A compound that contains only C and H was burned in excess O2 to give CO2 and H2

Question

A compound that contains only C and H was burned in excess O2 to give CO2 and H2O. When 0.270 g of the compound was burned, the amount of CO2 formed reacted completely with 20.0 mL of 2.00 M NaOH solution according to the equation:

2 OH-(aq) + CO2(g) ----> CO32-(aq) + H2O (l)

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 degrees C. ( Pure camphor freezes at 179.8 degrees C and has Kf= 37.7 degrees C/m.)

a) What is the empirical formula of the compound?

b) What is the molecular mass of the compound?

c) What is the molecular formula of the compound?

Explanation / Answer

moles NaOH = 0.0200 L x 2.00 M=0.0400
moles CO2 = 0.0400 / 2 = 0.0200

moles C in this compound = 0.0200
mass C = 0.0200 x 12.011 g/mol=0.240 g
mass H = 0.270 - 0.240=0.030 g
moles H = 0.030g/ 1.008 g/mol=0.0298

0.0298/ 0.0200 =1.5
C H 1.5
to get whole numbers multiply by 2
C2H3 would be the empirical formula ( molar mass =27.05 g/mol)

delta T = 1.9
1.9 = m x 37.7
m = 0.0504 = moles unknown / 0.0500 Kg
moles unknown = 0.00252
molar mass = 0.270/ 0.00252 =107.1 g/mol
107.1 / 27.05 =4
C8H12

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