Chemistry 5070/31 — October/November 2016

Cambridge O-Level · Practical Test · worked solutions for every part, with the mark scheme

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Topics Experimental Contexts · Analysis, Conclusions and Evaluation · Observations and Measurements · Qualitative Analysis

Q117MObservations and MeasurementsExperimental ContextsAnalysis, Conclusions and EvaluationFree sample

Mangalloy is made by alloying steel with manganese. The percentage by mass of manganese in mangalloy can be determined by converting all the manganese in the alloy into aqueous manganate(VII) ions and then using the solution to titrate aqueous iron(II) sulfate.

No indicator is needed for this titration as the products of the reaction are almost colourless and one drop of aqueous manganate(VII) ions in excess produces a permanent pale pink colour.

P is an aqueous solution of manganate(VII) ions, $\text{MnO}_4^-$ . The solution was prepared by converting all the manganese in a $2.12\text{ g}$ sample of mangalloy into manganate(VII) ions and making the final volume up to $250\text{ cm}^3$ by adding water.

Q is $0.0800\text{ mol / dm}^3$ iron(II) sulfate.

(a)

Put P into the burette.

The colour of P makes it difficult to see the bottom of the meniscus so you should take all your readings using the top of the meniscus.

Pipette a $25.0\text{ cm}^3$ (or $20.0\text{ cm}^3$ ) portion of Q into a flask.

Add P from the burette. At first the purple colour disappears quickly but as more P is added the colour disappears less quickly. At the end-point, one drop of P produces a pale pink colour that does not disappear on swirling.

Record your results in the table, repeating the titration as many times as you consider necessary to achieve consistent results.

Results

Burette readings

titration number	1	2
final reading / $\text{cm}^3$
initial reading / $\text{cm}^3$
volume of P used / $\text{cm}^3$
best titration results (✓)

Summary

Tick (✓) the best titration results.

Using these results, the average volume of P required was ______ $\text{cm}^3$ .

Volume of Q used was ______ $\text{cm}^3$ .

12M

DifficultyMedium-Easy

Worked solution

Answer

Record all initial and final readings to one decimal place, e.g. $0.0$ , $19.8$ .
For each titration, titre $=$ final reading $-\text{ initial reading}$ .
Tick the two or more concordant titres that agree within $0.2\text{ cm}^3$ .
Average volume of P $=$ mean of the ticked titres.
Volume of Q used $= 25.0\text{ cm}^3$ (or $20.0\text{ cm}^3$ ).

Final answer

Completed results table with readings to 1 decimal place, concordant titres ticked, and average titre calculated.

Detailed explanation

Walkthrough

This is a titration where the purple manganate(VII) solution is placed in the burette. Because P is dark, read the top of the meniscus, not the bottom. For each titration record the initial and final burette readings to one decimal place, then work out the titre by subtracting the initial reading from the final reading. Repeat the titration until at least two titres are concordant, i.e. within $0.2\text{ cm}^3$ of each other. Tick those best results and average only those ticked values. The volume of Q is the pipette volume, either $25.0\text{ cm}^3$ or $20.0\text{ cm}^3$ .

Key Takeaways

A burette is read to one decimal place. A titre is the difference between final and initial readings. Concordant results are those that agree closely, and the average is taken only from the ticked concordant results.

Common Mistakes

Writing burette readings to two decimal places or as whole numbers.
Forgetting to record an initial reading, or recording an initial reading of $50$ .
Averaging all titres instead of only the ticked concordant ones.
Making subtraction errors when calculating titres.
Leaving the table incomplete.

Things to Be Careful About

Because P is dark, read the top of the meniscus. Record a zero initial reading as $0.0$ , not $0$ . Reject final readings above $50$ and initial readings of $50$ . Concordance is judged on the uncorrected titres. The average volume of P is used in the later calculations.

Techniques used

record burette readings to one decimal placecalculate titre as final reading minus initial readingselect concordant resultsaverage the selected titres

(b)

Q is $0.0800\text{ mol / dm}^3$ iron(II) sulfate.

Calculate the number of moles of iron(II) sulfate present in the volume of Q used.

number of moles of iron(II) sulfate = ______

DifficultyEasy

Worked solution

Working

\text{moles of FeSO}_4 = \frac{\text{volume of Q in cm}^3}{1000} \times 0.0800

For $25.0\text{ cm}^3$ of Q:

\frac{25.0}{1000} \times 0.0800 = 0.00200\text{ mol}

Answer

$0.00200\text{ mol}$ (using $25.0\text{ cm}^3$ of Q)

Final answer

0.00200 mol (using 25.0 cm3 of Q)

Detailed explanation

Walkthrough

To find moles of iron(II) sulfate in the portion of Q, use $\text{moles} = \text{concentration} \times \text{volume in dm}^3$ . The concentration is $0.0800\text{ mol / dm}^3$ . The volume is the pipette volume, usually $25.0\text{ cm}^3$ , so convert to $\text{dm}^3$ by dividing by $1000$ : $25.0/1000 = 0.0250\text{ dm}^3$ . Then moles $= 0.0250 \times 0.0800 = 0.00200\text{ mol}$ . If the pipette volume was $20.0\text{ cm}^3$ , use $20.0/1000$ instead.

Key Takeaways

The core relationship is $\text{moles} = \text{concentration} \times \text{volume}$ . Volumes in $\text{cm}^3$ must be divided by $1000$ to become $\text{dm}^3$ .

Common Mistakes

Using the average volume of P instead of the volume of Q.
Forgetting to divide the volume in $\text{cm}^3$ by $1000$ .
Quoting the answer without the unit mol.

