Question

A gaseous compound of carbon, hydrogen and chlorine had the following composition: C = 24.78%; H = 2.08 % and Cl = 73.14 %. A 1.26 g sample of this gas occupies a volume of 435 ml. at a pressure of 0.736 atm and temperature of 27°C. a) What is the empirical formula of the unknown gas (3 points) show all setup & work Answer: empirical formula = b) What is the molar mass of the unknow gas (3 points) show all setup & work Answer: molar mass = c) What is the molecular formula of the unknown gas? (4 points) show all setup & work Answer: molecular formula =

Answer 1

Element    %           A. Wt       Relative number     simplest ratio
C         24.78         12           24.78/12 = 2.065    2.065/2.06 = 1
H          2.08          1            2.08/1 = 2.08     2.08/2.06 = 1
Cl        73.14         35.5         73.14/35.5 = 2.06   2.06/2.06 = 1
The empirical fromula = CHCl
The empirical formula mass = 12 + 1 + 35.5 = 48.5
b.
V = 435ml = 0.435L
P = 0.736atm
T = 27+273 = 300K
W = 1.26g
PV = nRT
n = W/M
PV = WRT/M
M   = WRT/PV
     =1.26*0.0821*300/(0.736*0.435)
     = 96.93g/mole
molar mass = 96.93g/mole
c.
molecular formula = ( empirical formula)n
   n = molar mass/empirical formula mass
       = 96.93/48.5   = 2
molecular formula = (CHCl)2
                   = CH2Cl2