Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 34

A compound is known to contain \(\mathrm{C}\) and \(\mathrm{H}\), and might also contain \(\mathrm{O} .\) It is analyzed for \(C\) and \(H\) only, yielding the mass percents \(C=54.53 \%\) and \(H=9.15 \%\). The molar mass of the compound is \(88.106 \mathrm{~g} / \mathrm{mol}\). What are the empirical and molecular formulas?

Short Answer

Expert verified
The empirical formula of the compound is \(C_2H_4O\) and the molecular formula is \(C_4H_8O_2\).

Step by step solution

01

Determine the mass percent of Oxygen

Since we know the mass percent of Carbon and Hydrogen, we can determine the mass percent of Oxygen by subtracting them from 100%: Mass percent of O \(= 100\% - 54.53\% - 9.15\% = 36.32\%\)
02

Convert mass percentages to moles

Now we will convert the mass percentages to moles using atomic masses. Let's assume we have 100 g of the compound, so the mass percentages will directly translate to grams: Moles of C \(= \frac{54.53\text{ g}}{12.01\text{ g/mol}} = 4.542\text{ mol}\) Moles of H \(= \frac{9.15\text{ g}}{1.008\text{ g/mol}} = 9.081\text{ mol}\) Moles of O \(= \frac{36.32\text{ g}}{16.00\text{ g/mol}} = 2.270\text{ mol}\)
03

Determine the mole ratio of C, H, and O

Now we'll divide each of the moles calculated above by the smallest mole value: Moles ratio of C \(= \frac{4.542}{2.270} = 2\) Moles ratio of H \(= \frac{9.081}{2.270} = 4\) Moles ratio of O \(= \frac{2.270}{2.270} = 1\)
04

Write the empirical formula

Using the mole ratios of C, H, and O determined in step 3, we can write the empirical formula: \(C_2H_4O_1\) or \(C_2H_4O\)
05

Calculate the empirical formula mass

Now let's calculate the mass of the empirical formula: Empirical formula mass \(= 2 \times 12.01 + 4 \times 1.008 + 1 \times 16.00 = 44.052\text{ g/mol}\)
06

Determine the molecular formula using the molar mass

Finally, we will determine the molecular formula by comparing the molar mass given in the problem and the empirical formula mass: Molecular formula \(= n \times\) (Empirical formula) Where \(n = \frac{\text{Molar mass of compound}}{\text{Empirical formula mass}}\) \(n = \frac{88.106}{44.052} = 1.998\), approximately \(2\) Molecular formula \(= 2 \times (C_2H_4O) = C_4H_8O_2\) Therefore, the empirical formula of the compound is \(C_2H_4O\) and the molecular formula is \(C_4H_8O_2\).

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

The compound ibuprofen, used in some pain relievers, has the molecular formula \(\mathrm{C}_{13} \mathrm{H}_{18} \mathrm{O}_{2}\). If \(0.250 \mathrm{~g}\) of ibuprofen is burned in oxygen, how many grams of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) will be produced?

A \(1.000-\mathrm{g}\) sample of a liquid is subjected to combustion analysis, yielding \(92.3 \% \mathrm{C}\) and \(7.7 \% \mathrm{H}\). It may or may not also contain oxygen. (a) What is the empirical formula for this compound? (b) The molar mass of this compound is determined to be about \(78 \mathrm{~g} / \mathrm{mol}\). What is the molecular formula for this compound?

Silicon nitride, \(\mathrm{Si}_{3} \mathrm{~N}_{4}\), is a ceramic material capable of withstanding high temperatures. It can be produced using the following unbalanced reaction: \(\mathrm{SiCl}_{4}+\mathrm{NH}_{3} \rightarrow \mathrm{Si}_{3} \mathrm{~N}_{4}+\mathrm{HCl}\) If \(64.2 \mathrm{~g}\) of \(\mathrm{SiCl}_{4}\) is reacted with \(20.0 \mathrm{~g}\) of \(\mathrm{NH}_{3}\), how many grams of \(\mathrm{Si}_{3} \mathrm{~N}_{4}\) are produced if the reaction has a \(96.0 \%\) yield?

Consider the unbalanced chemical equation \(\mathrm{CaC}_{2}+\mathrm{CO} \rightarrow \mathrm{C}+\mathrm{CaCO}_{3}\) When the reaction is complete, \(135.4 \mathrm{~g}\) of \(\mathrm{CaCO}_{3}\) produced and \(38.5 \mathrm{~g}\) of \(\mathrm{CaC}_{2}\) is left over. Assuming the reaction had a \(100 \%\) yield, what were the mass of the two reactants at the beginning of the reaction

(a) What is the molar mass of ribose \(\left(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5}\right)\) ? (b) What is the mass of \(3.87\) moles of ribose? (c) How many ribose molecules are there in \(3.87\) moles? (d) How many oxygen atoms are there in \(3.87\) moles of ribose? (e) What is the mass in grams of the oxygen atoms in part (d)?
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

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