   # An unknown compound has the formula C x H y O z . You burn 0.1523 g of the compound and isolate 0.3718 g of CO 2 and 0.1522 g of H 2 O. What is the empirical formula of the compound? if the molar mass is 72.1 g/mol, what is the molecular formula? ### 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 4, Problem 36PS
Textbook Problem
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## An unknown compound has the formula CxHyOz. You burn 0.1523 g of the compound and isolate 0.3718 g of CO2 and 0.1522 g of H2O. What is the empirical formula of the compound? if the molar mass is 72.1 g/mol, what is the molecular formula?

Interpretation Introduction

Interpretation:

The empirical and molecular formula of Unknown compound should be determined.

Concept Introduction:

• Empirical formula of a compound represents the smallest whole number relative ratio of elements in that compound.
• Equation for finding Molecular formula from the empirical formula,

MolarmassEmpiricalformula mass × Empirical formula

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).

### Explanation of Solution

When 0.1523g of unknown compound is burned in oxygen, 0.3718g of CO2and 0.1522gH2O can be isolated.

From the given masses the amount of CO2andH2O can be calculated.

Thus, amount of CO2andH2O isolated from the combustion of Unknown compound are,

0.3718gCO2×1molCO244.010gCO2=0.00844molCO20.1522gH2O×1molH2O18.015gH2O=0.00844molH2O

The given unknown compound has the formulaCxHyOz.

For every mole of CO2isolated, 1 mol of C must have been present in the unknown compound.

So,

0.00844molCO2×1molCinunknown1molCO2=0.00844molC

For every mole of H2Oisolated, 2 mol of H must have been present in the Unknown compound.

Therefore,

0.00844molH2O×2molHinunknown1molH2O=0.01688molH

Mass of carbon present in unknown compound is calculated as shown below;

MassofC=0.00844mol×12.01g/mol=0.1014g

Mass of hydrogen present in unknown compound is calculated as shown below;

MassofH=0

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