Chemistry 9701/23 — October/November 2010

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

questions

marks

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Topics Atoms, Molecules and Stoichiometry · Chemical Bonding · Halogen Compounds · Atomic Structure · Equilibria · Electrochemistry · +7 more

Q1Atomic StructureAtoms, Molecules and StoichiometryChemical BondingFree sample

The element magnesium, Mg, proton number 12, is a metal which is used in many alloys which are strong and light.

Magnesium has several naturally occurring isotopes.

(a)

What is meant by the term isotope?

DifficultyEasy

Worked solution

Answer

Atoms of the same element (with the same proton number / number of protons) but with different numbers of neutrons (or different nucleon / mass numbers).

Final answer

Atoms of the same element with the same proton number but different numbers of neutrons.

Detailed explanation

Background Concept

Isotopes are variants of a particular chemical element which differ in neutron number, and consequently in nucleon number (mass number). All isotopes of a given element have the same number of protons but different numbers of neutrons in each atom. Because they have the same number of protons and electrons, isotopes of the same element have virtually identical chemical properties, but their physical properties (such as mass and density) differ slightly.

Understanding the Question

The question asks for the definition of the term 'isotope'. This is a foundational concept in atomic structure, requiring a precise two-part statement: what is the same, and what is different.

Approach

Recall the standard IUPAC definition of an isotope. It must mention that they are atoms of the same element (implying the same atomic/proton number) and that they differ in their mass number (implying a different number of neutrons).

Step-by-Step Reasoning

Same element / same proton number: Isotopes belong to the same element, so they must have the same number of protons (atomic number). This defines the element's identity and its chemical behaviour.
Different neutrons / different mass number: To be an isotope and not just the same atom, the number of neutrons in the nucleus must differ. This results in a different mass number (nucleon number).

Key Takeaways

Isotopes have the same proton number but different neutron numbers.
The term 'isotope' refers to the atoms themselves, not the element as a whole (e.g., 'carbon isotopes' are correct; 'isotopes of carbon' is also acceptable, but 'isotopes' alone is vague).

Common Mistakes

Saying 'same mass number' (incorrect; they have different mass numbers).
Saying 'different number of electrons' (incorrect; neutral atoms of the same isotope have the same number of electrons, and ions are not isotopes).
Using the word 'molecule' instead of 'atom'.

Things to Be Careful About

Ensure both parts of the definition are present: what is the same (protons/element) and what is different (neutrons/mass number). Missing either part costs a mark.

Techniques used

define isotope

(b)

Complete the table below for two of the isotopes of magnesium.

Table 1.1

DifficultyEasy

Worked solution

Answer

isotope	number of protons	number of neutrons	number of electrons
$^{24}\text{Mg}$	12	12	12
$^{26}\text{Mg}$	12	14	12

Final answer

See table above.

Detailed explanation

Background Concept

The notation for an isotope is $^{A}_{Z}\text{X}$ , where $\text{X}$ is the chemical symbol, $Z$ is the proton number (atomic number), and $A$ is the mass number (nucleon number).

Proton number ( $Z$ ): The number of protons in the nucleus. This defines the element. For magnesium, $Z = 12$ .
Mass number ( $A$ ): The total number of protons and neutrons in the nucleus ( $A = Z + N$ , where $N$ is the number of neutrons).
Number of neutrons ( $N$ ): Calculated as $N = A - Z$ .
Number of electrons: In a neutral atom, the number of electrons equals the number of protons ( $Z$ ). If the species is an ion, electrons are added or subtracted based on the charge.

Understanding the Question

The question provides a table to complete for two isotopes of magnesium: $^{24}\text{Mg}$ and $^{26}\text{Mg}$ . The columns require the number of protons, neutrons, and electrons. The proton number for magnesium is given in the question stem as 12.

Approach

Identify the proton number ( $Z$ ) for Mg from the stem (12). This is the same for all Mg isotopes.
For a neutral atom, the number of electrons equals the proton number (12).
Calculate the number of neutrons for each isotope using $N = A - Z$ $N = A - Z$ .
- For $^{24}\text{Mg}$ : $N = 24 - 12 = 12$ .
- For $^{26}\text{Mg}$ : $N = 26 - 12 = 14$ .

Step-by-Step Reasoning

$^{24}\text{Mg}$ : Protons = 12. Electrons = 12 (neutral atom). Neutrons = $24 - 12 = 12$ .
$^{26}\text{Mg}$ : Protons = 12. Electrons = 12 (neutral atom). Neutrons = $26 - 12 = 14$ .

