Measuring Mixture Composition by Mass: Method and Error Checks

A 6.00 g sample contains sand and table salt. You separate the mixture and recover 3.60 g of sand and 2.10 g of salt. Have you measured the composition correctly?

The two recovered masses add to 5.70 g, leaving 0.30 g unaccounted for. You could divide each mass by 5.70 g to make the percentages add to 100%, but that would hide the missing material.

Before reading on, predict: if the experiment recovered only 95% of the original sample, can you still calculate each component’s percentage of the original mixture? Yes. Divide each recovered component’s mass by the original sample mass, then use the total mass recovery to check how reliable the result is.

For a mixture of sand and sodium chloride, the practical method is to dissolve the salt in water, filter out the sand, rinse the sand, and dry and weigh both components. The calculations are straightforward. The difficult part is recognising when the experiment has produced misleading numbers.

01The key idea: separate by a physical property

Suppose you have a mixture of sand and sodium chloride, the chemical name for table salt. Both are solids, so you cannot separate them simply by pouring the mixture into another container.

However, they behave differently in water:

• Sodium chloride dissolves in water.
• Sand remains as an insoluble solid.

That difference gives you a way to separate them without needing a chemical reaction.

Imagine inviting two people to a party where one is happy to join the dancing and the other refuses to leave the doorway. Adding water gives the salt a way to disperse through the liquid, while the sand remains in a separate solid phase.

The analogy has limits, of course. Salt does not disappear when it dissolves. Its sodium and chloride ions disperse throughout the water, and the salt can be recovered by removing that water.

This method works when the mixture’s components have different physical properties, such as solubility, and do not react under the conditions used. If both components dissolve in water, ordinary filtration will not separate them.

For the calculations below, assume the original sample contains only sand and sodium chloride.

02The practical method, step by step

Step 1: Measure the original sample

Start with a clean, dry sample. Measure its mass before adding water or beginning the separation.

Call this mass \(m_0\), measured in grams.

If you use a balance with a tare function, you can zero the balance with the weighing boat in place. Otherwise, weigh the empty boat and subtract its mass from the mass of the boat plus sample.

For example, if the boat and mixture have a combined mass of 18.42 g and the empty boat has a mass of 12.42 g, then

\[ m_0 = 18.42 - 12.42 \]

\[ m_0 = 6.00\text{ g} \]

Record the original mass carefully. Every percentage calculated later will use this value as its denominator.

Step 2: Dissolve the salt

Transfer the mixture into a beaker and add a suitable amount of distilled water. Stir until the salt has dissolved.

Use enough water to dissolve all the salt, but avoid adding much more than necessary. Excess water does not change the amount of salt, but it increases the time needed to recover it.

Check that no visible salt crystals remain undissolved. Remember that clear liquid is not evidence that no salt is present: dissolved salt is generally invisible in the solution.

Step 3: Filter the mixture

Place filter paper in a funnel and position the funnel over a clean container.

Pour the mixture through the filter paper. The sand remains on the paper, while the salt solution passes through into the container. This liquid is called the filtrate.

Two details matter here.

First, transfer the entire sample. Sand left in the original beaker is sand you cannot weigh later.

Second, rinse the sand on the filter paper with small portions of distilled water. Also rinse any sand left in the original beaker into the filter.

The rinsing removes salt solution clinging to the sand and carries the dissolved salt into the filtrate. Combine the rinsings with the original filtrate so that the salt is not discarded.

This is a crucial step. If you skip it, some salt may end up being weighed as part of the sand.

Step 4: Dry and weigh the sand

The sand on the filter paper is wet. You cannot treat its current mass as the mass of sand because some of that mass is water.

Dry the sand and filter paper using an appropriate laboratory drying method. A drying oven is suitable when used according to your laboratory’s instructions. Do not use a naked flame to dry filter paper.

Weigh the dry filter paper before use, or use another method that allows you to subtract its mass. After drying the sand, cool the material before weighing it.

