Mr Cole Chemistry

Relative Atomic Mass Questions

Relative Atomic Mass

Worksheet

Calculations & Spectra

Strontium exists as four commonly found isotopes, which are in the table below.

Isotope Mass Number Relative Abundance (%)
84Sr 84 0.56
86Sr 86 9.86
87Sr 87 7.00
88Sr 88 82.58
a) Use their abundances and m/z values to calculate the relative atomic mass.

87.7

b) Why is it a good approximation to assume that the M/Z ratio from the mass spectrometer is the same as the relative atomic mass of Sr?

The charge of the ions is 1+ so the M/Z ratio is the same as the mass of the ions. As the mass of the electron is negligible, the mass of the ion and the mass of the atom are the same.

The mass spectrum for bromine is given below.

Mass spectrum for bromine showing peaks at 79, 81, 158, 160, 162
a) Identify the chemical species responsible for each of the five main peaks in the spectrum.
  • 79: an atom of 79Br
  • 81: an atom of 81Br
  • 158: Br2 made of two 79Br atoms
  • 160: Br2 made of one 79Br and one 81Br
  • 162: Br2 made of two 81Br atoms
b) State the mass number of all the isotopes of bromine, and state their relative abundance (in comparison to each other).

79 and 81. 50:50 ratio.

c) Explain why the peaks at 158, 160 and 162 exist in the 1:2:1 pattern as seen in the spectra.

There is a 50% chance that any given atom will be either 79Br or 81Br. Treating each individual atom as independent, the four combinations are:

  • 79Br-79Br (Mass 158)
  • 79Br-81Br (Mass 160)
  • 81Br-79Br (Mass 160)
  • 81Br-81Br (Mass 162)

Because the middle two combinations both result in a mass of 160, the probability is double, resulting in a 1:2:1 ratio.

Chlorine exists as two main isotopes, Cl-35 and Cl-37, which have the rough relative abundance of 75% and 25% respectively.

a) Calculate the relative atomic mass of chlorine.

35.5

b) Predict and sketch the mass spectra for chlorine molecules.

Since the ratio of 35Cl to 37Cl is 3:1, the peaks for the diatomic molecule Cl2 will appear at masses 70, 72, and 74 in a 9:6:1 ratio.

Sketch of chlorine mass spectrum showing 9:6:1 ratio
4. A sample containing the element Zinc has a relative atomic mass of 64.66. The sample contains two isotopes, Zinc-64 and Zinc-66. Calculate the abundance of the two isotopes in the sample.
Working Out
Algebraic Method
Let abundance of 64Zn = x
Therefore, abundance of 66Zn = 100 – x
64.66 = [64x + 66(100 – x)] / 100
6466 = 64x + 6600 – 66x
6466 = 6600 – 2x
2x = 6600 – 6466
2x = 134
x = 67% (64Zn)
100 – 67 = 33% (66Zn)

Answer: 67% Zinc-64 and 33% Zinc-66

5. A sample containing strontium contains three isotopes: Strontium-84, Strontium-86, and Strontium-88. The sample has a relative atomic mass of 84.68 and contains 76.0% of the Strontium-84 isotope. Calculate the percentage abundance of each of the other two isotopes.
Working Out
Algebraic Method
84Sr = 76%
Remaining Abundance = 100 – 76 = 24%
Let 86Sr = x
Let 88Sr = 24 – x
84.68 = [(84 × 76) + 86x + 88(24 – x)] / 100
8468 = 6384 + 86x + 2112 – 88x
8468 = 8496 – 2x
2x = 8496 – 8468
2x = 28
x = 14% (86Sr)
24 – 14 = 10% (88Sr)

Answer: 14% Strontium-86 and 10% Strontium-88

Return to Topic