Why Losing Precipitate During Filtration Gives a Low Result

If 0.500 g of precipitate forms during a gravimetric analysis, but only 0.450 g makes it onto the balance, your calculated result will be too low. The missing 0.050 g still contains some of the analyte you are trying to measure. It has simply been lost before you could weigh it.

This is why losing precipitate during filtration causes an underestimate of the analyte concentration or percentage. You calculate the amount of substance from the mass you collect, not the amount that originally formed.

But there is an important detail: the precipitate is not necessarily the analyte itself. You must understand what the solid contains, how much of it was recovered, and whether the mass you weigh is actually dry and pure.

01Follow the analyte through the experiment

Imagine you are measuring the sulfate concentration in a water sample.

The substance you want to measure is called the analyte. In this case, the analyte is the sulfate ion, \(\mathrm{SO_4^{2-}}\).

You cannot simply place the dissolved sulfate ions on a balance. Instead, you use a chemical reaction to convert them into an insoluble solid that can be collected and weighed.

Step 1: Collect a representative sample

First, collect a suitable water sample and measure a known volume.

For example, you might measure exactly \(250.0\ \mathrm{mL}\) of water.

The sample needs to be representative of the water being investigated. If the original water is not mixed properly, the portion you analyse might contain a different sulfate concentration from the bulk sample.

This is a sampling problem, rather than a filtration problem. Accurate measurements later in the experiment cannot fix an unrepresentative sample.

Step 2: Convert the dissolved analyte into a precipitate

Add barium chloride solution, \(\ce{BaCl2}\), to supply barium ions.

The barium ions react with the dissolved sulfate ions to form insoluble barium sulfate, \(\ce{BaSO4}\).

The balanced ionic equation is:

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

A white solid appears in the solution.

Notice something important: the precipitate contains both barium and sulfate. The barium comes from the reagent, while the sulfate comes from the original water sample.

The mole ratio is 1:1. Therefore, every mole of barium sulfate collected represents one mole of sulfate ions originally present, assuming precipitation is complete.

Step 3: Separate the precipitate by filtration

Now you need to separate the solid from the remaining liquid.

Picture the apparatus:

• A filter funnel sits above a receiving flask.
• Filter paper is placed inside the funnel.
• The mixture is poured from the original beaker into the funnel.
• Liquid passes through the filter paper and collects in the flask.
• Solid particles should remain on the filter paper.

The liquid that passes through is called the filtrate. The solid remaining on the filter paper is the collected precipitate.

In an ideal experiment, all the barium sulfate stays on the filter paper.

But will it always happen?

Not necessarily.

What happens when precipitate is lost?

There are several ways to lose precipitate during filtration:

1. Particles pass through the filter paper. Very fine precipitate particles may be small enough to travel through the filter pores.
2. Solid sticks to the apparatus. Some precipitate may remain attached to the original beaker, stirring rod, or funnel instead of reaching the filter paper.
3. The filter paper is damaged. A tear or incorrectly positioned filter paper can allow solid particles to bypass the intended filtration pathway.
4. Solid is carried away during washing. If particles become dislodged and enter the filtrate, they will not contribute to the final collected mass.

The important point is that the analyte-containing solid is no longer included in the material you weigh.

Consider the following illustrative data:

Only \(0.450\ \mathrm{g}\) is available for the original calculation.

That is just 90% of the precipitate that actually formed.

Since the precipitate has a fixed chemical composition, the collected solid also represents only 90% of the original sulfate ions.

The calculated sulfate concentration will therefore be 10% too low.

Why does a clear filtrate not guarantee a successful filtration?

A filtrate can look perfectly clear while still containing very small suspended particles.

Your eyes are not a sufficiently sensitive measurement instrument to establish that every particle has been retained.

A useful experimental check is to pass some of the original filtrate through a finer filter. If additional solid is collected, the first filtration allowed precipitate to escape.

You can also test the resulting clear filtrate with a small additional amount of the precipitating reagent. If more precipitate forms, some analyte was still dissolved.

These tests investigate two different problems:

• Fine particles in the filtrate indicate mechanical loss during filtration.
• Additional precipitate formed from dissolved analyte indicates incomplete precipitation.

The distinction matters because changing the filter paper addresses the first problem, but does not automatically solve the second.

02Why does the calculated result become too low?

Let’s connect the lost solid to the calculation.

