Please show full work and reasoning, thanks! Molecular Formula from Elemental An

Question

Please show full work and reasoning, thanks!

Molecular Formula from Elemental Analysis and Molecular Mass Determination An unidentified covalent molecular compound contains only carbon, hydrogen, and oxygen. When 9.00 mg of this compound is burned, 26.74 mg of CO2 and 3.65 mg of H20 are produced The freezing point of camphor is lowered by 34.4°C when 2.867 g of the compound is dissolved in 15.00 g of camphor (Kf 40.0°C kg/mol). What is the molecular formula of the unidentified compound? Choose appropriate coefficients in the molecular formula below CHO Submit Answer Tries 0/5

Explanation / Answer

we have below equation to be used:

delta Tf = Kf*mb

34.4 = 40.0 *mb

mb= 0.86 molal

mass of solvent = 15 g

= 1.5*10^-2 kg [using conversion 1 Kg = 1000 g]

we have below equation to be used:

number of mol,

n = Molality * mass of solvent in Kg

= (0.86 mol/Kg)*(0.015 Kg)

= 1.29*10^-2 mol

mass of solute = 2.867 g

we have below equation to be used:

number of mol = mass / molar mass

1.29*10^-2 mol = (2.867 g)/molar mass

molar mass = 222.2 g/mol

Number of moles of CO2 = mass of CO2 / molar mass CO2

= 0.02674/44

= 0.0006077

Number of moles of H2O = mass of H2O / molar mass H2O

= 0.00365/18

= 0.0002028

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.0006077

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.0002028 = 0.0004056

Molar mass of O = 16 g/mol

mass O = total mass - mass of C and H

= 0.009 - 0.0006077*12 - 0.0004056*1

= 0.0013017

number of mol of O = mass of O / molar mass of O

= 0.0013017/16.0

= 0.0000814

Divide by smallest:

C: 0.0006077/0.0000814 = 7.5

H: 0.0004056/0.0000814 = 5

O: 0.0000814/0.0000814 = 1

multiply by 2 to get simplest whole number ratio:

C: 7.5*2 = 15

H: 5*2 = 10

O: 1*2 = 2

So empirical formula is:C15H10O2

Molar mass of C15H10O2 = 15*MM(C) + 10*MM(H) + 2*MM(O)

= 15*12.01 + 10*1.008 + 2*16.0

= 222.23 g/mol

Now we have:

Molar mass = 222.2 g/mol

Empirical formula mass = 222.23 g/mol

Multiplying factor = molar mass / empirical formula mass

= 222.2/222.23

= 1

So molecular formula is:CHO

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