Chemistry
Chemistry
4th Edition
ISBN: 9780078021527
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Chapter 25, Problem 100AP
Interpretation Introduction

Interpretation:

The amounts, in milligrams, of C, H, and O that were present in the original sample of Y are to be calculated, the empirical formula of Y is to be derived, and a plausible structure for Y is to be suggested in case the empirical formula is the same as the molecular formula.

Concept Introduction:

In an empirical formula, the mass of each element changes into moles by taking the molar mass from the periodic table.

Each mole value is divided by the smallest number of calculated moles.

Expert Solution & Answer
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Answer to Problem 100AP

Solution:

(a) The amounts of C, H, and O are calculated as follows:

15.81mg C, 1.32mg H, and 3.49mg O

(b) Empirical formula=C6H6O

(c)

Chemistry, Chapter 25, Problem 100AP , additional homework tip  1

Explanation of Solution

Given information: The mass of compound Y = 20.63mg.

The mass of CO2 formed = 57.94mg.

The mass of H2O formed = 11.85mg.

a)The mass of C,H,andO were present in the original sample of Y

The mass of C in 57.94mg of CO2 is calculated as follows:

Massof C=57.94×103gof CO2×1mol CO244.01g CO2×1mol C1mol CO2×12.01g C1mol C=15.81mg.

The mass of H in 11.85mg of H2O is calculated as follows:

Massof H=(11.85×103gof H2O)(1mol H2O18.02g H2O)(2mol H1mol H2O)(1.008g H1mol H)=1.32mg.

The mass of oxygen is calculated as follows:

Massof O=Massof sample(Massof C+Massof H)=20.63mg(15.81mg+1.32mg)=3.49mg.

b)Derive the empirical formula of Y.

The calculated value of compound Y is given as follows:

15.81mg C or 15.81×103g C;

1.32mg H or 1.32×103 g H;

3.49mg O or 3.49×103 g O.

The mass of each element changes into moles by taking the molar mass from the periodic table as follows:

(15.81 ×103g C×1mol C12.0g C)=1.31×103mol C

(1.32×103g H×1mol H1.01g H)=1.31×103mol H

(3.49×103g O×1mol O16.0g O)=0.21×103mol O.

Each mole is divided by the smallest number of calculated moles and taken round to the nearest whole number.

1.31×1030.21×103mol C6.276

1.31×1030.21×103mol H6.216

0.21×1030.21×103mol O1.

The value of each element is considered as the mole ratio of the elements, which is represented by subscripts in the empirical formula, as follows:

Empirical formula=C6H6O

c) Plausible structure for Y if the empirical formula is the same as the molecular formula.

The compound Y contains the molecular formula C6H6O, which means two structures can have different functional groups.

The structures are represented below.

Chemistry, Chapter 25, Problem 100AP , additional homework tip  2

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Chapter 25 Solutions

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