How to Find Where Material Was Lost in a Separation Experiment

Suppose you start with a dry mixture containing 2.00 g of sand and 3.00 g of sodium chloride. You dissolve the salt in water, filter the mixture, and evaporate the filtrate. At the end, you recover 1.92 g of dry sand and 2.85 g of dry salt.

Your original mixture weighed 5.00 g. Your recovered products weigh only 4.77 g.

Where did the missing 0.23 g go?

The answer isn’t necessarily that something escaped during heating. Some sand may have passed through the filter paper. Salt solution may have remained on the glassware, or dissolved salt may have been discarded with the filter paper. You need evidence to distinguish these possibilities.

To identify where material was lost in a separation experiment, track each component through every stage, collect the different fractions, and calculate a separate mass balance for each component. A single comparison between the starting mass and final product mass tells you how much is missing from your collected products, but not where it went.

01Follow each component through the apparatus

Start by picturing the salt-and-sand experiment.

You have two solids with different physical properties:

• Sodium chloride dissolves in water.
• Sand is insoluble in water.

These differences determine what happens during filtration and evaporation.

What should happen during filtration?

When you add water to the mixture and stir it, the sodium chloride dissolves.

At the particle level, sodium chloride separates into hydrated sodium and chloride ions. These ions are distributed throughout the water.

Sand behaves differently. It remains as solid particles because it does not dissolve appreciably in water.

When you pour the mixture into the filter funnel:

• Sand is trapped by the filter paper. The solid collected is called the residue.
• Water and dissolved sodium chloride pass through the filter paper. The liquid collected is called the filtrate.

Notice something important: the filtrate is not pure water. It contains dissolved sodium chloride.

Ordinary filter paper separates particles according to size. It does not remove dissolved ions.

What happens during washing?

The sand residue can still contain salt solution trapped between its grains and within the filter paper.

You can wash the residue with a small amount of distilled water.

The water dissolves the sodium chloride, which passes through the filter paper with the washings. The insoluble sand remains behind.

This improves the purity of the sand, but it creates another fraction containing sodium chloride.

If you discard those washings without recovering or measuring their contents, you lose track of some of your original salt.

What happens during evaporation?

The filtrate contains dissolved sodium chloride and water.

When you heat it in an evaporating basin:

• Water gains enough energy to escape as water vapour.
• Sodium chloride remains in the basin because it is effectively non-volatile under ordinary laboratory heating conditions.
• As the water evaporates, the solution becomes more concentrated.
• Eventually, sodium chloride forms a solid residue.

The important distinction is that water is intentionally removed, whereas sodium chloride is the substance you are trying to recover.

If the basin is heated too vigorously, droplets can spit out of it. Those droplets contain dissolved sodium chloride, so some of your desired product can be physically lost.

02Match the separation method to the possible loss

Different apparatus creates different places where material can remain or escape.

When investigating an unexpected mass deficit, consider the physical property responsible for the separation and the expected location of each component.

A useful habit is to ask two questions whenever you see separation apparatus:

1. Which physical property allows the substances to separate?
2. Where should each component be found after the process?

If you cannot answer the second question, you cannot confidently identify where material has been lost.

For a fuller explanation of how to select the appropriate apparatus, see How to Choose Separation Techniques in HSC Chemistry.

03Use a mass balance to measure the loss

Now we can put numbers to the investigation.

A mass balance compares the initial amount of a component with the amounts recovered in the different fractions.

For any component \(X\):

\[ \begin{aligned} m_{X,\mathrm{initial}} ={}& m_{X,\mathrm{target}}\\ &+m_{X,\mathrm{other\ fractions}}\\ &+m_{X,\mathrm{unaccounted}} \end{aligned} \]

Here:

• \(m_{X,\mathrm{initial}}\) is the initial mass of component \(X\).
• \(m_{X,\mathrm{target}}\) is the mass recovered in the intended product.
• \(m_{X,\mathrm{other\ fractions}}\) is the mass found in other collected fractions.
• \(m_{X,\mathrm{unaccounted}}\) is the mass that has not yet been located.

The word unaccounted is important.

It does not mean the material has been destroyed. It means you have not yet identified where it went.

