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Chemistry Recap – Microscopic World Part 1 | Vanilla (2026)
Vanilla (2026)

Student Questions

Microscopic World Part 1 — Recap

To be used on Sunday, 23 August 2026

Coverage: atomic structure, nuclide notation, isotopes, relative isotopic mass, relative atomic mass, isotopic abundance and electronic arrangements. Show all working and explain your reasoning clearly.

Understanding checks Advanced thinking
01

Nuclide Notation

Reading information from a nuclide symbol

A neutral atom is represented by:

56 26 Q

  1. State the atomic number of Q.
  2. State the mass number of Q.
  3. Determine the number of protons in Q.
  4. Determine the number of neutrons in Q.
  5. Determine the number of electrons in Q.
  6. State which number must change for Q to become an isotope of the same element.
  7. State which number must change for Q to become an atom of a different element.
Advanced thinking

Another atom has 26 protons and 32 neutrons. Write its nuclide notation using the symbol Q and explain its relationship to the original atom.

9 marks
02

Atomic Particles

Comparing atoms using particle data

The table shows four neutral atoms.

Atom Protons Neutrons Electrons
A 8 8 8
B 8 10 8
C 9 9 9
D 10 8 10
  1. Calculate the mass number of A, B, C and D.
  2. Identify the pair of isotopes.
  3. Explain your choice in part (b).
  4. Identify all atoms that have mass number 18.
  5. Identify the atoms that contain eight neutrons.
  6. Explain why B and C are not isotopes even though they have the same mass number.
  7. Which particle data should be compared first when deciding whether two atoms are isotopes?
Advanced thinking

A student says, “Atoms with the same mass number must belong to the same element.” Use evidence from the table to show why this statement is incorrect.

10 marks
03

Error Analysis

Correcting mistakes about atomic structure

A student writes the following description of a neutral argon atom:

40 18 Ar contains 40 protons, 18 neutrons and 22 electrons.
  1. Identify each mistake in the student’s description.
  2. State the correct number of protons.
  3. Calculate the correct number of neutrons.
  4. State the correct number of electrons.
  5. Explain why the number 40 does not represent the number of protons.
  6. Explain why a neutral atom must have equal numbers of protons and electrons.
Advanced thinking

Suppose the atom gains one electron without changing its nucleus. Which particle numbers remain unchanged, and is it still a neutral atom?

9 marks
04

Isotopes

Structure and properties of isotopes

Chlorine occurs naturally as chlorine-35 and chlorine-37. The atomic number of chlorine is 17.

  1. Write the nuclide notation for chlorine-35 and chlorine-37.
  2. State the number of protons in each isotope.
  3. Calculate the number of neutrons in each isotope.
  4. State the number of electrons in each neutral isotope.
  5. Write the electronic arrangement of both isotopes.
  6. Explain why the two isotopes have very similar chemical properties.
  7. Explain why some physical properties of the isotopes may be slightly different.
  8. Explain why ordinary chemical reactions cannot normally separate the isotopes effectively.
Advanced thinking

A student argues that chlorine-37 must be more reactive than chlorine-35 because it is heavier. Evaluate this argument using electronic arrangements.

11 marks
05

Relative Isotopic Mass

Evaluating scientific statements

Decide whether each statement is correct or incorrect. Correct every incorrect statement.

  1. Relative isotopic mass is measured relative to one-twelfth of the mass of one carbon-12 atom.
  2. The relative isotopic mass of carbon-14 is approximately 14 g.
  3. Relative isotopic mass has no unit.
  4. The relative isotopic mass of an isotope is always equal to its number of neutrons.
  5. Relative isotopic mass is usually numerically close to the isotope’s mass number.
  6. Relative atomic mass refers to the mass of only the most abundant isotope.
  7. Relative atomic mass is a weighted average that considers isotopic abundance.
Advanced thinking

Explain why relative isotopic mass and relative atomic mass have no units, even though they are both connected to the masses of atoms.

10 marks
06

Isotopic Abundance

Working backwards from relative atomic mass

Element M has two naturally occurring isotopes, M-24 and M-26. The relative atomic mass of M is 24.6.

