A liquid organic compound gives a positive Lassaigne's test for bromine and for sulphur, and a…

Step 1. Carbon. $\dfrac{12}{44} \times 0.352\ \text{g} = 0.0960\ \text{g}$. So $\%\ \text{C} = \dfrac{0.0960\ \text{g}}{0.326\ \text{g}} \times 100 = 29.45\%$.

Step 2. Hydrogen. $\dfrac{2}{18} \times 0.054\ \text{g} = 0.0060\ \text{g}$. So $\%\ \text{H} = \dfrac{0.0060\ \text{g}}{0.326\ \text{g}} \times 100 = 1.84\%$.

Step 3. Bromine. $\dfrac{80}{188} \times 0.376\ \text{g} = 0.1600\ \text{g}$. So $\%\ \text{Br} = \dfrac{0.1600\ \text{g}}{0.326\ \text{g}} \times 100 = 49.08\%$.

Step 4. Sulphur. $\dfrac{32}{233} \times 0.233\ \text{g} = 0.0320\ \text{g}$. So $\%\ \text{S} = \dfrac{0.0320\ \text{g}}{0.163\ \text{g}} \times 100 = 19.63\%$.

Step 5. Oxygen and the empirical formula. The four percentages add up to $100.00\%$, so there is no oxygen. Moles in $100\ \text{g}$: C $2.454$, H $1.84$, Br $0.6135$, S $0.6134$. Dividing by $0.6134$ gives $4 : 3 : 1 : 1$. The empirical formula is $\text{C}_4\text{H}_3\text{BrS}$, and its mass is $48 + 3 + 80 + 32 = 163\ \text{g mol}^{-1}$.

Step 6. Molecular formula. Molar mass $= 2 \times 81.5 = 163\ \text{g mol}^{-1}$, the same as the empirical formula mass. So the molecular formula is $\text{C}_4\text{H}_3\text{BrS}$, with $4$ carbon atoms.