Determine the empirical formula of a compound containing 48.38 grams of carbon, 6.74 grams of hydrogen, and 53.5 grams of oxygen. In an experiment, the molar mass of the compound was determined to be 180.15 g/mol. What is the molecular formula of the compound?

#Divide given mass with their molar masses.

C=48.38÷12.01= 4.03mols

H= 6.74÷1.0079= 6.69mols

O= 53.5÷16.00= 3.34mols

#Divide the moles with the lowest mole.

C = 4.03÷3.34 = 1.21

H = 6.69÷3.34 = 2.00

O = 3.34÷3.34 = 1

∴Emperical Formula = CH₂O

#Molar mass of CH₂O = 12.01 + (1.0079 x 2) + 16.00 = 30.02 g/mol

#Divide both moler masses: 180.15g/mol÷30.02g/mol = 6

#Multiply the answer with the Emperical formula to get the Molecular formula: 6 x (CH₂O) = C₆H₁₂O₆

Explanation:

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.