Answer : The empirical formula of a compound is, [tex]KMnO_4[/tex]
Solution :
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 = 24.56 g
Mass of Mn = 34.81 g
Mass of O = 40.50 g
Molar mass of K = 40 g/mole
Molar mass of Mn = 55 g/mole
Molar mass of O = 16 g/mole
Step 1 : convert given masses into moles.
Moles of K = [tex]\frac{\text{ given mass of K}}{\text{ molar mass of K}}= \frac{24.56g}{40g/mole}=0.614moles[/tex]
Moles of Mn = [tex]\frac{\text{ given mass of Mn}}{\text{ molar mass of Mn}}= \frac{34.81g}{55g/mole}=0.633moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{40.50g}{16g/mole}=2.53moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For K = [tex]\frac{0.614}{0.614}=1[/tex]
For Mn = [tex]\frac{0.633}{0.614}=1.03\approx 1[/tex]
For O = [tex]\frac{2.53}{0.614}=4.12\approx 4[/tex]
The ratio of K : Mn : O = 1 : 1 : 4
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]K_1Mn_1O_4=KMnO_4[/tex]
Therefore, the empirical formula of a compound is, [tex]KMnO_4[/tex]