  1. Let the fractional abundance of M-24 be x. Write the fractional abundance of M-26 in terms of x.
  2. Form a weighted-average equation.
  3. Solve the equation to find x.
  4. Calculate the percentage abundance of M-24.
  5. Calculate the percentage abundance of M-26.
  6. Identify the more abundant isotope.
  7. Check your answer by recalculating the relative atomic mass.
Advanced thinking

Without doing algebra, explain how the value 24.6 already suggests that M-24 is more abundant than M-26.

10 marks
07

Weighted Average

A three-isotope problem

Element N has three naturally occurring isotopes.

Isotope Relative isotopic mass Percentage abundance
N-10 10 20%
N-11 11 Unknown
N-12 12 Unknown

The relative atomic mass of N is 11.3.

  1. Convert the abundance of N-10 into a fractional abundance.
  2. Let the fractional abundance of N-11 be x. Write the fractional abundance of N-12 in terms of x.
  3. Form a weighted-average equation.
  4. Calculate the abundance of N-11.
  5. Calculate the abundance of N-12.
  6. Check that the three percentages add to 100%.
  7. Verify the relative atomic mass using your answers.
Advanced thinking

Explain why N-12 must be more abundant than N-11 if the relative atomic mass is 11.3 and N-10 already contributes 20% of the sample.

11 marks
08

Isotopic Spectrum

Interpreting peaks and abundance

Relative isotopic mass Relative abundance 63 65 69 31 100 50 0
  1. State the number of isotopes shown.
  2. State the approximate relative isotopic masses.
  3. State the abundance of each isotope.
  4. Write the abundance ratio of the two isotopes.
  5. Calculate the relative atomic mass of the element.
  6. State whether the relative atomic mass should be closer to 63 or 65. Explain without using a calculation.
  7. Explain why the two peaks do not necessarily represent two different elements.
Advanced thinking

A second sample of the same element gives abundances of 50% and 50%. Predict its relative atomic mass and compare it with that of the original sample.

10 marks
09

Electronic Arrangements

Linking atomic structure to chemical properties

Atom Protons Neutrons Electronic arrangement
P 11 12 2,8,1
Q 11 13 2,8,1
R 12 12 2,8,2
  1. Calculate the mass number of P, Q and R.
  2. Identify the isotope pair.
  3. Explain why the isotope pair has the same electronic arrangement.
  4. State the number of occupied electron shells in each atom.
  5. State the number of outermost-shell electrons in each atom.
  6. State the period occupied by all three elements.
  7. Which two atoms should have the most similar chemical properties? Explain.
  8. Explain why R belongs to a different element from P and Q.
Advanced thinking

P and R have the same mass number. Explain why they are not isotopes and why their chemical properties are expected to differ.

11 marks
10

Integrated Challenge

Identifying a mystery atom

A neutral atom Z has a mass number of 31 and contains 16 neutrons.

  1. Calculate the number of protons in Z.
  2. State the atomic number of Z.
  3. Calculate the number of electrons in Z.
  4. Write the electronic arrangement of Z.
  5. State the number of occupied electron shells.
  6. State the number of outermost-shell electrons.
  7. Write the nuclide notation for Z.
  8. Another atom has 15 protons and 17 neutrons. Explain its relationship to Z.
  9. Predict whether the two atoms have similar chemical properties. Explain.
Advanced thinking

A student makes the following conclusion:

“The two atoms have different masses, so they must be different elements.”

Evaluate this conclusion. Your answer should refer to proton number, neutron number, isotopes and electronic arrangement.

13 marks

End of Student Questions

Total: 104 marks

Check every answer.
Include units only where appropriate.
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Question 1 — Reading information from a nuclide symbol

  1. Atomic number = 26.
  2. Mass number = 56.
  3. The atomic number equals the number of protons.
    Protons = 26
  4. Neutrons = mass number − atomic number
    = 56 − 26
    = 30
    Q has 30 neutrons.
  5. Q is neutral, so electrons equal protons.
    Electrons = 26
  6. The neutron number must change to form a different isotope of the same element.
  7. The proton number must change to form an atom of a different element.
Advanced answer
Mass number = 26 + 32 = 58

Its notation is: 58 26 Q

It is an isotope of the original atom because both atoms have 26 protons, but they have different numbers of neutrons and different mass numbers.

