A 2.80-g sample of an oxide of bromine is converted to 4.698 g of AgBr.Calculate the empirical formula of the oxide.(molar mass for AgBr = 187.78 g/mol) A)BrO₃ B)BrO₂ C)BrO D)Br₂O E)none of these

Question

Quizplus · Accepted Answer

To find the empirical formula of the oxide, we need to first calculate the moles of AgBr produced from the given mass:

moles of AgBr = 4.698 g / 187.78 g/mol = 0.025 moles

Since AgBr is a 1:1 compound, this means that the moles of bromine in the oxide are also 0.025. We can find the mass of bromine in the oxide by using its molar mass:

mass of bromine = 0.025 moles x 79.9 g/mol = 1.998 g

Finally, we can calculate the mass of oxygen in the oxide:

mass of oxygen = 2.80 g - 1.998 g = 0.802 g

Now we can use these masses to find the empirical formula:

- Assume 100 g of the oxide exists.
- Mass of bromine is 1.998 g (or 1.998 g/79.9 g/mol = 0.025 moles).
- Mass of oxygen is 0.802 g (or 0.802 g/16.00 g/mol = 0.050 moles).
- Divide both masses by the smaller number of moles to get the mole ratio:

- Bromine: 0.025 moles / 0.025 moles = 1
- Oxygen: 0.050 moles / 0.025 moles = 2

- The empirical formula is therefore BrO2, which simplifies to BrO. Thus, the correct answer is B.

A 280-G Sample of an Oxide of Bromine Is Converted to to 4.698