Why ppm Is Not Always Equal to mg/L in Chemistry

A laboratory report gives the concentration of a dissolved substance as \(4.00\) ppm. Another report gives \(4.00\ \mathrm{mg/L}\). Are they describing the same concentration?

Not necessarily. The difference comes down to what each measurement is comparing.

• \(\mathrm{mg/L}\) measures the mass of solute per volume of solution.
• ppm measures a ratio, but that ratio could be based on mass, volume, or amount of substance.

If ppm means parts per million by mass, it is equivalent to \(\mathrm{mg/kg}\), not automatically \(\mathrm{mg/L}\). The two numerical values match only when the solution density is \(1.00\ \mathrm{kg/L}\).

Consider a sample with a density of \(1.20\ \mathrm{kg/L}\). Each litre has a total mass of \(1.20\ \mathrm{kg}\). If that litre contains \(4.00\ \mathrm{mg}\) of dissolved substance, its concentration is:

\[ \frac{4.00\ \mathrm{mg}}{1.20\ \mathrm{kg}} =3.33\ \mathrm{mg/kg} \]

That is \(3.33\) ppm by mass, not \(4.00\) ppm.

The mass concentration is still \(4.00\ \mathrm{mg/L}\). Nothing has changed about the amount of solute in the litre. The difference is that the two measurements use different denominators.

01What does ppm actually measure?

Parts per million describes a ratio in which the numerator represents one millionth of the denominator.

However, you must identify what those parts represent. The numerator and denominator need to use the same type of quantity.

There are three common possibilities.

1. ppm by mass (m/m)

Mass-based ppm compares the mass of the solute with the total mass of the solution.

\[ C_{\mathrm{ppm}} = \frac{m_{\mathrm{solute}}} {m_{\mathrm{solution}}} \times 10^6 \]

For example, \(1\) ppm by mass means:

\[ 1\ \mathrm{ppm}=1\ \mathrm{mg/kg} \]

One milligram of solute is present in every kilogram of the total solution.

Notice that the denominator includes both the solute and the solvent. It is not just the mass of the water.

2. ppm by volume (v/v)

Volume-based ppm compares the volume of one substance with the total volume of the mixture.

\[ 1\ \mathrm{ppm\ (v/v)} = 1\ \mathrm{\mu L/L} \]

This means one microlitre of a substance per litre of final mixture.

For example, a liquid mixture containing \(8\) ppm by volume has \(8\ \mathrm{\mu L}\) of the specified substance per litre of mixture.

3. ppm by amount (n/n)

Amount-based ppm compares the amount of substance, measured in moles, with the total amount of the mixture.

\[ 1\ \mathrm{ppm\ (n/n)} = 1\ \mathrm{\mu mol/mol} \]

This form is particularly useful when describing the composition of gas mixtures.

For an ideal gas mixture at the same temperature and pressure, volume fractions and mole fractions have the same numerical value. This is why gas concentrations are sometimes reported as ppmv.

Where does mg/L fit?

Milligrams per litre is a mass concentration.

It tells you how much mass of solute is present in a particular volume of the final solution.

For example:

\[ 5.0\ \mathrm{mg/L} \]

means that every litre of solution contains \(5.0\ \mathrm{mg}\) of solute.

Unlike mass-based ppm, the denominator is a volume rather than a mass.

This gives us an important distinction:

Do not confuse mole fraction with molarity, either. Mole fraction compares moles of solute with total moles of mixture, while molarity measures moles of solute per litre of solution.

02An experiment: why measuring volume is not enough

Imagine a laboratory analysing a trace dye in a concentrated sugar solution.

The analyst weighs \(0.400\ \mathrm{mg}\) of dye, dissolves it, and uses a volumetric flask to make the final solution volume \(100.0\ \mathrm{mL}\).

The calculation is straightforward:

\[ C=\frac{0.400\ \mathrm{mg}}{0.1000\ \mathrm{L}} \]

\[ C=4.00\ \mathrm{mg/L} \]

The measurement is valid. But can the analyst report the result as \(4.00\) ppm by mass?

Not yet.

The volumetric flask measures the final volume of the solution. It does not tell us the total mass of that solution.

The original assumption

A student might predict that \(100.0\ \mathrm{mL}\) of solution has a mass of \(100.0\ \mathrm{g}\).

That would give a density of \(1.00\ \mathrm{g/mL}\), approximately the density of water under ordinary laboratory conditions.

