How to Convert Between Mass Percentage and Molarity Using Density

A bottle of sodium chloride solution is labelled 12.0% by mass. Your chemistry teacher asks you to calculate its molarity.

You know the solution contains 12.0 g of NaCl per 100 g of solution. But molarity requires the number of moles in each litre of solution. How do you get from grams of solution to millilitres of solution?

The missing link is density.

Density allows you to convert between the mass of a solution and its volume. Once you have the volume, you can use the mass percentage and molar mass to calculate the number of moles per litre.

This gives us a direct method for converting between mass percentage and molarity, provided we know the density of the solution at the relevant temperature.

01Understand the three quantities involved

Before calculating, identify what each measurement actually represents. They describe different properties of the same solution.

Notice that the denominators are different.

Mass percentage compares the mass of solute with the total mass of solution. Molarity compares the amount of solute, measured in moles, with the total volume of solution.

Density connects mass and volume.

Mass percentage: a mass-based concentration

Mass percentage, also written as % w/w or % m/m, is calculated using:

\[ \text{Mass percentage}= \frac{m_{\text{solute}}}{m_{\text{solution}}} \times 100 \]

For example, a 12.0% w/w sodium chloride solution contains:

• 12.0 g of NaCl
• 88.0 g of water
• 100.0 g of solution in total

The denominator is the mass of the entire solution, not just the water.

Molarity: an amount-based concentration

Molarity, also called amount concentration, measures the number of moles of solute in each litre of solution.

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

Where:

• \(c\) = molarity (mol L⁻¹)
• \(n\) = amount of solute (mol)
• \(V\) = total volume of solution (L)

The volume refers to the final solution volume, not the volume of solvent added.

Density: the connection between mass and volume

Density is calculated using:

\[ \rho=\frac{m_{\text{solution}}}{V_{\text{solution}}} \]

Where:

• \(\rho\) = density
• \(m_{\text{solution}}\) = total mass of solution
• \(V_{\text{solution}}\) = total volume of solution

Rearranging:

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

This is the relationship we need.

02How density makes the conversion possible

Imagine you have exactly 1.000 L of solution.

If its density is 1.080 g/mL, what is its total mass?

Since 1.000 L is equal to 1000 mL:

\[ m=\rho V \]

\[ m=1.080\times1000 \]

\[ m=1080\text{ g} \]

You now know the mass of the entire litre of solution.

If the solution is 12.0% w/w, you can calculate the mass of dissolved solute:

\[ m_{\text{solute}}=0.120\times1080 \]

\[ m_{\text{solute}}=129.6\text{ g} \]

The final step is converting that mass into moles using the molar mass.

This is the entire logic of the conversion:

1. Use density to find the mass of the solution.
2. Use mass percentage to find the mass of solute.
3. Use molar mass to convert the solute mass into moles.
4. Divide the number of moles by the solution volume in litres.

The general formula: mass percentage to molarity

We can combine the steps into one equation.

Start with the mass percentage relationship:

\[ m_{\text{solute}}= \frac{w}{100}m_{\text{solution}} \]

Here, \(w\) is the numerical mass percentage. For example, use \(w=12.0\), not 0.120.

Using density, the mass of 1.000 L of solution is:

\[ m_{\text{solution}}=\rho\times1000 \]

This assumes density is measured in g/mL.

The amount of solute is:

\[ n=\frac{m_{\text{solute}}}{M_m} \]

Where \(M_m\) is the molar mass in g/mol.

Substituting these relationships into the molarity equation gives:

\[ \boxed{c=\frac{10w\rho}{M_m}} \]

This is the direct conversion formula.

Use this formula when:

• Mass percentage is given as % w/w.
• Density of the solution is known in g/mL.
• Molar mass of the solute is known in g/mol.

The resulting molarity is in mol/L.

You can also use density in kg/L because 1 kg/L has the same numerical value as 1 g/mL.

03Converting molarity back to mass percentage

The reverse conversion uses exactly the same information, but the reasoning runs in the opposite direction.

Suppose you know the molarity and density of a solution. You want to determine what percentage of its total mass is dissolved solute.

Start by considering 1.000 L of solution.

