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A 2.00 g sample of a bromine oxide (BrxOy) is converted to 2.94 g of AgBr. All the bromine in the original oxide compound ends up in the AgBr (molar mass for AgBr = 187.8). Determine the empirical formula of the oxide.

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Eduard22sly

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 of AgBr =?

Mole = mass / molar mass

Mole of AgBr = 2.94 / 188

Mole of AgBr = 0.0156 mole

  • Next, we shall determine the mole of Br in AgBr.

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

  • Next, we shall determine the mass of Br in the compound.

Mole of Br = 0.0156 mole

Molar mass of Br = 80 g/mol

Mass of Br =?

Mass = mole × molar mass

Mass of Br = 0.0156 × 80

Mass of Br = 1.248 g

  • Next, we shall determine the mass of Oxygen in the compound

Mass of Br = 1.248 g

Mass of compound = 2 g

Mass of O =?

Mass of O = mass of compound – mass of Br

Mass of O = 2 – 1.248

Mass of O = 0.752 g

  • Finally, we shall determine the empirical formula of the compound

Mass of Br = 1.248 g

Mass of O = 0.752 g

Empirical formula =?

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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