ppm, ppb and mg/L: When Can You Convert Between Them?

A water test reports \\(1.0\ \mathrm{mg/L}\\) of a dissolved substance. Is that \\(1.0\\) ppm? In dilute freshwater, approximately. In a denser solution, not necessarily. Before converting anything, predict this: two solutions each contain \\(1.0\ \mathrm{mg}\\) of solute per litre, but one litre of the first solution has a mass of \\(1.00\ \mathrm{kg}\\), while one litre of the second has a mass of \\(1.20\ \mathrm{kg}\\). Do they have the same ppm by mass?

No. The first contains \\(1.0\ \mathrm{mg}\\) of solute per \\(1.00\ \mathrm{kg}\\) of solution, giving \\(1.0\\) ppm by mass. The second contains the same amount of solute per \\(1.20\ \mathrm{kg}\\) of solution, giving \\(0.833\\) ppm by mass. The concentration in mg/L is identical, but the concentration by mass is different.

01Quick decision rule

Use this checklist before choosing a conversion.

• ppm or ppb by mass to another mass-based unit: Convert directly. These quantities describe the mass of solute relative to the total mass of solution.
• mg/L or µg/L to ppm or ppb by mass: Use the density of the solution. You can approximate the numerical values as equal only when the solution density is close to \\(1.00\ \mathrm{kg/L}\\).
• Percentage concentration: Check whether the label means mass/mass, mass/volume, or volume/volume. The conversion depends on which quantities are being compared.
• Density given in g/mL: Its numerical value is the same when expressed in kg/L. For example, \\(1.05\ \mathrm{g/mL}=1.05\ \mathrm{kg/L}\\).

The key question is always the same: What is in the denominator? Is it the mass of the solution, or its volume?

01Understand what ppm and ppb measure

ppm means parts per million. ppb means parts per billion. Both describe very small proportions, but the basis of the proportion must be specified.

For this reference, ppm and ppb refer to mass fractions unless stated otherwise.

Let:

• \\(m_s\\) = mass of the solute, the substance being measured.
• \\(m_{\mathrm{solution}}\\) = total mass of the solution, including both solute and solvent.

The mass-based concentrations are defined by

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

\\[ C_{\mathrm{ppb}}= \frac{m_s}{m_{\mathrm{solution}}} \times 10^9 \\]

The masses in each fraction must use the same units.

These definitions give two useful exact relationships:

• \\(1\\) ppm by mass \\(=1\ \mathrm{mg/kg}\\) of solution.
• \\(1\\) ppb by mass \\(=1\ \mathrm{\mu g/kg}\\) of solution.

Therefore,

\\[ 1\ \mathrm{ppm}=1000\ \mathrm{ppb} \\]

This conversion is exact when both quantities use the same mass basis.

Notice that the denominator is the mass of the whole solution, not just the water or other solvent. That distinction becomes important when the solution’s density differs significantly from that of water.

02Convert between mg/L and ppm by mass using density

mg/L and µg/L are mass concentrations expressed per unit volume. They tell you how much solute is present in a given volume of solution.

For example, \\(5.0\ \mathrm{mg/L}\\) means that each litre of solution contains \\(5.0\ \mathrm{mg}\\) of solute.

ppm by mass asks a different question: how much solute is present per unit mass of solution?

To move between these quantities, you need the solution’s density.

Let \\(\rho\\) represent the density of the solution in \\(\mathrm{kg/L}\\).

The conversion equations

For mg/L and ppm by mass:

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

The reverse conversion is

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

For µg/L and ppb by mass:

\\[ \boxed{ C_{\mathrm{ppb}}= \frac{C_{\mathrm{\mu g/L}}}{\rho} } \\]

And in the reverse direction,

\\[ C_{\mathrm{\mu g/L}}= C_{\mathrm{ppb}}\rho \\]

Here, the concentration values are numerical values in the stated units, and \\(\rho\\) is the numerical density in kg/L.

Why does division by density work? If one litre of solution has a mass of \\(\rho\\) kilograms, then a concentration of \\(C\ \mathrm{mg/L}\\) corresponds to \\(C/\rho\\) milligrams of solute per kilogram of solution. That is exactly the mass-based definition of ppm.

Why does 1 mg/L approximately equal 1 ppm in water?

At ordinary laboratory temperatures, the density of pure water is close to \\(1.00\ \mathrm{kg/L}\\). Dilute aqueous solutions also usually have densities close to this value.

Suppose a dilute solution has a density of \\(1.00\ \mathrm{kg/L}\\). One litre of solution then has a mass of \\(1.00\ \mathrm{kg}\\).

