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pentagon [3]
3 years ago
12

A 29.05 gram sample of cobalt is heated in the presence of excess oxygen. A metal oxide is formed with a mass of 40.88 g. Determ

ine the empirical formula of the metal oxide.
Chemistry
1 answer:
deff fn [24]3 years ago
4 0

The empirical formula of the oxide is Co₂O₃.

<em>Step 1</em>. Calculate the <em>mass of oxygen</em>

Your reaction is

 Cobalt + oxygen ⟶ cobalt oxide

29.05 g +    x g    ⟶     40.88 g

According to the <em>Law of Conservation of Mass</em>, the total mass of the reactants must equal the total mass of the products. Thus,

29.05 g + <em>x</em> g ⟶ 40.88 g

<em>x</em> = 40.88 – 29.05 = 11.83

<em>Step 2</em>. Calculate the <em>moles of each element</em>

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our job is to calculate the molar ratio of Co to O.

<em>Moles of Co</em> = 29.05 g Co × (1 mol Co /(58.93 g Co) = 0.492 96 mol Co

<em>Moles of </em>O = 11.83 g O × (1 mol O/16.00 g O) = 0.739 38 mol O

<em>Step 3</em>. Calculate the <em>molar ratio</em> of the elements

Divide each number by the smaller number of moles

Co:O = 0.429 26:0.739 38 = 1:1.4999

<em>Step 4</em>. Multiply each number by a factor that makes the <em>ratio close to whole numbers </em>

Multiply by 2. Then

Co:O = 2:2.998 ≈ 2:3

<em>Step 5</em>: Write the <em>empirical formula</em>

EF = Co₂O₃

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