Question

A 1.90-g sample of an oxide of bromine is converted to 3.188 g of AgBr. Calculate the empirical formula of the oxide. (molar mass for AgBr =187.78 g/mol)

a. BrO
b. Br2O
c. BrO
d. BrO3
e. None of these

Answer 1

Answer:

e. None of these

Explanation:

The oxide of Bromine (BrₐOₓ) produce "a" moles of Br that are equal to "a" moles of AgBr. The moles of AgBr are:

Moles AgBr:

3.188g AgBr * (1mol / 187.78g) = 0.01698 moles of AgBr = Moles of Br

These moles of Br weighs (Molar mass: 79.9g/mol):

0.01698 moles of Br * (79.9g / mol) = 1.36g of Br

Mass of Oxygen must be:

1.90g - 1.36g Br = 0.54g oxygen. In moles:

0.54g O * (1mol / 16g) = 0.03375 moles of O

Empirical formula is the simplest ratio of atoms in the molecule. Dividing in the moles of Br:

Br:  0.01698 moles of Br / 0.01698 moles = 1

O: 0.03375 moles of O / 0.01698 moles = 1.99 ≅ 2

That means empirical formula is:

BrO₂

And right option is:

e. None of these

Answer 2

The empirical formula of the oxide would be $BrO_2$

1.90 g of the oxide of bromine produces 3.188 g of AgBr

moles of 3.188 g AgBr = 3.188/187.78

                                        = 0.017 moles

AgBr ---> Ag+ + Br-

Thus, mole of Br in AgBr = 0.017 moles

Mass of 0.017 Br = mole x molar mass

                           = 0.017 x 79.904

                              = 1.357 g

Mass of oxygen in the oxide of bromine = 1.90 - 1.357

                                                                    = 0.543 g

mole of 0.543 oxygen = 0.543/16

                                       = 0.0339 moles

mole of 1.357 g Br = 0.017

Dividing by the smallest:

Br = 0.017/0.017 = 1

O = 0.0339/0.017 = 1.99

Empirical formula = BrO2

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