How to Convert Gravimetric Results into mg/L

A 250.0 mL water sample produces 0.4668 g of white solid after barium chloride is added. What is the sulfate concentration in mg/L?

It is tempting to divide 0.4668 g by 0.2500 L and convert the answer into milligrams per litre. But that would give the wrong result. The solid is barium sulfate, \(\ce{BaSO4}\), which contains barium supplied by the reagent as well as sulfate from the original water sample.

To convert a gravimetric result into mg/L, you need to work backwards from the mass of the precipitate to the mass of the original analyte. Then divide by the volume of original sample that was analysed.

01The conversion: from precipitate mass to mg/L

The calculation follows four steps:

1. Find the mass of the dry, pure precipitate.
2. Convert the precipitate mass into moles using its molar mass.
3. Use the balanced chemical equation to calculate the moles and mass of the original analyte.
4. Convert the analyte mass into milligrams and divide by the original sample volume in litres.

The general equation is:

\[
C=\frac{m_p}{M_p}\times r\times M_a
\times\frac{1000}{V_s}
\]

Where:

• \(C\) = analyte concentration in mg/L.
• \(m_p\) = mass of dry, pure precipitate in grams.
• \(M_p\) = molar mass of the precipitate in g/mol.
• \(r\) = mole ratio of analyte to precipitate from the balanced equation.
• \(M_a\) = molar mass of the analyte in g/mol.
• \(V_s\) = volume of original sample represented by the analysis, in litres.
• \(1000\) = conversion factor from grams to milligrams.

The mole ratio is particularly important. It comes directly from the balanced chemical equation, rather than being automatically equal to one.

For example, consider the precipitation of sulfate ions using barium ions:

\[
\ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)}
\]

The mole ratio is 1:1. Therefore, one mole of barium sulfate contains one mole of sulfate ions.

The equation tells us how to get from the measured precipitate to the original analyte.

02Follow the analyte through the experiment

Before doing any calculations, make sure you understand what actually happens to the analyte.

Imagine analysing a water sample for sulfate ions.

1. Collect and measure the sample

The chemist collects a representative water sample and measures a known volume, such as 250.0 mL.

This volume matters because it becomes the denominator when calculating mg/L.

If only a small portion of the original sample is analysed, that portion must be representative of the whole sample.

2. Add a precipitation reagent

Barium chloride is added to the sample. It dissolves and supplies barium ions, which react with sulfate ions to form insoluble barium sulfate.

\[
\ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)}
\]

The sulfate ions are now part of a solid that can be collected and weighed.

Notice that the barium atoms have also become part of the solid. This is why the precipitate mass cannot be treated as the sulfate mass.

3. Filter and wash the precipitate

The mixture is passed through a filter. The solid remains on the filter, while most of the liquid passes through.

The solid is washed to remove dissolved substances that might otherwise remain trapped between its particles.

4. Dry and weigh the precipitate

The collected solid is dried until it reaches a constant mass. It is then weighed using a balance.

The final measurement should represent the mass of the intended chemical compound, rather than water, filter material, or dissolved impurities.

This is the mass that can be used in the stoichiometric calculation.

03Wet precipitate versus pure, dry precipitate

This is one of the easiest places to lose marks in a gravimetric calculation.

Suppose you collect a white precipitate and immediately place it on a balance. The balance measures everything present, not just the chemical compound you want.

Why does the precipitate need to be dried?

Water trapped in a precipitate adds mass without adding any analyte.

If the wet mass is used, the calculated number of moles of precipitate will be too large. This produces an overestimate of the analyte concentration.

Drying to constant mass involves repeating the drying, cooling, and weighing process until successive measurements no longer change within the balance’s precision.

However, there is another important distinction.

Dry does not automatically mean pure.

Imagine that some dissolved sodium chloride remains trapped in the precipitate because it was not washed properly. Drying removes the water, but the sodium chloride remains behind as a solid.

The balance will still measure that extra mass.

This is why washing and drying are separate steps:

• Washing removes soluble impurities.
• Drying removes retained water.
• Weighing determines the mass of the resulting solid.

You need both washing and drying to obtain a reliable gravimetric result.

05What if the original sample was diluted?

This is another common source of mistakes.

Sometimes the original sample is too concentrated, or the chemist needs to dilute it before taking an aliquot for analysis.

An aliquot is a measured portion of a larger solution.

The important point is that the volume used in the final mg/L calculation must represent the volume of the original sample, not automatically the volume of the diluted aliquot.

You can calculate this equivalent original volume using:

\[
V_{\mathrm{equivalent}}
=
V_{\mathrm{original}}
\times
\frac{V_{\mathrm{aliquot}}}
{V_{\mathrm{diluted}}}
\]

All three volumes must use the same units.

