How to Calculate Solute Mass from Concentration and Volume

You need to prepare 250.0 mL of a solution with a mass concentration of 8.00 g L⁻¹. A quick calculation tells you that you need 2.00 g of dissolved solute. But should you add that solid to 250.0 mL of water, or dissolve it and make the final solution volume up to 250.0 mL?

The second method is correct.

To find the mass of a solute from its mass concentration and solution volume, use:

\[
\boxed{m=C_mV}
\]

Here, \(C_m\) is the mass concentration and \(V\) is the final volume of the solution. In this example:

\[
m=8.00\times0.2500=2.00\ \mathrm{g}
\]

The important detail is that 250.0 mL refers to the finished solution, not the water you started with.

There is one more thing to check before using this formula: not every concentration is expressed as mass per volume. If the question gives mol L⁻¹, percentage concentration, or a volume-based concentration, you need a different calculation.

01Identify the type of concentration first

A concentration tells you how much solute is present relative to a specified quantity of solution. The units tell you what that quantity is.

This distinction determines how you calculate the solute mass.

Here, \(p\) represents the numerical percentage. For example, if the concentration is 4.0%, use \(p=4.0\), not \(0.040\).

The main formulas to remember are:

• Mass concentration: multiply by solution volume.
• Molar concentration: multiply by solution volume, then by molar mass.
• Mass/volume percentage: multiply the percentage by the volume in mL, then divide by 100.
• Mass/mass percentage: you need the total mass of the solution.

If you want a broader explanation of these concentration units, see Ways to Measure Solution Concentration in HSC Chemistry.

02Finding solute mass from mass concentration (g/L)

This is the most direct situation.

Mass concentration describes the mass of dissolved solute in a particular volume of solution.

The relationship is:

\[
C_m=\frac{m}{V}
\]

Rearranging to make the solute mass the subject:

\[
\boxed{m=C_mV}
\]

Where:

• \(m\) = mass of solute, in grams (g)
• \(C_m\) = mass concentration, in grams per litre (g L⁻¹)
• \(V\) = final volume of solution, in litres (L)

The units matter. If your concentration is in g/L, your volume must be in litres to obtain the mass in grams.

Converting millilitres into litres

HSC Chemistry questions often provide solution volumes in millilitres.

Remember:

\[
1000\ \mathrm{mL}=1.000\ \mathrm{L}
\]

Therefore, divide a volume in millilitres by 1000 before using the mass concentration formula.

For example:

\[
250.0\ \mathrm{mL}=0.2500\ \mathrm{L}
\]

Do not multiply a concentration in g/L directly by a volume in mL. That introduces a factor-of-1000 error.

03Finding solute mass from molar concentration (mol/L)

Now consider a slightly different question.

A solution contains 0.250 mol/L of potassium nitrate. You know its volume, but the question asks for the mass of potassium nitrate in grams.

Can you simply multiply the concentration by the volume?

You can, but the result will be in moles, not grams.

Molar concentration, also called amount concentration, describes the number of moles of solute per litre of solution.

Its formula is:

\[
c=\frac{n}{V}
\]

Rearranging:

\[
n=cV
\]

Where:

• \(c\) = molar concentration, in mol L⁻¹
• \(n\) = amount of solute, in moles (mol)
• \(V\) = final solution volume, in litres (L)

We then need one additional relationship to convert moles into grams:

\[
m=nM
\]

Here, \(M\) is the molar mass of the solute, measured in g mol⁻¹.

Combining both equations gives:

\[
\boxed{m=cVM}
\]

This is the formula to use when a question gives molar concentration and solution volume but asks for solute mass.

04When do you need density?

Density becomes important when you need to convert between the mass and volume of a substance or solution.

However, you do not automatically need density whenever a question gives a solution volume.

For example:

• If concentration is in g/L, you can calculate mass directly.
• If concentration is in mol/L, you can calculate moles and then use molar mass.
• If concentration is in % m/v, you can calculate mass directly from volume.
• If concentration is in % m/m and you are given solution volume, you generally need the solution’s density to find its total mass.

