated compounds.

Use \(M(\ce{MgSO4})=120.4\ \mathrm{g\,mol^{-1}}\), \(M(\ce{H2O})=18.0\ \mathrm{g\,mol^{-1}}\), and \(N_A=6.02\times10^{23}\ \mathrm{mol^{-1}}\).

Solution 5

The original sample contains approximately \(2.81\times10^{23}\) oxygen atoms, and the measured mass loss is consistent with a mixture of anhydrous magnesium sulfate and its heptahydrate.

This is more interesting than simply multiplying the amount of magnesium sulfate by four or eleven. The original sample contains oxygen in both the sulfate groups and the water of crystallisation, but the two contributions must be calculated separately.

Step 1: Calculate the amount of magnesium sulfate remaining after heating.

The final residue is pure anhydrous \(\ce{MgSO4}\), so its mass tells us the total amount of magnesium sulfate units originally present.

\[ \begin{aligned} n_{\mathrm{MgSO4}}&=\frac{6.00}{120.4}\\ &=0.04983\ \mathrm{mol} \end{aligned} \]

Every original magnesium sulfate unit remains in the solid residue. Heating removes water, not the magnesium sulfate itself.

Step 2: Calculate the amount of water removed.

The original sample weighed 10.80 g, and the final residue weighs 6.00 g.

Therefore:

\[ \begin{aligned} m_{\mathrm{H2O}}&=10.80-6.00\\ &=4.80\ \mathrm g \end{aligned} \]

Convert this mass into moles:

\[ \begin{aligned} n_{\mathrm{H2O}}&=\frac{4.80}{18.0}\\ &=0.2667\ \mathrm{mol} \end{aligned} \]

This water was originally part of the hydrated magnesium sulfate.

Step 3: Calculate the amount of oxygen atoms contributed by the sulfate groups.

Each \(\ce{MgSO4}\) formula unit contains four oxygen atoms.

Therefore:

\[ \begin{aligned} n_{\mathrm O,\ sulfate} &=4\times0.04983\\ &=0.1993\ \mathrm{mol} \end{aligned} \]

Step 4: Calculate the amount of oxygen atoms contributed by the water.

Each water molecule contains one oxygen atom.

Therefore:

\[ n_{\mathrm O,\ water}=0.2667\ \mathrm{mol} \]

This is the oxygen that disappeared from the solid during heating. It must still be included because the question asks about the original sample.

Step 5: Add both contributions.

\[ \begin{aligned} n_{\mathrm O,\ total} &=0.1993+0.2667\\ &=0.4660\ \mathrm{mol} \end{aligned} \]

Step 6: Convert into the total number of oxygen atoms.

\[ \begin{aligned} N_{\mathrm O} &=0.4660\times6.02\times10^{23}\\ &=2.805\times10^{23} \end{aligned} \]

Therefore:

\[ \boxed{N_{\mathrm O}=2.81\times10^{23}\text{ oxygen atoms}} \]

Step 7: Check whether the mixture is physically possible.

There is one more important check.

Every hydrated magnesium sulfate formula unit contains seven water molecules. Anhydrous magnesium sulfate contains none.

Calculate the average number of water molecules per magnesium sulfate unit:

\[ \begin{aligned} \frac{n_{\mathrm{H2O}}}{n_{\mathrm{MgSO4}}} &=\frac{0.2667}{0.04983}\\ &=5.35 \end{aligned} \]

This is between zero and seven, which is consistent with a mixture of anhydrous magnesium sulfate and its heptahydrate.

In fact, approximately 76.4% of the magnesium sulfate formula units were hydrated:

\[ \frac{5.35}{7}\times100=76.4\% \]

The result makes physical sense.

Why not simply count four oxygen atoms per magnesium sulfate unit? Because that ignores the oxygen atoms in the water originally present.

Why not count eleven oxygen atoms per magnesium sulfate unit? Because that assumes every formula unit was hydrated, which would require seven water molecules per magnesium sulfate unit. The measured ratio is only 5.35.

The mass loss provides the missing information. It allows us to count the water oxygen atoms separately rather than assuming that every formula unit has the same hydration state.

This is a useful experimental lesson: a measurement taken after a change of state can reveal information about the original sample, but you must keep track of what was removed.

05What to remember

The central method is straightforward once you identify the correct entity and count its oxygen atoms.

If the amount of compound is already given in moles:

\[ \boxed{N_{\mathrm O}=n\times x\times N_A} \]

If the mass of compound is given:

\[ \boxed{N_{\mathrm O}=\frac{m}{M}\times x\times N_A} \]

Before calculating, check these four things:

• Have you counted every oxygen atom in the formula, including those inside brackets?
• Have you included water of crystallisation where applicable?
• Have you converted the sample mass into moles using the correct molar mass?
• Have you multiplied by the number of oxygen atoms per molecule or formula unit before converting to an actual atom count?

For a single atom or ion, the counting problem is different: you use atomic number, mass number, and charge to identify its subatomic particles. You can explore that distinction in How to Count Protons, Neutrons, and Electrons in Atoms and Ions.

The next useful step is to apply the same counting logic to balanced chemical equations. Once you can count oxygen atoms in a compound, you can track exactly where those atoms go when reactants rearrange to form products.