Chemistry 9701/22 — May/June 2014

Cambridge AS Level · AS Level Structured Questions · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Chemical Bonding · Atomic Structure · Atoms, Molecules and Stoichiometry · Hydrocarbons · Introduction to Organic Chemistry · Chemical Periodicity · +3 more

Q1Atomic StructureAtoms, Molecules and StoichiometryChemical PeriodicityChemical BondingFree sample

(a)

Explain what is meant by the term nucleon number.

DifficultyEasy

Worked solution

Answer

The (total) number of protons and neutrons in the nucleus of an atom.

Final answer

The total number of protons and neutrons in the nucleus of an atom

Detailed explanation

Background Concept

The atom has a small, dense, positively charged nucleus made of protons and neutrons, surrounded by electrons. The proton number (atomic number, $Z$ ) counts only the protons and fixes the element's identity. The nucleon number (mass number, $A$ ) counts all the particles in the nucleus — protons and neutrons together. The number of neutrons is therefore $A - Z$ . Because protons and neutrons each have a mass of about 1 unit while electrons are roughly $\frac{1}{1840}$ of that, the nucleon number is essentially the mass number of the atom.

Understanding the Question

The command word is explain, but for a one-mark definition question this simply means "give the meaning of the term." You must state that nucleon number is the count of both protons and neutrons, and make clear these are located in the nucleus.

Approach

Recall the definition and check that it mentions both particle types. A definition that names only protons (that is proton number) or only neutrons is incomplete and scores nothing.

Step-by-Step Reasoning

The mark scheme rewards: "the (total) number of protons and neutrons (in the nucleus of an atom)." The key idea is that nucleon number is a total — it sums the two types of nucleon. Saying "number of nucleons" is also acceptable because a nucleon is by definition a proton or a neutron. Saying "number of particles in the nucleus" captures the same idea. Saying "the mass number" alone is not enough — you must unpack what it counts.

Key Takeaways

Nucleon number $A$ = protons + neutrons.
Proton number $Z$ = protons only.
Neutrons = $A - Z$ .

Common Mistakes

Writing "number of protons" — that is proton number, not nucleon number.
Writing "number of electrons and neutrons" — electrons are not nucleons.
Omitting the nucleus and implying the count includes electrons.

Things to Be Careful About

The word "nucleon" already means a nuclear particle, so "total number of nucleons" is fully correct. Make sure you say total or sum, since the mark depends on capturing both particle types.

Techniques used

recall the definition of nucleon numberdistinguish nucleon number from proton number

(b)

Bromine exists naturally as a mixture of two stable isotopes, $^{79}\text{Br}$ and $^{81}\text{Br}$ , with relative isotopic masses of 78.92 and 80.92 respectively.

(i)

Define the term relative isotopic mass.

DifficultyMedium-Easy

Worked solution

Answer

The mass of an atom of a particular isotope relative to $\frac{1}{12}$ the mass of an atom of carbon-12.

Final answer

The mass of an atom of a particular isotope relative to 1/12 the mass of an atom of carbon-12

Detailed explanation

Background Concept

Masses of atoms are far too small to measure in grams conveniently, so chemists compare them on a relative scale. The standard is the carbon-12 isotope, whose mass is defined as exactly 12 units. The relative isotopic mass of an isotope is the mass of one atom of that isotope compared with $\frac{1}{12}$ of the mass of one atom of carbon-12. Because it is a ratio, it has no units. Note the distinction: relative isotopic mass refers to a single isotope, whereas relative atomic mass refers to the weighted average over the natural mixture of isotopes.

Understanding the Question

The command word is define. Two marks are available, so the definition must contain two components: (1) the mass of an atom/isotope, and (2) the comparison against the $\frac{1}{12}$ carbon-12 standard.

Approach

State the quantity being measured (the mass of an atom of the isotope) and the reference standard (one-twelfth the mass of a carbon-12 atom). Both halves must be present to earn both marks.

Step-by-Step Reasoning

First mark: "mass of an atom(s) or isotope." This establishes what is being measured.
Second mark: "relative to $\frac{1}{12}$ the mass of an atom of carbon-12" (or equivalently "relative to carbon-12 which is exactly 12 units"). This establishes the scale.