The last case is a common source of mistakes.

Comparing % m/v and % m/m

Consider two solutions, both labelled 4.0%.

They might look like the same concentration, but they describe different quantities.

Case A: 4.0% m/v

This means 4.0 g of solute per 100 mL of final solution.

Suppose you have 250 mL of this solution.

\[
m=\frac{4.0\times250}{100}
\]

\[
\boxed{m=10.0\ \mathrm{g}}
\]

No density is needed.

Case B: 4.0% m/m

This means 4.0 g of solute per 100 g of total solution.

Suppose you have 250 mL of this solution, and its density is 1.10 g/mL.

First, calculate the total solution mass:

\[
\begin{aligned}
m_{\text{solution}}&=\rho V\\
&=1.10\times250\\
&=275\ \mathrm{g}
\end{aligned}
\]

Now calculate the solute mass:

\[
\begin{aligned}
m_{\text{solute}}&=\frac{4.0}{100}\times275\\
&=11.0\ \mathrm{g}
\end{aligned}
\]

The solution contains 11.0 g of solute.

Notice that the answers are different: 10.0 g versus 11.0 g.

The difference comes from the concentration definitions, not from an arithmetic error.

Also, be careful about which density you are given:

• For % m/m, you need the density of the entire solution if you are converting its volume into mass.
• For % v/v, you first calculate the volume of liquid solute, then need the density of that pure solute to calculate its mass.

Do not automatically assume that a solution has the same density as pure water. Only use that approximation when the question explicitly allows it or provides a justified assumption.

05Why the final solution volume matters

Let’s return to the original preparation problem.

You want 250.0 mL of a solution containing 2.00 g of sodium chloride, \(\ce{NaCl}\).

You weigh 2.00 g of sodium chloride and add it to 250.0 mL of water.

Have you prepared the required solution?

Not necessarily.

The 250.0 mL measurement describes the water before the salt was dissolved. It does not describe the final solution volume.

When a substance dissolves, the resulting solution volume is not guaranteed to equal the original solvent volume. You cannot simply assume that volumes remain unchanged or add together.

This is a measurement problem: the original method does not control the quantity used in the concentration denominator.

Preparing a solution correctly

A volumetric flask allows you to measure the final solution volume directly.

For the sodium chloride example, the procedure is:

• Weigh exactly 2.00 g of sodium chloride using a balance.
• Transfer the solid into a beaker and dissolve it in a relatively small volume of deionised water.
• Transfer the dissolved solution into a 250.0 mL volumetric flask using a funnel.
• Rinse the beaker and funnel with deionised water, adding the rinses to the flask. This prevents dissolved solute from being left behind.
• Add more deionised water until the solution approaches the calibration line.
• Use a dropper to bring the bottom of the meniscus exactly to the line, viewed at eye level.
• Stopper the flask and invert it several times to mix the solution thoroughly.

Now the final solution volume is 250.0 mL, and the mass concentration is:

\[
\begin{aligned}
C_m&=\frac{m}{V}\\
&=\frac{2.00}{0.2500}\\
&=8.00\ \mathrm{g\,L^{-1}}
\end{aligned}
\]

The revised method controls the actual quantity needed in the concentration calculation.

This is why a volumetric flask is more appropriate than simply measuring out the required volume of water.

06A quick method for HSC Chemistry questions

Before calculating solute mass, work through these decisions.

First: Read the concentration units.

Identify whether the question gives g/L, mol/L, % m/v, % m/m, or % v/v.

Second: Identify the volume or mass being measured.

Check whether the question gives the final solution volume, the initial solvent volume, or the total solution mass.

Third: Convert the units.

For g/L and mol/L, convert millilitres into litres. For % m/v, use millilitres as specified by the percentage definition.

Fourth: Decide whether another quantity is needed.

You may need:

• Molar mass to convert moles into grams.
• Solution density to convert solution volume into solution mass.
• Solute density to convert a liquid solute’s volume into mass.