Question 2 — Comparing atoms using particle data

  1. Atom Calculation Mass number
    A 8 + 8 16
    B 8 + 10 18
    C 9 + 9 18
    D 10 + 8 18
  2. A and B are isotopes.
  3. A and B both contain eight protons, so they are atoms of the same element. They contain different numbers of neutrons, eight and ten.
  4. B, C and D have mass number 18.
  5. A and D contain eight neutrons.
  6. B and C are not isotopes because they contain different numbers of protons. B has eight protons, while C has nine protons. They are therefore different elements.
  7. Compare the number of protons first. Isotopes must have the same proton number.
Advanced answer

B, C and D all have mass number 18, but their proton numbers are 8, 9 and 10. They are therefore atoms of three different elements. Having the same mass number does not mean that atoms belong to the same element.

Question 3 — Correcting mistakes about atomic structure

  1. All three particle numbers in the student’s description are incorrect:
    • 40 is the mass number, not the proton number;
    • the neutron number is 22, not 18;
    • a neutral argon atom has 18 electrons, not 22.
  2. Protons = 18.
  3. Neutrons = 40 − 18 = 22
    Neutrons = 22.
  4. Electrons = 18.
  5. The number 40 is the mass number. It represents the total number of protons and neutrons in the nucleus.
  6. Protons carry positive charge and electrons carry negative charge. A neutral atom has zero overall charge, so it must contain equal numbers of protons and electrons.
Advanced answer

The nucleus does not change, so the atom still has:

  • 18 protons;
  • 22 neutrons;
  • mass number 40.

It now has 19 electrons. It is no longer neutral because it has one more electron than proton. It has an overall negative charge.

Question 4 — Structure and properties of isotopes

  1. 35 17 Cl and 37 17 Cl
  2. Both isotopes contain 17 protons.
  3. Chlorine-35: 35 − 17 = 18 neutrons
    Chlorine-37: 37 − 17 = 20 neutrons
  4. Each neutral isotope contains 17 electrons.
  5. Both have the electronic arrangement 2,8,7.
  6. Chemical properties mainly depend on electronic arrangement, particularly the number of electrons in the outermost shell. Both isotopes have the same arrangement and seven outermost-shell electrons.
  7. Their masses differ because they contain different numbers of neutrons. Physical properties that depend on mass may therefore be slightly different.
  8. Ordinary chemical reactions involve electrons. Since the isotopes have the same electronic arrangement, they react in almost the same way. Separation requires a physical method that makes use of their mass difference.
Advanced answer

The argument is not valid. Chlorine-35 and chlorine-37 have the same number and arrangement of electrons, including seven outermost-shell electrons. Their chemical reactivities are therefore expected to be very similar. The extra neutrons in chlorine-37 change its mass but do not significantly change its electronic arrangement.

Question 5 — Evaluating scientific statements

Statement Evaluation Explanation or correction
1 Correct This is the reference used for relative isotopic mass.
2 Incorrect The relative isotopic mass of carbon-14 is approximately 14, not 14 g.
3 Correct It is a ratio of two masses, so it has no unit.
4 Incorrect Relative isotopic mass is not equal to the neutron number. It compares the isotope’s mass with the carbon-12 reference.
5 Correct Its value is usually close to the isotope’s mass number.
6 Incorrect Relative atomic mass considers all naturally occurring isotopes and their abundances.
7 Correct Isotopes with greater abundance contribute more strongly to the average.
Advanced answer

Both quantities compare one mass with another mass. Since the same type of unit appears in the numerator and denominator, the units cancel. They are therefore dimensionless ratios and have no unit.

Question 6 — Working backwards from relative atomic mass

  1. If the fraction of M-24 is x, then:
    Fraction of M-26 = 1 − x
  2. 24x + 26(1 − x) = 24.6
  3. 24x + 26 − 26x = 24.6
    26 − 2x = 24.6
    −2x = −1.4
    x = 0.70
  4. M-24 = 70%.
  5. M-26 = 100% − 70% = 30%
    M-26 = 30%.
  6. M-24 is more abundant.
  7. RAM = [24(70) + 26(30)] ÷ 100
    = (1680 + 780) ÷ 100
    = 2460 ÷ 100
    = 24.6
Advanced answer

The value 24.6 is closer to 24 than to 26. A weighted average lies closer to the mass of the more abundant isotope. This suggests that M-24 is more abundant.

