How to Calculate Oxygen by Difference and Find Empirical Formulas

An unknown oxide weighs 1.590 g before an experiment and 1.270 g afterwards. The solid that remains is the metal. Where did the missing 0.320 g go?

If the metal has remained in the solid and nothing else has been lost, that missing mass must be oxygen. This is the basic idea behind calculating oxygen content by difference.

The method is straightforward:

• If you know the masses of all the other elements, subtract them from the original sample mass.
• If you know the mass percentages of all the other elements, subtract their percentages from 100%.
• Convert the resulting elemental masses into moles and simplify the ratio to obtain the empirical formula.

The important part is knowing when the subtraction is valid. A remainder is only oxygen if every other element has been accounted for.

All numerical data in the worked examples and questions below are illustrative.

01What does oxygen by difference actually mean?

Imagine a pure compound containing only copper and oxygen.

Its total mass must come from those two elements:

\[ m_{\text{compound}}=m_{\mathrm{Cu}}+m_{\mathrm O} \]

Rearranging gives:

\[ \boxed{m_{\mathrm O}=m_{\text{compound}}-m_{\mathrm{Cu}}} \]

This is called calculating oxygen by difference.

The same principle works for compounds containing more elements. For example, if a compound contains carbon, hydrogen, nitrogen, and oxygen:

\[ m_{\mathrm O}=m_{\text{total}}-m_{\mathrm C} -m_{\mathrm H}-m_{\mathrm N} \]

If the question gives percentages rather than actual masses, the calculation becomes:

\[ \boxed{\%\mathrm O=100-\%\mathrm C-\%\mathrm H-\%\mathrm N} \]

Notice that we subtract every other element, not just carbon and hydrogen.

Why would we calculate oxygen this way?

Oxygen is sometimes difficult to measure directly in an elemental composition experiment. Instead, an experiment can determine the amounts of the other elements, leaving oxygen as the unaccounted-for mass.

A useful experimental example is the reduction of a metal oxide.

02How an experiment can determine oxygen by difference

Suppose a chemist wants to determine the composition of an unknown copper oxide.

The problem is that the oxygen is chemically combined with copper. Simply weighing the black solid tells us its total mass, not how much of that mass comes from each element.

One solution is to remove the oxygen chemically while keeping the copper.

The reduction experiment

In a controlled laboratory experiment, a weighed sample of copper oxide is placed in a porcelain boat inside a horizontal combustion tube.

Dry hydrogen gas enters one end of the tube and flows over the sample. The tube is heated, allowing hydrogen to react with the copper oxide.

The reaction is:

\[ \ce{CuO(s) + H2(g) -> Cu(s) + H2O(g)} \]

The oxygen originally present in the copper oxide combines with hydrogen to form water vapour. The water vapour leaves through the outlet, while copper remains in the porcelain boat.

The boat and its contents can then be cooled under conditions that prevent the copper from reacting with air and weighed again.

What should happen to the solid’s mass?

If all the copper stays behind and all the oxygen is removed, the decrease in mass equals the original oxygen mass.

There is an important detail here. The water collected at the outlet is not pure oxygen from the original sample. It also contains hydrogen supplied by the hydrogen gas.

Therefore, the mass of collected water is not the mass of oxygen in the original copper oxide.

The difference between the initial solid mass and the final solid mass is what gives us the oxygen mass.

Why the experimental design matters

An initial attempt might simply heat the copper oxide in air and compare its mass before and afterwards.

That does not isolate oxygen removal. The solid might gain oxygen from the surrounding air rather than lose it.

Using a reducing gas changes the chemical environment. It allows oxygen to leave the solid through a defined reaction.

The interpretation still depends on several assumptions:

• The reduction is complete.
• All copper remains in the solid.
• No other material is lost from the sample.
• The final copper does not react with oxygen during cooling.

Hydrogen is flammable, so this is a supervised laboratory method, not a home experiment.

03The calculation: finding oxygen by difference

There are two common ways to approach a composition problem.