Suppose:

• Dry filter paper: 1.12 g
• Dry filter paper plus sand: 3.76 g

Then the recovered sand mass is

\[ m_{\text{sand}} = 3.76 - 1.12 \]

\[ m_{\text{sand}} = 2.64\text{ g} \]

The subtraction matters because the filter paper is not part of the original mixture.

Dry the material, cool it, and weigh it again. If successive measurements still decrease after further drying, water may remain. A useful endpoint is constant mass: successive measurements agree within the balance’s resolution after another drying and cooling cycle.

Step 5: Recover, dry, and weigh the salt

The filtrate contains the dissolved sodium chloride. Transfer it to a clean evaporating basin whose empty mass has been recorded.

Heat the solution gently to evaporate the water. Avoid vigorous boiling, which can cause liquid or salt to spit out of the basin.

Continue until the salt has been recovered, then ensure the product is dry and allow the basin to cool before weighing it. Use an appropriate heating method and follow your laboratory’s safety instructions.

Suppose:

• Empty evaporating basin: 42.35 g
• Basin plus dry salt: 45.17 g

Then

\[ m_{\text{salt}} = 45.17 - 42.35 \]

\[ m_{\text{salt}} = 2.82\text{ g} \]

Again, subtract the mass of the equipment. Otherwise, you would be counting the basin as part of the original mixture.

Wear safety glasses when heating solutions, and use suitable equipment to handle hot basins. Hot glassware can look identical to cool glassware.

Step 6: Check that both fractions are ready to weigh

Before calculating the composition, ask yourself:

• Have all the sand and salt been transferred into the appropriate containers?
• Was the sand rinsed to remove salt?
• Have both recovered fractions been dried adequately?
• Have the masses of the filter paper and evaporating basin been subtracted correctly?
• Were the materials cooled before their final measurements?

These checks matter because a balance measures the mass of whatever is on it. It cannot tell you whether that mass belongs entirely to the substance you intended to measure.

03Calculating the percentage composition

You now have the original sample mass and the recovered mass of each component.

The percentage of the original sample represented by a component is its recovered mass divided by the original sample mass, multiplied by 100.

For sand,

\[ \%\text{ sand} = \frac{m_{\text{sand}}}{m_0}\times100 \]

For salt,

\[ \%\text{ salt} = \frac{m_{\text{salt}}}{m_0}\times100 \]

Here, \(m_{\text{sand}}\) and \(m_{\text{salt}}\) are the recovered component masses in grams, and \(m_0\) is the original mixture mass in grams.

Notice that both calculations use the same original mass as the denominator.

The mass-recovery check

The mass recovery percentage tells you how much of the original sample has been accounted for by the recovered components.

For a two-component mixture,

\[ \%\text{ recovery} = \frac{m_{\text{sand}}+m_{\text{salt}}}{m_0} \times100 \]

If the separation is complete, the fractions are pure, and all the solids are dry, the recovered masses should add to the original sample mass, within experimental uncertainty.

Ideally, the mass recovery is 100%.

In practice, it may be lower because some material is lost. It may also be higher if the recovered solids contain water or other contaminants, or if there is a weighing error.

For a sample containing only sand and salt, the sum of the two calculated component percentages will equal the measured mass-recovery percentage, provided the same original mass is used throughout.

For example, if the sand accounts for 58.1% of the original mass and the salt accounts for 39.0%, the percentages sum to 97.1%. That is a 97.1% mass recovery, not a 100% recovery.

Do not force the percentages to add to 100% by dividing by the total recovered mass. That calculates the proportions of the recovered material, not the percentages of the original sample.

04Why the mass recovery can mislead you

A mass balance is a diagnostic tool, not a certificate that every step was correct.

Different experimental errors affect the result in different ways. Some reduce the recovered mass, some increase it, and others simply assign mass to the wrong component.

Incomplete separation: the fractions can be wrong even when the total is right

Suppose the entire sample is recovered, but some salt remains mixed with the sand.

The combined mass of the recovered fractions may still equal the original sample mass. However, the student has classified some salt as sand.