Calculate the recovery percentage

You can also calculate the percentage of a component recovered in its intended product:

\[ \%\text{ recovery} = \frac{\text{mass in target product}} {\text{initial mass of component}} \times100 \]

For example, if you begin with 2.00 g of sand and recover 1.92 g:

\[ \%\text{ recovery} = \frac{1.92}{2.00}\times100 =96.0\% \]

You have recovered 96.0% of the original sand in your intended product.

The remaining 4.0% might be in another fraction or genuinely unaccounted for.

These are two different things. A low product recovery does not automatically mean a large amount of material has escaped the apparatus.

The first experiment: why the final mass is insufficient

Return to the original salt-and-sand mixture.

The initial composition is:

After separation, the student collects:

The difference is:

\[ 5.00-4.77=0.23\text{ g} \]

The student knows that 0.23 g is missing from the intended products.

But the original experiment has a flaw: it measures only the final products.

It does not measure the material left in the filter paper, the liquid discarded during washing, or the substances adhering to the glassware.

The revised experiment needs to measure those intermediate fractions as well.

Before doing that, let’s calculate exactly what the first experiment tells us.

04Revise the experiment: collect every fraction

How could you improve the original experiment?

Instead of collecting only the intended products, retain the intermediate fractions separately.

For the salt-and-sand separation, investigate four possible locations:

The main filtrate

Pass it through a second, finer filter paper.

If sand particles appear on the second filter, some sand passed through the first filter.

The filter-cake washings

Collect the water used to wash the sand residue.

Evaporate the washings separately. Any sodium chloride remaining as a solid was previously dissolved in the liquid trapped within the filter cake.

The apparatus rinses

Rinse the original beaker, stirring rod, and filter funnel with a measured amount of distilled water.

Pass the rinse through a fine filter. This separates any remaining sand from the dissolved sodium chloride.

The evaporating basin

Inspect the basin for remaining solid. If you suspect salt has been lost during heating, examine the basin walls and surrounding apparatus for splashed material.

Illustrative experimental data

Suppose the revised investigation produces the following results.

The initial mixture still contains 2.00 g of sand and 3.00 g of sodium chloride.

All measurements are illustrative. Assume the recovered solids are dry and the fractions have been measured without double-counting.

Now the location of the missing material becomes much clearer.

05A different problem: material remaining in solution

There is another important situation where students incorrectly identify a separation loss.

Consider crystallisation.

Suppose you dissolve potassium nitrate in hot water and allow the solution to cool. Crystals form, and you collect them using filtration.

You might expect all the original potassium nitrate to appear as crystals.

That prediction is incorrect.

Some potassium nitrate remains dissolved in the liquid, even after the solution has cooled.

This remaining liquid is called the mother liquor.

Why the mother liquor matters

Solubility determines how much solute can remain dissolved at a particular temperature.

For a given amount of water, a saturated solution contains the maximum amount of dissolved solute possible at that temperature.

Cooling a saturated solution often decreases the solubility of the dissolved solid. Some of the solute then forms crystals.

However, the remaining solution still contains dissolved solute.

Consider these illustrative conditions:

• Initial potassium nitrate mass: 5.00 g.
• Water mass: 10.0 g.
• Final temperature: 20°C.
• Solubility at 20°C: 32.0 g of potassium nitrate per 100 g of water.

The maximum mass of potassium nitrate remaining dissolved is:

\[ m_{\mathrm{dissolved}} = \frac{32.0}{100}\times10.0 \]

\[ =3.20\text{ g} \]

Therefore, the mass of crystals formed is:

\[ m_{\mathrm{crystals}} = 5.00-3.20 \]

\[ =1.80\text{ g} \]

Only 1.80 g forms crystals, even though the original sample contained 5.00 g.

The remaining 3.20 g is still in the mother liquor.

The filter paper is working correctly. It traps the solid crystals but allows the dissolved potassium nitrate to pass through with the water.

Washing crystals can cause further loss

Suppose the student washes the crystals with 2.00 g of fresh water at 20°C.

Using the same solubility data, this water can dissolve:

\[ m_{\mathrm{dissolved}} = \frac{32.0}{100}\times2.00 \]

\[ =0.64\text{ g} \]

Assuming the wash water reaches saturation and enough crystals are available, another 0.64 g dissolves.