Finally: Check the units of your answer.

If the question asks for mass, your final answer should be in g, mg, or another appropriate mass unit.

A useful way to check your calculation is to substitute your answer back into the original concentration relationship.

07Questions and solutions

The following questions move from direct mass calculations to problems involving different concentration units, experimental design, and mixing solutions.

Question 1

A glucose solution has a mass concentration of 18.0 g/L and a final volume of 150.0 mL.

Calculate the mass of glucose dissolved in the solution.

Solution 1

The solution contains 2.70 g of glucose.

Since the concentration is given in g/L, use:

\[
m=C_mV
\]

First, convert the volume:

\[
150.0\ \mathrm{mL}=0.1500\ \mathrm{L}
\]

Substitute:

\[
\begin{aligned}
m&=18.0\times0.1500\\
&=2.70\ \mathrm{g}
\end{aligned}
\]

Therefore:

\[
\boxed{m=2.70\ \mathrm{g}}
\]

The common mistake here is using 150.0 directly instead of converting it into litres. Always check that the volume units match the concentration units.

Question 2

A magnesium chloride solution has a molar concentration of 0.0850 mol/L and a final volume of 425.0 mL.

The molar mass of magnesium chloride, \(\ce{MgCl2}\), is 95.21 g/mol.

Calculate the mass of magnesium chloride dissolved in the solution.

Solution 2

The solution contains 3.44 g of magnesium chloride.

Because the concentration is in mol/L, we must calculate the amount of solute in moles before converting to mass.

Use:

\[
m=cVM
\]

Step 1: Convert the solution volume into litres.

\[
V=\frac{425.0}{1000}
=0.4250\ \mathrm{L}
\]

Step 2: Calculate the amount of magnesium chloride.

\[
\begin{aligned}
n&=cV\\
&=0.0850\times0.4250\\
&=0.036125\ \mathrm{mol}
\end{aligned}
\]

Step 3: Convert moles into mass.

\[
\begin{aligned}
m&=nM\\
&=0.036125\times95.21\\
&=3.43946\ \mathrm{g}
\end{aligned}
\]

Step 4: Round to three significant figures.

\[
\boxed{m=3.44\ \mathrm{g}}
\]

The important distinction is that 0.036125 mol is an amount of substance, not a mass. The molar mass is what allows us to convert between these quantities.

Question 3

A potassium chloride solution has a concentration of 3.50% m/m and a volume of 250.0 mL.

The density of the solution is 1.08 g/mL.

Calculate the mass of potassium chloride in the solution.

Explain why simply multiplying 3.50% by 250.0 does not give the correct answer.

Solution 3

The solution contains 9.45 g of potassium chloride.

The concentration is expressed as % m/m, so we need the total mass of the solution rather than its volume.

Step 1: Calculate the total solution mass using density.

\[
m_{\text{solution}}=\rho V
\]

Substitute:

\[
\begin{aligned}
m_{\text{solution}}&=1.08\times250.0\\
&=270\ \mathrm{g}
\end{aligned}
\]

Step 2: Convert the percentage into a decimal fraction.

\[
3.50\%=\frac{3.50}{100}=0.0350
\]

Step 3: Calculate the mass of potassium chloride.

\[
\begin{aligned}
m_{\text{solute}}&=0.0350\times270\\
&=9.45\ \mathrm{g}
\end{aligned}
\]

Therefore:

\[
\boxed{m_{\ce{KCl}}=9.45\ \mathrm{g}}
\]

The tempting approach is to treat 250.0 mL as though it were 250.0 g. That would only be justified if the solution density were exactly 1.00 g/mL.

Here, the actual density is 1.08 g/mL, so the total solution mass is 270 g, not 250 g.

This is precisely why density matters when converting a mass/mass percentage into a solute mass from a solution volume.

Question 4

A student needs to prepare 500.0 mL of a glucose solution containing 8.00 g/L of glucose.

They weigh 4.00 g of glucose and dissolve it in 500.0 mL of water.

