2.0 g of a bromo hydrocarbon $$(X)$$ was subjected to Carius analysis, gave 3.36 g of AgBr. The percentage of carbon in the compound $$(X)$$ is 26.7%. Total number of carbon atoms in the empirical formula for compound $$(X)$$ is _____.
( Given molar mass in g $$mol^{-1}H:1,C:12,O:80,Cl:108$$)

Mass of compound $$(X) = 2.0\;g$$. Under Carius halogen analysis bromine is converted to $$AgBr$$, so the moles of $$AgBr$$ produced equal the moles of $$Br$$ present.

Molar mass of $$AgBr = 108 + 80 = 188\;g\,mol^{-1}$$.

Moles of $$AgBr$$ obtained $$= \frac{3.36}{188} = 0.017872\;mol$$.

Since the ratio $$AgBr : Br$$ is $$1:1$$, moles of bromine in the sample are also $$0.017872\;mol$$.

Mass of bromine in the sample $$= 0.017872 \times 80 = 1.4298\;g$$.

Percentage of bromine $$= \frac{1.4298}{2.0}\times100 = 71.49\%$$.

Percentage of carbon is given as $$26.7\%$$, so mass of carbon in the sample $$= 0.267 \times 2.0 = 0.534\;g$$.

Moles of carbon $$= \frac{0.534}{12} = 0.0445\;mol$$.

The remaining mass is hydrogen: $$2.0 - 0.534 - 1.4298 = 0.0362\;g$$.

Moles of hydrogen $$= 0.0362\;mol$$ (atomic mass of $$H = 1$$).

Now write the mole ratio:

$$C : H : Br = 0.0445 : 0.0362 : 0.017872$$.

Divide by the smallest value $$0.017872$$:

$$C : H : Br = 2.49 : 2.03 : 1$$.

These are very close to $$2.5 : 2 : 1$$. Multiply each by $$2$$ to clear the fractional $$0.5$$ and obtain the simplest integer ratio:

$$C : H : Br = 5 : 4 : 2$$.

Therefore the empirical formula of the bromo-hydrocarbon is $$C_5H_4Br_2$$.

Hence, the total number of carbon atoms present in the empirical formula is 5.