   Chapter 4, Problem 35PS

Chapter
Section
Textbook Problem

An unknown compound has the formula CxHyOz. You burn 0.0956 g of the compound and isolate 0.1356 g of CO2 and 0.0833 g of H2O. What is the empirical formula of the compound? If the molar mass is 62.1 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

When 0.0956g of unknown compound is burned in oxygen, 0.1356g of CO2 and 0.0833gH2O 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,

0.1356gCO2×1molCO244.010gCO2=0.00308molCO20.0833gH2O×1molH2O18.015gH2O=0.00462molH2O

For every mole of CO2 isolated, 1 mol of C must have been present in the unknown compound.

So,

0.00308molCO2×1molCinunknown1molCO2=0.00308molC

The given unknown compound has the formula CxHyOz.

For every mole of H2O isolated, 2 mol of H must have been present in the Unknown compound.

Therefore,

0.00462molH2O×2molHinunknown1molH2O=0

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