How to Calculate Mixture Composition After Removing a Component

A 100.0 g salt solution contains 12.0 g of sodium chloride and 88.0 g of water. You evaporate 38.0 g of water.

Has the percentage of sodium chloride changed?

The salt still has a mass of 12.0 g. However, the total mass of the remaining solution is now only 62.0 g. Its new percentage by mass is therefore:

\[
\frac{12.0}{62.0}\times100=19.4\%
\]

The original solution contained 12.0% sodium chloride. After removing water, it contains 19.4%.

This is the key to calculating mixture composition after removing a component: recalculate the percentage using the mass of the mixture that remains, not the original mixture mass.

The component you remove loses mass. Other components retain their original masses, provided none are accidentally lost or chemically changed. The total mass decreases, so the percentages of the remaining components change.

01The calculation method: track what remains

Start by separating the problem into three pieces of information:

• The original mass of each component.
• The mass of the component being removed.
• The mass of each component after removal.

Once you know these, calculating the new composition is straightforward.

Step 1: Calculate the final mass of each component

Suppose a mixture contains components A and B, and you remove some of component B.

The final mass of B is:

\[
m_{B,f}=m_{B,i}-m_{B,r}
\]

Where:

• \(m_{B,f}\) = final mass of component B, in grams.
• \(m_{B,i}\) = initial mass of component B, in grams.
• \(m_{B,r}\) = mass of component B removed, in grams.

If component A is unaffected:

\[
m_{A,f}=m_{A,i}
\]

If you remove component B completely, its final mass is zero.

Step 2: Calculate the final total mixture mass

Add together the masses of everything remaining:

\[
m_f=\sum m_{j,f}
\]

For a two-component mixture:

\[
m_f=m_{A,f}+m_{B,f}
\]

You can also use:

\[
m_f=m_i-m_r
\]

This second method works when only the target component is removed and no other material is lost.

Step 3: Calculate the new percentage composition

Use the standard percentage-by-mass equation:

\[
\%\text{ component}=
\frac{\text{mass of component remaining}}
{\text{final total mixture mass}}\times100
\]

Notice the denominator.

You must use the final mixture mass, not the original mixture mass.

If you need a refresher on the underlying percentage calculation, see Percentage Composition by Mass for HSC Chemistry.

The calculation in one table

A useful check is that the percentages of all components remaining in the mixture should add to approximately 100%.

02Choosing the correct separation method

Before doing any calculations, identify how the component is being removed.

Different separation methods rely on different physical properties. Choosing the wrong method can lead to an incorrect assumption about which substances have actually left the mixture.

Evaporation: removing a volatile solvent

Suppose you have sodium chloride dissolved in water.

Water is volatile, meaning it can evaporate relatively easily. Sodium chloride is effectively non-volatile under the conditions used for this separation.

When the solution is heated:

• Water absorbs energy and changes from liquid to gas.
• Water vapour escapes from the solution.
• Sodium chloride remains in the evaporating basin.
• The remaining solution becomes more concentrated.

If all the water is removed, the sodium chloride remains as a solid residue.

Use an evaporating basin when your aim is to remove the solvent and recover the non-volatile material.

One limitation is that the evaporated water is not collected. You infer its mass from the decrease in the mass of the apparatus and its contents.

Distillation: removing and collecting a volatile component

What if you want to remove water but also collect it?

Simple distillation is more appropriate.

The apparatus contains:

1. A heated round-bottom flask containing the mixture.
2. A still head connecting the flask to a condenser.
3. A thermometer positioned in the vapour pathway.
4. A water-cooled condenser.
5. A receiving flask that collects the condensed liquid.

As the mixture is heated, water evaporates and travels through the still head into the condenser.

Cooling water enters the condenser through its lower side-arm and leaves through its upper side-arm. This keeps the cooling jacket filled.

The water vapour loses energy, condenses back into liquid, and flows into the receiving flask. This collected liquid is called the distillate.

The non-volatile dissolved substances remain in the original flask.

The difference between evaporation and distillation is what happens to the vapour:

Distillation is particularly useful when you need to measure the mass of the removed liquid directly.