Method A: Starting and final masses

Define the mass loss as:

\[ \Delta m=m_{\text{initial}}-m_{\text{final}} \]

Here, \(\Delta m\) is positive when the solid loses mass.

For a pure metal oxide reduced completely to the metal:

\[ m_{\mathrm O}=\Delta m \]

The oxygen percentage is:

\[ \boxed{ \%\mathrm O= \frac{m_{\mathrm O}}{m_{\text{initial oxide}}} \times100 } \]

The denominator is the original oxide mass, not the final metal mass.

Method B: Given elemental percentages

If the question gives percentages of the other elements:

\[ \boxed{ \%\mathrm O=100-\sum\%\text{ of all other elements} } \]

For example, a pure compound contains:

• 40.0% carbon
• 6.7% hydrogen
• 53.3% oxygen

The oxygen percentage is:

\[ 100-40.0-6.7=53.3\% \]

This is oxygen by difference.

Both methods establish the mass of oxygen. Neither automatically gives the chemical formula.

For that, we need mole ratios.

05When oxygen is given as the missing percentage

Now consider a different type of question.

Instead of giving you an initial sample mass and a final residue mass, the question gives you the percentage composition of the elements.

The calculation of oxygen is even quicker.

However, you must still convert the percentages into moles to obtain an empirical formula.

06How do you know when to round a mole ratio?

Getting the oxygen percentage right is only half the job. You also need to recognise the simplest whole-number mole ratio.

Here are some useful patterns:

These are useful patterns, not automatic rounding rules.

For example, consider:

\[ 1:1.50:1 \]

The correct approach is:

\[ 2:3:2 \]

Not:

\[ 1:2:1 \]

The multiplier must be applied to every element.

What if the ratio is not close to a simple fraction?

Do not force it into a whole number.

Instead, check:

1. Did you calculate oxygen by subtracting every other element?
2. Did you use the correct atomic molar masses?
3. Did you divide every mole amount by the smallest one?
4. Did you round too early?
5. Is the sample pure, and have all elements been accounted for?

Experimental measurements are not always exact. A ratio of 1.99 may reasonably represent 2.00 when the measurements have limited precision.

Keep extra digits during your working and round the final formula only when the ratio is justified.

07The important distinction: empirical formula versus molecular formula

Finding oxygen by difference helps establish the composition of a compound. It does not necessarily identify the compound’s actual molecular formula.

An empirical formula represents the simplest whole-number ratio of atoms.

For example:

\[ \ce{C3H4O3} \]

could be an empirical formula for a substance whose molecular formula is:

\[ \ce{C6H8O6} \]

The second formula contains twice as many atoms of every element.

Both have the same elemental percentages by mass.

To determine the molecular formula of a molecular compound, you need an independent molar-mass measurement.

The relationship is:

\[ \text{Molecular formula} = (\text{Empirical formula})\times n \]

where \(n\) is a positive whole number.

There is another important distinction for ionic compounds.

A formula such as \(\ce{CuO}\) represents the simplest ratio of ions in an ionic lattice. It does not describe a discrete molecule containing one copper atom bonded to one oxygen atom.

The composition calculation gives you a ratio. You should not claim more about the substance’s identity than the evidence supports.

08Common mistakes when calculating oxygen by difference

One final limitation is worth remembering.

If a compound contains carbon, hydrogen, nitrogen, sulfur, and oxygen, but the question only provides carbon and hydrogen percentages, you cannot automatically call the remainder oxygen.

The remainder includes nitrogen, sulfur, and oxygen together.

Oxygen by difference works only when every other element has been measured or otherwise accounted for.

For a refresher on calculating mass fractions and identifying the correct denominator, see Percentage Composition by Mass for HSC Chemistry.

If your question instead gives carbon dioxide and water absorber mass gains from combustion analysis, the calculation requires an additional step to determine the original carbon and hydrogen masses. See How to Calculate Element Percentages from Chemical Formulas.

09Questions and solutions

The following questions progress from straightforward percentage calculations to experimental errors, impurities, and competing explanations of an empirical formula.