The final crystal mass becomes:

\[ 1.80-0.64=1.16\text{ g} \]

The liquid fractions now contain:

\[ 3.20+0.64=3.84\text{ g} \]

of potassium nitrate.

The complete mass balance is:

\[ 5.00=1.16+3.84 \]

The student has recovered only 1.16 g as crystals, but all 5.00 g of potassium nitrate is accounted for.

This is why you should never assume that a low crystal yield means the product escaped through the filter paper.

The mother liquor and washings may contain most of the missing substance.

06A practical method for locating material loss

When analysing an unfamiliar separation experiment, work through the following process.

Step 1: Identify the components.

Write down the initial mass of each substance, rather than relying only on the total mixture mass.

Step 2: Identify the physical property used for separation.

Is the process based on particle size, solubility, density, volatility, or attraction to a stationary phase?

Step 3: Predict where each component should go.

For example, during filtration of sand and salt solution, sand should remain on the filter paper, while dissolved sodium chloride should enter the filtrate.

Step 4: Identify the possible unwanted fractions.

Check the filtrate, washings, mother liquor, apparatus surfaces, and any material remaining in connecting tubes or collection vessels.

Step 5: Test the fractions.

Choose a test that reveals the missing component.

Examples include:

• Re-filtering a cloudy liquid to detect suspended solid particles.
• Evaporating a liquid fraction to detect dissolved non-volatile solids.
• Rinsing glassware to recover material adhering to its surfaces.
• Analysing a mother liquor to determine how much solute remains dissolved.

Step 6: Calculate a separate mass balance for each component.

Only after accounting for every fraction should you calculate the remaining unaccounted mass.

Experimental measurements that can mislead you

There are three particularly important limitations.

1. Wet products can give artificially high masses.

A filter residue may contain water even after it appears dry.

If you weigh it before drying to constant mass, you may mistake retained water for recovered product.

Dry the material, cool it, and weigh it again. Repeat until successive measurements show no meaningful change.

When using filter paper, subtract the original mass of the dry filter paper from the combined mass of the paper and residue.

2. Total mass recovery does not prove product purity.

You might recover the expected total mass but still have salt contaminating the sand, or sand contaminating the salt.

You need to consider both the mass and composition of each fraction.

3. A mass deficit does not identify a particular apparatus fault.

For example, if 0.10 g of sand is found in a beaker rinse, you know that sand remained on the apparatus.

However, you cannot necessarily distinguish between incomplete transfer from the original beaker and particles adhering to the stirring rod.

The mass balance identifies where the material was recovered, not always the exact physical event that caused it to remain there.

Also consider measurement uncertainty. Small differences between initial and recovered masses may result from balance precision, transfer uncertainty, or incomplete drying. Do not claim an exact loss mechanism when the measurements cannot support it.

For an ordinary physical separation, assume the substances retain their chemical identities. If heating causes decomposition or a chemical reaction, you must account for the resulting substances as well.

07Questions and solutions

The following questions move from basic recovery calculations to more involved investigations. Pay attention to the distinction between material missing from the intended product and material genuinely unaccounted for.

Question 1

A student filters a suspension containing 1.50 g of an insoluble solid in water.

After drying, the residue on the primary filter paper has a mass of 1.38 g.

The filtrate is passed through a second, finer filter paper. After drying, this second filter contains 0.07 g of solid.

Assume the solid does not dissolve in water and all filter-paper masses have been corrected.

Calculate:

a) The mass of solid that remains unaccounted for.

b) The percentage of the original solid recovered by the primary filter.

c) Whether the second filter explains the entire original deficit.

Solution 1

The unaccounted mass is 0.05 g, and the primary filter recovers 92.0% of the original solid. The second filter identifies some, but not all, of the initial deficit.

Step 1: Calculate the total mass of solid recovered.

The primary filter contains 1.38 g, while the second filter contains 0.07 g.

\[ m_{\mathrm{recovered}}=1.38+0.07 \]

\[ =1.45\text{ g} \]

Step 2: Calculate the unaccounted mass.

\[ m_{\mathrm{unaccounted}} = 1.50-1.45 \]

\[ =0.05\text{ g} \]

Therefore, 0.05 g remains unaccounted for.

Step 3: Calculate the primary filter recovery.

\[ \%\text{ recovery} = \frac{1.38}{1.50}\times100 \]

\[ =92.0\% \]

Step 4: Interpret the second filtration.

