Question

0.27gm of an organic compound gave on combustion 0.396gm of $C{O_2}$ and 0.216gm of ${H_2}O$. 0.36gm of the same substance gave 48.88mL of ${N_2}$ at 290K and 740mm pressure. Calculate the percentage composition of the compound.

Answer 1

Hint: We can first find the mass of C, H and N obtained in gm in order to find the % composition of them. If needed, then 760mm = 1atm pressure. Ideal gas equation is PV=nRT where R is the universal gas constant.

Complete step by step answer:
From the given data, we can find the %composition of C, N and H in the compound as carbon is obtained in the form of carbon dioxide upon combustion and hydrogen in the compound is obtained as water. Nitrogen present in the organic compound can be measured by nitrogen liberated upon its oxidation. So, we will first find the amount of the elements present in the compound and then we will find its composition.

Molecular weight of $C{O_2}$ = Atomic weight of C + 2(Atomic weight of O) = 12 + 2(16) =$44gmmo{l^{ - 1}}$
- So, we can say that if 44gm of carbon dioxide gets produced, the organic compound would contain 12gm of carbon, then if 0.369gm of carbon dioxide gas is produced, then the mass of carbon in the compound will be $\dfrac{{0.369 \times 12}}{{44}}$ = 0.100gm
Now, molecular weight of ${H_2}O$ = Atomic weight of O + 2(Atomic weight of H) = 16 + 2(1) = $18gmmo{l^{ - 1}}$

So, we can say that if 18 gm of water is produced then the compound would contain 2gm of hydrogen, so if 0.216gm of hydrogen is produced, then the weight of hydrogen in the organic compound will be $\dfrac{{0.216 \times 2}}{{18}}$ = 0.024gm

Now, we will find % of C and H.
     \[\% {\text{ of C = }}\dfrac{{{\text{Mass of C in the compound}} \times {\text{100}}}}{{{\text{Total mass of the compound}}}}\]
     \[\% C = \dfrac{{0.1 \times 100}}{{0.27}} = 37.03\% \]
We can also write that
     \[\% {\text{ of H = }}\dfrac{{{\text{Mass of H in the compound}} \times {\text{100}}}}{{{\text{Total mass of the compound}}}}\]
     \[\% H = \dfrac{{0.024 \times 100}}{{27}} = 8.88\% \]

Now, we are given that 48.88mL of nitrogen gas will be produced at 290K and 740mm pressure. So, we will need the amount of gas in weight. So, we can use the ideal gas equation
     \[PV = nRT\]

We are given pressure in mm units. So, we know that 760mm=1 atm. We need to put the value of pressure in the atm unit. So, we can say that 740mm=$\dfrac{{740}}{{760}}$ atm

Let’s put all the available values into the ideal gas equation.
     \[\dfrac{{740}}{{760}} \times 48 \times {10^{ - 3}} = n \times 0.082 \times 290\]
     \[\dfrac{{\dfrac{{740}}{{760}} \times 48 \times {{10}^{ - 3}}}}{{0.082 \times 290}} = n\]
So, $n = 0.002moles$

Now, we know that molecular weight of ${N_2}$ = 2(atomic weight of N) = 2(14) = 28 $gmmo{l^{ - 1}}$
So, mass of ${N_2}$ generated = $Moles \times {\text{molecular weight }}$ = (0.002)(28) = 0.056gm

So, we can say that if 0.056gm of nitrogen is there in 0.36gm of the compound.
     \[\% {\text{ of N = }}\dfrac{{{\text{Mass of N in the compound}} \times {\text{100}}}}{{{\text{Total mass of the compound}}}}\]
     \[\% N = \dfrac{{0.056 \times 100}}{{36}} = 15.55\% \]
Thus, we can conclude that % composition of C, H and N is 37.03%, 8.88% and 15.55% respectively.

Note: Remember that we cannot put the value of pressure in mm in the ideal gas equation. We need to convert it into an atm unit. Remember that we cannot measure the oxygen from the carbon dioxide and water generated because the oxygen atom in those molecules is not derived from the organic molecule but is the atmospheric oxygen.