For example, suppose:

• 100.0 mL of original sample is diluted to 500.0 mL.
• A 25.00 mL aliquot is taken from the diluted solution.

The aliquot represents:

\[
V_{\mathrm{equivalent}}
=
100.0\times\frac{25.00}{500.0}
\]

\[
V_{\mathrm{equivalent}}=5.000\ \mathrm{mL}
\]

Although the chemist analysed 25.00 mL of diluted solution, that portion contained analyte originally present in only 5.000 mL of the undiluted sample.

This is the volume you need when calculating the original sample concentration.

07Common errors when converting gravimetric results

These are the mistakes worth checking before submitting an HSC Chemistry calculation.

One particularly important point: drying to constant mass does not correct contamination. A precipitate can have a perfectly stable mass and still contain unwanted substances.

If a reagent blank produces a measurable solid residue, that background may also need to be accounted for. You can only subtract its mass directly when the blank contribution has the same known composition and is additive.

08Questions and solutions

The following practice questions use original illustrative data. They progress from straightforward gravimetric calculations to more complicated problems involving stoichiometry, impurities, dilution, and background measurements.

Question 1

A 100.0 mL water sample is analysed for sulfate ions. The experiment produces 0.2334 g of washed, dry, pure barium sulfate.

Calculate the sulfate concentration in mg/L.

Use:

• \(M(\ce{BaSO4})=233.4\ \mathrm{g\,mol^{-1}}\)
• \(M(\mathrm{SO_{4}}^{2-})=96.1\ \mathrm{g\,mol^{-1}}\)

Solution 1

The sulfate concentration is 961 mg/L.

First, calculate the moles of barium sulfate:

\[
n(\ce{BaSO4})=
\frac{0.2334}{233.4}
\]

\[
n(\ce{BaSO4})=0.001000\ \mathrm{mol}
\]

The balanced equation gives a 1:1 ratio:

\[
\ce{Ba^2+ + SO4^2- -> BaSO4}
\]

Therefore:

\[
n(\mathrm{SO_{4}}^{2-})=0.001000\ \mathrm{mol}
\]

Calculate the sulfate mass:

\[
m=nM
\]

\[
m=0.001000\times96.1
\]

\[
m=0.0961\ \mathrm{g}
\]

Convert to milligrams:

\[
0.0961\times1000=96.1\ \mathrm{mg}
\]

The sample volume in litres is:

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

Finally:

\[
C=\frac{96.1}{0.1000}
\]

\[
\boxed{C=961\ \mathrm{mg/L}}
\]

The important distinction is that the measured 0.2334 g belongs to barium sulfate, not sulfate alone. The mole ratio and molar mass of sulfate are what connect the precipitate to the original analyte.

Question 2

A 250.0 mL water sample is analysed for chloride ions using silver nitrate.

The empty filter crucible has a mass of 18.462 g. After precipitation, filtration, washing, and drying to constant mass, the crucible and silver chloride precipitate have a combined mass of 18.821 g.

Calculate the chloride concentration in mg/L.

Use:

• \(M(\ce{AgCl})=143.4\ \mathrm{g\,mol^{-1}}\)
• \(M(\mathrm{Cl}^{-})=35.5\ \mathrm{g\,mol^{-1}}\)

Solution 2

The chloride concentration is 356 mg/L.

The first step is to subtract the empty crucible mass. The balance measures the combined mass, not just the precipitate.

\[
m(\ce{AgCl})=18.821-18.462
\]

\[
m(\ce{AgCl})=0.359\ \mathrm{g}
\]

Calculate the moles of silver chloride:

\[
n(\ce{AgCl})=\frac{0.359}{143.4}
\]

\[
n(\ce{AgCl})=0.002504\ \mathrm{mol}
\]

The balanced equation is:

\[
\ce{Ag+ + Cl- -> AgCl}
\]

The mole ratio is 1:1, so:

\[
n(\mathrm{Cl}^{-})=0.002504\ \mathrm{mol}
\]

Calculate the chloride mass:

\[
m=nM
\]

\[
m=0.002504\times35.5
\]

\[
m=0.0889\ \mathrm{g}
\]

Convert to milligrams:

\[
0.0889\times1000=88.9\ \mathrm{mg}
\]

Convert the sample volume:

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

Therefore:

\[
C=\frac{88.9}{0.2500}
\]

\[
\boxed{C=356\ \mathrm{mg/L}}
\]

The tempting mistake is to use 18.821 g directly as the precipitate mass. That includes the crucible and would produce a completely unrealistic result. Always subtract the tare mass before using the precipitate mass in your mole calculation.

Question 3

A chemist wants to determine the concentration of silver ions in a 80.0 mL water sample.