If this prediction were correct, the two concentration values would match:

\[ \frac{0.400\ \mathrm{mg}}{0.1000\ \mathrm{kg}} = 4.00\ \mathrm{mg/kg} \]

However, a concentrated sugar solution can be considerably denser than water.

The assumption needs to be tested rather than accepted.

Revising the measurement

To determine the density, the analyst uses a calibrated \(100.0\ \mathrm{mL}\) volumetric flask and an analytical balance.

The procedure is:

1. Weigh the clean, dry, empty flask.
2. Fill it with the solution to the calibration mark.
3. Wipe the outside of the flask dry.
4. Weigh the filled flask.
5. Subtract the empty flask’s mass to find the solution mass.

Suppose the illustrative measurements are:

The density is therefore:

\[ \rho=\frac{120.00\ \mathrm{g}}{100.0\ \mathrm{mL}} \]

\[ \rho=1.200\ \mathrm{g/mL} \]

Since \(1\ \mathrm{g/mL}=1\ \mathrm{kg/L}\):

\[ \rho=1.200\ \mathrm{kg/L} \]

The original assumption was incorrect. This solution is \(20\%\) denser than water.

Now we can calculate the mass-based ppm:

\[ C_{\mathrm{ppm}} = \frac{0.400\ \mathrm{mg}} {0.12000\ \mathrm{kg}} \]

\[ \boxed{C_{\mathrm{ppm}}=3.33\ \mathrm{ppm}} \]

The same sample has a mass concentration of \(4.00\ \mathrm{mg/L}\), but a mass-based concentration of \(3.33\) ppm.

The original volumetric method did not fail. It correctly measured mass per volume. The problem was assuming that its result could automatically be labelled as mass-based ppm.

The revised measurement supplies the missing information: the density of the final solution.

All measurements in this laboratory scenario are illustrative.

03How density connects ppm and mg/L

We can now derive the relationship between mass-based ppm and \(\mathrm{mg/L}\).

Start with the definition:

\[ 1\ \mathrm{ppm\ (m/m)} = 1\ \mathrm{mg/kg} \]

Suppose a solution has a density of \(\rho\ \mathrm{kg/L}\).

This means one litre of solution has a total mass of \(\rho\) kilograms.

If the solution contains \(C_{\mathrm{ppm}}\) milligrams of solute per kilogram, then one litre contains:

\[ C_{\mathrm{ppm}}\times\rho \]

milligrams of solute.

Therefore:

\[ \boxed{ C_{\mathrm{mg/L}} = C_{\mathrm{ppm}}\times\rho } \]

Here:

• \(C_{\mathrm{mg/L}}\) is the mass concentration in milligrams per litre.
• \(C_{\mathrm{ppm}}\) is the concentration in ppm by mass.
• \(\rho\) is the density of the final solution in \(\mathrm{kg/L}\).

To convert in the opposite direction, rearrange:

\[ \boxed{ C_{\mathrm{ppm}} = \frac{C_{\mathrm{mg/L}}}{\rho} } \]

The units explain why these equations work:

\[ \frac{\mathrm{mg}}{\mathrm{kg}} \times \frac{\mathrm{kg}}{\mathrm{L}} = \frac{\mathrm{mg}}{\mathrm{L}} \]

The kilograms cancel, leaving milligrams per litre.

When can you treat ppm and mg/L as equal?

Only when the density assumption is justified.

For dilute aqueous solutions, the density is usually close to \(1.00\ \mathrm{kg/L}\). This is why students often see the shortcut:

\[ 1\ \mathrm{mg/L}\approx1\ \mathrm{ppm} \]

It is a useful approximation for dilute water-based solutions, not a universal conversion.

For a concentrated solution, a non-aqueous solvent, or a mixture with a significantly different density, use the actual density of the final solution.

Do not automatically use the density of pure water or the dissolved substance.

For a broader explanation of related concentration conversions, see ppm, ppb and mg/L: When Can You Convert Between Them?.

04What if ppm is based on volume or amount?

The density equation above applies specifically to mass-based ppm.

Other types of ppm require different information.

Converting volume-based ppm into mg/L

Suppose a liquid additive is present at \(8.0\) ppm by volume.

This means:

\[ 8.0\ \mathrm{\mu L/L} \]

To convert this into milligrams per litre, we need the density of the additive itself.

Assume its density is \(0.800\ \mathrm{g/mL}\), equivalent to \(0.800\ \mathrm{kg/L}\).

