Chemistry 9701/31 — May/June 2016

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation · Presentation of Data and Observations · Qualitative Analysis

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationFree sample

In this experiment you will determine the identity of the Group 2 metal, X, in the carbonate, $\text{XCO}_3$ . To do this you will react a known mass of $\text{XCO}_3$ with excess hydrochloric acid, $\text{HCl}$ , and measure the mass of carbon dioxide that is given off.

FA 1 is $\text{XCO}_3$ .
FA 2 is hydrochloric acid, $\text{HCl}$ .

(a)

Method

Weigh the stoppered tube containing FA 1 and record its mass.
Use the measuring cylinder to transfer $25\text{ cm}^3$ of FA 2 into the $250\text{ cm}^3$ beaker.
Weigh the beaker containing the acid and record the mass.
Carefully add all the sample of FA 1 to the acid in the beaker.
Stir the mixture until there is no further reaction.
Reweigh the beaker and its contents and record the mass.

KEEP THE CONTENTS OF THE BEAKER FOR USE IN QUESTION 2.

Reweigh the stoppered tube containing any residual FA 1 and record its mass.
Calculate the mass of FA 1 added to the acid and record this value.
Calculate the mass of carbon dioxide given off and record this value.

DifficultyMedium-Easy

Worked solution

Answer

Record all six masses clearly in a table. Give every quantity a heading and write the unit (g) with the heading or beside each value.

mass of stoppered tube + FA 1 (before reaction)
mass of beaker + acid
mass of beaker + contents after reaction
mass of stoppered tube + residual FA 1
mass of FA 1 added (calculated)
mass of $\text{CO}_2$ evolved (calculated)

Record the four measured masses to the same number of decimal places (at least 1 decimal place).

Calculate:

mass of FA 1 added = mass of stoppered tube + FA 1 – mass of stoppered tube + residual FA 1
mass of $\text{CO}_2$ evolved = mass of beaker + acid – mass of beaker + contents after reaction

Final answer

Candidate-dependent: four measured masses with headings, units and consistent decimal places, plus correctly calculated mass of FA 1 added and mass of CO2 evolved.

Detailed explanation

Background Concept

This experiment uses the mass-loss method. When the metal carbonate $\text{XCO}_3$ reacts with excess hydrochloric acid, carbon dioxide gas escapes from the open beaker, so the total mass of the beaker and its contents falls. The fall in mass is taken as the mass of $\text{CO}_2$ given off. The mass of solid actually added is found by weighing the sample tube before and after adding it; this is called weighing by difference and avoids errors from transferring the solid.

Understanding the Question

This part is the practical instruction. There is no single numerical answer to write on the paper because the masses are yours. The marks are awarded for how well you record and process your own readings: six identifiable masses, clear headings with units, consistent decimal places, and correct calculation of the two derived masses.

Approach

Use a table with columns such as “quantity” and “mass / g”. Record the two tube masses and the two beaker masses. Keep the same number of decimal places throughout. Then calculate the mass of FA 1 added by subtracting the final tube mass from the initial tube mass, and the mass of $\text{CO}_2$ evolved by subtracting the final beaker mass from the initial beaker mass.

Step-by-Step Reasoning

Weigh the stoppered tube containing FA 1 and record the mass. This is the initial tube mass.
Weigh the beaker containing the acid and record the mass.
After adding FA 1 and stirring, weigh the beaker and contents again. This is the final beaker mass.
Weigh the stoppered tube containing any residual solid. This is the final tube mass.
The mass of FA 1 added is initial tube mass – final tube mass.
The mass of $\text{CO}_2$ evolved is initial beaker mass – final beaker mass.
The six entries are the four measured masses and the two calculated masses. All headings must be unambiguous and the unit g must be present. All measured masses should be quoted to the same precision, for example 21.45 g, 36.92 g, 36.56 g, 19.83 g.

Key Takeaways

Weighing by difference allows you to know exactly how much solid reacted without transferring all of it. The mass of gas evolved is measured as a loss in mass of the open container. Good data presentation requires headings, units and consistent decimal places.

Common Mistakes

Forgetting to include units or writing the unit only once in a heading so that a column is ambiguous.
Writing one mass to 1 decimal place and another to 2 decimal places.
Calculating the mass of FA 1 added using the beaker masses instead of the tube masses.
Taking the mass of $\text{CO}_2$ as the change in the tube mass.

Things to Be Careful About

The mass of the beaker and contents decreases because $\text{CO}_2$ escapes wait, but any acid spray also escapes and contributes to that loss; this is exactly the error discussed in part (c). For part (a), simply perform the calculation as instructed: mass loss of the beaker equals the mass of carbon dioxide given off.

