Estimate the mass percentage of bromine in the compound using the Carius method. Given that 1.69 g of an organic compound gave 1.88 g of [tex]$AgBr$[/tex].

The Carius method is used to estimate the mass percentage of bromine in an organic compound by converting it into silver bromide (AgBr). In the given problem, 1.69g of an organic compound produced 1.88g of AgBr. To calculate the mass percentage of bromine, we need to use the molar mass of AgBr (187.77 g/mol) and the molar mass of bromine (Br is approximately 79.90 g/mol from the average of its isotopes).

First, we calculate the moles of AgBr produced:

Moles of AgBr = mass of AgBr / molar mass of AgBr
Moles of AgBr = 1.88g / 187.77 g/mol
Moles of AgBr = 0.0100 moles (rounded to 4 significant figures)

Since one mole of AgBr contains one mole of bromine, the moles of bromine are the same as the moles of AgBr:

Moles of Br = 0.0100 moles

Now, we convert the moles of bromine to grams using the molar mass of bromine:

Mass of Br = moles of Br * molar mass of Br
Mass of Br = 0.0100 moles * 79.90 g/mol
Mass of Br = 0.7990 g (rounded to 4 significant figures)

Finally, we calculate the mass percentage of bromine in the organic compound:

Mass percentage of Br = (mass of Br / mass of the organic compound) * 100%
Mass percentage of Br = (0.7990 g / 1.69g) * 100%
Mass percentage of Br = 47.28%

In conclusion, the estimated mass percentage of bromine in the compound using the Carius method is approximately 47.28%.