For a closer comparison of apparatus, including why reflux does not remove a component from a reaction mixture, see Reflux vs Distillation: How to Choose the Right Apparatus.

Other separation methods: don’t confuse what passes through

Not every separation involves evaporation.

A particularly common mistake is assuming that filtration removes dissolved salt from water.

It doesn’t.

Dissolved sodium chloride exists as dispersed sodium and chloride ions. These pass through ordinary filter paper along with the water.

Filtration can remove insoluble sand from salt water, but the salt remains in the filtrate.

03What does an experiment actually measure?

There is an important distinction between measuring how much mass has disappeared and identifying which substance has disappeared.

Imagine measuring an evaporating basin containing a solution before and after heating.

You obtain:

• Initial mass of basin and solution: 86.40 g.
• Final mass of basin and remaining contents: 66.40 g.

The decrease is:

\[
86.40-66.40=20.00\text{ g}
\]

This tells you that 20.00 g of material has left the apparatus.

But does it prove that exactly 20.00 g of water evaporated?

Only if water was the only material that could leave.

If the solution splattered during heating, some dissolved salt could have escaped as droplets. If another volatile substance was present, it could have evaporated too.

A balance measures mass. It does not identify chemical substances.

Improving the experimental method

An open evaporating basin is suitable when you want to remove a solvent and recover the residue. However, it cannot collect the vapour for direct measurement.

A revised experiment can use simple distillation.

Weigh the receiving flask before distillation, then weigh it again after collecting the distillate.

\[
m_{\text{distillate}}=
m_{\text{receiver and distillate}}
-m_{\text{empty receiver}}
\]

You can compare this value with the decrease in mass of the original flask and its contents.

If the two measurements agree within experimental uncertainty, this supports the model that the collected distillate accounts for the material lost from the original flask.

It does not automatically prove that the distillate is pure water. That conclusion also depends on the mixture containing no other volatile substances and avoiding the carryover of liquid droplets.

For a reliable experiment, heat gently, avoid splashing, and allow hot equipment to cool before weighing. If recovering a solid residue, dry it to constant mass so that residual solvent does not inflate the measured mass.

The important experimental principle is simple:

Mass lost from the original mixture must be accounted for before you can confidently assign that loss to a particular component.

06Common mistakes when calculating the new composition

Before attempting exam questions, check that you can distinguish these situations.

One final distinction is worth remembering: if a question asks for the composition of the collected distillate, you use the mass of the distillate as the denominator. If it asks for the composition of the remaining flask contents, you use the mass remaining in the flask.

These are two different mixtures, so they have different denominators.

07Questions and solutions

The following questions progress from a straightforward evaporation calculation to problems involving multiple components, distillation, reverse calculations, and solubility.

Question 1

A student prepares a 75.0 g glucose solution containing 15.0 g of glucose and 60.0 g of water.

The student evaporates 20.0 g of water.

Assuming no glucose is lost, calculate:

a) The final mass of the solution.

b) The final percentage by mass of glucose.

c) The final percentage by mass of water.

Solution 1

The final solution has a mass of 55.0 g and contains 27.3% glucose and 72.7% water by mass.

The glucose remains in the solution, while 20.0 g of water is removed.

Step 1: Calculate the remaining mass of water.

\[
m_{\text{water},f}=60.0-20.0
\]

\[
=40.0\text{ g}
\]

Step 2: Calculate the final solution mass.

\[
m_f=15.0+40.0
\]

\[
=55.0\text{ g}
\]

Step 3: Calculate the percentage of glucose.

\[
\%\text{ glucose}=
\frac{15.0}{55.0}\times100
\]

\[
=27.3\%
\]

Step 4: Calculate the percentage of water.

\[
\%\text{ water}=
\frac{40.0}{55.0}\times100
\]

\[
=72.7\%
\]

The two percentages add to 100.0%, confirming that the calculation is consistent.

The tempting mistake is to divide 15.0 g by the original 75.0 g and report 20.0%. That is the original composition, not the composition after evaporation.

Question 2

A 200.0 g aqueous mixture contains 16.0 g of sodium chloride, 24.0 g of copper sulfate, and water.