After preparation, the final solution volume is measured as 503.0 mL. Assume no glucose was lost during preparation.

(a) Calculate the actual mass concentration of the prepared solution.

(b) Explain why the student cannot claim that the solution has a concentration of exactly 8.00 g/L.

(c) Describe how the student should prepare the solution correctly to obtain the intended concentration.

Solution 4

(a) The actual mass concentration is 7.95 g/L, not 8.00 g/L.

(b) The student measured the initial water volume rather than the final solution volume.

(c) The student should dissolve the glucose in a smaller volume of water and make the final solution volume up to exactly 500.0 mL in a volumetric flask.

Part (a): Calculate the actual mass concentration.

Use:

\[
C_m=\frac{m}{V}
\]

The mass of glucose is 4.00 g, and the final solution volume is 503.0 mL.

Convert the volume into litres:

\[
503.0\ \mathrm{mL}=0.5030\ \mathrm{L}
\]

Substitute:

\[
\begin{aligned}
C_m&=\frac{4.00}{0.5030}\\
&=7.9523\ \mathrm{g\,L^{-1}}
\end{aligned}
\]

Therefore:

\[
\boxed{C_m=7.95\ \mathrm{g\,L^{-1}}}
\]

Part (b): Explain the experimental error.

The student calculated the required mass correctly:

\[
m=C_mV
\]

\[
m=8.00\times0.5000=4.00\ \mathrm{g}
\]

However, the 500.0 mL was the initial volume of water, not the final solution volume.

The final solution volume was 503.0 mL, so the actual concentration is lower than intended.

Part (c): Describe the corrected method.

The student should:

• Weigh 4.00 g of glucose.
• Dissolve it in a smaller volume of deionised water.
• Transfer the solution into a 500.0 mL volumetric flask.
• Rinse the original beaker and funnel into the flask.
• Add deionised water until the bottom of the meniscus reaches the calibration mark.
• Stopper and invert the flask to ensure an even concentration throughout.

The corrected method gives:

\[
C_m=\frac{4.00}{0.5000}
=8.00\ \mathrm{g\,L^{-1}}
\]

The key experimental principle is that the final solution volume must be measured directly. Measuring the starting water volume is not an equivalent substitute.

Question 5

Two sodium chloride solutions are combined to produce a single solution.

The original solutions have the following properties:

After mixing, the final solution volume is measured as 348.0 mL.

Assume there is no chemical reaction, evaporation, or loss of solute during transfer. Treat all supplied values as illustrative experimental data.

The molar mass of sodium chloride is 58.44 g/mol.

(a) Calculate the total mass of sodium chloride in the combined solution.

(b) Calculate the final molar concentration of sodium chloride.

(c) Explain why using 350.0 mL as the final solution volume would produce an incorrect result, even though the two initial volumes add to 350.0 mL.

Solution 5

(a) The combined solution contains 16.1 g of sodium chloride.

(b) Its final molar concentration is 0.792 mol/L.

(c) The correct volume for calculating the final concentration is 348.0 mL, because this is the measured final solution volume. The initial volumes cannot automatically be added together.

This problem requires us to combine two different concentration descriptions. The important idea is to calculate the amount of sodium chloride in each original solution before combining them.

Part (a): Calculate the total mass of sodium chloride.

Start with Solution A.

Step 1: Calculate the amount of sodium chloride in Solution A.

Its concentration is given in mol/L, so:

\[
n_A=cV
\]

Convert the volume:

\[
200.0\ \mathrm{mL}=0.2000\ \mathrm{L}
\]

Substitute:

\[
\begin{aligned}
n_A&=0.300\times0.2000\\
&=0.0600\ \mathrm{mol}
\end{aligned}
\]

Step 2: Convert the amount into mass.

\[
\begin{aligned}
m_A&=n_AM\\
&=0.0600\times58.44\\
&=3.5064\ \mathrm{g}
\end{aligned}
\]

Solution A contains 3.51 g of sodium chloride.

