How to Calculate Mass Percent from Solute and Solvent Masses
Learn how to calculate mass percent (% m/m) using solute and solvent masses. Follow step-by-step HSC Chemistry examples, understand dilution and evaporation, and practise more challenging mass percentage problems with worked solutions.
Suppose you dissolve 15.0 g of salt in 135.0 g of water. A student calculates \(15.0 \div 135.0 \times 100 = 11.1\%\) and calls this the mass percentage. Is that correct?
Not quite. The calculation compares the mass of the solute with the mass of the solvent. Mass percent compares the solute with the mass of the entire solution, including both the solute and the solvent.
The total solution mass is \(15.0 + 135.0 = 150.0\) g. Therefore:
\[ \text{Mass percent}=\frac{15.0}{150.0}\times100 \]
\[ \boxed{10.0\%\text{ m/m}} \]
This is the key idea: divide the mass of the solute by the total mass of the solution, not just the mass of the solvent.
01The weighing experiment: finding the correct masses
Before using the formula, it helps to understand where the numbers come from.
Imagine preparing a salt solution using an electronic balance, a clean, dry beaker, water, and solid sodium chloride.
The procedure is straightforward:
- Place the empty beaker on the balance and record its mass.
- Add water to the beaker and record the new reading.
- Add the salt and record the reading again.
- Stir until the salt dissolves, making sure no material is lost.
The following measurements are illustrative data from a proposed preparation.
| Measurement | Balance reading |
|---|---|
| Empty beaker | 52.40 g |
| Beaker + water | 187.40 g |
| Beaker + salt solution | 202.40 g |
We can use these readings to work out the mass of each component.
- Mass of water: \(187.40-52.40=135.00\) g.
- Mass of salt: \(202.40-187.40=15.00\) g.
- Mass of solution: \(202.40-52.40=150.00\) g.
Notice that the beaker’s mass must be subtracted. The balance measures everything sitting on its weighing pan, not just the solution.
What should happen when the salt dissolves?
Predict what the balance should read after stirring.
If no material is lost, it should still read 202.40 g.
Dissolving salt changes how its particles are distributed throughout the water. It does not make the salt’s mass disappear.
If the reading decreases, possible explanations include splashing, evaporation, or some salt remaining on the equipment rather than entering the beaker.
This also explains why simply measuring the volume of water and adding the volume of the solid is not a reliable way to calculate mass percent. The volume of a solution is not necessarily the sum of the separate component volumes.
A balance-based method avoids that problem by measuring mass directly.
02The mass percent formula
A solution contains two important components:
- Solute: the substance being dissolved, such as salt or glucose.
- Solvent: the substance that dissolves the solute, such as water.
Together, they form the solution.
For a solution containing one solute and one solvent:
\[ m_{\text{solution}} = m_{\text{solute}}+m_{\text{solvent}} \]
Mass percent is calculated using:
\[ \boxed{ \%\text{ m/m}= \frac{m_{\text{solute}}} {m_{\text{solution}}}\times100 } \]
Substituting the expression for total solution mass gives the formula you need when the question supplies the masses of the solute and solvent:
\[ \boxed{ \%\text{ m/m}= \frac{m_{\text{solute}}} {m_{\text{solute}}+m_{\text{solvent}}} \times100 } \]
Here:
- \(m_{\text{solute}}\) is the mass of the dissolved substance.
- \(m_{\text{solvent}}\) is the mass of the dissolving substance.
- \(m_{\text{solution}}\) is the combined mass of the solution.
The result is a percentage, so it has no mass unit.
Mass percent is also written as percentage by mass or % w/w (weight by weight). In this calculation, the quantities are masses, not forces or volumes.
A useful way to remember the denominator: the word solution means the entire mixture. Include both the solute and the solvent.
If a solution contains several different solutes, the denominator still includes the total mass of the entire solution. Only the numerator changes to the mass of the particular solute being investigated.
Worked example: calculate mass percent from solute and solvent masses
A student dissolves 6.00 g of sodium chloride in 114.0 g of water. Calculate the mass percentage of sodium chloride in the resulting solution.