Techniques used

record raw masses with clear headings and unitsweigh by difference to find the mass of solid addedcalculate the mass of carbon dioxide evolved from mass losskeep all measured masses to the same number of decimal places

(b)

Calculations

Show your working and appropriate significant figures in the final answer to each step of your calculations.

(i)

Calculate the number of moles of carbon dioxide given off when $\text{XCO}_3$ reacted with the acid.
Use the data in the Periodic Table on page 16.

\text{moles of CO}_2 = \text{......................... mol}

DifficultyEasy

Worked solution

Working

$M_r(\text{CO}_2) = 12.0 + 2(16.0) = 44.0$

n(\text{CO}_2) = \frac{\text{mass of CO}_2}{44.0}

Substitute the mass of $\text{CO}_2$ you obtained in part (a), in g.

Answer

$n(\text{CO}_2) = \dfrac{\text{mass of CO}_2}{44.0}$ mol

Final answer

n(CO2) = mass of CO2 / 44.0 mol

Detailed explanation

Background Concept

The number of moles of a substance is calculated using
$n = \frac{m}{M}$
where $m$ is the mass in g and $M$ is the molar mass in g mol $^{-1}$ . For carbon dioxide, the relative molecular mass is found by adding the relative atomic masses of one carbon atom and two oxygen atoms: 12.0 + 16.0 + 16.0 = 44.0.

Understanding the Question

You have already measured the mass of carbon dioxide evolved. This part asks you to convert that measured mass into an amount in moles so that it can be linked to the amount of $\text{XCO}_3$ in the next parts.

Approach

Write down the formula $n = m/M$ , substitute 44.0 for the molar mass of $\text{CO}_2$ , and divide your measured mass by 44.0. Give your answer to a sensible number of significant figures, usually the same as the mass reading.

Step-by-Step Reasoning

Identify the mass of $\text{CO}_2$ from part (a).
Use $M(\text{CO}_2) = 44.0$ g mol $^{-1}$ .
Calculate $n(\text{CO}_2) = m(\text{CO}_2) / 44.0$ .
Include the unit mol in your final answer.
For example, if the mass of $\text{CO}_2$ were 0.42 g, then $n = 0.42/44.0 = 0.0095$ mol. Your own value will differ.

Key Takeaways

Moles link a measured mass to the stoichiometry of a reaction. The molar mass is obtained from the Periodic Table.

Common Mistakes

Using 44 or forgetting to multiply the oxygen by 2.
Forgetting the unit mol.
Quoting too many significant figures, such as 0.00954545..., when the mass has only two or three significant figures.

Things to Be Careful About

Use the mass of carbon dioxide, not the mass of FA 1 added. Make sure the mass is in grams before dividing.

Techniques used

calculate moles from mass using molar massuse the relative molecular mass of carbon dioxide

(ii)

Write the equation for the reaction of FA 1, $\text{XCO}_3$ , with hydrochloric acid, $\text{HCl}$ . Include state symbols.

DifficultyEasy

Worked solution

Answer

\text{XCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{XCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})

Final answer

XCO3(s) + 2HCl(aq) -> XCl2(aq) + H2O(l) + CO2(g)

Detailed explanation

Background Concept

A metal carbonate reacts with an acid to produce a salt, water and carbon dioxide. Here X is a Group 2 metal, so it forms a 2+ ion, X $^{2+}$ , and the salt formed is XCl $_2$ . The acid is hydrochloric acid, HCl.

Understanding the Question

You need to write a fully balanced symbol equation for the reaction of solid $\text{XCO}_3$ with aqueous HCl, including state symbols. The equation is then used to relate the moles of $\text{CO}_2$ to the moles of $\text{XCO}_3$ in part (iii).

Approach

Start with the general pattern: carbonate + acid $\rightarrow$ salt + water + carbon dioxide. Substitute the specific salt XCl $_2$ , then balance the chloride ions and hydrogen atoms. Add state symbols last.

Step-by-Step Reasoning

Write the reactants: $\text{XCO}_3(\text{s}) + \text{HCl}(\text{aq})$ .
Write the products: $\text{XCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})$ .
Balance the chlorides: two HCl are needed because XCl $_2$ contains two Cl $^-$ ions.
Check hydrogen and oxygen balance: two HCl provide two H atoms, which form one H $_2$ O; the carbonate provides the oxygen for CO $_2$ and H $_2$ O.
Add state symbols: solid carbonate, aqueous acid, aqueous salt, liquid water and gaseous carbon dioxide.

