Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 106

The insecticide dieldrin contains carbon, hydrogen, oxygen, and chlorine. When burned in an excess of oxygen, a 1.510 g sample yields \(2.094 \mathrm{g} \mathrm{CO}_{2}\) and \(0.286 \mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) The compound has a molecular mass of 381 u and has half as many chlorine atoms as carbon atoms. What is the molecular formula of dieldrin?

Short Answer

Expert verified
The molecular formula of dieldrin is \(C_{10}H_{20}O_{4}Cl\).

Step by step solution

01

Calculation of moles of Carbon and Hydrogen

The yield of \(CO_2\) and \(H_2O\) can be used to calculate the moles of Carbon (C) and Hydrogen (H). The molar mass of \(CO_2\) is approximately 44.01g/mol and \(H_2O\) is approximately 18.02g/mol. So, the moles of \(CO_2\) (hence C) can be calculated as, \[moles = \frac{2.094g}{44.01g/mol} = 0.0476 mol\]. Each \(CO_2\) molecule contains one C atom, so the sample contains 0.0476 mol of C. Similarly, the moles of \(H_2O\) (hence H) can be calculated as, \[moles = \frac{0.286g}{18.02g/mol} = 0.0159 mol\]. Each \(H_2O\) molecule contains two H atoms, hence the sample contains \(0.0159 mol \times 2 = 0.0318 mol\) of H.
02

Calculation of mass of Carbon and Hydrogen

Next, we calculate the mass of Carbon and Hydrogen in the dieldrin sample. The mass of Carbon (C) can be calculated from its molar mass (12.01g/mol) as, \(0.0476 mol \times 12.01g/mol = 0.571 g\). Similarly, the mass of Hydrogen (H) can be calculated from its molar mass (1.01g/mol) as, \(0.0318 mol \times 1.01g/mol = 0.032 g\).
03

Calculation of mass of Oxygen and Chlorine

The rest of the mass of the sample is accounted for by Oxygen (O) and Chlorine (Cl). Their total mass can be calculated as, \(1.510 g - (0.571 g + 0.032 g) = 0.907 g\). The problem specifies that the compound has half as many chlorine atoms as carbon atoms. The molar mass of chlorine is 35.45 g/mol. From Step 1, we know there are 0.0476 mol of carbon in the compound, so there must be \(0.0476 mol/2 = 0.0238 mol\) of chlorine in the compound. From the molar mass of chlorine, this implies a mass of,\(0.0238 mol x 35.45 g/mol = 0.843 g\). The rest of the mass must therefore be due to oxygen, so the mass of oxygen is \((0.907 - 0.843) g = 0.064 g\).
04

Calculation of molecular formula

The molecular formula can be found by dividing each quantity of atoms in the empirical formula by the smallest quantity of \(2.984 \times 10^{-2} mol\), so the ratio of atoms becomes: {C:H:O:Cl}={5:10:2:0.5}. Multiplying all atomic ratios by 2 then gives {C:H:O:Cl}={10:20:4:1}. The molecular formula of dieldrin is then determined to be \(C_{10}H_{20}O_{4}Cl\).

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

Name these compounds: (a) \(\operatorname{SrO} ;\) (b) \(\mathrm{ZnS} ;\) (c) \(\mathrm{K}_{2} \mathrm{CrO}_{4}\) (d) \(\mathrm{Cs}_{2} \mathrm{SO}_{4} ;(\mathrm{e}) \mathrm{Cr}_{2} \mathrm{O}_{3} ;(\mathrm{f}) \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3} ;(\mathrm{g}) \mathrm{Mg}\left(\mathrm{HCO}_{3}\right)_{2}\) (h) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4} ; \quad\) (i) \(\quad \mathrm{Ca}\left(\mathrm{HSO}_{3}\right)_{2} ; \quad(\mathrm{j}) \quad \mathrm{Cu}(\mathrm{OH})_{2}\) (k) \(\mathrm{HNO}_{3} ;\) (1) \(\mathrm{KClO}_{4} ;\) (m) \(\mathrm{HBrO}_{3} ;\) (n) \(\mathrm{H}_{3} \mathrm{PO}_{3}\).

The hemoglobin content of blood is about \(15.5 \mathrm{g} / 100 \mathrm{mL}\) blood. The molar mass of hemoglobin is about \(64,500 \mathrm{g} / \mathrm{mol},\) and there are four iron (Fe) atoms in a hemoglobin molecule. Approximately how many Fe atoms are present in the 6 Lof blood in a typical adult?

Dimethylhydrazine is a carbon-hydrogen-nitrogen compound used in rocket fuels. When burned in an excess of oxygen, a \(0.312 \mathrm{g}\) sample yields \(0.458 \mathrm{g} \mathrm{CO}_{2}\) and \(0.374 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\). The nitrogen content of a \(0.486 \mathrm{g}\) sample is converted to \(0.226 \mathrm{g} \mathrm{N}_{2} .\) What is the empirical formula of dimethylhydrazine?

Determine the empirical formula of (a) benzo-[a]pyrene, a suspected carcinogen found in cigarette smoke, consisting of \(95.21 \%\) C and \(4.79 \%\) H, by mass; (b) hexachlorophene, used in germicidal soaps, which consists of \(38.37 \%\) C \(, 1.49 \%\) H, \(52.28 \%\) Cl, and \(7.86 \%\) O by mass.

A hydrate of copper(II) sulfate, when heated, goes through the succession of changes suggested by the photograph. In this photograph, (a) is the original fully hydrated copper(II) sulfate; (b) is the product obtained by heating the original hydrate to \(140^{\circ} \mathrm{C}\) (c) is the product obtained by further heating to \(400^{\circ} \mathrm{C}\) and (d) is the product obtained at \(1000^{\circ} \mathrm{C}\) A \(2.574 \mathrm{g}\) sample of \(\mathrm{CuSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O}\) was heated to \(140^{\circ} \mathrm{C},\) cooled, and reweighed. The resulting solid was reheated to \(400^{\circ} \mathrm{C},\) cooled, and reweighed. Finally, this solid was heated to \(1000^{\circ} \mathrm{C},\) cooled, and reweighed for the last time. $$ \text {Original sample } \quad \quad\quad\quad\quad \text {\(2.574 \mathrm{g}\) } $$ $$ \text {After heating to \(140^{\circ} \mathrm{C}\) } \quad \quad\quad\quad\quad \text {\(1.833 \mathrm{g}\) } $$ $$ \text {After reheating to \(400^{\circ} \mathrm{C}\)} \quad \quad\quad\quad\quad \text {\(1.647 \mathrm{g}\) } $$ $$ \text {After reheating to \(1000^{\circ} \mathrm{C}\)} \quad \quad\quad\quad\quad \text {\(0.812 \mathrm{g}\)} $$ (a) Assuming that all the water of hydration is driven off at \(400^{\circ} \mathrm{C},\) what is the formula of the original hydrate? (b) What is the formula of the hydrate obtained when the original hydrate is heated to only \(140^{\circ} \mathrm{C} ?\) (c) The black residue obtained at \(1000^{\circ} \mathrm{C}\) is an oxide of copper. What is its percent composition and empirical formula?
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

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