Question

A compound is 54.53 % C, 9.15% H, and 36.32% O by mass. What is its empirical formula? Insert subscripts as needed. empirical formula: CHO The molecular mass of the compound is 132 amu. What is its molecular formula? Insert subscripts as needed. molecular formula: CHO

Answer 1

The compound is 54.53% Carbon(), 9.15%. Hydrogen (H) and 36:32% oxygen (0) by mass- 3. The mass ratio of ciho o in the compound. = 54.63 : 976 56 3638 is 54.53: 9.15 : 36.32 Now, the ratio of atoms in the compound CiH:0 2:54.5.3: 9.5 : 36:32 dividing the mass of each element by their atomie weight respectively, a .. atomic mass of carbon a 12 amu .. . of Hydrogen a 1 amu Le 1 : oxygen a 16 amu. = 4.54 9.15 e 2.27 dividing by the lowest 3 2:4:1. (9.45 = 4.03 ~4). a. The empirical formula of the compound=C₂ H₂O .

Empirical and Molecular formula of a compound are related as - . Molecular formula = nx( Empirical formula) : n= positive integer Here Let, molecular formula n(C2H40) = Can Hanon a Molecular mass of the compound :: 's n[ 2x12) +(4x) +(1x16] 3. = n(24+ 4 + 16) 3 A4n . By the problem, 4402 132 na 132 - 3 S. Molecular formula of the compounds . : C6H12O3