The silver ions are precipitated using chromate ions, forming silver chromate:

\[
\ce{2Ag+(aq) + CrO4^2-(aq) -> Ag2CrO4(s)}
\]

After filtration and drying, 0.1327 g of pure silver chromate is collected.

Calculate the silver ion concentration in mg/L.

Use:

• \(M(\ce{Ag2CrO4})=331.8\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{Ag})=107.9\ \mathrm{g\,mol^{-1}}\)

Solution 3

The silver ion concentration is 1.08 × 10³ mg/L.

This question has an important difference from the previous examples: the mole ratio is 2:1, not 1:1.

First, calculate the moles of silver chromate:

\[
n(\ce{Ag2CrO4})=
\frac{0.1327}{331.8}
\]

\[
n(\ce{Ag2CrO4})=0.0004000\ \mathrm{mol}
\]

From the balanced equation:

\[
\ce{2Ag+ + CrO4^2- -> Ag2CrO4}
\]

Two moles of silver ions are required to form one mole of silver chromate.

Therefore:

\[
n(\mathrm{Ag}^{+})=2\times0.0004000
\]

\[
n(\mathrm{Ag}^{+})=0.0008000\ \mathrm{mol}
\]

Calculate the mass of silver:

\[
m=nM
\]

\[
m=0.0008000\times107.9
\]

\[
m=0.08632\ \mathrm{g}
\]

Convert to milligrams:

\[
0.08632\times1000=86.32\ \mathrm{mg}
\]

Convert the original sample volume:

\[
80.0\ \mathrm{mL}=0.0800\ \mathrm{L}
\]

Calculate the concentration:

\[
C=\frac{86.32}{0.0800}
\]

\[
\boxed{C=1.08\times10^3\ \mathrm{mg/L}}
\]

A student who assumes a 1:1 mole ratio would calculate only half the correct amount of silver.

The precipitate formula is what determines the mole relationship. Never assume the ratio without checking the balanced equation.

Question 4

A 100.0 mL water sample is analysed for sulfate ions using barium chloride.

The following measurements are recorded:

The chemist discovers that the precipitate was not washed properly. A separate analysis shows that 0.0400 g of soluble reagent salt remains in the dried solid.

Calculate the sulfate concentration in mg/L.

Then explain why using either the wet precipitate mass or the uncorrected dry precipitate mass would give an incorrect result.

Use:

• \(M(\ce{BaSO4})=233.4\ \mathrm{g\,mol^{-1}}\)
• \(M(\mathrm{SO_{4}}^{2-})=96.1\ \mathrm{g\,mol^{-1}}\)

Solution 4

The corrected sulfate concentration is 824 mg/L.

The crucial detail is that the dried precipitate still contains 0.0400 g of soluble salt. We must remove this contribution from the measured dry mass before calculating the sulfate amount.

First, calculate the total dry solid mass:

\[
m_{\mathrm{dry}}=12.740-12.500
\]

\[
m_{\mathrm{dry}}=0.240\ \mathrm{g}
\]

Subtract the mass of the contaminant:

\[
m(\ce{BaSO4})=0.240-0.0400
\]

\[
m(\ce{BaSO4})=0.200\ \mathrm{g}
\]

Now calculate the moles of pure barium sulfate:

\[
n(\ce{BaSO4})=\frac{0.200}{233.4}
\]

\[
n(\ce{BaSO4})=0.0008569\ \mathrm{mol}
\]

The balanced equation gives a 1:1 mole ratio:

\[
n(\mathrm{SO_{4}}^{2-})=0.0008569\ \mathrm{mol}
\]

Calculate the sulfate mass:

\[
m=nM
\]

\[
m=0.0008569\times96.1
\]

\[
m=0.08235\ \mathrm{g}
\]

Convert to milligrams:

\[
m=82.35\ \mathrm{mg}
\]

The original sample volume is:

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

Therefore:

\[
C=\frac{82.35}{0.1000}
\]

\[
\boxed{C=824\ \mathrm{mg/L}}
\]

Now consider the two tempting incorrect approaches.

Using the wet mass

The wet precipitate mass is:

\[
12.900-12.500=0.400\ \mathrm{g}
\]

This includes retained water and solution. If used as pure barium sulfate, it would give approximately 1647 mg/L, twice the correct result.

Using the uncorrected dry mass

The dry solid mass is 0.240 g, but 0.0400 g of this is contamination.

Using 0.240 g without correcting for the impurity would give approximately 988 mg/L.

This is still too high.

The deeper lesson is that drying solves the moisture problem, not the purity problem. A constant mass is not necessarily a chemically pure mass.

Question 5

A chemist analyses a groundwater sample for sulfate ions.

