Chemistry Recap Questions to be used on 18/8/2026

Question 1 — Interpreting nuclide notation

1. Atomic number = 20.
2. Mass number = 41.
3. Protons = 20.
4. Neutrons = mass number − atomic number.
   41 − 20 = 21
   The atom has 21 neutrons.
5. A neutral atom has equal numbers of protons and electrons.
   Electrons = 20
   The atom has 20 electrons.
6. The number of protons determines the element’s identity. Changing the neutron number only produces a different isotope.

Question 2 — Comparing atomic particles

1. Atom	Mass number
   A	12 + 12 = 24
   B	12 + 13 = 25
   C	13 + 12 = 25
   D	14 + 14 = 28

2. A and B.
3. They have the same number of protons, 12, but different numbers of neutrons, 12 and 13.
4. A and C both contain 12 neutrons.
5. B and C have different proton numbers, 12 and 13. They are therefore atoms of different elements, even though both have mass number 25.

Question 3 — Properties of isotopes

1. All three have six protons.
2. They have different numbers of neutrons:
   • carbon-12: 6 neutrons;
   • carbon-13: 7 neutrons;
   • carbon-14: 8 neutrons.
3. Each neutral isotope has six electrons because each has six protons. They therefore have the same electronic arrangement, 2,4.
4. Their chemical properties are very similar because chemical behaviour mainly depends on electronic arrangement, especially the outermost-shell electrons.
5. Their masses differ because they contain different numbers of neutrons. Properties depending on mass may therefore differ.
6. Not normally. Chemical reactions depend mainly on electrons, and the isotopes have the same electronic arrangement. Physical methods based on mass differences are required.

Question 4 — Relative isotopic mass

Relative isotopic mass has no unit because it is a ratio of two masses. The mass units cancel.

Question 5 — Finding isotopic abundance

1. Fraction of X-79 = x.
   Fraction of X-81 = 1 − x
2. 79x + 81(1 − x) = 79.9
3. 79x + 81 − 81x = 79.9
   
   −2x = −1.1
   
   x = 0.55
   X-79 = 55% and X-81 = 45%.
4. X-79 is more abundant.
5. Relative atomic mass is the weighted average of the masses of all naturally occurring isotopes. Different isotope masses and abundances usually produce a non-whole-number average.

Question 6 — Three-isotope calculation

1. The fractional abundance of Y-21 is 0.14. The total abundance of Y-20 and Y-22 is 0.86.
   Fraction of Y-22 = 0.86 − x
2. 20x + 21(0.14) + 22(0.86 − x) = 20.86
3. 20x + 2.94 + 18.92 − 22x = 20.86
   
   21.86 − 2x = 20.86
   
   2x = 1.00
   
   x = 0.50
   Y-20 = 50%.
4. Y-22 fraction = 0.86 − 0.50 = 0.36
   Y-22 = 36%.
5. RAM = [20(50) + 21(14) + 22(36)] ÷ 100
   
   = (1000 + 294 + 792) ÷ 100
   
   = 2086 ÷ 100
   
   = 20.86
   The calculated relative atomic mass is 20.86.

Question 7 — Interpreting an isotopic spectrum

1. Two isotopes.
2. Relative isotopic masses of approximately 35 and 37.
3. 75:25 = 3:1
   The abundance ratio is 3:1.
4. RAM = [35(75) + 37(25)] ÷ 100
   
   = (2625 + 925) ÷ 100
   
   = 35.5
   Relative atomic mass = 35.5.
5. Isotopes of the same element have the same number of protons but different numbers of neutrons. The two peaks represent different isotopic masses, not necessarily different elements.

Question 8 — Electronic arrangements and isotopes

1. Atom	Mass number
   U	6 + 6 = 12
   V	6 + 8 = 14
   W	7 + 7 = 14

2. U and V.
3. U and V both have six protons. Since they are neutral, both also have six electrons. Their electronic arrangement is therefore 2,4.
4. U, V and W each have two occupied electron shells.
5. • U: 4 outermost-shell electrons;
   • V: 4 outermost-shell electrons;
   • W: 5 outermost-shell electrons.
6. U and V. They have the same electronic arrangement, so they are expected to have very similar chemical properties.

Question 9 — Reconstructing an atom

1. Atomic number = 14.
2. Mass number = protons + neutrons
   
   = 14 + 15
   
   = 29
   Mass number = 29.
3. It is neutral, so electrons = protons.
   Electrons = 14
   The atom has 14 electrons.
4. Three occupied electron shells.
5. Four outermost-shell electrons.
6. 29 14 X
7. Both atoms have 14 protons and are therefore the same element. They have different neutron numbers, 15 and 16, so they are isotopes. The second atom has mass number 30.

Question 10 — Dalton’s atomic theory and modern evidence

1. Statements that remain acceptable in the intended context include:
   • Statement 2: elements are identified by their number of protons.
   • Statement 4: chemical reactions involve the rearrangement of atoms.
   • Statement 5: acceptable for ordinary chemical reactions because chemical reactions do not change atomic nuclei.
2. Isotopes challenge Statements 1 and 6:
   • Atoms of the same element need not be completely identical.
   • Isotopes have different neutron numbers and therefore different masses.
3. Statement 3 is challenged by the discovery of protons, neutrons and electrons. Atoms have an internal structure and are therefore not indivisible.
4. In a chemical reaction, electrons are transferred or shared and atoms are rearranged. Proton numbers do not change. In a nuclear process, the nucleus changes. If the proton number changes, the atom becomes an atom of a different element.