/*! 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 89 Researchers from the University ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 89

Researchers from the University of Cape Town in South Afric have discovered that penguins are able to detect a che ::: released by plankton as they are being consumed by schor fish. By following this scent they are able to find these sc of fish and feast on them. The chemical they smell is comp of \(38.65 \%\) carbon, \(9.74 \%\) hydrogen, and \(51.61 \%\) sulfur. D mine the empirical formula of this compound.

Short Answer

Expert verified
The empirical formula is \( C_2H_6S \).

Step by step solution

01

Convert Percentage to Mass

Assume a 100 g sample to simplify the calculations. This means you have 38.65 g of carbon (C), 9.74 g of hydrogen (H), and 51.61 g of sulfur (S).
02

Convert Mass to Moles

Use the molar mass of each element to convert mass to moles.For Carbon: \[ \text{Moles of C} = \frac{38.65 \, \text{g}}{12.01 \, \text{g/mol}} = 3.22 \, \text{mol} \]For Hydrogen: \[ \text{Moles of H} = \frac{9.74 \, \text{g}}{1.008 \, \text{g/mol}} = 9.67 \, \text{mol} \]For Sulfur: \[ \text{Moles of S} = \frac{51.61 \, \text{g}}{32.06 \, \text{g/mol}} = 1.61 \, \text{mol} \]
03

Determine the Simplest Ratio

Divide the number of moles of each element by the smallest number of moles calculated in the previous step.For Carbon: \[ \frac{3.22}{1.61} = 2 \]For Hydrogen: \[ \frac{9.67}{1.61} = 6 \]For Sulfur: \[ \frac{1.61}{1.61} = 1 \]
04

Write the Empirical Formula

Using the simplest whole number ratio derived from the previous step, the empirical formula of the compound is \( C_2H_6S \).

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 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

mass to mole conversion
In chemistry, converting mass to moles is a fundamental skill. It lets you work with chemical amounts more practically. Moles indicate the number of particles in a sample. To convert mass to moles, you use the molar mass. The molar mass is the weight of one mole of a substance.

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

Cosmone is a molecule used by fragrance manufacturers to provide a rich and elegant musky essence to many perfumes. Cosmone has the molecular formula \(\mathrm{C}_{15} \mathrm{H}_{26} \mathrm{O}\). (a) Calculate the molar mass of cosmone. (b) Calculate the mass of \(3.82\) moles of cosmone. (c) Calculate the number of molecules of cosmone in a sample containing \(8.36 \times 10^{-4} \mathrm{~mol}\) cosmone. (d) Calculate the number of moles of carbon in a \(8.35-\mathrm{mol}\) sample of cosmone. (e) Calculate the mass of oxygen in a 4.29-g sample of cosmone. (f) Calculate the mass of a sample of cosmone that contains \(6.58 \times 10^{19}\) atoms of hydrogen. (g) Calculate the mass of one molecule of cosmone. (h) Calculate the number of atoms of carbon in \(8.00-g\) of cosmone.

Which of the following chlorides has the highest and which ha the lowest percentage of chlorine, by mass, in its formula? (a) \(\mathrm{KCl}\) (c) \(\mathrm{SiCl}_{4}\) (b) \(\mathrm{BaCl}_{2}\) (d) \(\mathrm{LiCl}\)Which of the following chlorides has the highest and which ha the lowest percentage of chlorine, by mass, in its formula? (a) \(\mathrm{KCl}\) (c) \(\mathrm{SiCl}_{4}\) (b) \(\mathrm{BaCl}_{2}\) (d) \(\mathrm{LiCl}\)

Researchers at Anna Gudmundsdottir's laboratory at the University of Cincinnati have been studying extremely reactive chemicals known as radicals. One of the interesting phenomena they have discovered is that these radicals can be chemically attached to fragrance molecules, effectively tethering them to a solution. When light strikes these tethered molecules, the fragrance is released. This property would allow us to produce perfumes, cleansers, and other consumer products that release fragrance only when exposed to light. If limonene, \(\mathrm{C}_{10} \mathrm{H}_{16}\), the molecule that gives fruits their citrus scent, were able to be tethered to one of these radicals and every photon of light would release one molecule of limonene, calculate the time in seconds required to release \(1.00\) picogram of limonene if ambient light releases \(2.64 \times 10^{18} \mathrm{photons} / \mathrm{sec}\).

Determine the molar masses of these compounds: (a) \(\mathrm{KBr}\) (f) \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) (b) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) (g) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) (c) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) (h) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (i) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\) (e) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

Determine the molar masses of these compounds: (a) \(\mathrm{KBr}\) (f) \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) (b) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) (g) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) (c) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) (h) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (i) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\) (e) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

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.