Question

3. An elemental analysis of an organic compound showed the following results: C = 82.66 %, H= 17.34 %. A 1.34 g of that compound was vaporized and was found to occupy 520 ml at -1 °C and 750 torr. What is the molecular weight and molecular formula of the compound? be 4. Compute the approximate density of ethane, CH3CH3, at 27°C and 3.00 atm. The molecular weight of ethane is 30.07.

Answer 1

3) from ideal gas equation

   PV = (w/M.wt)RT

P = pressure = 750 torr= 0.99 atm

V = volume of gas = 0.52 L

n = no of mol of gas = W/M.wt = 1.34/M.wt

R = 0.0821 l.atm.k-1.mol-1

T = -1 + 273.15 = 272.15 k

M.wt = wRT/PV

   = (1.34*0.0821*272.15)/(0.99*0.52)

   = 58 g/mol

C = 82.66/12 , H = 17.34/1

C = 6.9       , H = 17.34

simplest ratio

C = 6.9/6.9 = 1 , H = 17.34/6.9 = 2.5

simplest whole number ratio

C = 2 , H = 5

Empirical formula = C2H5

n = 58/27 = 2.

Molecular formula , 2*C2H5 = C4H10

4)

from ideal gas equation

   P*M.wt = (w/v)RT

   d = P*M.wt/RT

P = pressure = 3 atm

M.wt = molecularweight of ethane = 30.07 g/mol

d = density of gas = w/v = ? g/l

R = 0.0821 l.atm.k-1.mol-1

T = 27 + 273.15 = 300.15 k

d = (3*30.07)/(0.0821*300.15)

   = 3.66 g/l