First, calculate the mass of the additive in one litre of mixture:

\[ m=0.0080\ \mathrm{mL} \times0.800\ \mathrm{g/mL} \]

\[ m=0.0064\ \mathrm{g} \]

Converting grams to milligrams:

\[ 0.0064\ \mathrm{g}=6.4\ \mathrm{mg} \]

Therefore:

\[ \boxed{C=6.4\ \mathrm{mg/L}} \]

Notice which density we used.

We used the density of the additive, not the density of the final mixture. This is because we started with a volume of additive and needed to convert that volume into a mass.

Converting amount-based ppm into mg/L

For a gas mixture, ppm may describe the mole fraction.

Consider a gas containing \(350\) ppm of a particular gas by amount.

This means:

\[ 350\ \mathrm{\mu mol/mol} \]

It does not mean \(350\ \mathrm{mg/L}\).

To convert it into mass concentration, we need:

• The molar mass of the gas.
• The temperature of the mixture.
• The pressure of the mixture.

Why temperature and pressure?

Because they determine how many moles of gas occupy one litre. The mole fraction tells us the proportion of those moles that belong to the gas being measured.

This is a different conversion problem from the liquid-solution density calculation.

It requires the ideal gas equation, \(PV=nRT\), rather than the density relationship for mass-based ppm.

07Questions and solutions

Question 1

A freshwater sample contains \(5.60\ \mathrm{mg/L}\) of a dissolved substance. Its density is \(1.00\ \mathrm{kg/L}\).

Express its concentration in ppm by mass.

Solution 1

The concentration is \(5.60\) ppm by mass.

Since the density is exactly \(1.00\ \mathrm{kg/L}\), the numerical values in \(\mathrm{mg/L}\) and ppm by mass are equal.

Using:

\[ C_{\mathrm{ppm}} = \frac{C_{\mathrm{mg/L}}}{\rho} \]

Substitute:

\[ C_{\mathrm{ppm}} = \frac{5.60}{1.00} \]

\[ \boxed{C=5.60\ \mathrm{ppm}} \]

The result is exact under the stated density assumption. Remember that this equality is not automatically valid for every solution.

Question 2

A laboratory analyses a concentrated aqueous solution using a \(100.0\ \mathrm{mL}\) volumetric flask.

The following measurements are recorded:

Calculate:

a) The density of the solution in \(\mathrm{kg/L}\).

b) The concentration in \(\mathrm{mg/L}\).

c) The concentration in ppm by mass.

A technician reports the answer as \(3.60\) ppm. Determine whether this is correct.

Solution 2

The correct concentration is \(3.60\ \mathrm{mg/L}\), or \(3.00\) ppm by mass. The technician has confused the two concentration bases.

Step 1: Calculate the mass of the solution.

Subtract the empty flask mass from the filled flask mass:

\[ \begin{aligned} m_{\mathrm{solution}} &=194.20-74.20\\ &=120.00\ \mathrm{g} \end{aligned} \]

Step 2: Calculate the density.

Density is mass divided by volume:

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

Substitute:

\[ \rho=\frac{120.00\ \mathrm{g}}{100.0\ \mathrm{mL}} \]

\[ \rho=1.200\ \mathrm{g/mL} \]

Therefore:

\[ \boxed{\rho=1.200\ \mathrm{kg/L}} \]

Step 3: Calculate the mass concentration.

Convert the volume into litres:

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

Now divide the solute mass by the final solution volume:

\[ \begin{aligned} C_{\mathrm{mg/L}} &=\frac{0.360\ \mathrm{mg}}{0.1000\ \mathrm{L}}\\ &=3.60\ \mathrm{mg/L} \end{aligned} \]

Step 4: Calculate ppm by mass.

First, convert the total solution mass into kilograms:

\[ 120.00\ \mathrm{g}=0.12000\ \mathrm{kg} \]

Then:

\[ \begin{aligned} C_{\mathrm{ppm}} &=\frac{0.360\ \mathrm{mg}}{0.12000\ \mathrm{kg}}\\ &=3.00\ \mathrm{mg/kg} \end{aligned} \]

Therefore:

\[ \boxed{C=3.00\ \mathrm{ppm}} \]

Step 5: Assess the technician’s result.

The technician reported \(3.60\) ppm.

That value is actually the concentration in \(\mathrm{mg/L}\). The mistake was treating the \(100.0\ \mathrm{mL}\) sample as though it had a mass of \(100.0\ \mathrm{g}\).

