Use the precipitate to find mass of bromine

In Carius method, bromine in the organic compound is converted to silver bromide, $\mathrm{AgBr}$.

Molar mass of $\mathrm{AgBr}$: $108 + 80 = 188$

So, $188\,$g of $\mathrm{AgBr}$ contains $80\,$g of Br.

Find bromine present in 1.88 g of AgBr

$\text{Mass of Br} = \frac{80}{188} \times 1.88$

$= 0.8\,\text{g}$

Find percentage of bromine in the organic compound

Mass of organic compound taken $= 1.6\,$g

$\%\text{Br} = \frac{0.8}{1.6} \times 100$

$= 50$

Final answer

The mass percentage of bromine in the compound is: $\boxed{50}$

Comparison with stored answer

Stored correct answer = $50$

My derived answer also is $50$, so they agree.