The mixture undergoes simple distillation, producing 50.0 g of pure water distillate.

Assume that both salts remain in the original flask.

a) Calculate the final mass of the mixture remaining in the flask.

b) Calculate the final percentage by mass of each component.

c) A student claims that the total percentage of dissolved solids is still 20.0% because the original mixture contained 40.0 g of dissolved solids. Explain the error and calculate the correct percentage.

Solution 2

The remaining mixture has a mass of 150.0 g and contains 26.7% dissolved solids by mass.

The important point is that the 50.0 g distillate consists entirely of water, so neither salt loses mass.

Step 1: Calculate the initial mass of water.

\[
m_{\text{water},i}=200.0-16.0-24.0
\]

\[
=160.0\text{ g}
\]

Step 2: Calculate the remaining water mass.

\[
m_{\text{water},f}=160.0-50.0
\]

\[
=110.0\text{ g}
\]

Step 3: Calculate the final mixture mass.

\[
m_f=16.0+24.0+110.0
\]

\[
=150.0\text{ g}
\]

Step 4: Calculate the percentage of sodium chloride.

\[
\%\text{ NaCl}=
\frac{16.0}{150.0}\times100
\]

\[
=10.7\%
\]

Step 5: Calculate the percentage of copper sulfate.

\[
\%\text{ CuSO}_4=
\frac{24.0}{150.0}\times100
\]

\[
=16.0\%
\]

Step 6: Calculate the percentage of water.

\[
\%\text{ water}=
\frac{110.0}{150.0}\times100
\]

\[
=73.3\%
\]

Step 7: Calculate the combined percentage of dissolved solids.

The total mass of dissolved solids is still:

\[
m_{\text{solids}}=16.0+24.0=40.0\text{ g}
\]

However, the denominator has changed.

\[
\%\text{ solids}=
\frac{40.0}{150.0}\times100
\]

\[
=26.7\%
\]

The student’s 20.0% calculation uses the original 200.0 g mixture as the denominator.

That calculation is correct for the original solution, but incorrect for the remaining solution.

Question 3

An aqueous mixture has an initial mass of 120.0 g. It contains sodium chloride, copper sulfate, and water.

The water is completely removed by evaporation. The resulting dry solid residue has a mass of 30.0 g and contains 40.0% sodium chloride by mass.

Assume both salts are non-volatile and neither decomposes during heating.

Calculate the original percentage by mass of:

a) Sodium chloride.

b) Copper sulfate.

c) Water.

Solution 3

The original mixture contained 10.0% sodium chloride, 15.0% copper sulfate, and 75.0% water by mass.

The important distinction is that the 40.0% sodium chloride figure describes the dry residue, not the original aqueous mixture.

Step 1: Calculate the mass of sodium chloride in the dry residue.

\[
m_{\text{NaCl}}=
\frac{40.0}{100}\times30.0
\]

\[
=12.0\text{ g}
\]

Step 2: Calculate the mass of copper sulfate.

The residue contains only sodium chloride and copper sulfate.

\[
m_{\text{CuSO}_4}=30.0-12.0
\]

\[
=18.0\text{ g}
\]

Step 3: Calculate the original mass of water.

All water was removed, so the difference between the original mixture and the dry residue is the original water mass.

\[
m_{\text{water}}=120.0-30.0
\]

\[
=90.0\text{ g}
\]

Step 4: Calculate the original percentage of sodium chloride.

\[
\%\text{ NaCl}=
\frac{12.0}{120.0}\times100
\]

\[
=10.0\%
\]

Step 5: Calculate the original percentage of copper sulfate.

\[
\%\text{ CuSO}_4=
\frac{18.0}{120.0}\times100
\]

\[
=15.0\%
\]

Step 6: Calculate the original percentage of water.

\[
\%\text{ water}=
\frac{90.0}{120.0}\times100
\]

\[
=75.0\%
\]

The final check is:

\[
10.0+15.0+75.0=100.0\%
\]

This question works backwards from the residue to reconstruct the original mixture.

The tempting mistake is to report 40.0% sodium chloride as the original percentage. That value applies only after the water has been removed. The original mixture contained a substantial amount of water, so its sodium chloride percentage was much lower.

