Chapter 10, Problem 97IL

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Iron forms a series of compounds of the type Fex(CO)y. In air, those compounds are oxidized to Fe2O3 and CO2 gas. After heating a 0.142-g sample of Fex(CO)y in air, you isolate the CO2 in a 1.50-L flask at 25 °C. The pressure of the gas is 44.9 mm Hg. What is the empirical formula of Fex(CO)y?

Interpretation Introduction

Interpretation:

Considering the given compound the empirical formula for the compound using given conditions should be determined.

Concept Introduction:

Ideal gas Equation:

Any gas is described by using four terms namely pressure, volume, temperature and the amount of gas. Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained. It is referred as ideal gas equation.

nTPV = RnTPPV = nRTwhere,n = moles of gasP = pressureT = temperatureR = gas constant

Under some conditions gases don not behave like ideal gas that is they deviate from their ideal gas properties. At lower temperature and at high pressures the gas tends to deviate and behave like real gases.

Boyle’s Law:

At given constant temperature conditions the mass of given ideal gas in inversely proportional to its volume.

Charles’s Law:

At given constant pressure conditions the volume of ideal gas is directly proportional to the absolute temperature.

Two equal volumes of gases with same temperature and pressure conditions tend to have same number of molecules with it.

Explanation

Given:

â€‚Â TotalÂ amountÂ ofÂ givenÂ sampleÂ =Â 0.142Â gPressureÂ ofÂ CO2â€Šâ€Š=â€Šâ€Šâ€Š44.9Â mmHgVolumeÂ ofÂ CO2â€Šâ€Š=â€Šâ€Šâ€Š1.5LTemperatureÂ =Â 25oCâ€Šâ€Š=â€Šâ€Š273.15+25Â â€Š=â€Šâ€Š298.15K

Now, the moles of CO2 is calculated from the given set of conditions,

â€‚Â PV=nRTn=â€Šâ€ŠPVRTâ€Šâ€Š=â€Šâ€Šâ€Š44.9760Ã—1.5L0.0821Ã—298.15=3.62Ã—10âˆ’3moles

â€‚Â moles=massMolarâ€Šâ€Šmassmass=molesÃ—Molarâ€Šâ€Šmassâ€Šâ€Š=â€Šâ€Š3.62Ã—10-3Ã—28g/molâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Š=â€Š0

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