This creates two errors in opposite directions:

• The measured sand mass is too high.
• The measured salt mass is too low.

The total may still give 100% recovery because the salt has not disappeared. It has simply ended up in the wrong fraction.

That is why rinsing the sand matters, and why mass recovery alone cannot establish the accuracy of the composition.

Wet solids: water can disguise a loss

Suppose you recover 4.00 g of dry sand and 6.20 g of salt crystals that still contain water. The original sample weighed 10.00 g.

The measured recovery would be

\[ \frac{4.00+6.20}{10.00}\times100 =102\% \]

The result is impossible as a true recovery of the original dry solids. The extra mass must come from somewhere outside the original dry sample, such as retained water, contamination, or measurement error.

Drying the salt could bring its mass closer to the true value. But do not assume that every recovery above 100% is caused by water: incorrect tare measurements and contamination can produce the same pattern.

Transfer losses: a low recovery does not identify the missing component

If some sand is left in the beaker, the recovered sand mass decreases. If salt solution is spilled, the recovered salt mass decreases.

Both errors can produce a low mass recovery, but they do not necessarily affect the component percentages equally.

A result such as 96% recovery tells you that the measured fractions account for only 96% of the original mass. It does not tell you whether the missing material was sand, salt, or a combination of both.

The practical response is to inspect the separation and transfer steps rather than automatically adjusting the percentages.

05Questions and solutions

Question 1

A 6.40 g sample contains only two components, A and B. After separation and drying, 3.84 g of A and 2.24 g of B are recovered.

Calculate the percentage of the original sample represented by each component, the mass recovery, and the mass not recovered.

Solution 1

The recovered fractions represent 60.0% for A and 35.0% for B, with a mass recovery of 95.0% and 0.32 g unaccounted for.

Step 1: Calculate the component percentages.

\[ \begin{aligned} \%\text{ A}&=\frac{3.84}{6.40}\times100=60.0\%\\ \%\text{ B}&=\frac{2.24}{6.40}\times100=35.0\% \end{aligned} \]

Step 2: Calculate the mass recovery.

\[ \begin{aligned} m_{\text{recovered}}&=3.84+2.24\\ &=6.08\text{ g} \end{aligned} \]

\[ \%\text{ recovery} =\frac{6.08}{6.40}\times100 =95.0\% \]

Step 3: Find the unaccounted mass.

\[ m_{\text{unaccounted}}=6.40-6.08 =0.32\text{ g} \]

The two percentages sum to 95.0%, not 100%. That is consistent with the measured mass recovery. Do not normalise the percentages to conceal the missing mass.

Question 2

A student separates a 7.50 g mixture containing a soluble component A and an insoluble component B.

The student records:

• Dry filter paper: 0.95 g
• Dry filter paper plus component B: 4.21 g
• Empty evaporating basin: 28.60 g
• Basin plus dry component A: 31.36 g

The student reports that A makes up 45.8% of the original sample and B makes up 54.2%, calculated using the total recovered mass as the denominator.

Are these the correct percentages of the original sample? Calculate the correct measured percentages, the mass recovery, and the unaccounted mass.

Solution 2

No. The student’s percentages describe the proportions of the recovered material, not the percentages of the original sample. The correct measured percentages are 36.8% for A and 43.5% for B, with 80.3% mass recovery.

Step 1: Find the mass of component B.

\[ m_B=4.21-0.95=3.26\text{ g} \]

Step 2: Find the mass of component A.

\[ m_A=31.36-28.60=2.76\text{ g} \]

Step 3: Calculate the percentages using the original mass.

\[ \begin{aligned} \%\text{ A} &=\frac{2.76}{7.50}\times100\\ &=36.8\% \end{aligned} \]

\[ \begin{aligned} \%\text{ B} &=\frac{3.26}{7.50}\times100\\ &\approx43.5\% \end{aligned} \]

Step 4: Calculate the mass recovery.

\[ m_{\text{recovered}}=2.76+3.26=6.02\text{ g} \]

\[ \%\text{ recovery} =\frac{6.02}{7.50}\times100 \approx80.3\% \]

Step 5: Calculate the unaccounted mass.

\[ m_{\text{unaccounted}}=7.50-6.02=1.48\text{ g} \]

The student’s percentages were obtained by dividing by 6.02 g. This forces the recovered fractions to add to 100%, even though 1.48 g of the original sample has not been accounted for.