Step 1
\[ m_{\text{solute}}=6.00\text{ g} \]
\[ m_{\text{solvent}}=114.0\text{ g} \]
Step 2
\[ m_{\text{solution}}=6.00+114.0 \]
\[ m_{\text{solution}}=120.0\text{ g} \]
Step 3
\[ \%\text{ m/m}= \frac{6.00}{120.0}\times100 \]
\[ \boxed{5.00\%\text{ m/m}} \]
The solution contains 5.00 g of sodium chloride for every 100 g of solution.
Notice that we never needed to calculate the volume of the solution.
Worked example: calculate mass percent from balance readings
A clean, dry beaker has a mass of 38.50 g. After adding water, the balance reads 138.50 g. A student then adds 10.00 g of glucose, which dissolves completely. The final balance reading is 148.50 g.
Calculate the mass percentage of glucose in the solution.
Step 1
Subtract the empty beaker reading from the reading after adding water.
\[ m_{\text{water}}=138.50-38.50 \]
\[ m_{\text{water}}=100.00\text{ g} \]
Step 2
Subtract the empty beaker reading from the final balance reading.
\[ m_{\text{solution}}=148.50-38.50 \]
\[ m_{\text{solution}}=110.00\text{ g} \]
Step 3
The increase in mass after adding glucose is:
\[ m_{\text{glucose}}=148.50-138.50 \]
\[ m_{\text{glucose}}=10.00\text{ g} \]
Step 4
\[ \%\text{ m/m}= \frac{10.00}{110.00}\times100 \]
\[ \boxed{9.09\%\text{ m/m}} \]
The glucose makes up 9.09% of the total solution mass.
The important experimental skill here is knowing which balance reading represents the solution itself. Always account for the container before substituting into the formula.
03Mass percent versus other concentration units
Not every concentration calculation uses mass.
Chemistry questions can describe concentration using mass, volume, or amount of substance. These quantities are related, but they are not interchangeable.
| Concentration type | Formula | What it measures |
|---|---|---|
| Mass percent (% m/m) | \(\frac{m_{\text{solute}}}{m_{\text{solution}}}\times100\) | Mass of solute relative to total solution mass |
| Mass concentration (g/L) | \(\frac{m_{\text{solute}}}{V_{\text{solution}}}\) | Mass of solute per litre of solution |
| Volume percent (% v/v) | \(\frac{V_{\text{solute}}}{V_{\text{solution}}}\times100\) | Volume of solute relative to total solution volume |
| Molar concentration (mol/L) | \(\frac{n_{\text{solute}}}{V_{\text{solution}}}\) | Amount of solute per litre of solution |
Do you need density?
For mass percent, density is not required when the masses of the solute and solvent are already provided.
This is because the calculation uses mass directly.
However, density becomes useful when a question gives you a volume and asks you to work with mass.
The relationship is:
\[ m=\rho V \]
where:
- \(m\) is mass in grams.
- \(\rho\) is density, commonly in g/mL.
- \(V\) is volume in mL.
Use the density of the actual substance whose volume is supplied. For example, if the question gives the volume of the solvent, use the solvent’s density. If it gives the volume of the finished solution, use the solution’s density.
Do not automatically assume that 1 mL of solution has a mass of 1 g. That approximation is associated with pure water under particular conditions, not every liquid or solution.
Also, do not assume that the volume of the solute and solvent simply add together. Masses are additive when no material is lost, but volumes do not necessarily behave that way.
04What happens when you change the solution?
Mass percent is not fixed if you add or remove material.
Understanding which component changes makes these questions much easier.
| Process | What happens to the solute? | What happens to the solvent? | Effect on mass percent |
|---|---|---|---|
| Add more solvent | Mass stays constant | Mass increases | Decreases |
| Evaporate solvent | Mass stays constant | Mass decreases | Increases |
| Remove a sample of homogeneous solution | Mass decreases | Mass decreases | Stays constant |
| Add more solute | Mass increases | Mass stays constant | Increases, assuming it dissolves |
The third row is particularly important.
If you remove 20 g of a homogeneous solution, you remove both solute and solvent in their existing proportions. The remaining solution therefore has the same mass percentage.
Evaporation is different. If only the solvent evaporates, the solute remains while the total solution mass decreases. The mass percentage consequently increases.