Key Takeaways

Metal carbonate + acid is a general reaction: carbonate + acid $\rightarrow$ salt + water + carbon dioxide. A Group 2 metal always forms a 2+ ion, so the chloride is XCl $_2$ .

Common Mistakes

Writing the salt as XCl instead of XCl $_2$ .
Forgetting to balance the HCl.
Omitting state symbols.
Writing H $_2$ CO $_3$ as a product instead of H $_2$ O and CO $_2$ .

Things to Be Careful About

State symbols are required. The acid is aqueous, the carbonate is solid, the salt is aqueous, water is liquid and carbon dioxide is a gas.

Techniques used

write the balanced equation for a carbonate reacting with acidassign correct state symbols to reactants and products

(iii)

Use your answers to (i) and (ii) to calculate the number of moles of $\text{XCO}_3$ that were added to the acid.

\text{moles of XCO}_3 = \text{............................ mol}

DifficultyEasy

Worked solution

Working

From the equation in (ii), 1 mol $\text{XCO}_3$ produces 1 mol $\text{CO}_2$ .

n(\text{XCO}_3) = n(\text{CO}_2)

Answer

$n(\text{XCO}_3) = \dfrac{\text{mass of CO}_2}{44.0}$ mol

Final answer

n(XCO3) = n(CO2) = mass of CO2 / 44.0 mol

Detailed explanation

Background Concept

The balanced equation in part (ii) shows the stoichiometric relationship between reactants and products. The coefficient in front of $\text{XCO}_3$ and $\text{CO}_2$ are both 1, so one mole of $\text{XCO}_3$ gives one mole of $\text{CO}_2$ .

Understanding the Question

You have calculated the moles of carbon dioxide in part (i). The reaction stoichiometry lets you state the moles of $\text{XCO}_3$ that reacted without any further measurement.

Approach

Look at the coefficients in the balanced equation. Since both $\text{XCO}_3$ and $\text{CO}_2$ have coefficient 1, the amount of $\text{XCO}_3$ equals the amount of $\text{CO}_2$ .

Step-by-Step Reasoning

From part (i), you know $n(\text{CO}_2)$ .
From part (ii), the ratio $n(\text{XCO}_3):n(\text{CO}_2)$ is 1:1.
Therefore $n(\text{XCO}_3) = n(\text{CO}_2)$ .
Substitute the expression from part (i) if needed.

Key Takeaways

The balanced equation is the key to converting moles of one species into moles of another. A 1:1 ratio means the same numerical amount.

Common Mistakes

Using a 2:1 ratio because of the 2HCl in the equation.
Trying to use the mass of FA 1 here instead of the moles of CO $_2$ .

Things to Be Careful About

The 2 in front of HCl does not affect the XCO $_3$ :CO $_2$ ratio. Only the coefficients of XCO $_3$ and CO $_2$ matter.

Techniques used

use stoichiometric ratio to convert moles of CO2 to moles of XCO3apply the 1:1 mole ratio from the balanced equation

(iv)

Use your answer to (iii) to calculate the relative atomic mass, $A_r$ , of X.
Identify X.

\begin{aligned} A_r\text{ of X} &= \text{............................} \\ \text{X is } &\text{..........................} \end{aligned}

DifficultyMedium-Easy

Worked solution

Working

From (iii), $n(\text{XCO}_3) = n(\text{CO}_2)$ .

M_r(\text{XCO}_3) = \frac{\text{mass of FA 1 added}}{n(\text{XCO}_3)}

The carbonate group, $\text{CO}_3$ , has $M_r = 12.0 + 3(16.0) = 60.0$ .

A_r(\text{X}) = M_r(\text{XCO}_3) - 60.0

Compare this value with:
Be 9.0, Mg 24.3, Ca 40.1, Sr 87.6, Ba 137.3.

Answer

X is the Group 2 metal whose $A_r$ is closest to the calculated value.

Final answer

Candidate-dependent: X is the Group 2 metal whose A_r is closest to the calculated value (Be, Mg, Ca, Sr or Ba).

Detailed explanation

Background Concept

The molar mass of a compound is the mass of one mole of its molecules or formula units. Since the mass of the sample and the number of moles of $\text{XCO}_3$ are both known, the molar mass can be found from $M = m/n$ . The carbonate group $\text{CO}_3$ contributes 60.0 to this molar mass, so subtracting 60.0 gives the relative atomic mass of X.

Understanding the Question

This part uses all the previous results to identify the unknown Group 2 metal. You know the mass of FA 1 added and the number of moles of $\text{XCO}_3$ , so you can calculate $M_r(\text{XCO}_3)$ , then obtain $A_r(\text{X})$ and match it to one of the Group 2 elements.