The density tells you the total mass of that litre. The molarity tells you how many moles of solute it contains.

You can then convert those moles into grams and calculate the mass percentage.

The general formula is:

\[ \boxed{w=\frac{cM_m}{10\rho}} \]

Where:

• \(w\) = mass percentage, expressed as a numerical percentage
• \(c\) = molarity (mol/L)
• \(M_m\) = molar mass (g/mol)
• \(\rho\) = density (g/mL)

04A common trap: mass percentage is not mass per volume

Consider a solution labelled 10.0% w/w.

A common mistake is to interpret this as 10.0 g of solute per 100 mL of solution.

That is incorrect.

A 10.0% w/w solution contains 10.0 g of solute per 100 g of solution. The volume occupied by those 100 g depends on density.

Let’s see how much difference this makes.

Suppose a hypothetical solution has:

• Mass percentage = 10.0% w/w
• Density = 1.20 g/mL
• Solute molar mass = 60.0 g/mol

All data in this example are illustrative.

Step 1: Calculate the mass of 100 mL of solution.

\[ m=1.20\times100 \]

\[ m=120\text{ g} \]

Step 2: Calculate the mass of solute in that volume.

\[ m_{\text{solute}}=0.100\times120 \]

\[ m_{\text{solute}}=12.0\text{ g} \]

Step 3: Calculate the solute mass in one litre.

Since 1 L contains ten times the volume:

\[ m_{\text{solute}}=120\text{ g} \]

Step 4: Convert the mass into moles.

\[ n=\frac{120}{60.0} \]

\[ n=2.00\text{ mol} \]

Therefore:

\[ \boxed{c=2.00\text{ mol/L}} \]

What would happen if you incorrectly treated the solution as containing 10.0 g per 100 mL?

You would calculate 100 g of solute per litre, giving:

\[ c=\frac{100}{60.0} \]

\[ c=1.67\text{ mol/L} \]

That is an underestimate of the actual molarity.

Know the difference between the percentage types

The conversion formula in this article applies to mass percentage expressed as % w/w.

If a question gives a percentage without specifying its basis, check the context before choosing a formula.

05Measuring density: where does the value come from?

In an experimental setting, density is not simply assumed. It can be determined by measuring the mass of a known volume of solution.

A common approach uses a volumetric flask and an electronic balance.

Consider a 100.0 mL volumetric flask calibrated at 20°C.

The measurement proceeds as follows:

1. Weigh the clean, dry, empty flask.
2. Fill the flask with the solution until the bottom of the meniscus reaches the calibration mark.
3. Ensure the solution is at the calibration temperature and that there are no air bubbles.
4. Dry the outside of the flask and weigh it again.
5. Subtract the empty flask mass from the filled flask mass.

Suppose the following illustrative measurements are obtained:

The density is:

\[ \rho=\frac{108.000}{100.0} \]

\[ \boxed{\rho=1.080\text{ g/mL}} \]

The observation tells us that each millilitre of this solution has a mass of 1.080 g at the measurement temperature.

There is an important limitation here.

Measuring density does not tell you the mass percentage of solute. It only tells you the mass of a known volume of the entire solution.

You still need independent information about the solution’s composition. This might be supplied in the question, provided by a manufacturer’s label, or determined using a suitable analytical method.

Why temperature matters

Density changes with temperature because most solutions expand when heated.

Imagine a sealed sample of solution. No solute or solvent is added or removed, but its temperature increases.

Its mass percentage remains unchanged because the masses of solute and total solution have not changed.

However, its volume generally increases. The density therefore decreases, and the molarity also decreases.

This is why you should use density measured at the same temperature as the molarity you are calculating.

For an HSC question, if the temperature is specified, use the corresponding density. If it is not specified, use the density provided and state any temperature assumption that matters.

06A reliable decision process for exam questions

When you encounter a question involving mass percentage, molarity, and density, first identify which quantity you need to calculate.

Before substituting values, check these five things:

• Percentage basis: Is the concentration w/w, w/v, or v/v?
• Density: Is it the density of the solution rather than the pure solvent?
• Units: Are density and molar mass in compatible units?
• Volume: Are you using the total solution volume rather than the solvent volume?
• Temperature: Does the density correspond to the temperature of the solution?

