Question

an compound on analysis give the following % composition ; K=31.84% , Cl=28.98% , O= 39.18% calculate the empirical formula of the compound

Answer 1

The empirical formula is $KClO_3$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of K= 31.84 g

Mass of Cl = 29.98 g

Mass of O = 39.18 g

Step 1 : convert given masses into moles.

Moles of K =$\frac{\text{ given mass of K}}{\text{ molar mass of K}}= \frac{31.84g}{39g/mole}=0.82moles$

Moles of Cl =$\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{29.98g}{35.5g/mole}=0.84moles$

Moles of O =$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{39.18g}{16g/mole}=2.45moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For K = $\frac{0.82}{0.82}=1$

For Cl = $\frac{0.84}{0.82}=1$

For O =$\frac{2.45}{0.82}=3$

The ratio of K: Cl: O= 1: 1: 3

Hence the empirical formula is $KClO_3$

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Answer 2

The empirical formula is KClO3

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of K= 31.84 g

Mass of Cl = 29.98 g

Mass of O = 39.18 g

Step 1 : convert given masses into moles.

Moles of K =given mass of K/ molar mass of K= 31.84g/39g/mole}=0.82moles

Moles of Cl =given mass of Cl/molar mass of Cl= 29.98g/35.5g/mole}=0.84moles

Moles of O =given mass of O/molar mass of O= 39.18g/16g/mole}=2.45moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For K = {0.82/0.82}=1

For Cl ={0.84/0.82}=1

For O ={2.45/0.82}=3

The ratio of K: Cl: O= 1: 1: 3

Hence the empirical formula is KClO3.