Isotopes & Average Atomic Mass — Practice · Grade 11 Science

Chapter: Atoms, Electrons & Periodicity

What this practice covers

By the end of Isotopes & Average Atomic Mass you should be able to:

10 questions, multiple choice, numeric and written answer, each with a worked explanation.

Vocabulary used: isotopes, percent abundance, fractional abundance, average atomic mass, weighted average, mass spectrometer.

Problem 1

Why is the atomic mass listed for most elements on the periodic table a decimal rather than a whole number?
  1. It is the weighted average of all the element's naturally occurring isotopes
  2. Electrons add a fractional amount of mass
  3. Atoms are constantly gaining and losing neutrons
  4. Measuring instruments are imprecise
Answer and explanation

It is the weighted average of all the element's naturally occurring isotopes

The listed value averages every natural isotope, weighted by abundance, so it is rarely a whole number.

Problem 2

Two isotopes of the same element must have the same number of ___ but different numbers of ___.
  1. neutrons; protons
  2. protons; neutrons
  3. electrons; protons
  4. protons; electrons
Answer and explanation

protons; neutrons

Isotopes share the proton count (same element) but differ in neutrons, which changes the mass number.

Problem 3

An element has two isotopes: one of mass 20.0 amu at 90.0% abundance and one of mass 22.0 amu at 10.0% abundance. Find the average atomic mass in amu.
Answer and explanation

20.2 amu

Average = 20.0 × 0.900 + 22.0 × 0.100 = 18.0 + 2.2 = 20.2 amu.

Problem 4

Copper has two isotopes: copper-63 (62.93 amu, 69.15%) and copper-65 (64.93 amu, 30.85%). Find copper's average atomic mass in amu.
Answer and explanation

63.55 amu

62.93 × 0.6915 = 43.52 amu; 64.93 × 0.3085 = 20.03 amu; sum = 63.55 amu.

Problem 5

Chlorine's average atomic mass, 35.45 amu, is much closer to 35 than to 37. What does that tell you about the abundances of its two isotopes?
Answer and explanation

Chlorine-35 must be far more abundant than chlorine-37 (roughly 75% versus 25%), because a weighted average is pulled toward the mass of the most common isotope.

Problem 6

An element has just two isotopes, of mass 10.0 amu and 11.0 amu, and its average atomic mass is 10.8 amu. What percent of its atoms are the mass-11 isotope?
Answer and explanation

80 %

Let the mass-10 fraction be x: 10x + 11(1 − x) = 10.8 → 11 − x = 10.8 → x = 0.20. So the mass-11 fraction is 0.80 = 80%.

Problem 7

Masses 10.0 and 11.0 u have abundances 20% and 80%. Find average. Show the reasoning, not just the conclusion.
Hints

An arithmetic average would incorrectly assume equal abundance.

Answer and explanation

10.0×0.20+11.0×0.80=10.8 u.

Problem 8

Could two samples of the same element have different average masses?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Yes, if their isotope abundances differ; proton count remains the same.

Problem 9

A two-isotope model can infer a fraction from its meanA two-isotope model can infer a fraction from its meanx = fraction of X-101 − x = fraction of X-1110.4 = 10.0x + 11.0(1 − x)10.4 = 11.0 − x → x = 0.60Check: 60% lighter, 40% heavier → mean below midpoint.
Original inverse calculation with invented masses and selected mean, keeping number fractions distinct from mass fractions.
Interpret the supplied visual. If the mean moved closer to 11.0 u, which modeled fraction would increase?
Answer and explanation

The X-11 fraction would increase, while the X-10 fraction x would decrease.

To infer an unknown abundance in a two-isotope model, define one fraction x and use one minus x for the other. The measured mean supplies a weighted-balance equation. Solving it is justified only if the two isotope masses and the sample model are appropriate; an unrecognized third isotope, mass bias or uncertainty changes the inference. A solution outside zero to one is a warning that the inputs or assumptions are inconsistent, not a physically negative abundance.

The X-11 fraction would increase, while the X-10 fraction x would decrease.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 10

Two modeled isotopes have masses 24.0 and 26.0 u. A sample mean is 24.6 u. Find the fraction of each and check the answer.
Answer and explanation

24.6=24.0x+26.0(1−x)=26.0−2.0x, so x=0.70. Fractions are 70% lighter and 30% heavier; 0.70(24.0)+0.30(26.0)=24.6 u.

24.6=24.0x+26.0(1−x)=26.0−2.0x, so x=0.70. Fractions are 70% lighter and 30% heavier; 0.70(24.0)+0.30(26.0)=24.6 u.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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