One more point: density-based conversion is not the same as converting mg/L or ppm into molarity. Those concentrations use different units and require their own conversion steps. For that distinction, see ppm, ppb and mg/L: When Can You Convert Between Them?.

07Questions and solutions

The following questions move from a standard HSC calculation towards problems involving dilution, different solutes, temperature changes, and conservation of mass.

Question 1

An aqueous potassium chloride solution has a mass percentage of 7.50% w/w and a density of 1.04 g/mL at 20°C.

The molar mass of KCl is 74.55 g/mol.

Calculate the molarity of the solution.

Solution 1

The molarity is 1.05 mol/L.

We are given mass percentage and density, so we can use the direct conversion formula.

Step 1: Write the known values.

\[ w=7.50 \]

\[ \rho=1.04\text{ g/mL} \]

\[ M_m=74.55\text{ g/mol} \]

Step 2: Substitute into the formula.

\[ c=\frac{10w\rho}{M_m} \]

\[ c=\frac{10\times7.50\times1.04}{74.55} \]

\[ c=\frac{78.0}{74.55} \]

\[ c=1.0463\text{ mol/L} \]

Step 3: Round to three significant figures.

\[ \boxed{c=1.05\text{ mol/L}} \]

The solution contains 1.05 moles of KCl per litre of final solution.

The main thing to check is that the density is expressed in g/mL. If it were supplied in another unit, you would need to convert it before using this particular formula.

Question 2

A laboratory needs to prepare 250.0 mL of a potassium nitrate solution with a concentration of 0.800 mol/L.

The available stock solution has:

• Mass percentage = 16.0% w/w
• Density = 1.10 g/mL
• Molar mass of KNO₃ = 101.10 g/mol

Calculate the volume of stock solution required.

Assume the stock solution is transferred quantitatively and diluted to a final volume of 250.0 mL.

Solution 2

The required stock-solution volume is 115 mL.

This question requires us to work backwards from the desired molarity to the amount of solute, then use mass percentage and density to find the volume of stock solution.

Step 1: Calculate the number of moles required in the final solution.

Convert the final volume into litres:

\[ V=0.2500\text{ L} \]

Using \(n=cV\):

\[ n=0.800\times0.2500 \]

\[ n=0.200\text{ mol} \]

Step 2: Calculate the mass of potassium nitrate required.

\[ m=nM_m \]

\[ m=0.200\times101.10 \]

\[ m=20.22\text{ g} \]

This is the mass of KNO₃ that must be transferred from the stock solution.

Step 3: Calculate the mass of stock solution needed.

The stock solution is 16.0% w/w, meaning that 16.0% of its total mass is KNO₃.

\[ 16.0=\frac{20.22}{m_{\text{solution}}}\times100 \]

Rearranging:

\[ m_{\text{solution}}=\frac{20.22\times100}{16.0} \]

\[ m_{\text{solution}}=126.375\text{ g} \]

Step 4: Convert the stock-solution mass into volume.

Using density:

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

\[ V=\frac{126.375}{1.10} \]

\[ V=114.886\text{ mL} \]

Step 5: Round the answer.

\[ \boxed{V=115\text{ mL}} \]

Approximately 115 mL of stock solution is required.

Notice that this is the volume of stock solution to transfer, not the volume of water to add. The final solution must be made up to a total volume of 250.0 mL.

Question 3

Two hypothetical aqueous solutions, X and Y, contain different dissolved solutes.

Both solutions have:

• Mass percentage = 12.0% w/w
• Density = 1.08 g/mL at 20°C

However, their solutes have different molar masses:

Calculate:

(a) The molarity of each solution.

(b) The number of moles of solute in 100.0 mL of each solution.

(c) Explain why the solutions have different molarities despite having identical mass percentages and densities.

The numerical data are illustrative and describe two different solute systems.

Solution 3

Solution X has a molarity of 2.16 mol/L, while solution Y has a molarity of 1.08 mol/L. Solution X contains twice as many moles of solute in the same volume.

Part (a): Calculate the molarity of each solution.

