Question

The osmotic pressure of a solution containing 1.02 g of an unknown compound dissolved in 175.0 mL of solution at 25 ∘C is 1.93 atm. The combustion of 23.40 g of the unknown compound produced 41.70 gCO2 and 17.07 gH2O. What is the molecular formula of the compound (which contains only carbon, hydrogen, and oxygen)?

Answer 1

First, let ys calculate the number of moles

PI = M*RT

and data given is:

PI = 1.93 atm

M = mol/L

L = 175 ml = 0.175 L

T = 25 C = 298 K

R = 0.082

then

PI = M*RT

M = PI/(RT) = 1.93/(0.082*298) = 0.07898183008 mol per liter

but we have only V = 175 ml = 0.175 L

mol = M*V = 0.079*0.175 = 0.013825 mol are present

therefore

Molar Mass = masS/mol = 1.02/0.013825 = 73.77 g/mol

Now, let us find an empirical equation

generic equation

CHO + O2 = CO2 + H2O

CHO = 23.4

mol = mass/MW = 23.4/73.77 = 0.31720 mol of CHO are burned

and produces

CO2 = 41.70 g

mol CO2 = m/MW = 41.70/44 = 0.9477 mol of CO2

H2O = 17.07 g

mol H2O = m/MW = 17.07/18 = 0.9483 mol of H2O

then

0.31720 CHO + O2 = 0.9477 CO2 + 0.9483 H2O

ratio of C

0.31720 mol of CHO = 0.9477 mol of CO2: therefore there are n = 0.9477/0.31720 = 2.99 = 3 mol of C

0.31720 mol of CHO = 0.31720 mol of H2O: therefore are 0.31720*2/0.31720 = 2 mol of H present

For O is complex:

0.31720 mol of CHO = 0.31720 H2O + 0.9477 mol of CO2

then

1 mol of O + 2/2 mol of O = 2 mol of O

then

C3H6O2

MW of product = 12*3+6+16*2 =74

which is pretty near to that of 73.77 calculated