The empirical formula of the compound (BrxOy) is BrO₃
We'll begin by calculating the mole of 2.94 g of AgBr
Mass of AgBr = 2.94 g
Molar mass of AgBr = 108 + 80 = 188 g/mol
Mole = mass / molar mass
Mole of AgBr = 2.94 / 188
From the question given above, we were told that all the bromine in the original oxide compound ends up in the AgBr.
Therefore, the mole of Br in 0.0156 mole of AgBr is 0.0156 mole since 1 mole of AgBr contains 1 mole of Br
Mole of Br = 0.0156 mole
Molar mass of Br = 80 g/mol
Mass = mole × molar mass
Mass of Br = 0.0156 × 80
Mass of Br = 1.248 g
Mass of compound = 2 g
Mass of O = mass of compound – mass of Br
Mass of O = 2 – 1.248
Mass of Br = 1.248 g
Mass of O = 0.752 g
Divide by their molar mass
Br = 1.248 / 80 = 0.0156
O = 0.752 / 16 = 0.047
Divide by the smallest
Br = 0.0156 / 0.0156 = 1
O = 0.047 / 0.0156 = 3
Therefore, the empirical formula of the compound is BrO₃
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