---
title: "A pure sample of chlorine is analyzed using mass spectrometry, producing the mass spectrum shown. Based on the mass spectrum, which of the following is the best estimate of the average atomic mass of chlorine, and what is the correct justification?"
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url: "https://nerd-notes.com/ubq/119629/"
date_modified: "2026-08-21T08:11:51+00:00"
---

# A pure sample of chlorine is analyzed using mass spectrometry, producing the mass spectrum shown. Based on the mass spectrum, which of the following is the best estimate of the average atomic mass of chlorine, and what is the correct justification?

A pure sample of chlorine is analyzed using mass spectrometry, producing the mass spectrum shown. Based on the mass spectrum, which of the following is the best estimate of the average atomic mass of chlorine, and what is the correct justification?

![A mass spectrum graph rendered strictly in grayscale. The horizontal axis is labeled Mass-to-Charge Ratio (m/z) with tick marks and integer labels at 34, 35, 36, 37, and 38. The vertical axis is labeled Relative Abundance (%) with tick marks and labels at 0, 25, 50, 75, and 100. Horizontal gridlines extend from each vertical axis tick mark. There are exactly two vertical solid black bars: the first bar is centered at m/z = 35 with a height reaching exactly 75%, and the second bar is centered at m/z = 37 with a height reaching exactly 25%. No other peaks, lines, labels, text, or annotations appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787299911-VPe2yQ.jpg)

- **A.** \(36.0\text{ u}\), because the average atomic mass is the unweighted arithmetic mean of the two mass numbers, \(\dfrac{35 + 37}{2}\).
- **B.** \(36.0\text{ u}\), because the two isotopes are present in equal natural abundances in the sample.
- **C.** \(35.5\text{ u}\), because the isotope with mass number \(37\) is approximately three times as abundant as the isotope with mass number \(35\).
- **D.** \(35.5\text{ u}\), because the isotope with mass number \(35\) is approximately three times as abundant as the isotope with mass number \(37\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119629/*