Start with the conversion formula:

\[ c=\frac{10w\rho}{M_m} \]

For solution X:

\[ c_X=\frac{10\times12.0\times1.08}{60.0} \]

\[ \boxed{c_X=2.16\text{ mol/L}} \]

For solution Y:

\[ c_Y=\frac{10\times12.0\times1.08}{120.0} \]

\[ \boxed{c_Y=1.08\text{ mol/L}} \]

Part (b): Calculate the number of moles in 100.0 mL.

Convert the volume:

\[ 100.0\text{ mL}=0.1000\text{ L} \]

For solution X:

\[ n_X=c_XV \]

\[ n_X=2.16\times0.1000 \]

\[ \boxed{n_X=0.216\text{ mol}} \]

For solution Y:

\[ n_Y=c_YV \]

\[ n_Y=1.08\times0.1000 \]

\[ \boxed{n_Y=0.108\text{ mol}} \]

Part (c): Explain the difference.

Both solutions contain the same mass of solute per litre because their mass percentages and densities are identical.

However, one mole of solute X has a mass of 60.0 g, while one mole of solute Y has a mass of 120.0 g.

The same mass of Y therefore represents fewer moles.

This is the important distinction: mass percentage and density determine the mass of solute per litre, but molar mass determines how many moles that mass represents.

A solution with the same mass concentration can have a different molarity simply because its solute has a different molar mass.

Question 4

A sealed sample of a solution contains a solute with a molar mass of 90.0 g/mol.

The solution has a mass percentage of 18.0% w/w.

Its density is measured at two temperatures:

Assume no solute or solvent is lost and no chemical reaction occurs.

(a) Calculate the molarity at 20°C.

(b) Calculate the molarity at 40°C.

(c) A student argues that the molarity must remain constant because no solute has been added or removed. Explain why this reasoning is incorrect.

Solution 4

The molarity decreases from 2.28 mol/L at 20°C to 2.22 mol/L at 40°C.

The student is correct that the amount of solute has not changed. However, molarity depends on the volume of the solution as well as the number of moles.

Part (a): Calculate the molarity at 20°C.

Use:

\[ c=\frac{10w\rho}{M_m} \]

Substitute:

\[ c_{20}=\frac{10\times18.0\times1.14}{90.0} \]

\[ \boxed{c_{20}=2.28\text{ mol/L}} \]

Part (b): Calculate the molarity at 40°C.

Use the density measured at 40°C.

\[ c_{40}=\frac{10\times18.0\times1.11}{90.0} \]

\[ \boxed{c_{40}=2.22\text{ mol/L}} \]

Part (c): Explain why the molarity changes.

The solution expands as its temperature increases.

This increases the volume occupied by the solution without changing the mass of solute or the total mass of the sample.

Consequently:

• The mass percentage remains at 18.0% w/w.
• The density decreases.
• The molarity decreases because the same amount of solute occupies a greater volume.

The percentage decrease in molarity is:

\[ \frac{2.28-2.22}{2.28}\times100 \]

\[ \boxed{2.63\%} \]

The student’s mistake is assuming that an unchanged number of moles guarantees an unchanged molarity.

It doesn’t. Molarity depends on both the number of moles and the volume of the solution.

Question 5

This final problem combines mass percentage, density, conservation of mass, and evaporation.

Two sodium chloride solutions are prepared at 20°C.

Their properties are:

Both solutions are transferred completely into a single container and mixed.

The mixture is then heated, allowing some water to evaporate. Assume:

• No sodium chloride is lost.
• Only water evaporates.
• No chemical reaction occurs.
• The final mixture is uniform.
• The final mass percentage is 15.0% w/w.
• The final density, measured at 20°C, is 1.10 g/mL.

The molar mass of NaCl is 58.44 g/mol.

Calculate:

(a) The initial mass of sodium chloride in the combined mixture.

(b) The mass of water evaporated.

(c) The final molarity of the mixture.

Solution 5

The final molarity is 2.82 mol/L, and 99.0 g of water evaporates.

The key is to recognise that evaporation removes solvent but conserves the mass of dissolved sodium chloride. We must track the total mass and solute mass separately.

