A 0.270 g of a compound that only contains hydrogen and carbon was burn in exces

Question

A 0.270 g of a compound that only contains hydrogen and carbon was burn in excess oxygen. The amount of CO2 formed reacted completely with 20.0 mL of 2.00 M NaOH solution according to the following reaction. When 0.270 g of the compound was dissolved in 50.0 g of camphor, the resulting solution had a freezing point of 177.9 °C Pure camphor has a freezing point of 179.8 oC and Kf 37.7 °C kg/mol. 20H-1(aq) + CO2(g) C03-2(aq) + H2O There are NaOH solution. The number of moles of CO2 reacted with the NaOH is moles (3 s.f.) of sodium hydroxide in 20.0 mL of a 2.00M moles. There are grams of C in the CO2 reacted. The molality of 0.270 grams of unknown compound dissolved in 50.0 grams of camphor is molal (2 s.f. The number of moles of the unknown compound in the solution is the compound is moles. (2 s.f.) The molecular weight of grams/mole. The ratio of C to H in the empirical formula is o molecular formula is . The ratio of C to H in the to

Explanation / Answer

1000ml= 1L, 1000kg= 1gm and atomic weights (g/mole): C=12, H= 1

moles of NaOH used= molarity* volume of NaOH in liters= 2*20/1000= 0.04

NaOH ----------> Na+OH-

as per the reaction, 2 moles of OH- ( from NaOH) requires 1 mole of CO2

0.04 moles of OH- gives 0.04/2=0.02 moles of CO2

moles of CO2= 0.02, C+O2-------->CO2

hence 1 mole of CO2 correspond to 1 mole of C

0.02 moles of CO2 correspond to .02 moles of C.

atomic weight of C= 12 g/mole, mass of C= 0.02*12=0.24 gm of C

mass of Hydrogen = total mass-mass of C= 0.27-0.24=0.03

moles of Hydrogen= mass/molar mass of H= 0.03/1=0.03

moles ratio of C: H= 0.02:0.03, = 2:3

So the empirical formula is C2H3.

Freezing point depression= molality* I( van;t Hoff factor)*Kf( Freezing point depresion constant)

Freezing ponit depression = Freezing point of camphor- freezing poinit of solution = 179.8-177.9= 1.9

I for Camphor= 1

hence molality= 1.9/(1*37.7)= 0.050

molality= moles of compound/ mass of camphor solvent in kg

0.050 = moles of compound/ (50/1000)

moles of compound= 0.050*50/1000 = 0.0025

moles= mass/molar mass of compound

molar mass of compound= mass/moles= 0.270/0.0025=108 g/mole

molar mass of empirical formula is =2*12+3*1= 27 g/mole

n* molar mass of empirical formula= molar mass of compound

n= no of empirical formula units= 108/27= 4

hence molecular formula is (C2H3)4= C8H12

molality = moles of compound/ kg of camphor=

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