The measured solution mass was \(120.00\ \mathrm{g}\), so the correct mass-based concentration is lower.

Question 3

Two laboratory samples contain the same dissolved substance.

Their results are:

Calculate the missing concentration values.

Then determine:

a) Which sample has the greater mass concentration?

b) Which sample has the greater concentration in ppm by mass?

Explain why the rankings are different.

Solution 3

Sample A has the greater mass-based concentration, while Sample B has the greater mass concentration.

The two rankings differ because the samples have different densities.

Step 1: Calculate the mass-based ppm for Sample A.

Use:

\[ C_{\mathrm{ppm}}= \frac{C_{\mathrm{mg/L}}}{\rho} \]

Substitute:

\[ \begin{aligned} C_{\mathrm{ppm,A}} &=\frac{3.60}{0.900}\\ &=4.00\ \mathrm{ppm} \end{aligned} \]

Step 2: Calculate the mass concentration for Sample B.

Use:

\[ C_{\mathrm{mg/L}}=C_{\mathrm{ppm}}\times\rho \]

Substitute:

\[ \begin{aligned} C_{\mathrm{mg/L,B}} &=3.20\times1.20\\ &=3.84\ \mathrm{mg/L} \end{aligned} \]

Step 3: Compare the results.

Sample A has the greater mass fraction because \(4.00\) ppm is greater than \(3.20\) ppm.

Sample B has the greater mass concentration because \(3.84\ \mathrm{mg/L}\) is greater than \(3.60\ \mathrm{mg/L}\).

This is possible because a litre of Sample B has a greater total mass.

The same mass of solute per kilogram does not necessarily give the same mass of solute per litre when the densities differ.

Question 4

A liquid mixture contains an additive at \(12.0\) ppm by volume.

The additive has a density of \(0.800\ \mathrm{g/mL}\), while the final mixture has a density of \(1.10\ \mathrm{kg/L}\).

Calculate the additive concentration in \(\mathrm{mg/L}\).

Explain which density should be used and why.

Solution 4

The additive concentration is \(9.60\ \mathrm{mg/L}\).

We must use the density of the additive because the original concentration is volume-based. We are converting the volume of the additive into its mass.

Step 1: Interpret the volume-based ppm.

By definition:

\[ 12.0\ \mathrm{ppm\ (v/v)} = 12.0\ \mathrm{\mu L/L} \]

Therefore, each litre of final mixture contains \(12.0\ \mathrm{\mu L}\) of additive.

Step 2: Convert the additive volume into millilitres.

\[ 12.0\ \mathrm{\mu L}=0.0120\ \mathrm{mL} \]

Step 3: Calculate the additive mass.

Use:

\[ m=\rho V \]

Substitute the additive density:

\[ \begin{aligned} m &=0.800\ \mathrm{g/mL}\times0.0120\ \mathrm{mL}\\ &=0.00960\ \mathrm{g}\\ &=9.60\ \mathrm{mg} \end{aligned} \]

Step 4: Calculate the mass concentration.

This mass is present in one litre of final mixture.

Therefore:

\[ \boxed{C=9.60\ \mathrm{mg/L}} \]

The final mixture density of \(1.10\ \mathrm{kg/L}\) was not needed.

It would be relevant if we were converting a mass-based ppm concentration into \(\mathrm{mg/L}\). Here, the starting concentration was volume-based, so we needed the density of the additive itself.

Question 5

A concentrated stock solution contains a tracer at \(40.0\) ppm by mass.

Its density is \(1.20\ \mathrm{kg/L}\).

A student transfers \(25.0\ \mathrm{mL}\) of this stock solution into a \(250.0\ \mathrm{mL}\) volumetric flask and adds solvent until the final solution reaches the calibration mark.

The final diluted solution has a measured density of \(0.995\ \mathrm{kg/L}\).

Assume that:

• No tracer is lost during the transfer or dilution.
• The final solution is homogeneous.
• The stated volumes and densities refer to the solutions at the same temperature.

Calculate:

a) The mass of tracer transferred from the stock solution.

b) The final concentration in \(\mathrm{mg/L}\).

c) The final concentration in ppm by mass.

A student argues that the answer to part (c) must be:

\[ 40.0\times\frac{25.0}{250.0} =4.00\ \mathrm{ppm} \]

Explain why this apparently reasonable dilution calculation is incorrect.

