Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 99

Silver has only two naturally occurring isotopes: \({ }^{107} \mathrm{Ag}\) with a mass of \(106.90509 \mathrm{amu}\) and an abundance of \(51.84 \%\) and \({ }^{109} \mathrm{Ag}\) with a mass of \(108.9047\) amu. Calculate the atomic mass of silver.

Short Answer

Expert verified
The atomic mass of silver is approximately \(107.84 \, \mathrm{amu}\).

Step by step solution

01

\({ }^{107} \mathrm{Ag}\) Abundance (decimal)

$$\text{Abundance (decimal)} = \frac{51.84}{100} = 0.5184$$
02

\({ }^{109} \mathrm{Ag}\) Abundance (decimal)

For \({ }^{109} \mathrm{Ag}\), the abundance adds up to 100%, so we can just subtract the abundance of \({ }^{107} \mathrm{Ag}\) from 100% or we can also divide the leftover percentage by 100. $$\text{Abundance (decimal)} = 1 - 0.5184 = 0.4816$$ #Step 2: Determine the weighted mass of each isotope# Next, we need to determine the weighted mass of each isotope by multiplying the mass of each isotope with their respective decimal abundance.
03

\({ }^{107} \mathrm{Ag}\) Weighted Mass

$$\text{Weighted Mass} = (106.90509 \, \mathrm{amu}) (0.5184) = 55.40832696 \, \mathrm{amu}$$
04

\({ }^{109} \mathrm{Ag}\) Weighted Mass

$$\text{Weighted Mass} = (108.9047 \, \mathrm{amu}) (0.4816) = 52.435572 \| \mathrm{amu}$$ #Step 3: Calculate the atomic mass of silver# Finally, to find the atomic mass of silver, we need to add up the weighted masses of both isotopes.
05

Silver Atomic Mass

$$\text{Atomic Mass} = 55.40832696 \, \mathrm{amu} + 52.435572 \, \mathrm{amu} = 107.84389896 \, \mathrm{amu}$$ The atomic mass of silver is approximately \(107.84 \, \mathrm{amu}\).

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!

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

Postulate a reason why the first ionization energy increases as we progress from left to right across a period.

It always takes energy to remove an electron from a neutral atom. Explain why, and name the minimum amount of energy it takes to do this.

Metals are excellent conductors of electricity. Nonmetals are poor conductors of electricity and are called insulators. Some elements have the ability to conduct electricity roughly between that of metals and nonmetals. Give some examples of these elements, and name the class they belong to.

Of the atoms \(\mathrm{Na}, \mathrm{Mg}\), and \(\mathrm{Al}\), which should be the most difficult to ionize? Which has the smallest first ionization energy?

A student claims that the way to convert a neutral atom of \(\mathrm{Cl}\) into a \(\mathrm{Cl}^{-}\) anion is to decrease its atomic number by one while leaving the number of electrons unchanged. (a) Will this give the atom a \(-1\) charge and convert it into an anion? Explain. (b) Will the atom still be an atom of the element chlorine? Explain. (c) What should the student have done?
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Inorganic 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