Question 4

A 100.0 g mixture contains 40.0 g of ethanol and 60.0 g of water.

The mixture undergoes distillation. A total of 20.0 g of distillate is collected.

An analysis shows that the distillate contains 70.0% ethanol by mass and 30.0% water by mass.

These are illustrative measured data supplied for this question.

Calculate:

a) The mass of ethanol removed.

b) The mass of water removed.

c) The final mass of each component remaining in the original flask.

d) The final percentage by mass of ethanol and water in the remaining mixture.

Explain why it would be incorrect to assume that all 20.0 g of distillate is ethanol simply because ethanol is more volatile than water.

Solution 4

The remaining mixture contains 26.0 g of ethanol and 54.0 g of water, giving 32.5% ethanol and 67.5% water by mass.

The crucial information is the measured composition of the distillate. Both ethanol and water are volatile, so both can enter the vapour.

Step 1: Calculate the mass of ethanol removed.

The distillate has a total mass of 20.0 g and contains 70.0% ethanol.

\[
m_{\text{ethanol removed}}=
\frac{70.0}{100}\times20.0
\]

\[
=14.0\text{ g}
\]

Step 2: Calculate the mass of water removed.

\[
m_{\text{water removed}}=
\frac{30.0}{100}\times20.0
\]

\[
=6.0\text{ g}
\]

Check:

\[
14.0+6.0=20.0\text{ g}
\]

Step 3: Calculate the remaining ethanol mass.

\[
m_{\text{ethanol},f}=40.0-14.0
\]

\[
=26.0\text{ g}
\]

Step 4: Calculate the remaining water mass.

\[
m_{\text{water},f}=60.0-6.0
\]

\[
=54.0\text{ g}
\]

Step 5: Calculate the final total mass.

\[
m_f=26.0+54.0
\]

\[
=80.0\text{ g}
\]

Step 6: Calculate the final ethanol percentage.

\[
\%\text{ ethanol}=
\frac{26.0}{80.0}\times100
\]

\[
=32.5\%
\]

Step 7: Calculate the final water percentage.

\[
\%\text{ water}=
\frac{54.0}{80.0}\times100
\]

\[
=67.5\%
\]

The tempting approach is to subtract the entire 20.0 g distillate from the ethanol mass.

That would assume the distillate is pure ethanol, which contradicts the supplied measurement showing that it contains 30.0% water.

Ethanol is more volatile under these conditions, so the distillate is enriched in ethanol relative to the original mixture. However, enrichment does not mean complete separation.

This is why the composition of the collected fraction matters whenever more than one component can evaporate.

Question 5

A student has a 100.0 g aqueous potassium nitrate (\(\ce{KNO3}\)) mixture at 20°C.

It contains:

• 20.0 g of potassium nitrate.
• 80.0 g of water.

For this question, use an illustrative solubility value of 32.0 g of potassium nitrate per 100 g of water at 20°C.

The student removes water by evaporation. After each stage, the mixture is returned to 20°C so that its equilibrium composition can be assessed.

A total of 30.0 g of water is eventually removed.

Assume that no potassium nitrate is lost.

Calculate:

a) The mass of water that must be removed before crystals first appear at 20°C.

b) The mass of potassium nitrate that crystallises after 30.0 g of water has been removed.

c) The percentage by mass of potassium nitrate and water in the entire contents of the vessel.

d) The percentage by mass of potassium nitrate in the remaining liquid solution only.

Solution 5

After removing 30.0 g of water, 4.0 g of potassium nitrate crystallises. The entire vessel contents contain 28.6% potassium nitrate by mass, while the remaining liquid solution contains 24.2% potassium nitrate by mass.

This is the most conceptually important question in the set because the solute does not necessarily remain entirely dissolved as the solvent is removed.

There are two separate things to track:

1. The total mass of potassium nitrate in the vessel.
2. The mass of potassium nitrate that remains dissolved in the liquid.

These are not necessarily equal.

Step 1: Calculate the water mass at which the solution first becomes saturated.

The solubility is 32.0 g of potassium nitrate per 100 g of water.

