Chlorine occurs in nature in two isotopic forms of masses 35 u and 37 u in the ratio of 3 : 1 respectively. What is the average atomic mass of the Chlorine atom?
- (a)36.1 u
- (b)35.5 u
- (c)36.5 u
- (d)35.1 u
Answer
Why
Correct — B, 35.5 u. The average atomic mass is the weighted mean of the isotopic masses in their natural ratio. With Cl-35 and Cl-37 in the ratio 3 : 1, the average is (3 × 35 + 1 × 37) ÷ 4 = (105 + 37) ÷ 4 = 142 ÷ 4 = 35.5 u. This is why chlorine's atomic mass is listed as about 35.5 rather than a whole number.
Why the others are wrong
- (a)36.1 u — This is not the 3 : 1 weighted mean; it would require a different abundance ratio than the one given.
- (c)36.5 u — 36.5 comes from weighting the isotopes wrongly; a simple 1 : 1 mix gives 36, while the 3 : 1 ratio gives 35.5.
- (d)35.1 u — This is below the correct weighted mean; the arithmetic 142 ÷ 4 gives 35.5, not 35.1.
Concept
Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons, and so different masses. The average atomic mass of an element is the weighted mean of its isotopic masses, each weighted by its fractional abundance.
Weight each isotope by its share, three parts of 35 and one part of 37 over four total parts. The heavier isotope's smaller share pulls the average only slightly above 35, landing at 35.5.
Key facts
- Average atomic mass equals the sum of each isotopic mass times its fractional abundance.
- For Cl-35 and Cl-37 in a 3 : 1 ratio, the average is (3×35 + 1×37) ÷ 4 = 35.5 u.
- This weighted mean explains why atomic masses are often non-integers.
- Isotopes share the same atomic number but differ in mass number.
Weighting Cl-35 and Cl-37 by the ratio 3 : 1 gives 35.5 u.
Study next
Common traps
- Taking a simple average of 36 instead of the 3 : 1 weighted mean.
- Swapping the weights of the two isotopes.
Asked to compute an element's average atomic mass from isotope masses and their abundance ratio; use the weighted mean.
Related PYQs
The relative atomic mass of boron (which exists in two isotopic forms 10B and 11B) is 10·81. What will be the abundance of 10B and 11B, respectively (consider a sample of 100 atoms) ?
- (a) 19% and 81%
- (b) 81% and 19%
- (c) 38% and 62%
- (d) 62% and 38%
Answer(a) 19% and 81%
The same weighted-mean relationship run backwards. That item gives the average atomic mass and asks for isotope abundances; this one gives the abundance ratio and asks for the average mass.
Practice
- practice — not a real PYQ
Two isotopes of masses 10 u and 11 u occur in the ratio 1 : 4. The average atomic mass is
- (a)10.2 u
- (b)10.5 u
- (c)10.8 u
- (d)11.0 u
Answer(c) 10.8 u — (1×10 + 4×11) ÷ 5 = 54 ÷ 5. - practice — not a real PYQ
Atomic masses of elements are often non-integers mainly because
- (a)electrons have mass
- (b)elements exist as a mixture of isotopes
- (c)protons vary in mass
- (d)of measurement error
Answer(b) elements exist as a mixture of isotopes.