For gravimetric analysis, you normally begin by calculating the number of moles of precipitate:

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

Where:

• \(n\) = amount of precipitate, in mol
• \(m\) = mass of precipitate, in g
• \(M\) = molar mass of precipitate, in \(\mathrm{g\,mol^{-1}}\)

For barium sulfate:

\[
M(\ce{BaSO4})=233.4\ \mathrm{g\,mol^{-1}}
\]

Suppose \(0.500\ \mathrm{g}\) of barium sulfate forms.

The actual amount is:

\[
n(\ce{BaSO4})=\frac{0.500}{233.4}
\]

\[
=0.002142\ \mathrm{mol}
\]

Because the mole ratio between sulfate and barium sulfate is 1:1:

\[
n(\mathrm{SO_{4}}^{2-})=0.002142\ \mathrm{mol}
\]

Now imagine that \(0.050\ \mathrm{g}\) of precipitate is lost during filtration.

You only collect \(0.450\ \mathrm{g}\).

Your calculation becomes:

\[
n(\ce{BaSO4})=\frac{0.450}{233.4}
\]

\[
=0.001928\ \mathrm{mol}
\]

You would therefore calculate only \(0.001928\ \mathrm{mol}\) of sulfate ions.

But the original sample actually contained \(0.002142\ \mathrm{mol}\).

The difference is caused by the missing precipitate, not by an error in the balanced chemical equation or molar mass.

The relationship between percentage loss and percentage error

If a fraction of the precipitate is lost, the calculated analyte amount decreases by the same fraction, provided the precipitate is pure and has a fixed composition.

We can express this as:

\[
m_{\mathrm{collected}}
=
m_{\mathrm{formed}}-m_{\mathrm{lost}}
\]

If \(f\) represents the fraction of precipitate lost:

\[
f=\frac{m_{\mathrm{lost}}}{m_{\mathrm{formed}}}
\]

Then:

\[
\boxed{C_{\mathrm{reported}}=(1-f)C_{\mathrm{true}}}
\]

Here, \(C\) represents the calculated analyte concentration.

This relationship assumes:

• Precipitation was complete.
• The precipitate has a known, constant chemical composition.
• The only error is the loss of precipitate.
• The original sample volume is recorded correctly.

For example, if 8% of the precipitate is lost, only 92% is recovered. The reported analyte concentration will be 8% below the true concentration.

Using the signed percentage error convention:

\[
\%\text{ error}
=
\frac{C_{\mathrm{reported}}-C_{\mathrm{true}}}
{C_{\mathrm{true}}}\times100
\]

A negative result indicates an underestimate.

04Wet precipitate versus pure dry precipitate

There is another important distinction to understand.

Losing precipitate makes the calculated result too low. However, weighing a wet precipitate can make the calculated result too high.

Why?

Imagine weighing a wet towel. The balance measures the towel and the water together.

A wet precipitate creates the same problem. The balance cannot distinguish the mass of the chemical solid from the water trapped between its particles.

Compare the measurements:

There are two separate jobs here:

Washing removes soluble impurities.

The precipitate can trap dissolved salts and excess reagent between its particles. Washing with a suitable liquid removes these substances.

Drying removes retained water.

The collected precipitate must be dried until it reaches a constant mass.

Constant mass means that successive drying, cooling, and weighing cycles produce no further significant change in mass.

For precipitates that undergo a chemical change when heated, you must also account for the formula of the final compound being weighed. The measured mass must correspond to a substance with a known composition.

For barium sulfate in our example, we assume that drying removes retained water without changing the chemical identity of the solid.

The key is that a wet mass cannot simply be substituted into a gravimetric calculation.

In fact, filtration loss and retained water can produce errors in opposite directions. If both occur, the overall error depends on their relative sizes.

06How to identify the source of a low result

When a gravimetric analysis produces a result that seems too low, don’t automatically blame the filtration.

Several different problems can produce the same direction of error.

A useful way to investigate the problem is to follow the material through the experiment.

1. Check whether the original sample was representative and the correct volume was measured.
2. Check whether the precipitation reaction converted the analyte into the intended solid.
3. Inspect the filtrate for suspended particles and, if necessary, use a finer filter.
4. Check whether dissolved analyte remains by testing the clear filtrate with additional precipitating reagent.
5. Ensure the collected precipitate is washed and dried to constant mass.
6. Confirm that the final mass includes all the precipitate recovered from the experiment, not just the portion remaining on the first filter.