Approach

Use the formula $M_r = m/n$ with the mass of FA 1 and the moles from part (iii). Subtract 60.0 for the carbonate group. Then compare the result with the known $A_r$ values of the Group 2 metals and select the closest.

Step-by-Step Reasoning

Use $n(\text{XCO}_3)$ from part (iii).
Use the mass of FA 1 added from part (a).
Calculate $M_r(\text{XCO}_3) = \text{mass of FA 1} / n(\text{XCO}_3)$ .
Calculate $A_r(\text{X}) = M_r(\text{XCO}_3) - 60.0$ .
Compare with Be 9.0, Mg 24.3, Ca 40.1, Sr 87.6, Ba 137.3.
Identify X as the metal with the $A_r$ closest to your calculated value.
For example, an $A_r$ close to 40.1 would indicate calcium, while a value near 24.3 would indicate magnesium.

Key Takeaways

$M_r = m/n$ is the reverse of $n = m/M$ . The carbonate group has a fixed molar mass, so it can be removed arithmetically from the compound molar mass. The answer is a Group 2 metal identified by the nearest known relative atomic mass.

Common Mistakes

Forgetting to subtract 60.0 from $M_r(\text{XCO}_3)$ .
Using the mass of $\text{CO}_2$ instead of the mass of FA 1.
Choosing a metal without comparing the calculated value to the known values.
Using incorrect $A_r$ values for the Group 2 metals.

Things to Be Careful About

Use the exact mass of FA 1 added, not the initial mass of the tube. Show your working because the mark for identifying X is awarded only if your calculation is visible. Quote the final $A_r$ to a sensible number of significant figures.

Techniques used

calculate molar mass from mass and molessubtract the molar mass of the carbonate groupidentify a Group 2 metal by matching its relative atomic mass

(c)

One of the sources of error in this experiment is that it is very difficult to reduce acid spraying out of the beaker when the metal carbonate is added to the acid.

(i)

Explain what effect this acid spray would have on the value you calculated for the relative atomic mass, $A_r$ , of X.

DifficultyMedium

Worked solution

Answer

Acid spray escaping from the beaker causes an additional loss of mass. This makes the calculated mass of $\text{CO}_2$ appear larger than it actually is, so $n(\text{CO}_2)$ and therefore $n(\text{XCO}_3)$ appear too large.

Because
$M_r(\text{XCO}_3) = \frac{\text{mass of FA 1}}{n(\text{XCO}_3)}$
a too-large value of $n(\text{XCO}_3)$ gives a value of $M_r(\text{XCO}_3)$ that is too small, and hence $A_r(\text{X})$ is too small.

Final answer

A_r(X) is smaller (too low).

Detailed explanation

Background Concept

In this experiment the mass of carbon dioxide is found by measuring the loss in mass of the beaker and its contents. The loss is caused by gas escaping. If liquid acid also escapes as spray, it is included in the total mass loss and is mistakenly treated as if it were carbon dioxide.

Understanding the Question

The question asks you to decide whether the calculated relative atomic mass of X becomes larger or smaller because of this error. You must link the error in mass to the moles, then to the molar mass, and finally to $A_r$ .

Approach

Work through the calculation chain: too much mass loss $\rightarrow$ too much calculated $\text{CO}_2$ $\rightarrow$ too many moles of $\text{CO}_2$ $\rightarrow$ too many moles of $\text{XCO}_3$ $\rightarrow$ too small $M_r(\text{XCO}_3)$ $\rightarrow$ too small $A_r(\text{X})$ .

Step-by-Step Reasoning

The measured loss in mass is (mass of beaker + acid) – (mass of beaker and contents after reaction).
Acid spray also leaves the beaker, so the measured loss is greater than the true mass of $\text{CO}_2$ .
Therefore the calculated mass of $\text{CO}_2$ is too large.
Dividing by 44.0 gives too large a value of $n(\text{CO}_2)$ .
Since one mole of $\text{XCO}_3$ gives one mole of $\text{CO}_2$ , $n(\text{XCO}_3)$ is also too large.
The mass of FA 1 is unaffected by the spray, so dividing this fixed mass by too large a number of moles gives too small a value of $M_r(\text{XCO}_3)$ .
Subtracting 60.0 still gives too small a value of $A_r(\text{X})$ .

Key Takeaways

An error in a measured quantity can propagate through a calculation. Identifying whether a result is too high or too low requires following each step, not just guessing. Here an overestimate of gas leads to an underestimate of molar mass.

Common Mistakes

Saying the $A_r$ would be larger.
Stopping at “the mass of CO $_2$ is too high” without explaining the effect on $A_r$ .
Confusing the mass of FA 1 with the mass of CO $_2$ in the denominator.

