Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 23

Average atomic masses listed by IUPAC are based on a study of experimental results. Bromine has two isotopes, \(^{2}\) Br and \(^{81}\) Br, whose masses \((78.9183 \text { and } 80.9163\) amu, respectively) and abundances (50.69\% and \(49.31 \%,\) respectively were determined in earlier experiments. Calculate the average atomic mass of bromine based on these experiments.

Short Answer

Expert verified
The average atomic mass of bromine is 79.8926 amu.

Step by step solution

01

Identify the given information

The problem provides the masses of the two isotopes of bromine: isotope 79Br with a mass of 78.9183 amu and isotope 81Br with a mass of 80.9163 amu. It also provides their respective natural abundances: 79Br at 50.69% and 81Br at 49.31%.
02

Convert percentages to fractional abundances

Convert the given isotopic abundances from percentage to decimal (fractional) form by dividing by 100. For 79Br: 50.69% becomes 0.5069, and for 81Br: 49.31% becomes 0.4931.
03

Calculate the weighted average atomic mass

Multiply the mass of each isotope by its fractional abundance and then sum the results to get the average atomic mass of bromine. Avg atomic mass = (mass of 79Br x abundance of 79Br) + (mass of 81Br x abundance of 81Br) = (78.9183 amu x 0.5069) + (80.9163 amu x 0.4931).
04

Perform the calculations

Now calculate the two parts of the equation separately and add them together. For 79Br: 78.9183 amu x 0.5069 = 39.9999 amu For 81Br: 80.9163 amu x 0.4931 = 39.8927 amu Average atomic mass of bromine = 39.9999 amu + 39.8927 amu = 79.8926 amu.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isotopic Abundance
Isotopic abundance refers to the percentage of a particular isotope naturally occurring in a sample of an element. It is crucial to understand that elements are not a single-atom type, but a blend of isotopes, each with its distinct atomic mass but chemically similar properties.

For the isotope 79Br, the isotopic abundance is 50.69%, which means that in a sample of bromine, 79Br makes up for approximately half of the atoms present. Similarly, 81Br makes up the remaining 49.31% of the bromine atoms. These percentages are typically found through experimental data and reflect how common or rare an isotope is in nature.
Atomic Mass Unit (amu)
The atomic mass unit, or amu, is a standard unit of mass that quantifies the mass of an atom or molecule. It is defined as one-twelfth the mass of a carbon-12 atom, which has 6 protons and 6 neutrons. 1 amu is approximately equal to 1.66053906660 x 10-24 grams.

Having a standardized unit such as the amu allows chemists and physicists to compare the masses of different atoms and molecules on a relative scale with precision. For bromine isotopes, the atomic masses are given in amu: 78.9183 amu for 79Br and 80.9163 amu for 81Br.
Isotopes of Bromine
Isotopes of an element have the same number of protons but different numbers of neutrons within their nuclei, resulting in different atomic masses. Bromine naturally occurs as two primary isotopes: 79Br and 81Br.

These isotopes have distinct atomic masses but identical chemical characteristics. Identifying the isotopic composition of an element is fundamental to understanding its atomic structure and behavior. For bromine, the isotope 79Br has a slightly lower mass than 81Br, which impacts the calculation of the average atomic mass of natural bromine.
Weighted Average
A weighted average is an arithmetic mean of quantities that are given different importance or 'weights.' In the context of average atomic mass, isotopes with greater abundances have a bigger impact on the value.

To calculate the average atomic mass of an element like bromine, you have to account for the mass of each isotope and its relative abundance. This results in a weighted mean of the isotopic masses rather than a simple average. The contribution of each isotope to the overall atomic mass of bromine correlates with its isotopic abundance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

How are electrons and protons similar? How are they different?

Determine the number of moles of the compound and determine the number of moles of each type of atom in each of the following: (a) 2.345 mol LiCl (b) 0.0872 mol acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\) (c) \(3.3 \times 10^{-2} \mathrm{mol} \mathrm{Na}_{2} \mathrm{CO}_{3}\) (d) \(1.23 \times 10^{3}\) mol fructose, \(C_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (e) 0.5758 mol \(\operatorname{FeSO}_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{7}\)

Calculate the empirical or molecular formula mass and the molar mass of each of the following minerals: (a) limestone, \(\mathrm{CaCO}_{3}\) (b) halite, NaCl (c) beryl, \(\mathrm{Be}_{3} \mathrm{Al}_{2} \mathrm{Si}_{6} \mathrm{O}_{18}\) (d) malachite, \(\mathrm{Cu}_{2}(\mathrm{OH})_{2} \mathrm{CO}_{3}\) (e) turquoise, \(\mathrm{CuAl}_{6}\left(\mathrm{PO}_{4}\right)_{4}(\mathrm{OH})_{8}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\)

Which postulate of Dalton's theory is consistent with the following observation concerning the weights of reactants and products? When 100 grams of solid calcium carbonate is heated, 44 grams of carbon dioxide and 56 grams of calcium oxide are produced.

Which of the following represents the least number of molecules? (a) \(20.0 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}(18.02 \mathrm{g} / \mathrm{mol})\) (b) \(77.0 \mathrm{g}\) of \(\mathrm{CH}_{4}(16.06 \mathrm{g} / \mathrm{mol})\) (c) \(68.0 \mathrm{g}\) of \(\mathrm{CaH}_{2}(42.09 \mathrm{g} / \mathrm{mol})\) (d) \(100.0 \mathrm{g}\) of \(\mathrm{N}_{2} \mathrm{O}(44.02 \mathrm{g} / \mathrm{mol})\) (e) 84.0 g of HF (20.01 g/mol)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free