For a broader example of tracking an analyte through dissolution, precipitation, filtration, and conversion into a known compound, see How to Calculate the Percentage of Metal in an Ore.

Once you have established a reliable dry precipitate mass, you can use the balanced equation to calculate the original analyte concentration. The conversion from precipitate mass to concentration is covered in How to Convert Gravimetric Results into mg/L.

07Questions and solutions

Question 1

A gravimetric experiment should produce \(0.320\ \mathrm{g}\) of a pure precipitate. However, only \(0.288\ \mathrm{g}\) is recovered after filtration.

Assuming the precipitate has a fixed composition and no other experimental errors occur:

a) Calculate the percentage recovery of the precipitate.

b) Determine the percentage by which the calculated analyte amount is underestimated.

Solution 1

The percentage recovery is 90.0%, and the calculated analyte amount is 10.0% too low.

Part a: Percentage recovery

Use:

\[
\%\text{ recovery}
=
\frac{m_{\mathrm{recovered}}}
{m_{\mathrm{original}}}\times100
\]

Substitute:

\[
=\frac{0.288}{0.320}\times100
\]

\[
\boxed{90.0\%}
\]

Part b: Percentage underestimate

The percentage lost is:

\[
100-90.0=10.0\%
\]

Therefore, 10.0% of the original precipitate was not recovered.

Since the precipitate has a fixed composition, the same percentage of analyte is missing from the calculation.

The reported analyte amount is therefore 10.0% too low.

Question 2

A student precipitates sulfate ions from a water sample using barium chloride.

After filtration, the filtrate appears clear. However, when the filtrate is passed through a finer filter, an additional \(0.012\ \mathrm{g}\) of dry white solid is collected.

The liquid passing through this finer filter produces no additional precipitate when a small amount of barium chloride is added.

Explain:

a) Why the original filtration produced an inaccurate result.

b) What the additional filtration tells you about the original filter paper.

c) Why the original filtrate could appear clear despite containing solid particles.

Solution 2

The original filtration lost fine barium sulfate particles, causing the calculated sulfate concentration to be too low.

Part a: Why was the original result inaccurate?

The original filter did not retain all the barium sulfate.

Some solid particles passed through the filter paper and entered the filtrate.

Because these particles were not included in the original weighed mass, the calculated number of moles of barium sulfate was too small.

Since one mole of barium sulfate represents one mole of sulfate ions, the calculated amount of sulfate was also too small.

Part b: What does the additional filtration show?

The second filter captured \(0.012\ \mathrm{g}\) of solid that escaped through the first filter.

This provides direct evidence of mechanical filtration loss.

The original filter paper was not suitable for retaining all the fine precipitate particles.

Using an appropriate finer filter would help prevent this particular error.

The additional test with barium chloride also provides useful evidence. No further precipitate forms, suggesting that no detectable sulfate remains dissolved in the final filtrate.

Part c: Why did the first filtrate appear clear?

The particles passing through the first filter were sufficiently small that they were not readily visible in the liquid.

Visual clarity does not establish that a solution is free of suspended particles.

The finer filtration provides stronger evidence than simply inspecting the liquid.

Question 3

A \(2.000\ \mathrm{g}\) fertiliser sample is dissolved and diluted to \(100.0\ \mathrm{mL}\).

A \(25.00\ \mathrm{mL}\) aliquot is removed and treated with excess barium chloride to precipitate sulfate ions as barium sulfate.

The following dry masses are recorded:

Assume sulfate is the only substance producing an insoluble barium salt, and both collected portions are pure, dry barium sulfate.

Calculate the percentage by mass of sulfate in the original fertiliser sample.

Use:

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

Solution 3

The original fertiliser contains 16.47% sulfate by mass.

The key is to include the additional precipitate recovered from the first filtrate before calculating the sulfate content.

Step 1: Determine the total mass of barium sulfate produced from the aliquot.

\[
m(\ce{BaSO4})=0.1800+0.0200
\]

\[
=0.2000\ \mathrm{g}
\]

Step 2: Calculate the moles of barium sulfate.

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

\[
n(\ce{BaSO4})=\frac{0.2000}{233.4}
\]

\[
=8.569\times10^{-4}\ \mathrm{mol}
\]

Step 3: Calculate the moles of sulfate ions.

From the balanced equation, the mole ratio is 1:1.

\[
n(\mathrm{SO_{4}}^{2-})
=
8.569\times10^{-4}\ \mathrm{mol}
\]

Step 4: Calculate the mass of sulfate in the aliquot.

\[
m=nM
\]

\[
=8.569\times10^{-4}\times96.1
\]

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

This is the sulfate mass in the \(25.00\ \mathrm{mL}\) aliquot.