The second filter caught 0.07 g, showing that some solid passed through the primary filter.

However, another 0.05 g remains unaccounted for.

It may have remained on the original beaker or stirring rod, adhered to the funnel, or been lost during transfer.

The second filtration identifies one location of material loss, but not the entire deficit.

Question 2

A student dissolves 5.00 g of potassium nitrate in 10.0 g of water at 60°C.

The solution is cooled to 20°C and filtered to collect the crystals.

For this question, use an illustrative solubility of 32.0 g of potassium nitrate per 100 g of water at 20°C.

The crystals are then washed with 2.00 g of fresh water at 20°C.

Assume:

• All the potassium nitrate initially dissolves.
• The original mother liquor is completely removed before washing.
• Equilibrium is reached during cooling and washing.
• No water evaporates.
• No crystals are mechanically lost.

Calculate:

a) The mass of crystals formed after cooling.

b) The mass of crystals remaining after washing.

c) The total mass of potassium nitrate remaining in the mother liquor and washings.

Explain why the final crystal mass is much lower than the initial potassium nitrate mass.

Solution 2

The student initially forms 1.80 g of crystals, but only 1.16 g remains after washing. The remaining 3.84 g is dissolved in the mother liquor and washings.

Step 1: Calculate the mass of potassium nitrate remaining dissolved after cooling.

At 20°C, the solubility is 32.0 g per 100 g of water.

The original solution contains 10.0 g of water.

\[ m_{\mathrm{dissolved}} = \frac{32.0}{100}\times10.0 \]

\[ =3.20\text{ g} \]

Step 2: Calculate the mass of crystals formed.

The initial potassium nitrate mass was 5.00 g.

\[ m_{\mathrm{crystals}} = 5.00-3.20 \]

\[ =1.80\text{ g} \]

Therefore, 1.80 g of crystals form during cooling.

Step 3: Calculate how much potassium nitrate dissolves during washing.

The wash water has a mass of 2.00 g.

Using the same solubility:

\[ m_{\mathrm{dissolved}} = \frac{32.0}{100}\times2.00 \]

\[ =0.64\text{ g} \]

The wash water dissolves 0.64 g of potassium nitrate.

Step 4: Calculate the final crystal mass.

\[ m_{\mathrm{final\ crystals}} = 1.80-0.64 \]

\[ =1.16\text{ g} \]

Step 5: Account for the dissolved potassium nitrate.

The original mother liquor contains 3.20 g.

The washings contain another 0.64 g.

\[ m_{\mathrm{liquid\ fractions}} = 3.20+0.64 \]

\[ =3.84\text{ g} \]

The complete mass balance is:

\[ 5.00=1.16+3.84 \]

The low crystal recovery is caused by the solubility of potassium nitrate, not necessarily by a fault in the filtration apparatus.

This example also demonstrates why using excessive wash water can reduce the yield of a soluble crystalline product.

Question 3

A student performs simple distillation on a salt solution containing:

• 8.00 g of sodium chloride.
• 42.0 g of water.

After distillation, the following results are obtained:

Assume no chemical reaction occurs and no sodium chloride is detected in the condenser or receiver rinse.

Calculate:

a) The unaccounted mass of water.

b) The unaccounted mass of sodium chloride.

c) Where you would investigate each deficit.

Explain why the sodium chloride deficit should not be attributed to evaporation under ordinary distillation conditions.

Solution 3

The unaccounted masses are 1.2 g of water and 0.10 g of sodium chloride. The two deficits require different investigations because water is volatile under the experimental conditions, whereas sodium chloride is effectively non-volatile.

Step 1: Calculate the recovered water mass.

The collected distillate contains 39.2 g of water.

The condenser and receiver rinse contains another 1.6 g.

\[ m_{\mathrm{water,recovered}} = 39.2+1.6 \]

\[ =40.8\text{ g} \]

Step 2: Calculate the unaccounted water mass.

\[ m_{\mathrm{water,unaccounted}} = 42.0-40.8 \]

\[ =1.2\text{ g} \]

The missing water may remain in the boiling flask, still head, or condenser. It may also have escaped as vapour through an imperfect connection or failed to condense completely.