A concentration of \\(1.0\ \mathrm{mg/L}\\) therefore becomes

\\[ \frac{1.0\ \mathrm{mg/L}} {1.00\ \mathrm{kg/L}} = 1.0\ \mathrm{mg/kg} \\]

Since \\(1.0\ \mathrm{mg/kg}=1.0\\) ppm by mass, the numerical values match.

The same reasoning gives the familiar approximation

\\[ 1\ \mathrm{\mu g/L} \approx 1\ \mathrm{ppb} \\]

for dilute aqueous solutions.

This is an approximation based on density, not a universal identity.

Consider what happens when the density changes:

For a solution less dense than water, each litre contains less total solution mass, so a fixed mass of solute represents a larger mass fraction. For a denser solution, each litre contains more total solution mass, so the same solute mass represents a smaller mass fraction.

This is why concentrated salt solutions, dense chemical solutions, and many non-aqueous solutions require more care than dilute freshwater.

03Convert percentage concentrations correctly

A percentage concentration is not complete information unless you know which quantities are being compared.

Chemistry commonly uses three forms:

• Mass/mass, or % m/m: grams of solute per \\(100\ \mathrm{g}\\) of solution.
• Mass/volume, or % m/v: grams of solute per \\(100\ \mathrm{mL}\\) of solution.
• Volume/volume, or % v/v: millilitres of solute per \\(100\ \mathrm{mL}\\) of solution.

These are different definitions. Do not swap between them without checking the units.

Converting % m/m to ppm and ppb

For a concentration of \\(x\%\ \mathrm{m/m}\\), where \\(x\\) is the numerical value written before the percent sign,

\\[ \mathrm{ppm\ by\ mass}=10\,000x \\]

\\[ \mathrm{ppb\ by\ mass}=10\,000\,000x \\]

For example, a solution containing \\(0.0032\%\ \mathrm{m/m}\\) has

\\[ 0.0032\times 10\,000=32\ \mathrm{ppm} \\]

Therefore, its concentration is \\(32\\) ppm by mass, or \\(32\,000\\) ppb by mass.

No density is needed because both percentage by mass and ppm by mass use the same mass fraction.

Converting % m/v to mg/L

For a concentration of \\(x\%\ \mathrm{m/v}\\),

\\[ C_{\mathrm{mg/L}}=10\,000x \\]

This relationship is exact under the stated mass/volume definition.

Why? A concentration of \\(1\%\ \mathrm{m/v}\\) means \\(1\ \mathrm{g}\\) of solute per \\(100\ \mathrm{mL}\\) of solution. That is \\(10\ \mathrm{g/L}\\), or \\(10\,000\ \mathrm{mg/L}\\).

For example,

\\[ 0.025\%\ \mathrm{m/v} =250\ \mathrm{mg/L} \\]

To convert this result to ppm by mass, however, you still need the solution density.

Converting % m/m to mg/L

If you know both the mass percentage and the solution density, you can convert directly:

\\[ C_{\mathrm{mg/L}}=10\,000x\rho \\]

Here, \\(x\\) is the numerical value of the percentage in % m/m, and \\(\rho\\) is the solution density in kg/L.

For example, a \\(0.040\%\ \mathrm{m/m}\\) solution with a density of \\(1.05\ \mathrm{kg/L}\\) has

\\[ C_{\mathrm{mg/L}} =10\,000(0.040)(1.05) =420\ \mathrm{mg/L} \\]

The density matters because a litre of this solution has a mass of \\(1.05\ \mathrm{kg}\\), not \\(1.00\ \mathrm{kg}\\).

What about % v/v?

A volume percentage cannot be converted to mass-based ppm or mg/L from the percentage alone.

For example, \\(2.0\%\ \mathrm{v/v}\\) means \\(2.0\ \mathrm{mL}\\) of solute per \\(100\ \mathrm{mL}\\) of final solution. You need the solute’s density to convert that volume of solute into a mass.

To obtain ppm by mass, you also need the density of the final solution, because the denominator must be the total solution mass.

A volume-based concentration can be expressed in parts per million on a volume basis, but that is not the same as ppm by mass.

04Unit assumptions at a glance

Use this table when you are unsure whether a conversion is exact or approximate.

Remember that \\(\rho\\) always means the density of the final solution in these equations, not automatically the density of the pure solvent or the dissolved substance.