Things to Be Careful About

The volume of Q is the pipette volume, not a titre. Keep the answer to an appropriate number of significant figures; $0.00200$ has three significant figures, matching the concentration.

Techniques used

convert volume in cm3 to dm3calculate moles from concentration and volume

(c)

Using your answer from (b), calculate the number of moles of manganate(VII) ions present in the average volume of P required.
[Five moles of iron(II) sulfate react with one mole of manganate(VII) ions.]

number of moles of manganate(VII) ions = ______

DifficultyEasy

Worked solution

Working

\text{moles of MnO}_4^- = \frac{\text{answer to (b)}}{5}

For answer to (b) $= 0.00200\text{ mol}$ :

\frac{0.00200}{5} = 0.000400\text{ mol}

Answer

$0.000400\text{ mol}$

Final answer

0.000400 mol

Detailed explanation

Walkthrough

The reaction uses five moles of iron(II) sulfate for every one mole of manganate(VII) ions. So the moles of manganate(VII) in the titre is the moles of iron(II) sulfate divided by 5. If the answer to (b) is $0.00200\text{ mol}$ , then $0.00200/5 = 0.000400\text{ mol}$ .

Key Takeaways

A balanced mole ratio lets you convert moles of one reactant to moles of another. Here the ratio is 5:1, so divide by 5.

Common Mistakes

Multiplying by 5 instead of dividing.
Using the ratio the wrong way round.
Not using the answer from part (b).

Things to Be Careful About

Use the exact answer from (b), not a rounded value. The unit is mol.

Techniques used

apply the 5:1 mole ratio from the equation

(d)

Using your answer from (c), calculate the number of moles of manganate(VII) ions in $250\text{ cm}^3$ of P.

number of moles of manganate(VII) ions in $250\text{ cm}^3$ of P = ______

DifficultyMedium-Easy

Worked solution

Working

\text{moles in }250\text{ cm}^3 = \text{answer to (c)} \times \frac{250}{\text{average volume of P in cm}^3}

For average volume $= 19.8\text{ cm}^3$ :

0.000400 \times \frac{250}{19.8} = 0.00505\text{ mol}

Answer

$0.00505\text{ mol}$

Final answer

0.00505 mol

Detailed explanation

Walkthrough

The titre volume of P contains the moles found in (c). The whole $250\text{ cm}^3$ of solution P is larger by a factor of $250/\text{average titre}$ . Multiply the moles from (c) by this factor. With an average titre of $19.8\text{ cm}^3$ : $0.000400 \times 250/19.8 = 0.00505\text{ mol}$ .

Key Takeaways

Scaling a sample amount to a total volume uses the ratio total volume / sample volume.

Common Mistakes

Using $250/25.0$ instead of $250/\text{average volume of P}$ .
Forgetting to use the average volume of P.
Not using the answer from (c).

Things to Be Careful About

The average volume of P is in $\text{cm}^3$ , so the units cancel correctly. Keep enough significant figures before rounding.

Techniques used

scale moles from the titre volume to the total 250 cm3 volume

(e)

Using your answer from (d), calculate the mass of manganese in the $2.12\text{ g}$ sample of mangalloy.
[ $A_r$ : Mn, 55]

mass of manganese in $2.12\text{ g}$ of mangalloy = ______ $\text{g}$

DifficultyEasy

Worked solution

Working

\text{mass of Mn} = \text{answer to (d)} \times 55

0.00505 \times 55 = 0.278\text{ g}

Answer

$0.278\text{ g}$

Final answer

0.278 g

Detailed explanation

Walkthrough

Each manganate(VII) ion contains one manganese atom, so the moles of manganese equal the moles of manganate(VII) found in (d). Mass is moles times $A_r$ : $\text{mass} = \text{moles} \times 55$ . With $0.00505\text{ mol}$ : $0.00505 \times 55 = 0.278\text{ g}$ .

Key Takeaways

$\text{mass} = \text{moles} \times A_r$ . One mole of Mn atoms has a mass of $55\text{ g}$ .

Common Mistakes

Using the $M_r$ of $\text{MnO}_4^-$ instead of the $A_r$ of Mn.
Forgetting to multiply by 55.
Missing the unit g.

Things to Be Careful About

There is one Mn atom per $\text{MnO}_4^-$ ion, so the mole ratio is 1:1. Round to three significant figures at the end.

Techniques used

convert moles of manganate(VII) to mass of manganese using Ar

(f)

Using your answer from (e), calculate the percentage by mass of manganese in mangalloy.

percentage by mass of manganese in mangalloy = ______ %

DifficultyEasy

Worked solution

Working

\text{percentage by mass} = \frac{\text{answer to (e)}}{2.12} \times 100

\frac{0.278}{2.12} \times 100 = 13.1\%

Answer

$13.1\%$

Final answer

13.1%

Detailed explanation

Walkthrough

Percentage by mass is the mass of manganese divided by the mass of the alloy sample, multiplied by 100. Using $0.278\text{ g}$ of Mn in a $2.12\text{ g}$ sample: $0.278/2.12 \times 100 = 13.1\%$ .

Key Takeaways

$\text{percentage by mass} = (\text{mass of component} / \text{mass of sample}) \times 100$ .

Common Mistakes

Dividing by the wrong mass, such as the volume of solution.
Forgetting to multiply by 100.
Rounding intermediate values too early.

Things to Be Careful About

The sample mass is $2.12\text{ g}$ . Give the final percentage to three significant figures.

Techniques used

calculate percentage by massdivide mass of manganese by sample mass and multiply by 100

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