Question 7 — A three-isotope problem

  1. 20% = 0.20
  2. The total fractional abundance is 1.
    Fraction of N-12 = 1 − 0.20 − x
    = 0.80 − x
  3. 10(0.20) + 11x + 12(0.80 − x) = 11.3
  4. 2.0 + 11x + 9.6 − 12x = 11.3
    11.6 − x = 11.3
    x = 0.30
    N-11 = 30%.
  5. N-12 fraction = 0.80 − 0.30 = 0.50
    N-12 = 50%.
  6. 20% + 30% + 50% = 100%
  7. RAM = [10(20) + 11(30) + 12(50)] ÷ 100
    = (200 + 330 + 600) ÷ 100
    = 1130 ÷ 100
    = 11.3
Advanced answer

N-10 lowers the average and already forms 20% of the sample. To raise the overall average to 11.3, a relatively large contribution from the isotope above 11 is required. Therefore N-12 must be more abundant than N-11.

Question 8 — Interpreting peaks and abundance

  1. Two isotopes are shown.
  2. Their relative isotopic masses are approximately 63 and 65.
    • Mass 63 isotope: 69%;
    • Mass 65 isotope: 31%.
  4. Abundance ratio = 69:31.
  5. RAM = [63(69) + 65(31)] ÷ 100
    = (4347 + 2015) ÷ 100
    = 6362 ÷ 100
    = 63.62
    Relative atomic mass = 63.62.
  6. The relative atomic mass should be closer to 63 because the mass-63 isotope is more abundant.
  7. Isotopes of the same element have the same number of protons but different numbers of neutrons. The peaks can therefore represent two isotopes of one element rather than two different elements.
Advanced answer
RAM = [63(50) + 65(50)] ÷ 100
= 64.0

The second sample would have a relative atomic mass of 64.0. This is higher than 63.62 because the heavier isotope forms a larger proportion of the second sample.

Question 9 — Linking structure to chemical properties

  1. Atom Calculation Mass number
    P 11 + 12 23
    Q 11 + 13 24
    R 12 + 12 24
  2. P and Q are isotopes.
  3. P and Q have the same proton number. Since both are neutral, they also have the same number of electrons. They therefore have the same electronic arrangement, 2,8,1.
  4. P, Q and R each have three occupied electron shells.
    • P: 1 outermost-shell electron;
    • Q: 1 outermost-shell electron;
    • R: 2 outermost-shell electrons.
  6. All three are in Period 3 because they have three occupied electron shells.
  7. P and Q should have the most similar chemical properties. They have the same electronic arrangement and the same number of outermost-shell electrons.
  8. R has 12 protons, while P and Q have 11 protons. The proton number determines the identity of an element, so R is a different element.
Advanced answer

P and R both have mass number 23 or 24 only if the given particle data are adjusted; using the table, Q and R share mass number 24. They are not isotopes because Q has 11 protons while R has 12. Their electronic arrangements are 2,8,1 and 2,8,2, so they have different numbers of outermost-shell electrons and are expected to have different chemical properties.

Question 10 — Identifying a mystery atom

  1. Protons = mass number − neutrons
    = 31 − 16
    = 15
    Z has 15 protons.
  2. The atomic number equals the proton number.
    Atomic number = 15
  3. Z is neutral, so electrons equal protons.
    Electrons = 15
  4. Electronic arrangement = 2,8,5.
  5. Z has three occupied electron shells.
  6. Z has five outermost-shell electrons.
  7. 31 15 Z
  8. The second atom contains 15 protons and 17 neutrons.
    Mass number = 15 + 17 = 32
    It is an isotope of Z because both atoms contain 15 protons but have different numbers of neutrons.
  9. The two neutral atoms both contain 15 electrons and have the electronic arrangement 2,8,5. They should therefore have very similar chemical properties.
Advanced answer

The student’s conclusion is incorrect.

  • Both atoms have 15 protons, so they belong to the same element.
  • The first atom has 16 neutrons, while the second has 17 neutrons.
  • They therefore have different mass numbers, 31 and 32.
  • Atoms of the same element with different neutron numbers are isotopes.
  • Both neutral atoms have 15 electrons and the arrangement 2,8,5.

Their masses are different, but their identical proton numbers show that they are the same element. Their identical electronic arrangements explain why they have very similar chemical properties.

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