Question

A 15.58 g sample of a compound contains 4.97 g potassium (k), 4.51 g chlorine (cl), and oxygen (o). Calculate the empirical formula.

Answer 1

The number of atoms of each element present in the compound is shown by the formula in the form of simplest whole number is known as empirical formula.

Mass of sample = 15.58 g (given)

Mass of potassium in sample = 4.97 g (given)

Mass of chlorine in sample = 4.51 g (given)

Mass of oxygen in sample = (Mass of sample) - (Mass of potassium in sample + Mass of chlorine in sample)

Mass of oxygen in sample = $15.58 - (4.97+4.51) = 6.1 g$

Calculating the number of moles of each element in the sample by using formula:

$number of moles = \frac{mass of element}{Molar mass of element}$

The number of moles of each element in the sample:

• Potassium, $K$

Atomic mass of potassium = 39.098 u

Number of moles of Potassium, $K$ = $number of moles = \frac{4.97}{39.098} = 0.127 mole$

• Chlorine, $Cl$

Atomic mass of chlorine = 35.453 u

Number of moles of Chlorine, $Cl$ = $number of moles = \frac{4.51}{35.453} = 0.127 mole$

• Oxygen, $O$

Atomic mass of chlorine = 15.99 u

Number of moles of Oxygen, $O$ = $number of moles = \frac{6.1}{15.99} = 0.381 mole$

To determine the mole ratio, will divide with the smallest mole number.

$K_{\frac{0.127}{0.127}}Cl_{\frac{0.127}{0.127}}K_{\frac{0.381}{0.127}}$

$KClO_3$

Hence, the empirical formula is $KClO_3$.