Unless otherwise stated, assume complete reactions and use the atomic molar masses supplied.

Question 1

A pure calcium oxide sample contains 71.5% calcium by mass.

Calculate:

(a) The percentage by mass of oxygen.

(b) The empirical formula of calcium oxide.

Use Ca = 40.08 g mol⁻¹ and O = 16.00 g mol⁻¹.

Solution 1

The oxygen content is 28.5% by mass, and the empirical formula is \(\ce{CaO}\).

Part (a): Calculate oxygen by difference.

The sample contains only calcium and oxygen.

\[ \begin{aligned} \%\mathrm O &=100-71.5\\ &=28.5\% \end{aligned} \]

Part (b): Determine the empirical formula.

Assume a 100 g sample.

The elemental masses are:

• Calcium = 71.5 g
• Oxygen = 28.5 g

Convert to moles.

\[ n(\mathrm{Ca})=\frac{71.5}{40.08} =1.784\text{ mol} \]

\[ n(\mathrm O)=\frac{28.5}{16.00} =1.781\text{ mol} \]

Divide by the smallest amount:

\[ \mathrm{Ca:O}=1.002:1 \]

Within the precision of the supplied data, this is a 1:1 ratio.

Therefore:

\[ \boxed{\ce{CaO}} \]

The key point is that 28.5% oxygen is a mass percentage, not the number of oxygen atoms. You must convert the masses into moles before finding the formula.

Question 2

An unknown pure compound contains only carbon, hydrogen, nitrogen, and oxygen.

Its elemental composition is:

Calculate:

(a) The percentage by mass of oxygen.

(b) The empirical formula of the compound.

Use:

• C = 12.01 g mol⁻¹
• H = 1.008 g mol⁻¹
• N = 14.01 g mol⁻¹
• O = 16.00 g mol⁻¹

Solution 2

The oxygen percentage is 28.0%, and the empirical formula is \(\ce{C2H3NO}\).

Part (a): Find oxygen by difference.

There are four elements, so we must subtract all three known percentages.

\[ \begin{aligned} \%\mathrm O &=100-42.1-5.30-24.6\\ &=28.0\% \end{aligned} \]

Part (b): Find the empirical formula.

Assume a 100 g sample.

The elemental masses are:

• C = 42.1 g
• H = 5.30 g
• N = 24.6 g
• O = 28.0 g

Convert each mass into moles.

\[ n(\mathrm C)=\frac{42.1}{12.01} =3.505 \]

\[ n(\mathrm H)=\frac{5.30}{1.008} =5.258 \]

\[ n(\mathrm N)=\frac{24.6}{14.01} =1.756 \]

\[ n(\mathrm O)=\frac{28.0}{16.00} =1.750 \]

Divide all amounts by the smallest value, 1.750.

\[ \begin{aligned} \mathrm C&=2.003\\ \mathrm H&=3.005\\ \mathrm N&=1.003\\ \mathrm O&=1.000 \end{aligned} \]

The simplest whole-number ratio is:

\[ \mathrm{C:H:N:O}=2:3:1:1 \]

Therefore:

\[ \boxed{\ce{C2H3NO}} \]

The tempting shortcut would be to subtract carbon and hydrogen from 100% and call the remainder oxygen.

That would give 52.6%, which is incorrect because nitrogen also contributes to the sample’s mass.

Always check how many elements are present before calculating oxygen by difference.

Question 3

A student performs the reduction of a 1.590 g sample of pure copper(II) oxide.

The expected final mass of copper after complete reduction is 1.270 g.

However, the student turns off the hydrogen supply while the tube is still hot and allows the copper to cool in contact with air. Some of the copper reacts with oxygen during cooling.

The final measured solid mass is 1.300 g.

Explain:

(a) Why the final mass is higher than expected.

(b) Whether the calculated oxygen mass will be too high or too low.

(c) How the experimental design should be improved.

Solution 3

The calculated oxygen mass will be too low because some oxygen from the air has been added back to the copper during cooling.