When a question gives a mass loss, treat it as a positive amount and subtract it from the component that actually leaves. Do not automatically subtract all mass losses from the solvent.
The distinction between evaporation and spilling a sample becomes especially important in multi-step calculations.
05Questions and solutions
Try each question before reading its solution. The difficulty increases as you progress.
Question 1
A student dissolves 9.0 g of glucose in 141 g of water.
Calculate the mass percentage of glucose in the resulting solution.
Solution 1
The solution has a mass percentage of 6.0% m/m.
First, calculate the total solution mass:
\[ m_{\text{solution}}=9.0+141 \]
\[ m_{\text{solution}}=150\text{ g} \]
Next, substitute into the formula:
\[ \%\text{ m/m}= \frac{9.0}{150}\times100 \]
\[ \boxed{6.0\%\text{ m/m}} \]
The common mistake is dividing 9.0 g by 141 g. That would compare glucose with water rather than with the entire solution.
Question 2
A laboratory has 240.0 g of a homogeneous salt solution with a mass percentage of 12.5% m/m.
A student transfers 80.0 g of this solution into a second beaker and adds 20.0 g of pure water.
Calculate:
a) The mass of salt transferred into the second beaker.
b) The final mass percentage of salt in the second beaker.
Solution 2
The final solution has a mass percentage of 10.0% m/m, and the transferred sample contains 10.0 g of salt.
Step 1: Calculate the mass of salt in the original solution.
The original solution contains 12.5 g of salt per 100 g of solution.
\[ m_{\text{salt, original}} =0.125\times240.0 \]
\[ m_{\text{salt, original}}=30.0\text{ g} \]
Step 2: Calculate the mass of salt in the 80.0 g sample.
The transferred sample is one-third of the original solution mass:
\[ m_{\text{salt, sample}} =0.125\times80.0 \]
\[ \boxed{m_{\text{salt, sample}}=10.0\text{ g}} \]
Step 3: Calculate the new total solution mass.
The student adds 20.0 g of water, but no additional salt.
\[ m_{\text{new solution}}=80.0+20.0 \]
\[ m_{\text{new solution}}=100.0\text{ g} \]
Step 4: Calculate the new mass percentage.
\[ \%\text{ m/m}= \frac{10.0}{100.0}\times100 \]
\[ \boxed{10.0\%\text{ m/m}} \]
The tempting mistake is to keep using the original 240.0 g as the denominator. However, only 80.0 g of that original solution entered the second beaker.
The 20.0 g of added water increases the denominator without increasing the solute mass. This is why the concentration decreases.
Question 3
A 300.0 g solution contains 14.0% m/m of a dissolved substance.
The solution undergoes one of two possible processes:
Process A: The solution is heated, causing 45.0 g of water to evaporate. Assume no solute is lost.
Process B: Instead of evaporation, 45.0 g of the homogeneous solution is accidentally spilled.
Calculate the final mass percentage for each process and explain why the results differ.
Solution 3
After evaporation, the mass percentage increases to 16.5% m/m. After spilling a homogeneous sample, it remains 14.0% m/m.
The difference comes from which components leave the solution.
Process A: Evaporation
Step 1: Calculate the initial solute mass.
\[ m_{\text{solute}}=0.140\times300.0 \]
\[ m_{\text{solute}}=42.0\text{ g} \]
Step 2: Calculate the initial solvent mass.
\[ m_{\text{water}}=300.0-42.0 \]
\[ m_{\text{water}}=258.0\text{ g} \]
Step 3: Subtract the evaporated water.
\[ m_{\text{water, final}}=258.0-45.0 \]
\[ m_{\text{water, final}}=213.0\text{ g} \]
The new solution mass is:
\[ m_{\text{solution, final}}=42.0+213.0 \]
\[ m_{\text{solution, final}}=255.0\text{ g} \]
Step 4: Calculate the new mass percentage.
\[ \%\text{ m/m}= \frac{42.0}{255.0}\times100 \]
\[ \boxed{16.5\%\text{ m/m}} \]
The solute mass stayed constant, but the total solution mass decreased. Therefore, the solute makes up a larger percentage of the remaining solution.
Process B: Spilling homogeneous solution
Step 1: Calculate the mass of solute spilled.