Key Takeaways

The proton number is constant for all isotopes of an element.
The mass number is the sum of protons and neutrons.
Neutral atoms have equal numbers of protons and electrons.

Common Mistakes

Calculating neutrons as $A + Z$ instead of $A - Z$ .
Assuming the number of electrons changes for different isotopes (electrons only change if the atom is ionised).

Things to Be Careful About

Ensure the values are placed in the correct rows for the correct isotopes.

Techniques used

calculate neutrons from mass and proton numbersdetermine electrons in neutral atom

(c)

A sample of magnesium had the following isotopic composition:
$^{24}\text{Mg}$ , 78.60%; $^{25}\text{Mg}$ , 10.11%; $^{26}\text{Mg}$ , 11.29%.

Calculate the relative atomic mass, $A_r$ , of magnesium in the sample.
Express your answer to an appropriate number of significant figures.

DifficultyMedium-Easy

Worked solution

Working

A_r = \frac{(24 \times 78.60) + (25 \times 10.11) + (26 \times 11.29)}{100}

A_r = \frac{1886.40 + 252.75 + 293.54}{100}

A_r = \frac{2432.69}{100} = 24.33

Answer

24.33

Final answer

24.33

Detailed explanation

Background Concept

The relative atomic mass ( $A_r$ ) of an element is the weighted average mass of an atom of the element relative to $\frac{1}{12}$ of the mass of a carbon-12 atom. Because elements exist as a mixture of isotopes, the $A_r$ is calculated by taking the sum of the mass numbers (or more precisely, isotopic masses) multiplied by their relative abundances (percentage or fraction), divided by the total abundance (100% or 1).

Formula:

A_r = \frac{\sum (\text{isotope mass} \times \% \text{ abundance})}{100}

Understanding the Question

The question gives the isotopic composition of a magnesium sample: $^{24}\text{Mg}$ (78.60%), $^{25}\text{Mg}$ (10.11%), and $^{26}\text{Mg}$ (11.29%). We are to calculate the $A_r$ and express it to an appropriate number of significant figures.

Approach

Multiply each isotope's mass number by its percentage abundance.
Sum these products.
Divide by 100 to get the weighted average.
Round the final answer to the correct number of significant figures (usually matching the precision of the input data, which is 4 sig figs here).

Step-by-Step Reasoning

Contribution from $^{24}\text{Mg}$ : $24 \times 78.60 = 1886.40$
Contribution from $^{25}\text{Mg}$ : $25 \times 10.11 = 252.75$
Contribution from $^{26}\text{Mg}$ : $26 \times 11.29 = 293.54$
Total sum: $1886.40 + 252.75 + 293.54 = 2432.69$
$A_r = 2432.69 / 100 = 24.3269$
Rounding to 4 significant figures (as the percentages are given to 4 sig figs): $24.33$ .

Key Takeaways

$A_r$ is a weighted average, not a simple average.
The number of significant figures in the final answer should reflect the precision of the given data (usually 3 or 4 sig figs for $A_r$ values in exams).

Common Mistakes

Forgetting to divide by 100 (if using percentages directly).
Rounding to too few or too many significant figures.
Using atomic mass unit values (e.g., 24.305) instead of the integer mass numbers provided in the question (unless specified; CIE typically accepts integer mass numbers for these calculations unless high-precision data is given).

Things to Be Careful About

Check the significant figures of the input data. Here, 78.60% has 4 sig figs, so the answer should be given to 4 sig figs (24.33).

Techniques used

calculate weighted average relative atomic mass

(d)

Antimony, Sb, proton number 51, is another element which is used in alloys.

Magnesium and antimony each react when heated separately in chlorine.

Construct a balanced equation for the reaction between magnesium and chlorine.

DifficultyEasy

Worked solution

Answer

\text{Mg}(\text{s}) + \text{Cl}_2(\text{g}) \rightarrow \text{MgCl}_2(\text{s})

(Note: state symbols are often not strictly required unless specified, but the balanced equation $\text{Mg} + \text{Cl}_2 \rightarrow \text{MgCl}_2$ is the core mark.)

Final answer

Mg + Cl2 -> MgCl2

Detailed explanation

Background Concept

Magnesium is a Group 2 metal and chlorine is a Group 17 non-metal (halogen). When heated together, they undergo a combination (synthesis) reaction to form an ionic compound, magnesium chloride. Magnesium loses 2 electrons to form $\text{Mg}^{2+}$ , and each chlorine atom gains 1 electron to form $\text{Cl}^-$ .

Understanding the Question

The question asks for a balanced equation for the reaction between magnesium and chlorine. Magnesium is a solid metal, and chlorine is a diatomic gas at room temperature.

