/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 How many molecules of ethane \(\... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

How many molecules of ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) are present in \(0.334 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{6} ?\)

Short Answer

Expert verified
The number of molecules present in 0.334 g of \(\mathrm{C}_{2} \mathrm{H}_{6}\) is approximately \(\6.63 x 10^22\) molecules.

Step by step solution

01

Determine the Molar Mass of Ethane

The molar mass of Ethane (`\(\mathrm{C}_{2} \mathrm{H}_{6}\)`) is `30.07 \mathrm{~g/mol}`. It's calculated by adding the molar masses of Carbon (C) and Hydrogen (H) in the atomic units. Ethane has 2 Carbon atoms (12.01 g/mol each) and 6 Hydrogen atoms (1.01 g/mol each) so the calculation becomes `(2*12.01) + (6*1.01) = 30.07 g/mol`.
02

Calculate the number of Moles

Moles of a substance is calculated by dividing the given mass of the substance by the molar mass of that substance. Ergo, the moles of Ethane can be calculated as `0.334 g ÷ 30.07 g/mol = 0.011 moles`.
03

Calculate the Number of Molecules

Using Avogadro’s number (`6.023 x 10^23 molecules/mol`), the number of molecules of Ethane can be calculated as `0.011 moles x 6.023 x 10^23 molecules/mol = 6.63 x 10^22 molecules`.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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

See all solutions

Recommended explanations on Chemistry Textbooks

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.