An acceptable alternative phrasing uses the carbon-12 standard directly: "the mass of an atom of the isotope relative to carbon-12, which is taken as exactly 12 units." A correct expression such as $\dfrac{\text{mass of one atom of the isotope}}{\frac{1}{12} \times \text{mass of one } ^{12}\text{C atom}}$ also scores.

Key Takeaways

Relative isotopic mass is a ratio to the carbon-12 standard, so it is dimensionless.
It applies to one isotope; relative atomic mass is the weighted average.
The standard is $\frac{1}{12}$ of a carbon-12 atom.

Common Mistakes

Omitting the carbon-12 reference and giving only "the mass of an atom."
Saying "relative to carbon-12" without specifying $\frac{1}{12}$ — the mark scheme accepts the wording "relative to carbon-12 which is exactly 12" but a bare "compared to carbon" is too vague.
Attaching units such as g or u to a relative mass.

Things to Be Careful About

The definition must reference one atom, not a mole. Also make sure you say the mass of the isotope (or atom), since relative isotopic mass is defined per isotope.

Techniques used

recall the definition of relative isotopic massreference the carbon-12 standard

(ii)

Using the relative atomic mass of bromine, 79.90, calculate the relative isotopic abundances of $^{79}\text{Br}$ and $^{81}\text{Br}$ .

DifficultyMedium

Worked solution

Working

Let $x$ = percentage abundance of $^{79}\text{Br}$ , so $^{81}\text{Br}$ has abundance $(100 - x)\%$ .

\frac{78.92x + 80.92(100 - x)}{100} = 79.90

78.92x + 8092 - 80.92x = 7990

2x = 102 \quad\Rightarrow\quad x = 51

So $^{79}\text{Br} = 51\%$ and $^{81}\text{Br} = 100 - 51 = 49\%$ .

Answer

$^{79}\text{Br} : ^{81}\text{Br} = 51 : 49$

Final answer

79Br = 51%, 81Br = 49%

Detailed explanation

Background Concept

The relative atomic mass of an element is the weighted average of the relative isotopic masses of its isotopes, weighted by their natural abundances. If an isotope has relative isotopic mass $m$ and fractional abundance $a$ , its contribution to the average is $m \times a$ . Summing these contributions over all isotopes gives the relative atomic mass. Because the abundances are given as percentages, we divide by 100 to convert to fractions, or equivalently work directly in percentages and divide the final sum by 100.

Understanding the Question

We are told bromine is a mixture of two isotopes with relative isotopic masses 78.92 ( $^{79}\text{Br}$ ) and 80.92 ( $^{81}\text{Br}$ ), and that the overall relative atomic mass is 79.90. We must find the percentage abundance of each isotope. The command word is calculate, so full working is expected.

Approach

Let $x$ be the percentage abundance of $^{79}\text{Br}$ . Then the percentage abundance of $^{81}\text{Br}$ is $(100 - x)$ , because the two abundances must sum to 100%. Form the weighted-average expression, set it equal to the known relative atomic mass, and solve for $x$ .

Step-by-Step Reasoning

Set up the average. The weighted average is

\frac{78.92x + 80.92(100 - x)}{100} = 79.90

This earns the first mark (the correct expression with the two isotopic masses and the unknown $x$ ).
2. Clear the denominator. Multiply both sides by 100:

78.92x + 80.92(100 - x) = 7990

Expand and simplify.

78.92x + 8092 - 80.92x = 7990

8092 - 2x = 7990

2x = 102

x = 51

This earns the second mark.
4. Find the second abundance. $^{81}\text{Br} = 100 - 51 = 49\%$ . The ratio $^{79}\text{Br} : ^{81}\text{Br} = 51 : 49$ earns the third mark.

Check: The average sits much closer to 79 than to 81, so the lighter isotope must be the more abundant — 51% versus 49% is consistent with a value of 79.90, just above 79.

Key Takeaways

Relative atomic mass is an abundance-weighted average of isotopic masses.
Two abundances always sum to 100%, so one unknown suffices.
The answer can be reported as percentages or as a ratio.

Common Mistakes

Forgetting to divide by 100, giving an answer 100 times too large.
Assuming a 50:50 split and not actually solving the equation.
Mixing up which isotope is more abundant — the average 79.90 lies nearer 79, so $^{79}\text{Br}$ dominates.

