Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 97

Determine the empirical formula of the compound that is \(26.4 \%\) by mass \(\mathrm{Na}, 36.8 \%\) by mass \(\mathrm{S}\), and also contains oxygen.

Short Answer

Expert verified
The empirical formula of the compound is NaSO₂.

Step by step solution

01

Find the grams of each element in 100g compound

Assume you have a 100g sample of the compound. Given the mass percentages of Na, S, and O, we can determine the grams of each element in the 100g sample. For Na: \(26.4\% \times 100g = 26.4g\) For S: \(36.8\% \times 100g = 36.8g\) For O: The remaining mass percentage is for O, so \(100\% - 26.4\% - 36.8\% = 36.8\%\). Therefore, the mass of O in the 100g sample is \(36.8\% \times 100g = 36.8g\).
02

Convert grams to moles

Next, we will convert the mass of each element in grams to moles. To do this, we will use the molar masses of each element: Na (22.99 g/mol), S (32.07 g/mol), and O (16.00 g/mol). Moles of Na: \(\frac{26.4 g}{22.99 g/mol} = 1.15\) moles Moles of S: \(\frac{36.8 g}{32.07 g/mol} = 1.15\) moles Moles of O: \(\frac{36.8 g}{16.00 g/mol} = 2.30\) moles
03

Find the simplest whole-number ratio

Divide the moles of each element by the smallest number of moles to find the simplest whole-number ratio: For Na: \(\frac{1.15}{1.15} = 1\) For S: \(\frac{1.15}{1.15} = 1\) For O: \(\frac{2.30}{1.15} = 2\) The empirical formula is NaSO₂.

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

If you have \(44.6 \mathrm{~g}\) of carbon tetrachloride, how many atoms of chlorine do you have?

A \(1.540-g\) sample of a liquid is subjected to combustion analysis, yielding \(40.00 \% \mathrm{C}\) and \(6.71 \% \mathrm{H}\). It may also contain oxygen. (a) What is the empirical formula for this compound? (b) The molar mass of this compound is determined to be about \(30 \mathrm{~g} / \mathrm{mol}\). What is the molecular formula for this compound?

\(5.00 \mathrm{~g}\) of solid sodium (Na) and \(30.0 \mathrm{~g}\) of liquid bromine \(\left(\mathrm{Br}_{2}\right)\) react to form solid \(\mathrm{NaBr}\). (a) Write a balanced chemical equation for this reaction. (b) Which reactant is limiting? (c) What is the theoretical yield for this reaction in grams? (d) How many grams of excess reactant are left over at the end of the reaction? (e) When this reaction is actually performed, \(14.7 \mathrm{~g}\) of \(\mathrm{NaBr}\) is recovered. What is the percent yield of the reaction?

A compound is \(91.77 \%\) by mass \(\mathrm{Si}\) and \(8.23 \%\) by mass \(\mathrm{H}\) and has a molar mass of approximately \(122 \mathrm{~g} / \mathrm{mol}\). What is its molecular formula?

Consider the unbalanced chemical equation \(\mathrm{Cl}_{2} \mathrm{O}_{7}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HClO}_{4}\) The reaction is carried out at \(82.0 \%\) yield and gives \(52.8 \mathrm{~g}\) of \(\mathrm{HClO}_{4}\) (a) What is the theoretical yield of \(\mathrm{HClO}_{4}\) ? (b) How many grams of \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) and \(\mathrm{H}_{2} \mathrm{O}\) were consumed in the reaction?
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

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