Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 48

The empirical formula of a compound is CH. If the molar mass of this compound is about \(78 \mathrm{~g},\) what is its molecular formula?

Short Answer

Expert verified
The compound's molecular formula is \(C_6H_6\).

Step by step solution

01

Determine the molar mass of the empirical formula

You calculate the molar mass of the empirical formula \(CH\). Carbon (C) has an atomic mass of approximately \(12 \mathrm{~g/mol}\) and Hydrogen (H) has an atomic mass of approximately \(1 \mathrm{~g/mol}\). Add these together to find the molar mass of the empirical formula, which equates to \(13 \mathrm{~g/mol}\).
02

Determine the ratio of the actual molar mass to the empirical formula mass

You divide the given molar mass of the compound \(78 \mathrm{~g/mol}\) by the molar mass of the empirical formula \(13 \mathrm{~g/mol}\). The result is 6. Therefore, the molecular formula contains 6 times the atoms as the empirical formula.
03

Write out the Molecular Formula

Since we found out that the molecular formula contains six times the atoms as the empirical formula, we can write out the molecular formula of the compound as \(C_6H_6\).

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

Without doing any detailed calculations, arrange the following substances in the increasing order of number of moles: \(20.0 \mathrm{~g} \mathrm{Cl}, 35.0 \mathrm{~g} \mathrm{Br},\) and\(94.0 \mathrm{~g} \mathrm{I}\)

A research chemist used a mass spectrometer to study the two isotopes of an element. Over time, she recorded a number of mass spectra of these isotopes. On analysis, she noticed that the ratio of the taller peak (the more abundant isotope) to the shorter peak (the less abundant isotope) gradually increased with time. Assuming that the mass spectrometer was functioning normally, what do you think was causing this change?

One of the reactions that occurs in a blast furnace, where iron ore is converted to cast iron, is$$\mathrm{Fe}_{2} \mathrm{O}_{3}+3 \mathrm{CO} \longrightarrow 2 \mathrm{Fe}+3 \mathrm{CO}_{2}$$Suppose that \(1.64 \times 10^{3} \mathrm{~kg}\) of \(\mathrm{Fe}\) are obtained from a \(2.62 \times 10^{3} \mathrm{~kg}\) sample of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). Assuming that the reaction goes to completion, what is the percent purity of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the original sample?

Why must a chemical equation be balanced? What law is obeyed by a balanced chemical equation?

A mixture of methane \(\left(\mathrm{CH}_{4}\right)\) and ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) of mass \(13.43 \mathrm{~g}\) is completely burned in oxygen. If the total mass of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) produced is \(64.84 \mathrm{~g},\) calculate the fraction of \(\mathrm{CH}_{4}\) in the mixture.
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

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