Things to Be Careful About

Keep the equation balanced: both terms must be divided by 100 (or both left in percentage terms).
Report both abundances; giving only one loses the final mark.
The mark scheme accepts the answer as a ratio (51:49) or as percentages.

Techniques used

set up a weighted-average equation for isotopic abundancesolve a linear equation for percentage abundanceconvert fractional abundance into a ratio

(c)

Bromine reacts with the element A to form a compound with empirical formula ABr $_3$ . The percentage composition by mass of ABr $_3$ is A, 4.31; Br, 95.69.

Calculate the relative atomic mass, $A_r$ , of A.
Give your answer to three significant figures.

DifficultyMedium

Worked solution

Working

The empirical formula $\text{ABr}_3$ means the mole ratio $\text{A} : \text{Br} = 1 : 3$ .

\frac{4.31}{A_r} : \frac{95.69}{79.9} = 1 : 3

\frac{95.69 / 79.9}{4.31 / A_r} = 3

A_r = \frac{3 \times 4.31 \times 79.9}{95.69} = 10.796

Answer

$A_r = 10.8$ (to 3 s.f.)

Final answer

10.8

Detailed explanation

Background Concept

When an element's percentage composition by mass is known, dividing each percentage by the relevant relative atomic mass converts the mass percentages into relative numbers of moles. These mole ratios give the empirical formula. Here the empirical formula is already given as $\text{ABr}_3$ , so the mole ratio of A to Br is fixed at $1 : 3$ ; we can use this ratio in reverse to find the unknown $A_r$ of A.

Understanding the Question

The compound $\text{ABr}_3$ is 4.31% A and 95.69% Br by mass. The relative atomic mass of bromine is 79.9 (the value from part (b)). We must find $A_r$ of A and give the answer to three significant figures. The command word is calculate.

Approach

Convert each mass percentage into moles by dividing by the relative atomic mass. The ratio of these mole values must equal the ratio of atoms in the formula, $1 : 3$ . Set up the proportion and solve for $A_r$ .

Step-by-Step Reasoning

Moles of Br per 100 g: $\dfrac{95.69}{79.9} = 1.1976$ mol.
Moles of A per 100 g: $\dfrac{4.31}{A_r}$ mol.
Apply the $1 : 3$ ratio. Since the formula is $\text{ABr}_3$ , the moles of Br are three times the moles of A:

\frac{95.69 / 79.9}{4.31 / A_r} = 3

This earns the first mark (correct ratio set up).
4. Solve for $A_r$ . Rearranging:

A_r = \frac{3 \times 4.31 \times 79.9}{95.69} = 10.796

This earns the second mark.
5. Round to three significant figures: $A_r = 10.8$ . The third mark is for the correct number of significant figures.

Key Takeaways

Percentage composition $\div$ relative atomic mass gives mole ratios.
A stated empirical formula fixes the atom ratio and can be used to find an unknown $A_r$ .
Always check the required significant figures at the end.

Common Mistakes

Inverting the ratio (using $1 : 3$ the wrong way round), which gives $A_r \approx 0.39$ .
Using the relative atomic mass of bromine as 80 instead of 79.9.
Giving the unrounded value 10.796 instead of 10.8, losing the significant-figures mark.

Things to Be Careful About

The question explicitly demands three significant figures — 10.8 is correct, 10.796 is not.
Use the bromine value consistent with the data (79.9).
Alternative correct methods (e.g. treating 100 g of compound and using the mass of Br present) are allowed and give the same result.

Techniques used

use percentage composition to set up a mole ratioapply the empirical formula ratio to find relative atomic massround to three significant figures

(d)

The elements in Period 3 of the Periodic Table show different behaviours in their reactions with oxygen.

(i)

Describe what you would see when separate samples of magnesium and sulfur are reacted with oxygen.

Write an equation for each reaction.

magnesium

sulfur

DifficultyMedium-Easy

Worked solution

Answer

Magnesium: burns with a bright white flame / bright white light, forming a white solid (white smoke).

\text{Mg} + \tfrac{1}{2}\text{O}_2 \rightarrow \text{MgO}

Sulfur: burns with a blue flame, giving white/steamy fumes; the yellow solid disappears.