Step 5: Calculate the sulfate mass in the entire digest.

The original digest volume is \(100.0\ \mathrm{mL}\).

The aliquot represents:

\[
\frac{100.0}{25.00}=4
\]

Therefore:

\[
m_{\mathrm{total}}(\mathrm{SO_{4}}^{2-})
=0.08235\times4
\]

\[
=0.3294\ \mathrm{g}
\]

Step 6: Calculate the percentage of sulfate in the original fertiliser.

\[
\%\mathrm{SO_{4}}^{2-}
=
\frac{m_{\mathrm{SO_{4}}^{2-}}}
{m_{\mathrm{fertiliser}}}\times100
\]

\[
=\frac{0.3294}{2.000}\times100
\]

\[
\boxed{16.47\%}
\]

If the additional \(0.0200\ \mathrm{g}\) had been ignored, the calculated sulfate percentage would have been only 14.82%.

The original result would therefore have been 10.0% too low.

Question 4

During a gravimetric experiment, a student collects a barium sulfate precipitate on a filter.

The following measurements are obtained:

Assume both dry portions are pure barium sulfate.

a) Calculate the mass of water retained in the original wet precipitate.

b) Calculate the total mass of dry precipitate formed.

c) Calculate the percentage underestimate if only the dry mass from the first filter is used.

d) Explain why using the wet mass of \(0.500\ \mathrm{g}\) would appear to give the correct total mass, despite being an invalid measurement.

Solution 4

The wet precipitate contains 0.050 g of water, the total dry precipitate mass is 0.500 g, and using only the first dry mass produces a 10.0% underestimate.

Part a: Mass of retained water

The difference between the wet and dry measurements is the mass removed during drying.

\[
m_{\mathrm{water}}=0.500-0.450
\]

\[
\boxed{m_{\mathrm{water}}=0.050\ \mathrm{g}}
\]

Part b: Total dry precipitate

Both dry portions must be included.

\[
m_{\mathrm{total}}=0.450+0.050
\]

\[
\boxed{m_{\mathrm{total}}=0.500\ \mathrm{g}}
\]

Part c: Percentage underestimate

The first filter retained only:

\[
\frac{0.450}{0.500}\times100=90.0\%
\]

Therefore, the percentage lost from the first filter was:

\[
100-90.0=10.0\%
\]

The calculated analyte amount would also be 10.0% too low.

Part d: Why does the wet mass appear correct?

The wet mass of the main filter cake is \(0.500\ \mathrm{g}\).

Interestingly, this is numerically equal to the total dry precipitate mass collected across both filters.

However, this agreement is accidental.

The main filter contains:

• \(0.450\ \mathrm{g}\) of dry barium sulfate.
• \(0.050\ \mathrm{g}\) of retained water.

The second filter contains the missing \(0.050\ \mathrm{g}\) of barium sulfate.

The water happens to add exactly the same mass that was lost as solid.

If the student uses the wet mass, the calculation appears correct, but it is not a valid gravimetric measurement. The water is not part of the barium sulfate and its amount could change between experiments.

This example demonstrates why two experimental errors can partially or completely cancel without making the procedure reliable.

Question 5

A chemist analyses a homogeneous \(500.0\ \mathrm{mL}\) water sample for sulfate ions by precipitating them as barium sulfate.

Assume that the sulfate concentration is uniform throughout the original sample, precipitation is complete, and the precipitate is pure and has a fixed composition.

Consider three different experimental situations.

Situation A

Before any aliquot is measured, \(50.0\ \mathrm{mL}\) is accidentally spilled from the original \(500.0\ \mathrm{mL}\) sample bottle.

The remaining sample is mixed, and a fresh \(100.0\ \mathrm{mL}\) aliquot is measured. All the precipitate from this aliquot is successfully recovered.

Situation B

A \(100.0\ \mathrm{mL}\) aliquot is measured accurately. However, \(10.0\ \mathrm{mL}\) spills from the measured aliquot before the precipitation reagent is added.

The student still uses \(100.0\ \mathrm{mL}\) as the original sample volume when calculating the concentration.

Situation C

A \(100.0\ \mathrm{mL}\) aliquot is measured. Then \(10.0\ \mathrm{mL}\) spills before precipitation.

