   # Analysis of a gaseous chlorofluorocarbon, CCl x F y , shows that it contains 11.79% C and 69.57% Cl. In another experiment, you find that 0.107 g of the compound fills a 458-mL flask at 25 °C with a pressure of 21.3 mm Hg. What is the molecular formula of the compound? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 10, Problem 72GQ
Textbook Problem
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## Analysis of a gaseous chlorofluorocarbon, CClxFy, shows that it contains 11.79% C and 69.57% Cl. In another experiment, you find that 0.107 g of the compound fills a 458-mL flask at 25 °C with a pressure of 21.3 mm Hg. What is the molecular formula of the compound?

Interpretation Introduction

Interpretation: The molecular formula for the given gaseous chlorofluorocarbon with given reactions conditions, percent of carbon and chlorine should be determined.

Concept introduction:

Empirical formula: It is the simplest value that denotes the ratio of atoms present in chemical compound but not all the exact number of the types of atoms present in the compound.

Molecular formula: It is defined as the chemical formula which defines all the type and number of atoms present in the compound.

### Explanation of Solution

Determine the mole of C, Cl and F

moles =massmolar massmoles of C =11.7912.01=0.982molesmoles of Cl =69.5735.453=1.962molesmoles of F =18.6419=0.981molessinceMass of F = 100-(mass of carbon + mass of chlorine)=10011.7969.57=18.64

The obtained moles should be divided by the smallest number obtained in order to obtain the whole number.

moles of C =0.982moles =0.9820.981=1moles of Cl =1.962moles =1.9620.981=2moles of F =0.981moles =0.9810.981=1

Therefore, the empirical formula for the given compound is C1Cl2F1

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