\text{S} + \text{O}_2 \rightarrow \text{SO}_2

Final answer

Mg: bright white flame, white solid/smoke; Mg + 1/2O2 -> MgO. S: blue flame, white/steamy fumes, yellow solid disappears; S + O2 -> SO2

Detailed explanation

Background Concept

Period 3 elements react with oxygen to form oxides, but the vigour and appearance of the reaction vary across the period. Magnesium is a reactive metal that burns vigorously, while sulfur is a non-metal that burns with a characteristic blue flame. The observations (flame colour, colour of the product, fumes) are a classic examinable feature of Period 3 chemistry.

Understanding the Question

The command word is describe for the observations and write an equation for each reaction. Four marks are available: one observation and one equation for magnesium, and one observation and one equation for sulfur. The question asks specifically what you would see.

Approach

Recall the standard observations for each element burning in oxygen, then write a correctly balanced equation with the correct product: magnesium forms the oxide MgO, sulfur forms the dioxide SO $_2$ .

Step-by-Step Reasoning

Magnesium. It burns in oxygen with a bright white flame (bright white light), producing a white solid — often seen as white smoke or white ash. The product is magnesium oxide:

\text{Mg} + \tfrac{1}{2}\text{O}_2 \rightarrow \text{MgO}

Correct multiples (e.g. $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ ) are allowed.

Sulfur. It burns with a blue flame, giving off white/steamy fumes of sulfur dioxide, and the yellow solid disappears as it is consumed. The product is sulfur dioxide:

\text{S} + \text{O}_2 \rightarrow \text{SO}_2

Again, multiples are allowed.

Key Takeaways

Metal + oxygen $\rightarrow$ metal oxide (basic oxide); non-metal + oxygen $\rightarrow$ non-metal oxide (acidic oxide).
Observations must be specific: flame colour and product appearance.
Equations must be balanced with the correct product formula.

Common Mistakes

Writing "burns brightly" without naming the colour (white) — loses the observation mark.
Giving the sulfur product as SO $_3$ instead of SO $_2$ .
Omitting state symbols where the mark scheme expects a balanced equation; here they are not required, but the equation must balance.

Things to Be Careful About

"White smoke" and "white solid" are both accepted for magnesium; the key word is white.
For sulfur, any one of the three observations (blue flame, white/steamy fumes, yellow solid disappears) earns the mark.
Make sure the equation balances in both atoms and, if you use multiples, keeps the same ratio.

Techniques used

recall the visual observations of Period 3 elements burning in oxygenwrite balanced equations for combustion in oxygen

(ii)

Write equations for the reactions of aluminium oxide, $\text{Al}_2\text{O}_3$ , with

sodium hydroxide,

hydrochloric acid.

DifficultyMedium

Worked solution

Answer

With sodium hydroxide:

\text{Al}_2\text{O}_3 + 2\text{NaOH} + 3\text{H}_2\text{O} \rightarrow 2\text{NaAl(OH)}_4

With hydrochloric acid:

\text{Al}_2\text{O}_3 + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2\text{O}

Final answer

Al2O3 + 2NaOH + 3H2O -> 2NaAl(OH)4; Al2O3 + 6HCl -> 2AlCl3 + 3H2O

Detailed explanation

Background Concept

Aluminium oxide, $\text{Al}_2\text{O}_3$ , is amphoteric: it reacts both with acids (acting as a base) and with alkalis (acting as an acid). This is a defining property of aluminium and its oxide/hydroxide, and it distinguishes aluminium from the purely basic oxides of the s-block metals. With acid, the oxide behaves as a base and is neutralised to an aluminium salt plus water. With alkali, it behaves as an acid and dissolves to form an aluminate (tetrahydroxoaluminate) species.

Understanding the Question

The command word is write equations. Two marks: one for the reaction with sodium hydroxide and one for the reaction with hydrochloric acid. Both equations must be balanced.

Approach

Recall the amphoteric behaviour. For the acid reaction, treat $\text{Al}_2\text{O}_3$ as a base forming $\text{AlCl}_3$ and water. For the alkali reaction, form the soluble aluminate, e.g. sodium tetrahydroxoaluminate, $\text{NaAl(OH)}_4$ .

Step-by-Step Reasoning

With hydrochloric acid. Aluminium oxide neutralises the acid:

\text{Al}_2\text{O}_3 + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2\text{O}

Balance: 2 Al on each side; 6 Cl on each side; 3 O and 6 H on each side. Correct.