The following procedure is used:

• 40.00 mL of the original groundwater sample is transferred into a 200.0 mL volumetric flask and diluted to the calibration mark.
• A 50.00 mL aliquot of the diluted solution is removed.
• Excess barium chloride is added to this aliquot.
• The resulting precipitate is filtered, washed, and dried to constant mass.
• The measured mass of the dry solid is 0.12834 g.

A reagent blank is also prepared using the same reagent volumes but without any groundwater sample. It produces 0.01167 g of dry barium sulfate due to sulfate contamination in the reagent.

The sample filtrate has a final measured volume of 58.0 mL because some liquid is retained during filtration.

Assume:

• All sulfate in the analysed aliquot is precipitated.
• The blank contribution is additive and consists entirely of barium sulfate.
• There are no analyte losses during transfer or filtration.
• The precipitate has the formula \(\ce{BaSO4}\).

Calculate the original groundwater sulfate concentration in mg/L.

Explain why neither 50.00 mL nor 58.0 mL is the correct denominator in the final concentration calculation.

Use:

• \(M(\ce{BaSO4})=233.4\ \mathrm{g\,mol^{-1}}\)
• \(M(\mathrm{SO_{4}}^{2-})=96.1\ \mathrm{g\,mol^{-1}}\)

Solution 5

The original groundwater sulfate concentration is 4.81 × 10³ mg/L.

This problem requires three separate corrections:

1. Correct the measured precipitate mass using the reagent blank.
2. Convert the corrected precipitate mass into the sulfate mass.
3. Determine the original sample volume represented by the diluted aliquot.

Step 1: Correct the precipitate mass using the reagent blank.

The measured sample solid contains both the sulfate-derived precipitate and the background precipitate from the reagent.

Subtract the blank:

\[
m_{\mathrm{corrected}}
=0.12834-0.01167
\]

\[
m_{\mathrm{corrected}}=0.11667\ \mathrm{g}
\]

This is the mass of barium sulfate attributable to the groundwater sample.

Step 2: Convert the corrected precipitate mass into moles.

\[
n(\ce{BaSO4})=
\frac{0.11667}{233.4}
\]

\[
n(\ce{BaSO4})=0.0005000\ \mathrm{mol}
\]

Step 3: Calculate the moles of sulfate ions.

The balanced equation is:

\[
\ce{Ba^2+ + SO4^2- -> BaSO4}
\]

The mole ratio is 1:1.

Therefore:

\[
n(\mathrm{SO_{4}}^{2-})=0.0005000\ \mathrm{mol}
\]

Step 4: Calculate the mass of sulfate ions.

\[
m=nM
\]

\[
m=0.0005000\times96.1
\]

\[
m=0.04805\ \mathrm{g}
\]

Convert to milligrams:

\[
m=48.05\ \mathrm{mg}
\]

Step 5: Find the original sample volume represented by the aliquot.

This is the most important reasoning step.

The chemist initially diluted 40.00 mL of original sample to 200.0 mL.

Only 50.00 mL of that diluted solution was analysed.

Therefore, the original sample volume represented is:

\[
V_{\mathrm{equivalent}}
=
40.00\times\frac{50.00}{200.0}
\]

\[
V_{\mathrm{equivalent}}=10.00\ \mathrm{mL}
\]

Convert to litres:

\[
V_{\mathrm{equivalent}}=0.01000\ \mathrm{L}
\]

Step 6: Calculate the original sulfate concentration.

\[
C=\frac{48.05}{0.01000}
\]

\[
\boxed{C=4.81\times10^3\ \mathrm{mg/L}}
\]

Now consider why the two other volumes are incorrect.

The 50.00 mL volume is the volume of diluted solution analysed. It represents only 10.00 mL of the original sample.

Using 50.00 mL would give the concentration of the diluted solution, not the original groundwater.

The 58.0 mL filtrate volume is also irrelevant to the original sample concentration. It includes the effect of added reagent and the liquid retained during filtration. It does not represent the volume of original groundwater that supplied the sulfate ions in the precipitate.

The precipitate contains sulfate that originally came from exactly 10.00 mL of groundwater. That is the denominator we must use.

This is the key idea behind the entire conversion: the mass must be traced back to the original analyte, and the volume must be traced back to the original sample.

Once you can convert a gravimetric result into mg/L, the next useful decision is whether you need to report that result as ppm instead.

Remember that mg/L and ppm by mass are not universally interchangeable. Their numerical values are approximately equal for dilute aqueous solutions because their density is close to 1.00 kg/L.

For a closer look at that distinction, see ppm, ppb and mg/L: When Can You Convert Between Them?.

If you need to revisit the underlying precipitation reactions, mole ratios, and conversion from precipitate mass to analyte amount, see Gravimetric analysis: calculate analyte amounts from precipitates.