Question

A sample of a compound of Cl and O reacts with an excess of H2 to give 0.233g of HCl and 0.403g of H2O. Determine the empirical formula of the compound

Answer 1

Answer:

Cl₂O₇

Explanation:

For the reaction:

ClₓOₙ + H₂ → HCl + H₂O

Moles of HCl and moles of H₂O are:

HCl: 0.233g HCl ₓ (1mol / 36.46g) = 6.39x10⁻³ mol HCl

H₂O: 0.403g H₂O ₓ (1mol / 18.02g) = 2.236x10⁻² mol H₂O

As you can see, moles of HCl are equivalent to moles of Cl in the compound and moles of H₂O are equivalent to moles of O in the compound, that means:

6.39x10⁻³ mol Cl

2.236x10⁻² mol O

Empirical formula is the simplest ratio of atoms presents in a molecule. If Cl is 1, Oxygen will be:

2.236x10⁻² mol / 6.39x10⁻³ = 3.5

As empirical formula must be given in natural numbers, the empirical formula is:

Cl₂O₇

Answer 2

Answer:

Cl2O7

Explanation:

Molar mass of HCl= 36.5gmol-1

Molar mass of water =18gmol-1

To get the ratio of chlorine present in the original substance

0.233/36.5= 0.00638

To get ratio of oxygen present in the original compound

0.403/18= 0.0224

Then divide through by the lowest ratio

Cl- 0.00638/0.00638 O- 0.0224/0.00638

Cl- 1 O- 3.5

Multiplying through by 2 to give whole numbers

Cl-2. O-7

Hence the empirical formula of the compound is Cl2O7