An organic compound is estimated through Dumas method and was found to evolve 6 moles of $$CO_2$$, 4 moles of $$H_2O$$ and 1 mole of nitrogen gas. The formula of the compound is:

From the data of the Dumas combustion we know the exact numbers of moles of the three products that are formed:

$$n_{\,CO_2}=6,\qquad n_{\,H_2O}=4,\qquad n_{\,N_2}=1$$

First we convert these moles of the products into the moles of the original elements present in the unknown compound.

Each mole of $$CO_2$$ contains exactly one mole of carbon atoms because the formula is $$CO_2$$. Therefore

$$n_{\,C}=n_{\,CO_2}=6\; \text{mol}$$

Each mole of $$H_2O$$ contains two moles of hydrogen atoms since the subscripts of hydrogen in water are 2. Thus

$$n_{\,H}=2\,n_{\,H_2O}=2\times4=8\; \text{mol}$$

Each mole of $$N_2$$ contains two moles of nitrogen atoms because nitrogen gas is di-atomic ($$N_2$$). Hence

$$n_{\,N}=2\,n_{\,N_2}=2\times1=2\; \text{mol}$$

So the mole (atom) ratio of the elements in the burnt sample is

$$C:H:N = 6:8:2$$

We obtain the simplest whole-number ratio (the empirical ratio) by dividing every number by the greatest common divisor, which is 2:

$$\frac{6}{2} : \frac{8}{2} : \frac{2}{2} = 3 : 4 : 1$$

Therefore the empirical formula is

$$C_3H_4N$$

The options offered all have even numbers of carbon and hydrogen atoms, so we scale the empirical formula by the smallest integer that converts it to one of the listed formulas. Multiplying by 2 gives

$$2\,(C_3H_4N)=C_6H_8N_2$$

This molecular formula exactly matches Option C.

Hence, the correct answer is Option C.