Verified Answer :

Answer:

The empirical formula of the compound is  \(SO_3\).

Explanation:

Suppose in 100 grams of compound there is 40% of sulfur and 60% of oxygen  by weight.

Mass of sulfur = 40% of 100 g = \(\frac{40}{100}\times 100=40 g\)

Mass of oxygen = 60% of 100 g = \(\frac{60}{100}\times 100=60 g\)

Moles of sulfur = \(\frac{40 g}{32 g/mol}=1.25 mol\)

Moles of oxygen = \(\frac{60 g}{16 g/mol}=3.75 mol\)

In order to determine its empirical formulas divide smallest number of moles by each number of moles of elements:

Sulfur = \(\frac{1.25 mol}{1.25 mol}=1\)

Oxygen = \(\frac{3.75 mol}{1.25 mol}=3\)

Empirical formula = \(S_1O_3\)

The empirical formula of the compound is  \(SO_3\).

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The empirical formula of the compound is \(SO_3\)

What is the Empirical formula?

Empirical formula is the simplest whole number ratio of atoms of each element in a compound.

Calculation:

Given, the amount of sulfur is 40%

The amount of oxygen is 60%

  • Step 1:, we calculate the mass of the elements

Mass of sulfur \(= \dfrac{40}{100} \times100= 40 g\)

Mass of Oxygen\(= \dfrac{60}{100} \times100= 60 g\)

  • Step 2:, we calculate moles:  mass/ atomic mass

Moles of sulfur   \(\dfrac{40g}{32g/mol}= 1.25\; mol\)

Moles of Oxygen \(\dfrac{60g}{16g/mol}= 3.75\; mol\)

  • Step 3:, To calculate the empirical formula, divide the mole of each element by the smallest number of mole :

Sulfur \(= \dfrac{1.25mol}{1.25mol} = 1\)

Oxygen\(= \dfrac{3.75mol}{1.25mol} = 3\)

Thus, the empirical formula of the compound is \(SO_3\).

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