A sample of oxygen contains two isotopes of oxygen with masses 16 u an

A sample of oxygen contains two isotopes of oxygen with masses 16 u and 18 u respectively. The proportion of these isotopes in the sample is 3 : 1. What will be the average atomic mass of oxygen in this sample?

17.5 u
17 u
16 u
16.5 u
This question was previously asked in
UPSC NDA-2 – 2018
The average atomic mass of oxygen in this sample is 16.5 u.
– Average atomic mass is calculated as the weighted average of the masses of its isotopes based on their relative abundance.
– The sample contains isotopes with masses 16 u and 18 u in a proportion of 3:1.
– This means out of 4 parts, 3 parts have mass 16 u and 1 part has mass 18 u.
– Relative abundance of 16u isotope = 3 / (3+1) = 3/4 = 0.75
– Relative abundance of 18u isotope = 1 / (3+1) = 1/4 = 0.25
– Average atomic mass = (Mass₁ × Abundance₁) + (Mass₂ × Abundance₂)
– Average atomic mass = (16 u × 0.75) + (18 u × 0.25)
– Average atomic mass = 12 u + 4.5 u = 16.5 u.
Naturally occurring oxygen is predominantly isotope ¹⁶O (around 99.76%), with smaller amounts of ¹⁷O (around 0.038%) and ¹⁸O (around 0.205%). The standard average atomic mass of oxygen listed on the periodic table is approximately 15.999 u, reflecting the natural isotopic abundance. This question presents a hypothetical sample with a specific, non-natural isotopic ratio.