Question

A) Analysis of a volatile liquid shows that it contains 24.79 percent carbon, 2.04 percent hydrogen, and 73.17 percent chlorine by mass. At 268.0 °C and 1.6 atm, 652.0 mL of the vapor has a mass of 2.275 g. What is the molar mass of the compound? B) For the compound above, determine the molecular formula. Enter your formula as plain text, with carbon first, then hydrogen, the chlorine. So if the formula were C10H21Cl, you would enter C10H21Cl.

Answer 1

Assume 100 g basis

mass of C = 24.79

mol of C = mass/MW = 24.79/12 = 2.0658

mass of H = 2.04

mass/MW = 2.04 /1 = 2.04

mass of Cl = 73.17

mass/MW = 73.17 /70.9 = 1.03201

ratio is : 2:2:1

C2H2Cl

ate least, this is the empirical formula

for... molar mass

MM = mass/mol

find mols

PV = nRT

n = PV/(RT) = 1.6*(652*10^-3)/ ((0.082)(268+273)) = 0.02351 mol

MM = 2.275 / 0.02351 = 96.767 g/mol

then..

MM of empirical formula = C2H2Cl --> 2*12 + 2*1 + 1*35.5 = 61.5

n = MM calculated / MM empirical = 96.767/61.5 = 1.5734 --> 1.5x

C2H2Cl --> 1.5x

C6H6Cl1.5

nearest will be

C12H12Cl2