05Worked applications

07Common mistakes

• Assuming mg/L always equals ppm. The shortcut works approximately for dilute aqueous solutions because their density is close to \\(1\ \mathrm{kg/L}\\). Outside that situation, use density.
• Using the mass of the solvent instead of the solution. ppm and ppb by mass use the total mass of the solution as the denominator.
• Treating every percentage as % m/m. A value in % m/v converts directly to mg/L, not automatically to ppm by mass. Check the label first.
• Using the wrong density. For volume-to-mass conversions, use the density of the final solution. The density of pure water or of the solute alone is not necessarily suitable.
• Mixing up factors of 1000. There are 1000 µg in 1 mg and 1000 ppb in 1 ppm, provided the ppm and ppb values use the same basis.

08Questions and solutions

Question 1

A dilute aqueous sample contains \\(4.60\ \mathrm{mg/L}\\) of a dissolved substance. Assume its density is \\(1.00\ \mathrm{kg/L}\\).

Convert the concentration to ppm by mass and ppb by mass.

Solution 1

The concentration is \\(4.60\\) ppm by mass, or \\(4.60\times10^3\\) ppb by mass.

Because the density is \\(1.00\ \mathrm{kg/L}\\), the numerical mg/L and ppm-by-mass values are equal:

\\[ \begin{aligned} C_{\mathrm{ppm}} &=\frac{4.60}{1.00}\\ &=4.60\ \mathrm{mg/kg}\\ &=4.60\ \mathrm{ppm} \end{aligned} \\]

Convert ppm to ppb using the exact mass-basis relationship:

\\[ 4.60\times1000 =4.60\times10^3\ \mathrm{ppb} \\]

The result is \\(4.60\\) ppm by mass and \\(4600\\) ppb by mass. The density assumption makes the first conversion straightforward; it is not a general rule for all solutions.

Question 2

An industrial rinse solution has a density of \\(1.16\ \mathrm{kg/L}\\) and contains \\(58.0\ \mathrm{\mu g/L}\\) of a metal ion.

Calculate its concentration in ppb by mass and ppm by mass. Explain why simply writing \\(58.0\\) ppb is not justified.

Solution 2

The correct concentrations are \\(50.0\\) ppb by mass and \\(0.0500\\) ppm by mass.

The given unit, µg/L, is mass per volume. To convert it to ppb by mass, divide by the solution density:

\\[ \begin{aligned} C_{\mathrm{ppb}} &=\frac{58.0}{1.16}\\ &=50.0\ \mathrm{\mu g/kg}\\ &=50.0\ \mathrm{ppb} \end{aligned} \\]

Now convert ppb to ppm:

\\[ C_{\mathrm{ppm}}=\frac{50.0}{1000} =0.0500\ \mathrm{ppm} \\]

The tempting answer of \\(58.0\\) ppb uses the approximation that \\(1\ \mathrm{\mu g/L}\approx1\\) ppb. That approximation assumes a density close to \\(1.00\ \mathrm{kg/L}\\). Here, the solution is denser, so each litre contains more total solution mass, and the mass fraction is smaller than the numerical µg/L value suggests.

Question 3

A solution contains \\(0.0060\%\ \mathrm{m/m}\\) of a dissolved substance and has a density of \\(0.920\ \mathrm{kg/L}\\).

Calculate its concentration in:

• ppm by mass,
• ppb by mass, and
• mg/L.

Solution 3

The concentration is \\(60\\) ppm by mass, \\(60\,000\\) ppb by mass, and \\(55\ \mathrm{mg/L}\\) to two significant figures.

First, convert the mass percentage directly to ppm:

\\[ \begin{aligned} C_{\mathrm{ppm}} &=10\,000x\\ &=10\,000(0.0060)\\ &=60\ \mathrm{ppm} \end{aligned} \\]

The percentage and ppm are both mass-based, so no density is needed.

Next, convert ppm to ppb:

\\[ 60\times1000=60\,000\ \mathrm{ppb} \\]

Finally, convert ppm by mass to mg/L using the density:

\\[ \begin{aligned} C_{\mathrm{mg/L}} &=C_{\mathrm{ppm}}\rho\\ &=60\times0.920\\ &=55.2\ \mathrm{mg/L} \end{aligned} \\]

To two significant figures, this is \\(55\ \mathrm{mg/L}\\).

Notice that the mg/L value is lower than the numerical ppm value. That makes sense because one litre of this solution has a mass of only \\(0.920\ \mathrm{kg}\\). Each litre contains less total solution mass than a litre of a solution with density \\(1.00\ \mathrm{kg/L}\\).

