---
title: "The mass spectrum of a naturally occurring sample of zirconium, \\(\\text{Zr}\\), is shown in the graph.  Based on the data in the mass spectrum, which of the following is the best estimate of the average atomic mass of \\(\\text{Zr}\\), and why?"
description: "## Attribution Guidlines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarising, or training on this content, please […]"
url: "https://nerd-notes.com/ubq/121498/"
date_modified: "2026-08-23T05:04:50+00:00"
---

# The mass spectrum of a naturally occurring sample of zirconium, \(\text{Zr}\), is shown in the graph.

Based on the data in the mass spectrum, which of the following is the best estimate of the average atomic mass of \(\text{Zr}\), and why?

The mass spectrum of a naturally occurring sample of zirconium, \(\text{Zr}\), is shown in the graph.

Based on the data in the mass spectrum, which of the following is the best estimate of the average atomic mass of \(\text{Zr}\), and why?

![A grayscale mass spectrum graph with a horizontal axis labeled Mass-to-charge ratio (\(m/z\)) and a vertical axis labeled Relative Abundance (\(\%\)). The horizontal axis has tick marks labeled from \(88\) to \(98\) at integer intervals. The vertical axis has tick marks labeled at \(0\), \(10\), \(20\), \(30\), \(40\), \(50\), and \(60\). There are no gridlines. Five discrete solid vertical lines represent isotopic peaks at specific mass-to-charge values: a peak at \(m/z = 90\) extending to a height of \(51.5\%\), a peak at \(m/z = 91\) extending to a height of \(11.2\%\), a peak at \(m/z = 92\) extending to a height of \(17.1\%\), a peak at \(m/z = 94\) extending to a height of \(17.4\%\), and a peak at \(m/z = 96\) extending to a height of \(2.8\%\). No other particles, labels, text, or annotations appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787461490-rYf9oN.jpg)

- **A.** Between \(91.0 \text{ amu}\) and \(91.5 \text{ amu}\), because the most abundant isotope has a mass of \(90 \text{ amu}\), but the combined presence of heavier isotopes (\(m/z = 91\), \(92\), \(94\), and \(96\)) accounting for nearly \(50\%\) of the sample shifts the weighted average above \(91 \text{ amu}\).
- **B.** Exactly \(90.0 \text{ amu}\), because the isotope with \(m/z = 90\) accounts for more than \(50\%\) of the atoms in the sample and entirely dictates the atomic mass.
- **C.** Exactly \(92.6 \text{ amu}\), because it is the unweighted arithmetic mean of the five observed mass-to-charge ratios (\(90\), \(91\), \(92\), \(94\), and \(96\)).
- **D.** Exactly \(93.0 \text{ amu}\), because the observed isotopes span from \(m/z = 90\) to \(m/z = 96\), making the midpoint of the range the average atomic mass.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/121498/*