The correct denominator for percentages of the original sample is 7.50 g. The low recovery also indicates that the experimental method needs investigation before the measured percentages can be treated as reliable estimates of the original composition.

Question 3

A 5.00 g mixture contains only sand and sodium chloride. A student recovers 3.10 g of dry solid from the filter paper and 1.90 g of dry salt crystals from the filtrate.

The student forgot to rinse the sand. Independent evidence shows that 0.35 g of salt remained mixed with the recovered sand.

Assume all the original material was transferred, both fractions are dry, and there were no other losses. Calculate:

(a) The apparent percentages of sand and salt reported by the student.

(b) The true percentages of sand and salt in the original sample.

(c) The mass recovery.

Explain why the mass recovery does not reveal the separation error.

Solution 3

The student would report 62% sand and 38% salt, but the true composition is 55% sand and 45% salt. The mass recovery is still 100% because the salt was misclassified rather than lost.

Step 1: Calculate the apparent percentages.

The student treats the entire filter residue as sand:

\[ \%\text{ apparent sand} =\frac{3.10}{5.00}\times100 =62\% \]

The recovered crystals are treated as salt:

\[ \%\text{ apparent salt} =\frac{1.90}{5.00}\times100 =38\% \]

Step 2: Correct the sand mass.

The filter residue contains 0.35 g of salt. Therefore,

\[ m_{\text{sand}}=3.10-0.35 =2.75\text{ g} \]

The true sand percentage is

\[ \%\text{ sand} =\frac{2.75}{5.00}\times100 =55\% \]

Step 3: Correct the salt mass.

The salt that remained on the sand must be added to the recovered crystals:

\[ m_{\text{salt}}=1.90+0.35 =2.25\text{ g} \]

Therefore,

\[ \%\text{ salt} =\frac{2.25}{5.00}\times100 =45\% \]

The corrected percentages sum to 100%, as expected for the two-component sample.

Step 4: Check the mass recovery.

\[ \%\text{ recovery} =\frac{3.10+1.90}{5.00}\times100 =100\% \]

The recovery is perfect because all 5.00 g of material was collected. However, the fractions were not pure: some salt was counted as sand.

This demonstrates an important limitation of the mass-recovery check. It detects a total mass discrepancy, but it cannot detect every error in the allocation of mass between components.

Question 4

An 8.00 g mixture contains only sand and sodium chloride. After separation, the student measures:

• Dry sand: 3.60 g
• Recovered salt crystals: 4.70 g

A repeat drying test shows that the salt crystals contain 0.40 g of water.

Calculate the corrected salt mass, the measured percentages of the original sample represented by the dry sand and salt, and the corrected mass recovery. Explain why the initial mass recovery was misleading.

Solution 4

The corrected salt mass is 4.30 g. The dry sand and salt represent 45.0% and 53.75% of the original sample, respectively, and the corrected mass recovery is 98.75%.

Step 1: Remove the water contribution from the salt mass.

\[ m_{\text{salt}}=4.70-0.40 =4.30\text{ g} \]

Step 2: Calculate the component percentages.

For sand,

\[ \%\text{ sand} =\frac{3.60}{8.00}\times100 =45.0\% \]

For salt,

\[ \%\text{ salt} =\frac{4.30}{8.00}\times100 =53.75\% \]

These are the measured percentages represented by the corrected dry fractions.

Step 3: Calculate the corrected recovery.

\[ m_{\text{recovered}}=3.60+4.30 =7.90\text{ g} \]

\[ \%\text{ recovery} =\frac{7.90}{8.00}\times100 =98.75\% \]

The corrected unaccounted mass is

\[ m_{\text{unaccounted}}=8.00-7.90 =0.10\text{ g} \]

Step 4: Compare the initial and corrected recovery.