Things to Be Careful About

The mass of FA 1 does not change when acid sprays out. It is the moles of $\text{XCO}_3$ that become too large, so the molar mass becomes too small.

Techniques used

trace the effect of a systematic error through the calculationrelate increased mass loss to moles and to M_r

(ii)

Why is a small amount of acid spray not likely to cause an error in the identification of X?

DifficultyEasy

Worked solution

Answer

A small amount of acid spray causes only a small loss of mass, so the calculated $A_r$ is only slightly smaller. The value is still closest to the same Group 2 metal, so it does not cause confusion in the identification of X.

Final answer

Small acid spray causes only a small decrease in A_r, so X is still identified as the same Group 2 metal.

Detailed explanation

Background Concept

Identification in this experiment uses the nearest known $A_r$ among the Group 2 metals. A small systematic error changes the calculated $A_r$ only slightly, so unless the true value is very close to the midpoint between two metals, the nearest value remains unchanged.

Understanding the Question

Part (i) showed that acid spray makes $A_r$ smaller. This part asks why a small amount of spray does not matter. The key is the size of the effect: the change is small compared with the gap between the $A_r$ values of adjacent Group 2 metals.

Approach

Compare the magnitude of the error with the separation between the known $A_r$ values. A small error will not shift the calculated value away from the correct metal.

Step-by-Step Reasoning

A small acid spray means very little extra mass is lost.
The calculated mass of $\text{CO}_2$ is only slightly too large.
The calculated $A_r$ is only slightly too small.
The nearest Group 2 metal remains the same because the shift is small compared with the differences between Be, Mg, Ca, Sr and Ba.
Therefore the identity of X is still clear.

Key Takeaways

Not all errors change the conclusion. Small errors may leave the identification unchanged if the metal's $A_r$ is not close to a boundary between two elements.

Common Mistakes

Saying there is no effect at all; there is a small effect, just not enough to change the identity.
Not mentioning that the value remains closest to the same metal.

Things to Be Careful About

The mark is for recognising that the error is too small to change the nearest match, not for claiming the result is exact.

Techniques used

evaluate the significance of an errorjudge whether an error affects identification

(iii)

How could you minimise acid spraying out of the beaker?

DifficultyEasy

Worked solution

Answer

Add the carbonate slowly, in small portions. Alternatively, use a taller beaker or a conical flask, put a cotton wool plug in the mouth of the beaker, use less solid or less concentrated acid, use lumps of solid rather than powder, or carry out the reaction at a lower temperature.

Final answer

Add the carbonate slowly in small portions / use a taller beaker or conical flask / use a cotton wool plug.

Detailed explanation

Background Concept

When a carbonate is added to acid, carbon dioxide is produced rapidly and can carry droplets of acid out of the container. The rapid effervescence is worse with powder, concentrated acid and warming. Slowing the reaction or physically preventing spray reduces the error.

Understanding the Question

The question asks for a practical change that reduces acid spray. You only need one correct method; the mark is for any sensible approach that slows the reaction or contains the spray.

Approach

Think of two ways to reduce spray: slow down the production of gas, or physically trap the spray. Slower addition, larger container, cotton wool plug, less concentrated acid, lumps rather than powder, and lower temperature all meet this aim.

Step-by-Step Reasoning

Adding the solid slowly, a little at a time, means less gas is produced at any instant, so less spray is thrown out.
A taller beaker or conical flask gives droplets more distance to travel before reaching the opening.
A cotton wool plug allows gas to escape but traps liquid droplets.
Using lumps instead of powder decreases the surface areaasi and slows the reaction.
Using less concentrated acid slows the reaction.
A lower temperature reduces the rate of effervescence.

Key Takeaways

Controlling the rate of gas evolution and containing the spray are practical ways to reduce the mass-loss error. Several methods are acceptable as long as they directly reduce splashing or trap droplets.

Common Mistakes

Saying “stir more quickly” or “use more acid”, which would increase spraying.
Suggesting a stopper that prevents gas escaping; the carbon dioxide must be allowed to leave the beaker.
Giving a vague answer like “be more careful” without a concrete method.

Things to Be Careful About

The improvement must be specific enough to be reproduced. The cotton wool plug is particularly good because it stops spray while still allowing carbon dioxide to escape.

Techniques used

modify procedure to reduce splashingcontrol the rate of gas evolution

The rest of this paper

2 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation12M
Q3Qualitative Analysis · Analysis, Conclusions and Evaluation · Manipulation, Measurement and Observation13M

Loading the full paper…