Calculate the percent composition of the elements in 3-chloro-2-methylbutanoic acid.?

3-chloro-2-methylbutanoic acid:  CH₃-CHCl-CH(CH₃)-COOH
Molecular formula:  C₅H₉O₂Cl
Molar mass = (12.0×5 + 1.0×9 + 16.0×2 + 35.5) g/mol = 136.5 g/mol

Mass percent of C in C₅H₉O₂Cl = (12.0×5/136.5) × 100% = 44.0%
Mass percent of H in C₅H₉O₂Cl = (1.0×9/136.5) × 100% = 6.6%
Mass percent of O in C₅H₉O₂Cl = (16.0×2/136.5) × 100% = 23.4%
Mass percent of Cl in C₅H₉O₂Cl = (35.5/136.5) × 100% = 26.0%