Before the water was removed, the measured total was

\[ 3.60+4.70=8.30\text{ g} \]

The apparent recovery was

\[ \frac{8.30}{8.00}\times100 =103.75\% \]

The initial recovery exceeded 100% because water was counted as salt. After correcting for the water, the recovery falls below 100%, revealing a 0.10 g mass deficit.

The original measurement contained two different effects: an excess mass from water and a shortfall in recovered dry solids. Correcting one does not automatically eliminate the other.

Question 5

A 10.00 g mixture contains only sand and sodium chloride. After separation, the student records the following results:

• The dry filter residue has a mass of 6.20 g, after subtracting the filter paper’s mass. It contains 0.50 g of sodium chloride that was not washed off the sand.
• The recovered salt crystals have a mass of 3.70 g, but they contain 0.20 g of water.
• During transfer, 0.30 g of dry sodium chloride was spilled and not recovered.

Assume all values are exact, the spilled material was entirely sodium chloride, and there were no other losses or contaminants.

Calculate the true mass and percentage of each original component. Then calculate the apparent mass recovery using the student’s recorded masses and explain why a recovery close to 100% is not enough to validate the result.

Solution 5

The original mixture was 57.0% sand and 43.0% sodium chloride. The apparent mass recovery is 99.0%, but that figure hides a combination of misclassification, retained water, and material loss.

Step 1: Determine the true sand mass.

The filter residue contains 6.20 g in total, including 0.50 g of sodium chloride. Therefore,

\[ m_{\text{sand}}=6.20-0.50 \]

\[ m_{\text{sand}}=5.70\text{ g} \]

Step 2: Determine the dry salt mass recovered as crystals.

The recorded crystal mass includes 0.20 g of water:

\[ m_{\text{salt, crystals}}=3.70-0.20 \]

\[ m_{\text{salt, crystals}}=3.50\text{ g} \]

Step 3: Reconstruct the original salt mass.

The original salt must include three contributions:

• 3.50 g recovered as dry crystals
• 0.50 g incorrectly retained with the sand
• 0.30 g spilled during transfer

Therefore,

\[ \begin{aligned} m_{\text{salt}}&=3.50+0.50+0.30\\ &=4.30\text{ g} \end{aligned} \]

Check the reconstructed original masses:

\[ 5.70+4.30=10.00\text{ g} \]

The total agrees with the original sample mass.

Step 4: Calculate the true percentage composition.

For sand,

\[ \%\text{ sand} =\frac{5.70}{10.00}\times100 =57.0\% \]

For sodium chloride,

\[ \%\text{ salt} =\frac{4.30}{10.00}\times100 =43.0\% \]

The true percentages sum to 100%, consistent with the known two-component mixture.

Step 5: Calculate the apparent mass recovery.

Using the student’s recorded masses without correcting the water or the salt retained with the sand,

\[ m_{\text{measured}}=6.20+3.70 =9.90\text{ g} \]

Thus,

\[ \%\text{ apparent recovery} =\frac{9.90}{10.00}\times100 =99.0\% \]

The recovery looks convincing, but consider the three errors.

First, 0.50 g of salt has been assigned to the sand fraction. This changes the apparent composition without changing the total measured mass.

Second, the 0.20 g of water increases the measured crystal mass.

Third, the 0.30 g of spilled salt decreases the recovered mass.

The water partly masks the transfer loss: the recorded masses are only 0.10 g below the original mass because the 0.20 g excess from water offsets part of the 0.30 g loss. Meanwhile, the salt left with the sand creates a separate error in the component percentages.

The tempting conclusion is that 99.0% recovery proves the separation was accurate. It does not. The experiment has nearly recovered the original total mass, but the individual fractions are contaminated and some material has been lost.

The next useful step is to test repeatability: repeat the separation, improve the rinsing and drying, and compare the measured compositions. A credible mass balance is a useful starting point for evaluating experimental uncertainty, not a substitute for checking the method itself.