A tempting approach is to average the two starting percentages or assume the final volume is simply 100.0 + 200.0 = 300.0 mL.

Neither approach is valid for this problem.

We need to use mass conservation first.

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

Start with solution A.

Its total mass is:

\[ m_A=\rho_AV_A \]

\[ m_A=1.15\times100.0 \]

\[ m_A=115.0\text{ g} \]

The mass of sodium chloride is:

\[ m_{\mathrm{NaCl,A}}=0.200\times115.0 \]

\[ m_{\mathrm{NaCl,A}}=23.0\text{ g} \]

Now calculate the total mass of solution B:

\[ m_B=1.03\times200.0 \]

\[ m_B=206.0\text{ g} \]

The sodium chloride mass in solution B is:

\[ m_{\mathrm{NaCl,B}}=0.0500\times206.0 \]

\[ m_{\mathrm{NaCl,B}}=10.30\text{ g} \]

Add the two solute masses:

\[ m_{\mathrm{NaCl,total}}=23.0+10.30 \]

\[ \boxed{m_{\mathrm{NaCl,total}}=33.30\text{ g}} \]

The initial combined mass of solution is:

\[ m_{\mathrm{total}}=115.0+206.0 \]

\[ m_{\mathrm{total}}=321.0\text{ g} \]

Part (b): Calculate the mass of water evaporated.

The final mixture has a mass percentage of 15.0% w/w.

We know the mass of sodium chloride remains 33.30 g.

Start with the mass percentage equation:

\[ 15.0=\frac{33.30}{m_{\mathrm{final}}}\times100 \]

Rearrange to find the final total solution mass:

\[ m_{\mathrm{final}}= \frac{33.30\times100}{15.0} \]

\[ m_{\mathrm{final}}=222.0\text{ g} \]

The initial solution mass was 321.0 g.

The final solution mass is 222.0 g.

Since only water evaporates:

\[ m_{\mathrm{evaporated}}= 321.0-222.0 \]

\[ \boxed{m_{\mathrm{evaporated}}=99.0\text{ g}} \]

This result makes physical sense. Removing solvent increases the mass percentage of solute because the solute mass stays constant while the total solution mass decreases.

Part (c): Calculate the final molarity.

We now know the final solution mass and density.

First, calculate the final volume:

\[ V_{\mathrm{final}}= \frac{m_{\mathrm{final}}}{\rho_{\mathrm{final}}} \]

\[ V_{\mathrm{final}}= \frac{222.0}{1.10} \]

\[ V_{\mathrm{final}}=201.8\text{ mL} \]

Convert to litres:

\[ V_{\mathrm{final}}=0.2018\text{ L} \]

Next, calculate the moles of sodium chloride.

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

\[ n=\frac{33.30}{58.44} \]

\[ n=0.5698\text{ mol} \]

Finally, calculate the molarity:

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

\[ c=\frac{0.5698}{0.2018} \]

\[ \boxed{c=2.82\text{ mol/L}} \]

The final mixture has a molarity of 2.82 mol/L.

Why the tempting shortcut fails

It might seem reasonable to use the original combined volume of 300.0 mL. However, this is not the final solution volume.

There are two separate issues:

1. The volumes of the original solutions are not guaranteed to be additive when mixed.
2. Evaporation changes the total solution mass and volume.

The final density gives us the actual relationship between the final solution mass and volume. We therefore calculate the final volume from the final mass and density rather than relying on the starting volumes.

The other important constraint is conservation of solute. Because no NaCl is lost, its mass remains 33.30 g throughout the process, even though the mass percentage and molarity change.

This is the central idea behind the entire conversion: mass percentage tells you how much solute is present relative to the total solution mass, while molarity tells you how many moles are present relative to the total solution volume. Density connects those two measurements.

08Where to go next

Once you can move confidently between mass percentage and molarity, the next useful skill is working with dilution and solution preparation.

In particular, practise tracking the amount of solute when a stock solution is diluted. The amount of solute stays constant during dilution, but the total volume changes, so the molarity changes.

That distinction, combined with the density conversions in this article, gives you a reliable method for solving multi-step HSC Chemistry concentration problems.