Question 4

An ethanol solution is labelled \\(2.0\%\ \mathrm{v/v}\\). A student converts this to \\(20\,000\\) ppm and \\(20\,000\ \mathrm{mg/L}\\), without specifying a concentration basis.

Evaluate both conversions. What additional information is required to calculate mg/L and ppm by mass?

Solution 4

The value \\(20\,000\\) is valid as parts per million by volume, but neither \\(20\,000\ \mathrm{mg/L}\\) nor \\(20\,000\\) ppm by mass follows from the label alone.

The label \\(2.0\%\ \mathrm{v/v}\\) means \\(2.0\ \mathrm{mL}\\) of ethanol per \\(100\ \mathrm{mL}\\) of final solution. Its volume fraction is \\(0.020\\), so

\\[ 0.020\times10^6 =20\,000\ \mathrm{ppm\ by\ volume} \\]

That calculation is valid because the percentage and ppm use the same volume basis.

However, mg/L requires the mass of ethanol per litre of solution. The label tells us that one litre contains \\(20\ \mathrm{mL}\\) of ethanol, but it does not tell us the mass of that ethanol. We need the density of pure ethanol.

If the density of ethanol is \\(\rho_e\\) in \\(\mathrm{g/mL}\\), then

\\[ C_{\mathrm{mg/L}} =20\,000\rho_e\ \mathrm{mg/L} \\]

To convert this result to ppm by mass, we also need the density of the final solution:

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

where \\(\rho\\) is the solution density in \\(\mathrm{kg/L}\\).

The error is not the arithmetic. It is treating volume-based concentration as though it were automatically mass-based concentration.

Question 5

A laboratory report gives two concentrations for the same homogeneous solution:

• \\(0.040\%\ \mathrm{m/m}\\),
• \\(420\ \mathrm{mg/L}\\).

Assume both measurements are accurate and refer to the same solute.

(a) Determine the solution density required for the two results to agree.

(b) A chemist then adds pure water with a mass equal to the original solution mass. No solute is lost or reacts. Calculate the new concentration in ppm by mass.

(c) Can the new concentration in mg/L be determined exactly from the information given? Explain.

Solution 5

The reports are consistent if the original solution density is \\(1.05\ \mathrm{kg/L}\\). After adding an equal mass of pure water, the concentration becomes \\(200\\) ppm by mass. The new mg/L concentration cannot be determined exactly without the final solution volume or density.

Part (a): Find the density.

First, convert the percentage by mass to ppm by mass:

\\[ \begin{aligned} C_{\mathrm{ppm}} &=10\,000(0.040)\\ &=400\ \mathrm{ppm} \end{aligned} \\]

Therefore, the solution contains \\(400\ \mathrm{mg}\\) of solute per kilogram of solution.

Use the relationship between ppm by mass and mg/L:

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

Substitute the reported concentration:

\\[ \begin{aligned} 420&=400\rho\\ \rho&=\frac{420}{400}\\ &=1.05\ \mathrm{kg/L} \end{aligned} \\]

The two measurements are consistent with a solution density of \\(1.05\ \mathrm{kg/L}\\). There is no contradiction: the numerical values differ because the concentration units use different denominators.

Part (b): Calculate the new ppm by mass.

Let the original solution have total mass \\(M\\).

Its solute mass fraction is

\\[ \frac{400}{10^6}=0.000400 \\]

Therefore, the original mass of solute is \\(0.000400M\\).

Adding pure water with mass \\(M\\) doubles the total solution mass to \\(2M\\), while the solute mass remains unchanged. The new mass fraction is

\\[ \frac{0.000400M}{2M} =0.000200 \\]

Convert this fraction to ppm by mass:

\\[ 0.000200\times10^6 =200\ \mathrm{ppm} \\]

The new concentration is exactly \\(200\\) ppm by mass under the stated assumptions. The key constraint is that the total mass doubles while the solute mass stays constant.

Part (c): Determine whether the new mg/L value is known.

No. The new mg/L concentration cannot be calculated exactly without knowing the final solution volume or density.

It is tempting to halve the original \\(420\ \mathrm{mg/L}\\) and claim \\(210\ \mathrm{mg/L}\\). That would require the final solution volume to be exactly twice the original volume. Adding an equal mass of water does not guarantee that the volume doubles.

Mass-based concentration halves here because the total mass doubles. Volume-based concentration depends on the final volume, which has not been provided.

This is the next decision to make in dilution problems: identify whether the concentration is defined by total solution mass or final solution volume, then determine which of those quantities is known to change.