A compound of carbon, hydrogen, and oxygen was burned in oxygen, and 1 g of the compound produced 1.418 g #CO_2# and 0.871 g #H_2O#. (Question below?)

In another experiment, 0.1103 g of the compound was dissolved in 45 g of water. This solution had a freezing point of -0.0734 C. What is the molecular formula of this compound?

1 Answer
Jan 13, 2018

We gots....#C_2H_6O_2#...this is a 2nd year inorganic question I take it....?

Explanation:

We assess the molar quantities of carbon, and hydrogen, and oxygen.....

#"Moles of carbon"=(1.418*g)/(44.01*g*mol^-1)=0.0322*mol.#

#"Moles of hydrogen"=(2xx0.871*g)/(18.01*g*mol^-1)=0.0967*mol.#

Now we check that we have accounted for all the mass....

#"Mass of carbon"=12.011*g*mol^-1xx0.0322*mol=0.387*g. " Mass of hydrogen"=1.00794*g*mol^-1xx0.0967*mol=0.0975*g.#

Now the mass of carbon PLUS the mass of hydrogen is not equal to the starting mass of compound. The balance was due to oxygen.....because an analyst would never give you percentage oxygen....you would be expected to get this by difference....

And so ................

#"mass of oxygen"=(1.00-0.387-0.0975)*g=0.516*g#

And thus with respect to oxygen there were...

#(0.516*g)/(16.00*g*mol^-1)=0.0322*mol#

And now (FINALLY!) we can address the empirical formula, by dividing thru by the lowest molar quantity...

#C_((0.0322*mol)/(0.0322*mol))H_((0.0967*mol)/(0.0322*mol))O_((0.0322*mol)/(0.0322*mol))-=CH_3O#

And after all that arithmetic we gots to make another calculation to address the molecular mass....

#"Freezing point depression"=k_f*"molality"#...

...and so...#"molality"=(0.0734*K)/(1.853*K*kg*mol^-1)=0.0396*mol*kg^-1#.

And so #((0.1103*g)/("molar mass of stuff"))/(45xx10^-3*kg)=0.0396*mol*kg^-1#..

#"Molar mass of stuff"=(0.1103xx10^-3*kgxx45xx10^-3*kg)/(0.0396*mol*kg^-1)=61.2*g*mol^-1#...

And since the molecular formula is a muliple of the empirical formula, we gots #2xxCH_3O=C_2H_6O_2#...this corresponds to ethylene glycol or a peroxide. I doubt that this question was drawn from actual experimental data....which in my opinion is a bit of a no-no....