Approach

Write the unbalanced equation: $\text{Mg} + \text{Cl}_2 \rightarrow \text{MgCl}_2$ .
Check if it is balanced: 1 Mg on both sides, 2 Cl on both sides. It is already balanced.
Add state symbols if required (though the mark scheme often accepts without for simple synthesis, it's good practice: Mg(s), Cl2(g), MgCl2(s)).

Step-by-Step Reasoning

Reactants: $\text{Mg}$ (solid) and $\text{Cl}_2$ (gas, diatomic).
Product: $\text{MgCl}_2$ (ionic solid).
Balancing: One Mg atom reacts with one $\text{Cl}_2$ molecule to form one formula unit of $\text{MgCl}_2$ .
Equation: $\text{Mg} + \text{Cl}_2 \rightarrow \text{MgCl}_2$ .

Key Takeaways

Halogens are diatomic molecules ( $\text{Cl}_2$ , $\text{Br}_2$ , $\text{I}_2$ ).
Group 2 metals react with halogens to form 1:2 ionic halides ( $\text{MgX}_2$ ).

Common Mistakes

Writing chlorine as $\text{Cl}$ instead of $\text{Cl}_2$ .
Writing the product as $\text{MgCl}$ instead of $\text{MgCl}_2$ (ignoring the +2 oxidation state of Mg).

Things to Be Careful About

Ensure the equation is balanced. In this case, it is naturally balanced with 1:1:1 stoichiometry.

Techniques used

write and balance chemical equation

(e)

When a 2.45 g sample of antimony was heated in chlorine under suitable conditions, 4.57 g of a chloride A were formed.

(i)

Calculate the amount, in moles, of antimony atoms that reacted.

DifficultyEasy

Worked solution

Working

n(\text{Sb}) = \frac{\text{mass}}{\text{molar mass}} = \frac{2.45}{122} = 0.0200 \text{ mol}

Answer

0.0200 mol

Final answer

0.0200 mol

Detailed explanation

Background Concept

The amount of substance in moles ( $n$ ) is calculated by dividing the mass of the substance ( $m$ ) by its molar mass ( $M_r$ or $M$ ). For an element, the molar mass is numerically equal to the relative atomic mass ( $A_r$ ) in g/mol.

Understanding the Question

We are given 2.45 g of antimony (Sb) and asked to calculate the amount in moles. The $A_r$ of Sb is 122 (from the periodic table or data booklet).

Approach

Use the formula $n = \frac{m}{M_r}$ .

Step-by-Step Reasoning

Mass of Sb = 2.45 g.
Molar mass of Sb = 122 g/mol.
$n(\text{Sb}) = \frac{2.45}{122} = 0.02008... \approx 0.0200 \text{ mol}$ (to 3 sig figs).

Key Takeaways

Always use the correct molar mass for the element or compound.
Keep track of significant figures (3 sig figs is appropriate here as 2.45 has 3 sig figs).

Common Mistakes

Using the wrong molar mass for Sb.
Forgetting to convert mass to moles.

Things to Be Careful About

Antimony's $A_r$ is 121.76, often rounded to 122 in CIE data booklets. Use the value provided in the question's data booklet.

Techniques used

calculate moles from mass and molar mass

(ii)

Calculate the amount, in moles, of chlorine atoms that reacted.

DifficultyMedium-Easy

Worked solution

Working

Mass of chlorine in A = mass of A - mass of Sb

\text{Mass of Cl} = 4.57 - 2.45 = 2.12 \text{ g}

n(\text{Cl}) = \frac{2.12}{35.5} = 0.0597 \approx 0.0600 \text{ mol}

Answer

0.0600 mol

Final answer

0.0600 mol

Detailed explanation

Background Concept

In a reaction where an element combines with chlorine to form a chloride, the mass of the chloride is the sum of the mass of the element and the mass of the chlorine that reacted. By finding the mass of chlorine by difference, we can then calculate the moles of chlorine atoms.

Understanding the Question

A 2.45 g sample of Sb reacts with chlorine to form 4.57 g of chloride A. We need to find the moles of chlorine atoms that reacted (not molecules).

Approach

Calculate the mass of chlorine that reacted: $m(\text{Cl}) = m(\text{A}) - m(\text{Sb})$ .
Calculate the moles of Cl atoms using $n = \frac{m}{A_r}$ .

Step-by-Step Reasoning

Mass of A = 4.57 g.
Mass of Sb = 2.45 g.
Mass of Cl = $4.57 - 2.45 = 2.12$ g.
Molar mass of Cl = 35.5 g/mol.
$n(\text{Cl}) = \frac{2.12}{35.5} = 0.059718... \approx 0.0600 \text{ mol}$ .