Now consider Solution B.

Step 3: Calculate the total mass of Solution B.

Its concentration is given as 8.00% m/m, so we need the total solution mass.

Use:

\[
m_{\text{solution}}=\rho V
\]

Substitute:

\[
\begin{aligned}
m_{\text{solution}}&=1.05\times150.0\\
&=157.5\ \mathrm{g}
\end{aligned}
\]

Step 4: Calculate the sodium chloride mass in Solution B.

\[
\begin{aligned}
m_B&=\frac{8.00}{100}\times157.5\\
&=12.6\ \mathrm{g}
\end{aligned}
\]

Solution B contains 12.6 g of sodium chloride.

Step 5: Add the two solute masses.

Since no sodium chloride is lost or consumed in a reaction, the total solute mass is conserved.

\[
\begin{aligned}
m_{\text{total}}&=m_A+m_B\\
&=3.5064+12.6\\
&=16.1064\ \mathrm{g}
\end{aligned}
\]

Therefore:

\[
\boxed{m_{\text{total}}=16.1\ \mathrm{g}}
\]

Part (b): Calculate the final molar concentration.

We now need the total number of moles of sodium chloride and the final solution volume.

Step 6: Convert the mass of sodium chloride in Solution B into moles.

\[
\begin{aligned}
n_B&=\frac{m_B}{M}\\
&=\frac{12.6}{58.44}\\
&=0.2156\ \mathrm{mol}
\end{aligned}
\]

Step 7: Calculate the total amount of sodium chloride.

\[
\begin{aligned}
n_{\text{total}}&=n_A+n_B\\
&=0.0600+0.2156\\
&=0.2756\ \mathrm{mol}
\end{aligned}
\]

Step 8: Convert the final solution volume into litres.

\[
348.0\ \mathrm{mL}=0.3480\ \mathrm{L}
\]

Step 9: Calculate the final molar concentration.

Use:

\[
c=\frac{n}{V}
\]

Substitute:

\[
\begin{aligned}
c&=\frac{0.2756}{0.3480}\\
&=0.7919\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

Therefore:

\[
\boxed{c=0.792\ \mathrm{mol\,L^{-1}}}
\]

Part (c): Explain why 350.0 mL is incorrect.

The tempting approach is to add the original volumes:

\[
200.0+150.0=350.0\ \mathrm{mL}
\]

However, this assumes that solution volumes are perfectly additive.

Mixing two solutions does not guarantee that their combined volume equals the sum of their original volumes. The actual final volume must be measured or provided.

In this question, the measured final volume is 348.0 mL, so that is the value we must use when calculating the final concentration.

There is also an important distinction between the volumes used at different stages:

• Use the original volume of Solution A with its original molar concentration.
• Use the original volume and density of Solution B to find its solute mass.
• Use the measured final volume only when calculating the concentration of the combined solution.

The conservation of solute mass allows us to add the quantities of sodium chloride. It does not mean we can automatically add the solution volumes.

As a final check, we can calculate the final mass concentration:

\[
C_m=\frac{16.1064}{0.3480}
=46.3\ \mathrm{g\,L^{-1}}
\]

Converting the final molar concentration back into mass concentration gives the same result:

\[
C_m=cM
\]

\[
C_m=0.792\times58.44
\approx46.3\ \mathrm{g\,L^{-1}}
\]

The two calculations agree, supporting our result.

08Where to go next

Once you can find the mass of a dissolved solute, the next step in many HSC Chemistry calculations is converting that mass into moles.

Remember:

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

For a solution with a known molar concentration, you can also calculate moles directly:

\[
n=cV
\]

These two methods connect solution concentration to reacting quantities in chemical equations.

This becomes particularly useful in limiting reagent questions, where you may need to calculate the amount of acid available before determining how much solid can react.

For a complete application, see How to Calculate CO₂ Volume from a Carbonate Reaction.

The key decision in every calculation is the same: identify what the concentration measures, use the matching volume or mass, and only introduce density or molar mass when the units require it.