AQA A-Level Chemistry Paper 1, 2021: Question 2
6 marks · Medium difficulty · State/Explain/Numerical
Define mass number, determine subatomic particle numbers for titanium species, and calculate the percentage abundance of titanium-46 from isotopic data.
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Question text
02 This question is about atomic structure.
02.1 Define the mass number of an atom.
[1 mark]
02.2 Complete Table 3 to show the numbers of neutrons and electrons in the species
shown.
[2 marks]
Table 3
Number of Number of Number of
neutrons electrons
protons
46Ti 22
49Ti2+ 22
02.3 A sample of titanium contains four isotopes, 46Ti, 47Ti, 48Ti and 49Ti
This sample has a relative atomic mass of 47.8
In this sample the ratio of abundance of isotopes 46Ti, 47Ti and 49Ti is 2:2:1
Calculate the percentage abundance of 46Ti in this sample.
[3 marks]
Abundance of 46Ti %
Mark scheme
Show the mark scheme
Question Answers Additional comments/Guidelines Mark
Do not allow reference to mass or average
02.1 Number of protons + neutrons (in the nucleus of the atom) 1
Ignore references to C-12 being 12
Mark as rows
Number of Number of Number of
protons neutrons electrons 1
02.2 46Ti 22 24 22
49Ti2+ 22 27 20
Allow
Let 49Ti be y
M1 47.8 = (46 x 2) + ( 47 x 2) + (48 x n) + 49 1
M1 47.8 = (46 x 2y) + ( 47 x 2y) + (48 x (100 – 5y))+ (49 x y) (5 + n)
02.3 47.8 = 235y + 4800 – 240y 1
M2 0.2n =4 or n=20
M2 5y = 20 OR y = 4 1
46 2
M3 % Ti = x 100 = 8%
M3 abundance of 46Ti = 8%
How to answer it
Atomic Structure & Isotope Abundance Calculations
This question examines fundamental physical chemistry foundations:
- Precise Definitions: Stating the strict definition of mass number without confusing it with relative atomic mass.
- Subatomic Particle Accounting: Calculating protons, neutrons, and electrons in neutral atoms versus transition metal positive ions ( M²⁺ ).
- Algebraic Isotopic Calculations: Setting up and solving weighted-average equations for multi-isotope systems involving algebraic ratios and remaining percentages.
Definition of Mass Number
Define the mass number of an atom.
✅ Correct Answer
Total number of protons + neutrons (in the nucleus of an atom).
• [1 mark] for stating the sum or total number of protons and neutrons (or nucleons).
❌ Common Errors & Traps
- Mentioning mass: Saying "the mass of protons and neutrons" or "the average mass of an atom" scores 0 marks. Mass number is a discrete integer count, not a mass in grams or atomic mass units.
- Confusing with Ar: Adding standard definitions like "relative to 1/12th the mass of carbon-12" is irrelevant and risks contradictions.
🧠 Exam Technique
Notice the word "number". Whenever asked for atomic number or mass number, your answer must lead with "the number of...".
Subatomic Particles in Atoms and Ions
Complete Table 3 to show the numbers of neutrons and electrons in ⁴⁶Ti and ⁴⁹Ti²⁺.
✅ Completed Table
| Species | Number of protons | Number of neutrons | Number of electrons |
|---|---|---|---|
| ⁴⁶Ti | 22 | 24 | 22 |
| ⁴⁹Ti²⁺ | 22 | 27 | 20 |
• Row 1: 24 neutrons AND 22 electrons [1 mark]
• Row 2: 27 neutrons AND 20 electrons [1 mark]
💡 Key Knowledge
- Protons: Given as 22 (the atomic number Z of titanium).
- Neutrons: Calculated as Mass Number (A) - Proton Number (Z) .
• For ⁴⁶Ti: 46 - 22 = 24
• For ⁴⁹Ti: 49 - 22 = 27 - Electrons:
• In a neutral atom, electrons = protons = 22.
• In a 2+ ion, the species has lost 2 electrons: 22 - 2 = 20.
❌ Common Errors
- Adding electrons for positive charge: Doing 22 + 2 = 24 electrons for Ti²⁺ instead of subtracting.
- Incorrect subtraction: Simple arithmetic slip-ups under timed exam pressure when subtracting 22 from 46 or 49.
Isotope Abundance Calculation
A sample contains ⁴⁶Ti, ⁴⁷Ti, ⁴⁸Ti, and ⁴⁹Ti. Ar = 47.8. Abundance ratio of ⁴⁶Ti : ⁴⁷Ti : ⁴⁹Ti is 2 : 2 : 1. Calculate the percentage abundance of ⁴⁶Ti.
📐 Step-by-Step Calculation (Percentage Method)
Let the percentage abundance of ⁴⁹Ti = y %.
From the 2 : 2 : 1 ratio:
• Abundance of ⁴⁶Ti = 2y %
• Abundance of ⁴⁷Ti = 2y %
The remaining isotope is ⁴⁸Ti, which must make the total percentage 100%:
• Abundance of ⁴⁸Ti = 100 - (2y + 2y + y) = (100 - 5y) %
Ar = [(46 × 2y) + (47 × 2y) + (48 × (100 - 5y)) + (49 × y)] / 100 = 47.8
(Awarded M1)
4780 = 92y + 94y + 4800 - 240y + 49y
4780 = 4800 - 5y
5y = 4800 - 4780 = 20
y = 4
(Awarded M2 for finding 5y = 20 or y = 4)
The question asks for the percentage abundance of ⁴⁶Ti, which is 2y :
Abundance of ⁴⁶Ti = 2 × 4 = 8%
(Awarded M3)
✅ Final Answer
8%
• M1: Correct expression set equal to 47.8
• M2: Correct simplification leading to y = 4 (or n = 20 if using parts)
• M3: Final answer 8%
❌ Common Errors & Pitfalls
- Premature Stopping: Solving for y = 4 and writing 4% as the final answer! Remember that y is the abundance of ⁴⁹Ti, whereas the question asked for ⁴⁶Ti ( 2y ).
- Forgetting ⁴⁸Ti: Students often omit the fourth isotope ⁴⁸Ti from the numerator or fail to express its abundance as (100 - 5y) .
- Sign Errors: Expanding 48 × (100 - 5y) to 4800 + 240y instead of 4800 - 240y .
🧠 Top-Tier Check Strategy
Always substitute your final percentages back into the Ar formula to verify:
- ⁴⁶Ti = 8%
- ⁴⁷Ti = 8%
- ⁴⁸Ti = 100 - (5 × 4) = 80%
- ⁴⁹Ti = 4%
Ar = (46×8 + 47×8 + 48×80 + 49×4) / 100 = (368 + 376 + 3840 + 196) / 100 = 4780 / 100 = 47.8 — Perfect match!
Topics
Physical Chemistry · 3.1.1 Atomic Structure
Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.