Of the barium sulfate subsequently formed from the remaining liquid, another 10.0% is lost during filtration.

The student again uses \(100.0\ \mathrm{mL}\) as the original sample volume.

Answer the following:

a) What fraction of the true sulfate concentration would be reported in Situation A?

b) What fraction of the true sulfate concentration would be reported in Situation B?

c) What fraction of the true sulfate concentration would be reported in Situation C?

d) A student claims that the two 10.0% losses in Situation C produce a 20.0% underestimate. Explain why this reasoning is incorrect.

e) How would the result in Situation C change if the student correctly used the actual \(90.0\ \mathrm{mL}\) of liquid remaining after the spill as the sample volume?

Solution 5

Situation A reports 100% of the true concentration, Situation B reports 90%, and Situation C reports 81%. Situation C therefore underestimates the true concentration by 19.0%, not 20.0%.

This question tests an important distinction: losing part of a bulk sample is not always equivalent to losing part of an already measured aliquot or its precipitate.

Part a: Situation A

The reported concentration is 100% of the true concentration.

The original sample is homogeneous, so spilling \(50.0\ \mathrm{mL}\) from the bulk bottle does not change the concentration of the remaining water.

The chemist then measures a fresh \(100.0\ \mathrm{mL}\) aliquot.

Both the amount of sulfate and the sample volume are appropriate for the concentration calculation.

All the precipitate is recovered, so the result is accurate.

Part b: Situation B

The reported concentration is 90% of the true concentration.

Initially, the chemist measures \(100.0\ \mathrm{mL}\).

After spilling \(10.0\ \mathrm{mL}\), only \(90.0\ \mathrm{mL}\) remains.

Because the original sample is homogeneous, the remaining liquid contains 90% of the sulfate originally present in the measured aliquot.

However, the student still divides by the original \(100.0\ \mathrm{mL}\).

The calculation therefore uses a smaller amount of sulfate with an unchanged volume denominator.

The reported concentration is:

\[
C_{\mathrm{reported}}=0.90C_{\mathrm{true}}
\]

This is a 10.0% underestimate.

If the student had correctly used \(90.0\ \mathrm{mL}\) as the sample volume, the concentration would have been correct, assuming complete recovery.

Part c: Situation C

The reported concentration is 81% of the true concentration.

There are two separate losses.

First, the \(10.0\ \mathrm{mL}\) spill leaves only 90% of the original sulfate in the aliquot.

Second, filtration loses 10.0% of the precipitate that forms from this remaining sulfate.

Therefore, only 90% of the remaining 90% is recovered.

\[
\text{Overall recovery}=0.90\times0.90
\]

\[
=0.81
\]

The reported concentration is:

\[
\boxed{81.0\%\text{ of the true concentration}}
\]

Part d: Why is the underestimate 19%, not 20%?

The tempting approach is to add the two losses:

\[
10\%+10\%=20\%
\]

But this assumes both losses are calculated from the original amount.

They are not.

The second 10% loss applies to the precipitate remaining after the first loss, not the original amount of sulfate.

For an initial amount of 100 arbitrary units:

• The spill removes 10 units, leaving 90.
• Filtration loses 10% of those 90 units, which is another 9 units.
• The amount finally recovered is 81 units.

Therefore:

\[
100-81=19
\]

The total underestimate is:

\[
\boxed{19.0\%}
\]

This is why successive percentage losses must be applied multiplicatively rather than simply added.

Part e: Using the correct remaining sample volume

If the student correctly uses \(90.0\ \mathrm{mL}\) instead of \(100.0\ \mathrm{mL}\), the initial sample-spill error is removed from the concentration calculation.

However, the filtration loss remains.

The amount recovered is still only 90% of the sulfate actually present in the remaining \(90.0\ \mathrm{mL}\).

Therefore:

\[
C_{\mathrm{reported}}=0.90C_{\mathrm{true}}
\]

The reported result is now only 10.0% too low.

This demonstrates why the stage at which material is lost matters.

Losing some of a homogeneous bulk sample before measuring an aliquot does not necessarily change its concentration. Losing analyte-containing precipitate after measuring the sample volume reduces the calculated analyte amount without changing the volume denominator.

The central principle is that gravimetric analysis depends on recovering a known mass of a pure, dry compound that represents the original analyte. Once sampling, precipitation, and filtration losses have been accounted for, the next step is to use the balanced equation to convert that reliable precipitate mass into the original analyte concentration or percentage.