Part (a): Explain the increase in mass.

During reduction, oxygen leaves the copper oxide and reacts with hydrogen to form water vapour.

However, once the hydrogen supply is stopped, air can enter the tube.

The hot copper can then react with oxygen from the air, forming copper oxide again.

This adds mass to the final solid.

Part (b): Determine the direction of error.

If the copper had remained completely unoxidised, the final mass would have been 1.270 g.

The true oxygen mass removed from the original sample would therefore be:

\[ \begin{aligned} m_{\mathrm O} &=1.590-1.270\\ &=0.320\text{ g} \end{aligned} \]

But the student measures 1.300 g.

The apparent oxygen mass is:

\[ \begin{aligned} m_{\mathrm O,\ apparent} &=1.590-1.300\\ &=0.290\text{ g} \end{aligned} \]

The student underestimates the oxygen mass by:

\[ 0.320-0.290=0.030\text{ g} \]

The calculated oxygen percentage would also be too low.

Part (c): Improve the experiment.

The copper must be protected from air while it cools.

The experiment should maintain an appropriate oxygen-free atmosphere until the solid has cooled sufficiently to avoid reoxidation.

The sample should also be weighed consistently and, where appropriate, heated and cooled until a constant mass is obtained.

This question illustrates an important experimental principle: a correct mass-difference calculation cannot compensate for an uncontrolled chemical reaction.

Question 4

A 2.200 g sample contains an oxide of an unknown metal X and 0.200 g of an inert impurity.

The sample is reduced completely.

After the reaction:

• All the metal remains in the solid.
• The inert impurity remains unchanged.
• All oxygen is removed.
• The final solid residue weighs 1.400 g.

The relative atomic mass of X is 40.0.

Determine:

(a) The mass of oxygen removed.

(b) The mass percentage of oxygen in the original oxide, excluding the inert impurity.

(c) The empirical formula of the oxide.

Solution 4

The oxide has an empirical formula of \(\ce{X3O5}\), and its oxygen content is 40.0% by mass.

Part (a): Calculate the oxygen mass.

The inert impurity and metal both remain in the solid.

Therefore, the entire mass loss must be oxygen.

\[ \begin{aligned} m_{\mathrm O} &=2.200-1.400\\ &=0.800\text{ g} \end{aligned} \]

Part (b): Calculate the oxygen percentage.

Be careful with the denominator.

The original sample includes 0.200 g of inert impurity, so the mass of the actual oxide is:

\[ \begin{aligned} m_{\text{oxide}} &=2.200-0.200\\ &=2.000\text{ g} \end{aligned} \]

Therefore:

\[ \begin{aligned} \%\mathrm O &=\frac{0.800}{2.000}\times100\\ &=40.0\% \end{aligned} \]

Part (c): Determine the empirical formula.

First, find the mass of metal in the final residue.

The residue contains both metal and inert impurity.

\[ \begin{aligned} m_X&=1.400-0.200\\ &=1.200\text{ g} \end{aligned} \]

Convert the elemental masses into moles.

For X:

\[ n(X)=\frac{1.200}{40.0} =0.0300\text{ mol} \]

For oxygen:

\[ n(\mathrm O)=\frac{0.800}{16.00} =0.0500\text{ mol} \]

Divide by the smaller amount:

\[ \begin{aligned} X&=\frac{0.0300}{0.0300}=1\\ O&=\frac{0.0500}{0.0300}=1.667 \end{aligned} \]

The ratio is:

\[ 1:1.667 \]

This is approximately:

\[ 1:\frac{5}{3} \]

Multiply both values by 3:

\[ 3:5 \]

Therefore:

\[ \boxed{\ce{X3O5}} \]

The important trap was the inert impurity.

It did not affect the total mass loss, because it remained in the solid. However, it did affect the amount of metal present in the final residue.

Using the entire 1.400 g residue as the metal mass would produce an incorrect mole ratio.

