Question

How to determine the molecular formula of this organic compound
An organic compound containing only carbon, hydrogen, and oxygen was analyzed gravimetrically. When completely oxidized in air, 0.900g of the compound produced 1.80g of carbon dioxide and 0.736g of water. A separate 2.279g sample, when vaporized in a 1.00 $d{m}^3$ vessel at 100${}^{\circ }C$ had a pressure of 84kPa. Determine the molecular formula of the compound.

Answer 1

Hint: Before solving this question, we should first know the formula to find the molecular formula. $MolecularFormula={\displaystyle \frac{Molecularmass}{Empiricalformulamass}}$, In this, first, we have to find the molecular mass of the compound and then the empirical formula. Then we can easily find the Molecular formula of the organic compound.

Complete answer:
We have to first calculate the empirical formula:
Mass of C = 1.80g $C{O}_2$$\times {\displaystyle \frac{12.01gC}{44.01gC{O}_2}}$= 0.4192 g
Mass of H = $0.736g{H}_{2}O$$\times {\displaystyle \frac{2.016gH}{18.02g{H}_{2}O}}$= 0.0823 g
Mass of O = Mass of compound – Mass of C – Mass of H
                  = 0.900g – 0.4192g – 0.0823g
                  = 0.3265 g
Now, the masses need to be converted into moles and then find ratios.
$Moles={\displaystyle \frac{Mass}{MolarMass}}$
Moles of C = 0.0409
Moles of H = 0.0411
Moles of O = 0.0204
Integer would be the ratio of mass and moles
Integer: C = 2
               H = 4
               O = 1
Empirical formula comes out to be $C_2{H}_4{O}_1$
Now we will use Ideal Gas Law : PV= nRT
Here, $n={\displaystyle \frac{mass}{molarmass}}$
So, Formula can also be written as $PV={\displaystyle \frac{m}{M}}RT$
                                                             $M={\displaystyle \frac{m}{PV}}RT$
Here the mass is 2.279g
R is 8.314
T is 100${}^{\circ }C$i.e (100+273) = 373.15 L
P is 84 kPa
M = ${\displaystyle \frac{2.279\times 8.314\times 373.15}{84}}$
     = 84.2 g/mol
So, the molecular mass is 84.2 g/mol
Now, the Empirical formula is $C_2{H}_4{O}_1$= 44.05
Molecular mass = 84.2
$MolecularFormula={\displaystyle \frac{Molecularmass}{Empiricalformulamass}}$
                                    =${\displaystyle \frac{84.2}{44.05}}$= 2
Molecular formula = ${(C_2{H}_{4}O)}_2=C_4{H}_8{O}_2$

Note:
The empirical formula and molecular formula are two different terms. One should not get confused about it. The empirical formula tells about the relative abundance of the atoms in a compound whereas the molecular formula tells us about the actual number of atoms in a compound.