With sodium hydroxide. The oxide dissolves in the alkali to form a soluble aluminate:

\text{Al}_2\text{O}_3 + 2\text{NaOH} + 3\text{H}_2\text{O} \rightarrow 2\text{NaAl(OH)}_4

Balance: 2 Al, 2 Na; the oxygen and hydrogen balance with the added water. The mark scheme also accepts other equivalent forms such as $\text{Al}_2\text{O}_3 + 2\text{NaOH} \rightarrow 2\text{NaAlO}_2 + \text{H}_2\text{O}$ and the corresponding ionic equations.

Key Takeaways

Aluminium oxide is amphoteric — it reacts with both acids and alkalis.
The alkaline product is an aluminate; the acidic product is an aluminium salt.
Equations must be balanced in atoms and, for ionic versions, in charge.

Common Mistakes

Writing only the acid reaction and leaving the alkali blank, or vice versa.
Producing $\text{Al(OH)}_3$ with NaOH as if no further reaction occurred — the oxide dissolves to an aluminate.
Unbalanced equations (e.g. forgetting the water on the left of the alkali reaction).

Things to Be Careful About

Ionic equations are accepted, provided they balance in both atoms and charge.
Several product formulations are allowed for the alkali reaction; pick one and balance it correctly.
Do not confuse the amphoteric oxide of aluminium with the basic oxide of magnesium.

Techniques used

write balanced equations for an amphoteric oxide with alkali and with acidapply amphoteric behaviour of aluminium oxide

(e)

Phosphorus reacts with chlorine to form $\text{PCl}_5$ .

State the shape of and two different bond angles in a molecule of $\text{PCl}_5$ .

shape of $\text{PCl}_5$

bond angles in $\text{PCl}_5$

DifficultyMedium-Easy

Worked solution

Answer

Shape of $\text{PCl}_5$ : (trigonal) bipyramidal

Bond angles: $120^\circ$ and $90^\circ$

Final answer

Trigonal bipyramidal; bond angles 120° and 90°

Detailed explanation

Background Concept

VSEPR theory predicts molecular shape from the number of electron pairs around the central atom. Phosphorus in $\text{PCl}_5$ has five bonding pairs and no lone pairs, so the electron-pair geometry is trigonal bipyramidal. This geometry is special because it contains two inequivalent positions: three equatorial positions in a plane and two axial positions above and below that plane. The equatorial bonds are separated by $120^\circ$ , while the angle between an axial bond and an equatorial bond is $90^\circ$ . Phosphorus expands its octet to ten electrons here, which is possible for Period 3 elements.

Understanding the Question

The command word is state. Two marks: one for the shape, one for two different bond angles. The question stresses two different angles, which is the key clue that the geometry has inequivalent positions.

Approach

Count the bonding pairs around phosphorus (five) and lone pairs (none), deduce the trigonal bipyramidal shape, then read off the two distinct bond angles present in that geometry.

Step-by-Step Reasoning

Phosphorus has 5 valence electrons; each of the five P–Cl bonds uses one, giving five bonding pairs and no lone pairs.
Five bonding pairs with no lone pairs $\rightarrow$ trigonal bipyramidal shape.
In a trigonal bipyramid there are two sets of angles: the equatorial–equatorial angle of $120^\circ$ and the axial–equatorial angle of $90^\circ$ .

Key Takeaways

Five bonding pairs, no lone pairs $\rightarrow$ trigonal bipyramidal.
Trigonal bipyramidal geometry has two distinct bond angles: $120^\circ$ and $90^\circ$ .
Period 3 elements can expand their octet.

Common Mistakes

Calling the shape "trigonal planar" (that is three bonding pairs) or "octahedral" (six bonding pairs).
Giving only one bond angle, or giving $109.5^\circ$ (tetrahedral) instead of the correct pair.
Forgetting that the two axial positions are distinct from the three equatorial ones.

Things to Be Careful About

The question asks for two different bond angles — both $120^\circ$ and $90^\circ$ are required.
Do not confuse $\text{PCl}_5$ (five bonds, bipyramidal) with $\text{PCl}_3$ (three bonds plus a lone pair, trigonal pyramidal).
State the shape clearly as "trigonal bipyramidal."

Techniques used

count bonding pairs and lone pairs around the central atomapply VSEPR to deduce molecular shapeidentify the two distinct bond angles

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