Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 28

A compound contains 7 carbon atoms, 2 oxygen atoms and \(9.96 \times 10^{-24} \mathrm{~g}\) of other elements. The molecular mass of compound is \(\left(N_{\mathrm{A}}=6 \times 10^{23}\right)\) (a) 122 (b) 116 (c) 148 (d) 154

Short Answer

Expert verified
Answer: (b) 116

Step by step solution

01

Calculate mass of carbon atoms

The given compound contains 7 carbon atoms. The atomic mass of carbon is 12 g/mole. Since the atomic mass is the mass of one mole of a substance (1 mole contains Avogadro's number of particles), we can calculate the total mass of carbon atoms in the compound. $$ \text{Total mass of carbon atoms} = \text{number of carbon atoms} \times \text{atomic mass of carbon} $$ $$ \text{Total mass of carbon atoms} = 7 \times 12 = 84 \mathrm{~g/mol} $$
02

Calculate mass of oxygen atoms

Now, let's calculate the total mass of oxygen atoms in the given compound. The compound contains 2 oxygen atoms, and the atomic mass of oxygen is 16 g/mole. $$ \text{Total mass of oxygen atoms} = \text{number of oxygen atoms} \times \text{atomic mass of oxygen} $$ $$ \text{Total mass of oxygen atoms} = 2 \times 16 = 32 \mathrm{~g/mol} $$
03

Calculate mass of other elements

We are given the mass of other elements in the compound, which is \(9.96 \times 10^{-24}\mathrm{~g}\). In order to find the mass of the other elements per mole, we need to use Avogadro's number, which is \(6 \times 10^{23}\). $$ \text{Mass of other elements per mole} = \frac{\text{mass of other elements}}{\text{Avogadro's number}} $$ $$ \text{Mass of other elements per mole} = \frac{9.96 \times 10^{-24}\mathrm{~g}}{6 \times 10^{23}} = 1.66 \times 10^{-03} \mathrm{~g/mol} $$
04

Calculate the molecular mass of the compound

Now, we can find the molecular mass of the compound by adding the total mass of carbon atoms, oxygen atoms, and other elements. $$ \text{Molecular mass of compound} = \text{Total mass of carbon atoms} + \text{Total mass of oxygen atoms} + \text{Mass of other elements per mole} $$ $$ \text{Molecular mass of compound} = 84 + 32 + 1.66 \times 10^{-03} = 116.00166 \mathrm{~g/mol} $$ Since 116.00166 is closest to 116 among the given options, the correct molecular mass of the compound is: (a) 122 (b) 116 (c) 148 (d) 154 The answer is (b) 116.

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

Two successive reactions, \(\mathrm{A} \rightarrow \mathrm{B}\) and \(\mathrm{B} \rightarrow \mathrm{C}\), have yields of \(90 \%\) and \(80 \%\) respectively. What is the overall percentage yield for conversion of \(\mathrm{A}\) to \(\mathrm{C}\) ? (a) \(90 \%\) (b) \(80 \%\) (c) \(72 \%\) (d) \(85 \%\)

An aqueous solution of glucose is \(10 \%\) \((\mathrm{w} / \mathrm{v}) .\) The volume in which \(1 \mathrm{~mole}\) of glucose is dissolved, will be (a) 181 (b) 91 (c) \(0.91\) (d) \(1.81\)

The mass composition of universe may be given as \(90 \% \mathrm{H}_{2}\) and \(10 \% \mathrm{He}\). The average molecular mass of universe should be (a) \(2.20\) (b) \(2.10\) (c) \(3.80\) (d) \(3.64\)

A volume of \(10 \mathrm{ml}\) of an oxide of nitrogen was taken in a eudiometer tube and mixed with hydrogen until the volume was \(28 \mathrm{ml} .\) On sparking, the resulting mixture occupied \(18 \mathrm{ml}\). To this mixture, oxygen was added when the volume came to \(27 \mathrm{ml}\) and on explosion again, the volume fall to \(15 \mathrm{ml}\). Find the molecular weight of the oxide of nitrogen originally taken in eudiometer tube. All measurements were made at STP.

If rocket were fuelled with kerosene and liquid oxygen, what mass of oxygen would be required for every litre of kerosene? Assume kerosene to have the average composition \(\mathrm{C}_{14} \mathrm{H}_{30}\) and density, \(0.792 \mathrm{~g} / \mathrm{ml}\). (a) \(5.504 \mathrm{~kg}\) (b) \(2.752 \mathrm{~kg}\) (c) \(1.376 \mathrm{~kg}\) (d) \(3.475 \mathrm{~kg}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Physical 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