The spilled solution has the same composition as the original solution.
\[ m_{\text{solute, spilled}}=0.140\times45.0 \]
\[ m_{\text{solute, spilled}}=6.30\text{ g} \]
Step 2: Calculate the remaining solute mass.
\[ m_{\text{solute, remaining}}=42.0-6.30 \]
\[ m_{\text{solute, remaining}}=35.7\text{ g} \]
Step 3: Calculate the remaining solution mass.
\[ m_{\text{solution, remaining}}=300.0-45.0 \]
\[ m_{\text{solution, remaining}}=255.0\text{ g} \]
Step 4: Calculate the new mass percentage.
\[ \%\text{ m/m}= \frac{35.7}{255.0}\times100 \]
\[ \boxed{14.0\%\text{ m/m}} \]
The concentration does not change because the spill removes solute and solvent in exactly the same proportion as the original solution.
The key distinction is that evaporation removes solvent selectively, whereas spilling a homogeneous solution removes both components together.
Question 4
Two solutions contain the same dissolved substance.
- Solution A has a mass of 150 g and a mass percentage of 8.0% m/m.
- Solution B has a mass of 50 g and a mass percentage of 32.0% m/m.
The solutions are combined without any material being lost.
Calculate the mass percentage of the combined solution.
A student argues that the answer must be 20.0% because this is the arithmetic average of 8.0% and 32.0%.
Explain why this reasoning is incorrect.
Solution 4
The combined solution has a mass percentage of 14.0% m/m, not 20.0%.
The percentages cannot simply be averaged because the two original solutions have different masses.
Step 1: Calculate the mass of solute in Solution A.
\[ m_{\text{solute,A}}=0.080\times150 \]
\[ m_{\text{solute,A}}=12\text{ g} \]
Step 2: Calculate the mass of solute in Solution B.
\[ m_{\text{solute,B}}=0.320\times50 \]
\[ m_{\text{solute,B}}=16\text{ g} \]
Step 3: Calculate the total mass of solute.
\[ m_{\text{solute,total}}=12+16 \]
\[ m_{\text{solute,total}}=28\text{ g} \]
Step 4: Calculate the total solution mass.
\[ m_{\text{solution,total}}=150+50 \]
\[ m_{\text{solution,total}}=200\text{ g} \]
Step 5: Calculate the combined mass percentage.
\[ \%\text{ m/m}= \frac{28}{200}\times100 \]
\[ \boxed{14.0\%\text{ m/m}} \]
Why does the arithmetic average fail?
The 8.0% solution contributes 150 g to the mixture, while the 32.0% solution contributes only 50 g.
The first solution therefore has three times the mass of the second. Its lower concentration has a much greater influence on the final composition.
An arithmetic average would only work if the two solutions had equal masses.
For mixtures of different masses, always calculate the total solute mass and divide by the total solution mass.
Question 5
A homogeneous solution has a total mass of 200.0 g and a mass percentage of 20.0% m/m.
It is divided equally between two beakers, A and B, each initially containing 100.0 g of solution.
The following operations are performed:
- Beaker A: Remove 20.0 g of homogeneous solution, then replace it with 20.0 g of pure water.
- Beaker B: Evaporate exactly 20.0 g of water. No solute is lost.
The remaining contents of both beakers are then combined.
Calculate the mass percentage of the final solution.
Explain why the two beakers have different concentrations immediately before mixing, yet the combined solution has the same mass percentage as the original solution.
Also explain why simply averaging the two final percentages would give an incorrect result.
Solution 5
The recombined solution is still 20.0% m/m.
This is a more subtle mass-balance problem because the two beakers have different concentrations before mixing. The important detail is that their final masses are also different.
Step 1: Calculate the original masses of solute and solvent.
The original solution has a mass of 200.0 g and contains 20.0% solute.
\[ m_{\text{solute}}=0.200\times200.0 \]
\[ m_{\text{solute}}=40.0\text{ g} \]
Therefore:
\[ m_{\text{water}}=200.0-40.0 \]
\[ m_{\text{water}}=160.0\text{ g} \]
Step 2: Determine the initial composition of each beaker.
The solution is divided equally, so each beaker receives half of the original solute and water.