Key Takeaways

Use the law of conservation of mass to find the mass of the second reactant.
Be careful to calculate moles of atoms (Cl), not molecules ( $\text{Cl}_2$ ), unless specified. The question asks for 'chlorine atoms'.

Common Mistakes

Using the mass of A (4.57 g) directly to calculate moles of Cl.
Calculating moles of $\text{Cl}_2$ instead of Cl atoms (would give half the value).

Things to Be Careful About

The question specifically asks for 'chlorine atoms'. If it asked for 'chlorine molecules', you would divide by 71 (or divide the atom moles by 2). Here, $A_r$ of Cl = 35.5 is used.

Techniques used

determine mass of element by differencecalculate moles of atoms

(iii)

Use your answers to (i) and (ii) to determine the empirical formula of A.

DifficultyMedium-Easy

Worked solution

Working

Mole ratio $\text{Sb} : \text{Cl} = 0.0200 : 0.0600$
Divide by the smallest number of moles (0.0200):

\text{Sb} : \frac{0.0200}{0.0200} = 1

\text{Cl} : \frac{0.0600}{0.0200} = 3

Simplest ratio is 1 : 3.

Answer

$\text{SbCl}_3$

Final answer

SbCl3

Detailed explanation

Background Concept

The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. To find it, calculate the moles of each element, then divide by the smallest number of moles to get a ratio. If the ratio is not whole numbers (e.g., 1.5), multiply by a small integer (e.g., 2) to make them whole.

Understanding the Question

Using the moles of Sb (0.0200 mol) and Cl (0.0600 mol) calculated in parts (i) and (ii), determine the empirical formula of chloride A.

Approach

Write the mole ratio of Sb to Cl.
Divide both by the smaller value to get the simplest ratio.
Write the empirical formula.

Step-by-Step Reasoning

Moles of Sb = 0.0200 mol.
Moles of Cl = 0.0600 mol.
Ratio $\text{Sb} : \text{Cl} = 0.0200 : 0.0600$ .
Divide by 0.0200: $1 : 3$ .
Empirical formula is $\text{SbCl}_3$ .

Key Takeaways

The empirical formula is derived from the mole ratio, not the mass ratio.
Always divide by the smallest number of moles to simplify the ratio.

Common Mistakes

Using the mass ratio instead of the mole ratio.
Failing to simplify the ratio to whole numbers (though here it is already simple).

Things to Be Careful About

Ensure the ratio is simplified correctly. In this case, $0.06 / 0.02 = 3$ exactly.

Techniques used

determine empirical formula from mole ratio

(iv)

The empirical and molecular formulae of A are the same.

Construct a balanced equation for the reaction between antimony and chlorine.

DifficultyMedium-Easy

Worked solution

Answer

2\text{Sb}(\text{s}) + 3\text{Cl}_2(\text{g}) \rightarrow 2\text{SbCl}_3(\text{s})

Working

Since the empirical and molecular formulae are the same, the formula is $\text{SbCl}_3$ .
Reactants: $\text{Sb}$ and $\text{Cl}_2$ (diatomic).
Unbalanced: $\text{Sb} + \text{Cl}_2 \rightarrow \text{SbCl}_3$ .
Balance Cl: need 3 $\text{Cl}_2$ and 2 $\text{SbCl}_3$ .
Balance Sb: need 2 $\text{Sb}$ .
Balanced: $2\text{Sb} + 3\text{Cl}_2 \rightarrow 2\text{SbCl}_3$ .

Final answer

2Sb + 3Cl2 -> 2SbCl3

Detailed explanation

Background Concept

Antimony is a metalloid (Group 15) and forms covalent chlorides. The empirical formula is $\text{SbCl}_3$ , and the question states the molecular formula is the same. The reaction is between solid antimony and chlorine gas (diatomic, $\text{Cl}_2$ ) to form antimony trichloride.

Understanding the Question

Construct a balanced equation for the reaction between antimony and chlorine, given that the product is $\text{SbCl}_3$ .

Approach

Write the unbalanced equation with correct formulas for reactants and products.
Balance the atoms, starting with the most complex molecule or the element that appears in the fewest places.

Step-by-Step Reasoning

Reactants: $\text{Sb}$ (solid) and $\text{Cl}_2$ (gas).
Product: $\text{SbCl}_3$ .
Unbalanced: $\text{Sb} + \text{Cl}_2 \rightarrow \text{SbCl}_3$ .
To balance Cl, we need a common multiple of 2 and 3, which is 6. So use $3\text{Cl}_2$ and $2\text{SbCl}_3$ .
This gives 2 Sb on the right, so we need $2\text{Sb}$ on the left.
Balanced equation: $2\text{Sb} + 3\text{Cl}_2 \rightarrow 2\text{SbCl}_3$ .

