Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 11

Chromium (atomic mass \(=51.9961\) amu) has four isotopes. Their masses are \(49.94605 \mathrm{amu}, 51.94051 \mathrm{amu}, 52.94065 \mathrm{amu}\), and \(53.93888\) amu. The first two isotopes have a total abundance of \(87.87 \%\), and the last isotope has an abundance of \(2.365 \%\). What is the abundance of the third isotope? Estimate the abundances of the first two isotopes.

Short Answer

Expert verified
Answer: The estimated abundances of the first and second isotopes are 81.68% and 6.19%, respectively. The calculated abundance of the third isotope is 9.765%.

Step by step solution

01

Calculate the abundance of the third isotope

To find the abundance of the third isotope, we will first calculate the total abundance of the first, second, and fourth isotopes, and then subtract it from 100%. Total abundance of first two isotopes = 87.87% Abundance of the fourth isotope = 2.365% Total abundance of first, second, and fourth isotopes = 87.87% + 2.365% = 90.235% Now we can calculate the abundance of the third isotope: Abundance of the third isotope = 100% - 90.235% = 9.765% The abundance of the third isotope is 9.765%. Now let's estimate the abundances of the first two isotopes.
02

Estimate the abundances of the first two isotopes

We are given that the total abundance of the first and second isotopes is 87.87%, and we need to estimate their individual abundances. We can use the given information about their masses and the atomic mass of chromium to estimate the abundances. Let the x represent the abundance of the first isotope as a decimal (i.e., 0.50 for 50%). Then, the abundance of the second isotope can be represented as (0.8787 - x). So, we can set up an equation using the given atomic mass of chromium: 51.9961 = (49.94605 * x) + (51.94051 * (0.8787 - x)) Now, we need to solve for x.
03

Solve for x

To solve for x, we will first simplify the equation: 51.9961 = 49.94605x + 45.6069 - 51.94051x 51.9961 = -1.99446x + 45.6069 Now, isolate the x term: 1.99446x = 51.9961 - 45.6069 x ≈ 0.8168 Now we can calculate the abundance of the first isotope: Abundance of the first isotope = x * 100% = 81.68% We can find the abundance of the second isotope by subtracting the abundance of the first isotope from the total abundance of the first two isotopes: Abundance of the second isotope = 87.87% - 81.68% = 6.19% So, the estimated abundances of the first and second isotopes are 81.68% and 6.19% respectively.

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

Determine whether the statements given below are true or false. (a) The mass of an atom can have the unit mole. (b) In \(\mathrm{N}_{2} \mathrm{O}_{4}\), the mass of the oxygen is twice that of the nitrogen. (c) One mole of chlorine atoms has a mass of \(35.45 \mathrm{~g}\). (d) Boron has an average atomic mass of \(10.81\) amu. It has two isotopes, \(\mathrm{B}-10(10.01\) amu \()\) and \(\mathrm{B}-11(11.01 \mathrm{amu}) .\) There is more naturally occurring B-10 than B-11. (e) The compound \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{2} \mathrm{~N}\) has for its simplest formula \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{ON}_{1 / 2}\). (f) A 558.5-g sample of iron contains ten times as many atoms as \(0.5200 \mathrm{~g}\) of chromium. (g) If \(1.00\) mol of ammonia is mixed with \(1.00\) mol of oxygen the following reaction occurs, $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ All the oxygen is consumed. (h) When balancing an equation, the total number of moles of reactant molecules must equal the total number of moles of product molecules.

The isotope Si-28 has a mass of \(27.977\) amu. For ten grams of Si-28, calculate (a) the number of moles. (b) the number of atoms. (c) the total number of protons, neutrons, and electrons.

A gaseous mixture containing \(4.15 \mathrm{~mol}\) of hydrogen gas and \(7.13 \mathrm{~mol}\) of oxygen gas reacts to form steam. (a) Write a balanced equation for the reaction. (b) What is the limiting reactant? (c) What is the theoretical yield of steam in moles? (d) How many moles of the excess reactant remain unreacted?

Write a balanced equation for (a) the combustion (reaction with oxygen gas) of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), to give carbon dioxide and water. (b) the reaction between xenon tetrafluoride gas and water to give xenon, oxygen, and hydrogen fluoride gases. (c) the reaction between aluminum and iron(III) oxide to give aluminum oxide and iron. (d) the formation of ammonia gas from its elements. (e) the reaction between sodium chloride, sulfur dioxide gas, steam, and oxygen to give sodium sulfate and hydrogen chloride gas.

Calculate the mass in grams of \(2.688 \mathrm{~mol}\) of (a) chlorophyll, \(\mathrm{C}_{55} \mathrm{H}_{72} \mathrm{~N}_{4} \mathrm{O}_{5} \mathrm{Mg}\), responsible for the green color of leaves. (b) sorbitol, \(\mathrm{C}_{9} \mathrm{H}_{14} \mathrm{O}_{6}\), an artificial sweetener. (c) indigo, \(\mathrm{C}_{16} \mathrm{H}_{10} \mathrm{~N}_{2} \mathrm{O}_{2}\), a blue dye.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

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