Question 5

A metal M has a relative atomic mass of 40.0 and can form two oxides:

• \(\ce{MO}\), containing metal in the +2 oxidation state.
• \(\ce{M2O3}\), containing metal in the +3 oxidation state.

A 2.400 g sample containing only M and oxygen is reduced completely to 1.600 g of pure metal.

Assume:

• All the metal remains in the solid.
• All oxygen is removed.
• A single-phase oxide with formula unit \(\ce{M4O5}\) is chemically possible.
• The only alternative being considered is a mixture of \(\ce{MO}\) and \(\ce{M2O3}\).

A student calculates the empirical ratio \(\mathrm{M:O}=4:5\) and confidently concludes that the original sample must have been a single compound, \(\ce{M4O5}\).

Is the student’s conclusion justified?

If the sample were instead a mixture of \(\ce{MO}\) and \(\ce{M2O3}\), calculate the mole ratio of the two oxide formula units.

Explain why the mass measurements alone cannot distinguish the two possibilities.

Solution 5

No. The student’s conclusion is not justified.

The bulk composition is consistent with a single-phase oxide \(\ce{M4O5}\), but it is also consistent with a mixture containing two different oxides.

If the sample is a mixture, it contains 0.0200 mol of \(\ce{MO}\) and 0.0100 mol of \(\ce{M2O3}\), giving a 2:1 mole ratio.

Step 1: Calculate the oxygen mass.

\[ \begin{aligned} m_{\mathrm O} &=2.400-1.600\\ &=0.800\text{ g} \end{aligned} \]

Step 2: Calculate the mole amounts of metal and oxygen.

For metal:

\[ n(M)=\frac{1.600}{40.0} =0.0400\text{ mol} \]

For oxygen:

\[ n(\mathrm O)=\frac{0.800}{16.00} =0.0500\text{ mol} \]

The overall ratio is:

\[ \begin{aligned} \mathrm{M:O} &=0.0400:0.0500\\ &=4:5 \end{aligned} \]

This gives the empirical composition \(\ce{M4O5}\).

But an empirical ratio describes the overall composition. It does not prove that every part of the sample belongs to one compound.

Step 3: Test the mixture explanation.

Let:

• \(x\) = moles of \(\ce{MO}\)
• \(y\) = moles of \(\ce{M2O3}\)

The total amount of metal is:

\[ x+2y=0.0400 \]

The total amount of oxygen is:

\[ x+3y=0.0500 \]

Subtract the first equation from the second:

\[ \begin{aligned} y&=0.0500-0.0400\\ &=0.0100\text{ mol} \end{aligned} \]

Substitute into the metal equation:

\[ \begin{aligned} x+2(0.0100)&=0.0400\\ x&=0.0200\text{ mol} \end{aligned} \]

Therefore:

\[ \boxed{ n(\ce{MO}):n(\ce{M2O3})=2:1 } \]

Step 4: Explain why both models give the same result.

Consider the mixture:

• 2 mol of \(\ce{MO}\) contain 2 mol of M and 2 mol of O.
• 1 mol of \(\ce{M2O3}\) contains 2 mol of M and 3 mol of O.

Together, they contain:

\[ 4\text{ mol M}:5\text{ mol O} \]

That is exactly the same overall elemental ratio as \(\ce{M4O5}\).

Even the average oxidation state is the same.

A single-phase \(\ce{M4O5}\) has an average metal oxidation state of +2.5. The proposed mixture also has an average oxidation state of +2.5 because it contains equal amounts of M in the +2 and +3 states.

Consequently, neither the overall mass composition nor the average oxidation state distinguishes these models.

A separate investigation that identifies the solid phases, such as X-ray diffraction, would be needed to determine whether the sample is one compound or a mixture.

This is the deeper limitation of oxygen by difference: an accurate elemental ratio tells you what is present overall, but not necessarily how the material is organised.

Once you have calculated oxygen content and established a defensible empirical formula, the next decision is whether the question provides enough information to determine a molecular formula or identify a particular compound. That requires evidence beyond elemental percentages alone.