Each beaker initially contains:
\[ m_{\text{solute}}=20.0\text{ g} \]
\[ m_{\text{water}}=80.0\text{ g} \]
Each therefore starts at 20.0% m/m.
Step 3: Calculate the composition of Beaker A after its treatment.
Beaker A loses 20.0 g of homogeneous solution.
Because the original concentration is 20.0%, the removed sample contains:
\[ m_{\text{solute, removed}} =0.200\times20.0 \]
\[ =4.0\text{ g} \]
The mass of water removed is:
\[ 20.0-4.0=16.0\text{ g} \]
The remaining contents are:
- Solute: \(20.0-4.0=16.0\) g.
- Water: \(80.0-16.0=64.0\) g.
Now add 20.0 g of pure water.
The new composition is:
- Solute: 16.0 g.
- Water: \(64.0+20.0=84.0\) g.
- Total solution: 100.0 g.
Therefore:
\[ \%\text{ m/m}_A= \frac{16.0}{100.0}\times100 \]
\[ \boxed{16.0\%\text{ m/m}} \]
Step 4: Calculate the composition of Beaker B after evaporation.
Beaker B initially contains 20.0 g of solute and 80.0 g of water.
Only water evaporates, so:
\[ m_{\text{water, final}}=80.0-20.0 \]
\[ m_{\text{water, final}}=60.0\text{ g} \]
The solute remains at 20.0 g.
The final solution mass is:
\[ m_{\text{solution,B}}=20.0+60.0 \]
\[ m_{\text{solution,B}}=80.0\text{ g} \]
Its mass percentage is:
\[ \%\text{ m/m}_B= \frac{20.0}{80.0}\times100 \]
\[ \boxed{25.0\%\text{ m/m}} \]
Step 5: Recombine both beakers.
Add the solute masses:
\[ m_{\text{solute,total}}=16.0+20.0 \]
\[ m_{\text{solute,total}}=36.0\text{ g} \]
Add the water masses:
\[ m_{\text{water,total}}=84.0+60.0 \]
\[ m_{\text{water,total}}=144.0\text{ g} \]
The total mass is:
\[ m_{\text{solution,total}}=36.0+144.0 \]
\[ m_{\text{solution,total}}=180.0\text{ g} \]
Therefore:
\[ \%\text{ m/m}= \frac{36.0}{180.0}\times100 \]
\[ \boxed{20.0\%\text{ m/m}} \]
Step 6: Explain the surprising result.
There are two tempting approaches, and both need careful thought.
First, averaging the final concentrations gives:
\[ \frac{16.0+25.0}{2}=20.5\% \]
But this is incorrect because Beaker A contains 100.0 g of solution, while Beaker B contains only 80.0 g.
The correct average must account for the different final masses:
\[ \frac{(0.160\times100.0)+(0.250\times80.0)} {180.0}\times100 =20.0\% \]
Second, notice something even more interesting about the original operations.
Across both beakers:
- 20.0 g of water is added to A.
- 20.0 g of water evaporates from B.
These two changes cancel when we consider the combined system.
The only net change is that 20.0 g of homogeneous solution has been removed from A.
That removed sample contained 4.0 g of solute and 16.0 g of water. It therefore had exactly the same composition as the original solution.
Removing a sample with the original composition leaves the remaining mixture at the original mass percentage.
The total mass has decreased from 200.0 g to 180.0 g, but the ratio of solute to total solution mass has not changed.
This is the important distinction: individual treatments can change concentration substantially, while the combined system can retain its original composition.
06What to remember for your next chemistry calculation
When calculating mass percent from solute and solvent masses, use:
\[ \boxed{ \%\text{ m/m}= \frac{\text{mass of solute}} {\text{mass of solute}+\text{mass of solvent}} \times100 } \]
Before substituting, check three things:
- Are you using the mass of the entire solution in the denominator?
- Are all masses expressed in the same units?
- Have you correctly accounted for any material added, removed, or evaporated?
Most importantly, keep mass percent separate from molar concentration. Mass percent describes the composition of a solution by mass, whereas molar concentration describes the amount of dissolved substance per litre.
This distinction becomes useful when moving into equilibrium calculations. For example, calculating solubility from \(K_{sp}\) in HSC Chemistry requires molar concentrations of dissolved ions, not simply their mass percentages.