Solution 5

The correct final concentrations are:

\[ \boxed{4.80\ \mathrm{mg/L}} \]

and

\[ \boxed{4.82\ \mathrm{ppm\ by\ mass}} \]

The student’s \(4.00\) ppm answer is incorrect because the student has applied a volume dilution ratio directly to a mass-based concentration.

We need to track the actual mass of tracer and the total mass of the final solution.

Step 1: Convert the stock concentration into mass-based units.

The stock solution contains:

\[ 40.0\ \mathrm{ppm} = 40.0\ \mathrm{mg/kg} \]

This means every kilogram of stock solution contains \(40.0\ \mathrm{mg}\) of tracer.

Step 2: Calculate the mass of the stock solution transferred.

The transferred volume is:

\[ 25.0\ \mathrm{mL}=0.0250\ \mathrm{L} \]

Use the stock solution density:

\[ \begin{aligned} m_{\mathrm{stock}} &=\rho V\\ &=1.20\times0.0250\\ &=0.0300\ \mathrm{kg} \end{aligned} \]

The transferred stock solution has a mass of \(0.0300\ \mathrm{kg}\), or \(30.0\ \mathrm{g}\).

Step 3: Calculate the mass of tracer transferred.

The stock contains \(40.0\ \mathrm{mg}\) of tracer per kilogram.

Therefore:

\[ \begin{aligned} m_{\mathrm{tracer}} &=40.0\ \mathrm{mg/kg}\times0.0300\ \mathrm{kg}\\ &=1.20\ \mathrm{mg} \end{aligned} \]

The dilution does not change the tracer mass because no tracer is lost.

Step 4: Calculate the final mass concentration.

The final solution volume is:

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

Therefore:

\[ \begin{aligned} C_{\mathrm{mg/L}} &=\frac{1.20\ \mathrm{mg}}{0.2500\ \mathrm{L}}\\ &=4.80\ \mathrm{mg/L} \end{aligned} \]

\[ \boxed{C=4.80\ \mathrm{mg/L}} \]

Step 5: Calculate the total mass of the final solution.

This is where the final solution density becomes important.

\[ \begin{aligned} m_{\mathrm{final}} &=\rho V\\ &=0.995\times0.2500\\ &=0.24875\ \mathrm{kg} \end{aligned} \]

The final solution has a total mass of \(0.24875\ \mathrm{kg}\).

Step 6: Calculate the final ppm by mass.

Use the definition of mass-based ppm:

\[ C_{\mathrm{ppm}} = \frac{m_{\mathrm{tracer}}} {m_{\mathrm{final}}} \times10^6 \]

Equivalently, use \(\mathrm{mg/kg}\):

\[ \begin{aligned} C_{\mathrm{ppm}} &=\frac{1.20\ \mathrm{mg}} {0.24875\ \mathrm{kg}}\\ &=4.824\ \mathrm{mg/kg} \end{aligned} \]

Therefore:

\[ \boxed{C=4.82\ \mathrm{ppm}} \]

Step 7: Identify the student’s mistake.

The student used:

\[ 40.0\times\frac{25.0}{250.0} =4.00 \]

This calculation assumes that the volume ratio is also the mass ratio.

But the stock and final solutions have different densities.

The stock solution has a density of \(1.20\ \mathrm{kg/L}\), while the final solution has a density of \(0.995\ \mathrm{kg/L}\).

Consequently, \(25.0\ \mathrm{mL}\) of stock does not represent the same fraction of the total solution mass as \(25.0\ \mathrm{mL}\) represents of the final solution volume.

The actual mass dilution factor is:

\[ \frac{0.0300}{0.24875}=0.1206 \]

So the final mass-based concentration is:

\[ 40.0\times0.1206=4.82\ \mathrm{ppm} \]

The tempting calculation would only work if the stock and final solution densities were equal, making their volume ratio equal to their mass ratio.

This is an important distinction in dilution problems: a volume dilution factor cannot automatically be applied to a mass-based concentration.

08The decision to make before every conversion

When you encounter ppm in a Chemistry question, do not immediately reach for a conversion formula.

First identify what the ppm represents.

The central idea is simple: concentration numbers only become comparable when you know what they measure.

Once you can distinguish mass fraction, mass concentration, volume fraction, and amount concentration, you can choose the correct conversion rather than relying on a shortcut that happens to work for dilute water.

The next useful step is connecting mass concentration to molar concentration. That requires a different piece of information: the molar mass of the solute. See How to Convert Mass Concentration into Molarity: g/L and mg/L.