Key Takeaways

Halogens are diatomic ( $\text{Cl}_2$ ), not monatomic ( $\text{Cl}$ ).
Balancing equations with diatomic molecules often requires fractional coefficients or multiplying through to clear fractions.

Common Mistakes

Writing $\text{Cl}$ instead of $\text{Cl}_2$ as a reactant.
Balancing incorrectly (e.g., $\text{Sb} + \text{Cl}_2 \rightarrow \text{SbCl}_2$ ).

Things to Be Careful About

Ensure the equation is fully balanced with whole-number coefficients.

Techniques used

write and balance chemical equation from empirical formula

(f)

The chloride A melts at 73.4 °C while magnesium chloride melts at 714 °C.

(i)

What type of bonding is present in magnesium chloride?

DifficultyEasy

Worked solution

Answer

Ionic

Final answer

Ionic

Detailed explanation

Background Concept

Ionic bonding occurs between metals and non-metals, where electrons are transferred from the metal to the non-metal, forming a lattice of oppositely charged ions. Ionic compounds typically have high melting and boiling points due to the strong electrostatic forces of attraction between the ions in the lattice.

Understanding the Question

Magnesium chloride ( $\text{MgCl}_2$ ) has a high melting point (714 °C). We are asked to identify the type of bonding present.

Approach

Mg is a metal, Cl is a non-metal. Compounds formed between metals and non-metals are typically ionic.
The high melting point (714 °C) is characteristic of a giant ionic lattice with strong electrostatic forces.

Step-by-Step Reasoning

Element types: Mg (Group 2 metal) + Cl (Group 17 non-metal) -> Ionic compound.
Physical property: High melting point (714 °C) -> Strong bonds in a giant structure -> Ionic bonding.

Key Takeaways

High melting points and metal+non-metal combination indicate ionic bonding.

Common Mistakes

Saying 'metallic' (only for pure metals or alloys).
Saying 'covalent' (typically lower melting points, though network covalent can be high, MgCl2 is not).

Things to Be Careful About

Just state 'ionic'. No need for further explanation unless asked.

Techniques used

identify bonding type from melting point and element types

(ii)

Suggest what type of bonding is present in A.

DifficultyMedium-Easy

Worked solution

Answer

Covalent

Final answer

Covalent

Detailed explanation

Background Concept

Covalent bonding involves the sharing of electron pairs between atoms. Simple molecular covalent substances have low melting and boiling points because the intermolecular forces (van der Waals forces, dipole-dipole, hydrogen bonding) between molecules are weak, even though the covalent bonds within the molecules are strong.

Understanding the Question

Chloride A ( $\text{SbCl}_3$ ) has a much lower melting point (73.4 °C) compared to magnesium chloride (714 °C). We are to suggest the type of bonding in A.

Approach

The low melting point indicates weak intermolecular forces, characteristic of simple molecular covalent substances.
Sb is a metalloid and Cl is a non-metal; their electronegativity difference is not large enough to form a purely ionic bond. They share electrons.
The bonding within the $\text{SbCl}_3$ molecules is covalent.

Step-by-Step Reasoning

Melting point of A = 73.4 °C (low). This suggests a simple molecular structure.
In simple molecular substances, the atoms within the molecule are held together by covalent bonds.
Therefore, the bonding present in A (referring to the intramolecular bonding that defines the molecule) is covalent.
Note: The question asks for 'bonding present in A'. While there are van der Waals forces between molecules, the primary chemical bonding holding the atoms together in the formula unit is covalent. The mark scheme explicitly rejects 'van der Waals' forces as the answer for 'bonding'.

Key Takeaways

Low melting points indicate simple molecular structures with covalent bonding within molecules.
Distinguish between intramolecular bonding (covalent) and intermolecular forces (van der Waals). The question asks for 'bonding', which typically refers to the chemical bonds (covalent/ionic/metallic).

Common Mistakes

Answering 'van der Waals' forces or 'intermolecular forces'. These are forces between molecules, not the chemical bonding within the molecule. The mark scheme specifically rejects this.
Answering 'ionic' (contradicted by the low melting point).

Things to Be Careful About

Read the question carefully: 'What type of bonding'. In CIE marking, 'covalent' is the expected answer for molecular compounds, not the intermolecular forces. The mark scheme note 'not van der Waals' forces' is a crucial hint.

Techniques used

identify bonding type from melting point

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