Question

a 2.56 g of an oxide of bromine is converted to 3.76 g of AgBr. what is the empirical formula of oxide.
please answer asap please​

Answer 1

Answer:

this is an example you can try to learn

Explanation:

First figure out what mass of bromine and oxygen you have, convert those figures to moles, and find a whole number ratio.

MW Ag = 107.87

MW Br = 79.90

MW O = 16.00

%Br = (MW Br/MW AgBr) x 100 = (79.90/187.77) x 100 = 42.55%

mass Br = (.4255)(2.685) = 1.142 g

mass O = 1.60 - 1.142 = 0.458 g

mol Br = 1.142 g Br x 1 mol Br/79.90 g Br = 0.01429 mol

mol O = 0.458 g O x 1 mol O/16.00 g O = 0.02863 mol

mol O/mol Br = .02863/.01429 ≅ 2/1

So the empirical formula is BrO2.