a compound contain 5.03% hydrogen 70.69% chlorine 24.27% carbon by mass.its molar mass is 98.96g.what is its empircal formula? atomic masses of hydrogen, carbon and chlorine are 1.008,12.000 & 35.453u respectively
Answers
The empirical formula of the compound is
C2 H4 Cl2.
Explanation:
Let the total mass of the molecule =100 g
Mass of C in the molecule = 24.27 g
Mass of H in the molecule= 4.07 g
Mass of Cl in the molecule =71.65 g
Number of moles of Carbon = Mass of C in the molecule / molar mass of carbon = 24.7 / 12 = 2.02
Number of moles of Hydrogen = Mass of h in the molecule / molar mass of hydrogen = 4.07 / 1 = 4.07
Number of moles of Cl = Mass of Cl in the molecule / molar mass of chlorine =71.65 / 35.45 = 2.02
Ratio of the number of moles of C,H and Cl will be C : H : C L
Ratio = 2.02 : 4.07 : 2.02 = 1 : 2 : 1
So the Empirical formula will be CH2Cl
Empirical formula mass = 12.01 + 2 + 35.45 = 49.5 g
n = Molecular mass / Empirical formula mass = 98.98 / 49.5 = 2
Molecular formula = (Empirical formula) n = (CH2Cl)2 = C2H4Cl2