This means that 100 g of water can dissolve a maximum of 32.0 g of potassium nitrate at 20°C.

The initial potassium nitrate mass is 20.0 g.

Let the required mass of water at saturation be \(m_w\).

\[
\frac{32.0}{100}=
\frac{20.0}{m_w}
\]

Rearranging:

\[
m_w=\frac{20.0\times100}{32.0}
\]

\[
=62.5\text{ g}
\]

Therefore, the solution reaches saturation when 62.5 g of water remains.

Step 2: Calculate how much water must be removed to reach saturation.

Initially, there are 80.0 g of water.

\[
m_{\text{water removed}}=80.0-62.5
\]

\[
=17.5\text{ g}
\]

At this point, the solution is saturated but does not yet need to contain excess solid.

Crystals begin to appear when further water removal makes the solution unable to keep all 20.0 g of potassium nitrate dissolved at 20°C.

Step 3: Calculate the final mass of water.

A total of 30.0 g of water has been removed.

\[
m_{\text{water},f}=80.0-30.0
\]

\[
=50.0\text{ g}
\]

Step 4: Calculate how much potassium nitrate can remain dissolved.

At 20°C, 100 g of water dissolves a maximum of 32.0 g of potassium nitrate.

Therefore, 50.0 g of water can dissolve:

\[
m_{\text{dissolved}}=
\frac{32.0}{100}\times50.0
\]

\[
=16.0\text{ g}
\]

Step 5: Calculate the mass of potassium nitrate that crystallises.

The original potassium nitrate mass was 20.0 g.

Only 16.0 g can remain dissolved at the final temperature.

Therefore:

\[
m_{\text{crystallised}}=20.0-16.0
\]

\[
=4.0\text{ g}
\]

Step 6: Calculate the percentage composition of the entire vessel contents.

The entire vessel contains:

• 20.0 g of potassium nitrate, including dissolved and crystallised material.
• 50.0 g of water.

The total mass is:

\[
m_{\text{total}}=20.0+50.0
\]

\[
=70.0\text{ g}
\]

The overall percentage of potassium nitrate is:

\[
\%\text{ KNO}_3=
\frac{20.0}{70.0}\times100
\]

\[
=28.6\%
\]

The overall percentage of water is:

\[
\%\text{ water}=
\frac{50.0}{70.0}\times100
\]

\[
=71.4\%
\]

Step 7: Calculate the percentage of potassium nitrate in the liquid solution only.

This is where many students make a mistake.

The liquid solution contains only 16.0 g of dissolved potassium nitrate and 50.0 g of water.

Its total mass is:

\[
m_{\text{solution}}=16.0+50.0
\]

\[
=66.0\text{ g}
\]

Therefore:

\[
\%\text{ KNO}_3=
\frac{16.0}{66.0}\times100
\]

\[
=24.2\%
\]

The solid crystals are not included in the mass of the liquid solution.

The final results are:

The tempting approach is to calculate \(20.0/70.0\times100\) and report 28.6% as the concentration of the remaining solution.

That answer describes the entire contents of the vessel, including the solid crystals. It does not describe the liquid phase.

There is another important detail: the solubility value is expressed per 100 g of water, not per 100 g of solution. That is why the saturated liquid contains 32.0 g of potassium nitrate per 132.0 g of total solution, giving 24.2% by mass.

This distinction between the overall mixture and the liquid phase becomes essential whenever removing a solvent causes a dissolved substance to crystallise.

08Connecting the method to other HSC Chemistry questions

When a question describes the removal of a mixture component, use this sequence:

1. Identify the physical property that allows the component to be separated.
2. Choose the appropriate separation apparatus.
3. Identify exactly what enters the collected fraction and what remains behind.
4. Calculate the final mass of every component.
5. Add the remaining masses to obtain the correct denominator.
6. Calculate the new percentage composition.
7. Check that the percentages are consistent with the final mixture and the stated assumptions.

The next useful concept is understanding how changing the amount of solvent affects solution concentration and solubility. That is where mass balance, percentage composition, and crystallisation begin to work together in more demanding HSC Chemistry problems.