Chemistry 9701/11 — October/November 2021

Background Concept

The relative atomic mass ($A_r$) of an element is the weighted average mass of its naturally occurring isotopes, relative to one-twelfth the mass of a carbon-12 atom. In mass spectrometry, a sample is ionised and accelerated, then deflected by a magnetic field. Ions with different masses (isotopes) follow different paths and are detected at different mass-to-charge ratios (m/z). For singly charged ions ($z = +1$), the m/z value is numerically equal to the isotopic mass. The height of each peak (relative abundance) represents the proportion of that isotope present in the sample. To find $A_r$, we calculate the mean of these isotopic masses, weighting each by its relative abundance.

Understanding the Question

The question provides a mass spectrum of neon with three peaks at m/z = 20, 21, and 22. The numbers in brackets above each peak are their relative abundances: 100, 0.3, and 8. We are asked to calculate the relative atomic mass ($A_r$) of this specific sample of neon and select the correct value from the given options (A: 20.15, B: 20.20, C: 21.00, D: 21.82).

Approach

The strategy is to apply the formula for the weighted average:

$$A_r = \frac{\sum (\text{isotopic mass} \times \text{relative abundance})}{\sum \text{relative abundance}}$$

Since the ions are singly charged ($z=+1$), the m/z values (20, 21, 22) are used directly as the isotopic masses. We multiply each mass by its corresponding relative abundance, sum these products to get the numerator, sum the relative abundances to get the denominator, and divide to find the final $A_r$.

Step-by-Step Reasoning

1. Identify isotopic masses and abundances: From the spectrum, the isotopes have masses 20, 21, and 22 with relative abundances 100, 0.3, and 8 respectively.
2. Calculate the numerator (sum of mass × abundance):
   $(20 \times 100) + (21 \times 0.3) + (22 \times 8) = 2000 + 6.3 + 176 = 2182.3$
3. Calculate the denominator (total relative abundance):
   $100 + 0.3 + 8 = 108.3$
4. Divide to find $A_r$:
   $A_r = \frac{2182.3}{108.3} \approx 20.1505...$
5. Match with options: Rounding to two decimal places gives 20.15, which matches option A.

Key Takeaways

• The relative atomic mass is always a weighted average, not a simple arithmetic mean. The denominator must be the sum of the relative abundances, not 100 or the number of isotopes.
• In mass spectrometry, for singly charged ions, the m/z value is the isotopic mass.

Common Mistakes

• Forgetting the denominator: Simply summing the products ($2182.3$) and forgetting to divide by the total abundance ($108.3$). This would lead to an incorrect answer like 2182.3.
• Using a simple average: Adding the masses and dividing by 3 ($\frac{20+21+22}{3} = 21.00$), which ignores the relative abundances. This matches distractor C.
• Ignoring the smaller peaks: If the 0.3 peak is ignored, the calculation becomes $\frac{2000 + 176}{108} = 20.15$, which coincidentally gives the right answer here, but is conceptually wrong. If the 8 peak is ignored, $\frac{2000 + 6.3}{100.3} = 19.95$, which is incorrect.

Things to Be Careful About

• Significant figures: The final answer should be given to an appropriate number of decimal places, typically matching the precision of the options or the input data. Here, two decimal places (20.15) is correct.
• State symbols and charges: Ensure you correctly identify the charge of the ions. If the charge were +2, the isotopic mass would be $2 \times \text{m/z}$, but for standard mass spectrometry of noble gases, the charge is +1.
• Total abundance: Always verify that the denominator is the sum of all relative abundances, especially when they do not sum to 100 (as is the case here with 108.3).