The measurements alone cannot distinguish these possibilities.

The student should inspect the joints, check that the condenser has adequate cooling, and rinse the remaining apparatus sections separately.

Step 3: Calculate the recovered sodium chloride mass.

The boiling flask contains 7.70 g of dry sodium chloride.

The flask and still-head rinse produces another 0.20 g.

\[ m_{\mathrm{NaCl,recovered}} = 7.70+0.20 \]

\[ =7.90\text{ g} \]

Step 4: Calculate the sodium chloride deficit.

\[ m_{\mathrm{NaCl,unaccounted}} = 8.00-7.90 \]

\[ =0.10\text{ g} \]

This material cannot reasonably be explained by sodium chloride evaporating under ordinary distillation conditions.

The student should investigate splashing or bumping in the boiling flask, solid deposits in other parts of the apparatus, and material lost during transfer.

Step 5: Compare the two types of loss.

Water can travel through the apparatus as vapour, so an incomplete condensation or leak is a plausible explanation for its deficit.

Sodium chloride should remain in the liquid or solid residue. Its deficit therefore points towards physical transfer losses or material remaining on surfaces.

The same mass-balance method applies to both components, but the physical properties determine which explanations are reasonable.

Question 4

A separating funnel contains 20.0 g of an immiscible organic liquid A and 30.0 g of water.

Liquid A is less dense than water.

The student separates the mixture and collects:

A composition test shows that the interface contains:

• 0.9 g of water.
• 0.6 g of organic liquid A.

Calculate:

a) The unaccounted mass of water.

b) The unaccounted mass of organic liquid A.

c) The total unaccounted mass.

Explain why it would be incorrect to conclude that all the missing material was trapped in the interface.

Solution 4

The unaccounted masses are 0.3 g of water and 0.2 g of organic liquid A, giving a total unaccounted mass of 0.5 g.

Step 1: Identify the position of each liquid.

Organic liquid A is less dense than water.

Therefore:

• Water forms the lower layer.
• Organic liquid A forms the upper layer.

The student should drain the lower aqueous layer through the tap first, then collect the upper organic layer separately.

Step 2: Calculate the unaccounted water mass.

The initial water mass was 30.0 g.

The collected aqueous layer contains 28.8 g, while the interface contains 0.9 g.

\[ m_{\mathrm{water,unaccounted}} = 30.0-28.8-0.9 \]

\[ =0.3\text{ g} \]

Step 3: Calculate the unaccounted organic liquid mass.

The initial organic liquid mass was 20.0 g.

The collected upper layer contains 19.2 g, while the interface contains 0.6 g.

\[ m_{\mathrm{A,unaccounted}} = 20.0-19.2-0.6 \]

\[ =0.2\text{ g} \]

Step 4: Calculate the total unaccounted mass.

\[ m_{\mathrm{unaccounted}} = 0.3+0.2 \]

\[ =0.5\text{ g} \]

Step 5: Explain the likely location.

The interface contains 1.5 g, but its composition accounts for only 0.9 g of water and 0.6 g of organic liquid.

The remaining 0.5 g cannot also be assigned to the interface.

It may remain on the funnel walls, around the tap, or in the connecting stem. A spill during transfer is another possibility.

The appropriate next investigation is to rinse the funnel and tap into separate collection vessels and determine the composition of each rinse.

The key lesson is that knowing the total mass of an uncollected fraction is not enough. You need to know which components it contains.

Question 5

A student separates a dry mixture containing 2.00 g of sand and 3.00 g of sodium chloride using water, filtration, and evaporation.

After the first filtration, the student records the following masses:

• Unwashed filter cake: 2.12 g.
• Dry solid obtained by evaporating the main filtrate: 2.70 g.

The student calculates:

\[ 5.00-(2.12+2.70)=0.18\text{ g} \]

They conclude that only 0.18 g of the original mixture has been lost.

A second student questions this conclusion and performs additional measurements.

The following results are obtained:

Assume:

• No chemical reaction occurs.
• No material is spilled.
• The products do not become contaminated.
• All stated masses are measured accurately.
• The filter-cake washings, main filtrate, and apparatus rinses are separate fractions.
• The original filter cake contains only sand, sodium chloride, and the stated retained water.

