 # A compound of manganese, carbon, and oxygen contains 28.17% Mn and 30.80% C. When 0.125 g of this compound is dissolved in 5.38 g of cyclohexane, the solution freezes at 5.28°C. What is the molecular formula of this compound? ### General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
Publisher: Cengage Learning
ISBN: 9781305580343 ### General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
Publisher: Cengage Learning
ISBN: 9781305580343

#### Solutions

Chapter
Section
Chapter 12, Problem 12.105QP
Textbook Problem

## A compound of manganese, carbon, and oxygen contains 28.17% Mn and 30.80% C. When 0.125 g of this compound is dissolved in 5.38 g of cyclohexane, the solution freezes at 5.28°C. What is the molecular formula of this compound?

Expert Solution
Interpretation Introduction

Interpretation:

Given the mass percent of elements in a compound and freezing point of the solution of this compound, molecular formula of the compound has to be deduced.

Concept Introduction:

Depression in freezing point is a colligative property which refers to decrease in freezing point of the solution due to the addition of non-volatile solute. It is expressed as,

ΔTf = Kf.cm

Where,

ΔTf = depression in freezing pointKf = freezing point depression constantcm = molal concentration

Mass percent is one of the parameters that is used to express concentration of a substance (solute) in a solution.  It is expressed as,

Mass percent of a solute = mass of solutemass of solution × 100%

Molality or molal concentration is one of the many parameters that is used to express concentration of a solution.  It is expressed as,

Molality = number of moles of solutemass of solvent in kg

### Explanation of Solution

Given that mass of solute compound is 0.125 g and mass of solvent cyclohexane is 5.38 g which is equivalent to 0.00538 kg .

From the data given in the table 12.3 in text book, for cyclohexane, Kf= 20.0°C/m

Freezing point of pure cyclohexane is 6.55°C

Therefore, molality of ethylene glycol in solution is,

ΔTf = Kf.cm

Rewriting the above equation,

cm  ΔTfKf Tf(solvent)-Tf(solution)20.0°C/m 6.55°C-5.28°C20.0°C/m = 0.06350 m

Moles of the compound are,

no. of moles = molality × mass of solvent = 0.06350 mol/kg × 0.00538 kg = 3.416 ×104 mol

Molar mass of the compound is calculated as,

molar mass massno.of moles = 0.125 g3.416×104mol = 365.9 g/mol

Consider 100 g of the compound contains 28

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