Acompound is found to have a percent composition (by mass) of 85.63% c and 14.37% h. the molar mass of the compound was found to be 42.0 g/mol. which of the following is the empirical formula and the molecular formula, respectively a. c2h6 and c3h9 b. ch2 and c3h6 c. ch and c3h3 d. ch3 and c2h6 e. c2h3 and c4h6

Empirical formular = CH₂

Molecular formular = C₃H₆

Explanation:

Percent Mass of C = 85.63%

Percent Mass of H = 14.37%

Molar mass = 42.0 g/mol

Steps in Calculating the Empirical Formular;

1 - Divide the percent mass of the Elements with their respective atomic masses. Carbon = 12 and Hydrogen = 1

Carbon = 85.63 / 12 = 7.14

Hydrogen = 14.37 / 1 = 14.37

2 - Divide the values by the smallest one. In this case Carbon has the lowest value of 7.14

Carbon = 7.14 / 7.14 = 1

Hydrogen = 14.37 / 7.14 =  2.013

3. Approximate the values to the nearest whole number.

Hydrogen ≈ 2

Hence the empirical formular is given as; CH₂

Molecular formular is given as;

[12 + 2(1)]n = 42

14n = 42

n = 3

Molecular formular = C₃H₆