Answer the following:

a) Explain why the first student’s mass-deficit calculation is misleading.

b) Calculate the total mass of sand recovered across all fractions.

c) Calculate the total mass of sodium chloride recovered across all fractions.

d) Identify where each component was retained during the experiment.

e) Does complete mass accounting prove that the separation was successful? Explain.

Solution 5

The first student’s conclusion is incorrect. The 0.18 g deficit is artificially small because the unwashed filter cake contains 0.20 g of water. After correcting for moisture, 0.38 g of the original solid material is absent from the two initially weighed fractions. The additional investigations locate this material, and the complete component-by-component mass balance accounts for all 5.00 g.

Step 1: Determine the actual composition of the original filter cake.

The unwashed filter cake has a mass of 2.12 g.

Drying removes 0.20 g of retained water.

Therefore, the dry solid mass is:

\[ m_{\mathrm{dry\ cake}} = 2.12-0.20 \]

\[ =1.92\text{ g} \]

However, this 1.92 g is not entirely sand.

After washing and drying, only 1.82 g of sand remains.

The remaining solid mass must therefore have been sodium chloride:

\[ m_{\mathrm{NaCl\ in\ cake}} = 1.92-1.82 \]

\[ =0.10\text{ g} \]

This agrees with the 0.10 g of sodium chloride recovered from the filter-cake washings.

The original filter cake therefore contained:

• 1.82 g of sand.
• 0.10 g of sodium chloride.
• 0.20 g of water.

Step 2: Calculate the total sand recovery.

Sand was found in three locations:

• Primary filter cake: 1.82 g.
• Second filter: 0.08 g.
• Apparatus rinse: 0.10 g.

Therefore:

\[ m_{\mathrm{sand,total}} = 1.82+0.08+0.10 \]

\[ =2.00\text{ g} \]

All the original sand has been accounted for.

The second filter shows that 0.08 g passed through the primary filter.

The apparatus rinse shows that another 0.10 g remained on the beaker, stirring rod, or funnel.

Step 3: Calculate the total sodium chloride recovery.

Sodium chloride was found in three locations:

• Main filtrate: 2.70 g.
• Filter-cake washings: 0.10 g.
• Apparatus-rinse filtrate: 0.20 g.

Therefore:

\[ m_{\mathrm{NaCl,total}} = 2.70+0.10+0.20 \]

\[ =3.00\text{ g} \]

All the original sodium chloride has also been accounted for.

The main filtrate contains most of the salt, but some remained dissolved in the liquid trapped by the filter cake and some remained on the apparatus.

Step 4: Explain the misleading original calculation.

The first student added:

\[ 2.12+2.70=4.82\text{ g} \]

This included 0.20 g of water that was not part of the original dry mixture.

The actual solid mass in those two initial collection fractions was:

\[ 4.82-0.20=4.62\text{ g} \]

Therefore, the solid material absent from those fractions was:

\[ 5.00-4.62=0.38\text{ g} \]

This 0.38 g is accounted for by:

\[ 0.08+0.10+0.20=0.38\text{ g} \]

These amounts represent sand passing through the primary filter, sand retained on the apparatus, and sodium chloride retained on the apparatus.

The 0.10 g of sodium chloride in the filter cake is already included in the original 2.12 g cake measurement. Washing simply transfers it into a separately measured fraction. It must not be counted twice.

Step 5: Evaluate whether complete mass accounting proves successful separation.

No. Complete mass accounting proves that the original components have been located, but it does not prove that the intended products are pure or that recovery was high.

The intended pure sand product contains 1.82 g, while the intended sodium chloride product contains 2.70 g.

Their combined mass is:

\[ 1.82+2.70=4.52\text{ g} \]

The intended product recovery is therefore:

\[ \frac{4.52}{5.00}\times100 =90.4\% \]

The remaining 9.6% is accounted for in the other fractions.

This is the crucial distinction: a complete mass balance tells you where the material is, while product recovery and purity tell you how successful the separation was.

08What to investigate next

Once you can identify where each component went, the next useful step is understanding how removing one component changes the composition of the remaining mixture.

For example, removing water from a salt solution changes the percentage composition of the remaining solution, even though the mass of sodium chloride may stay constant.

See How to Calculate Mixture Composition After Removing a Component for the calculations that follow from a correctly understood separation and mass balance.