Question

What is the molecular formula of a compound containing 24.27% carbon, 4.07% hydrogen and 71.65 chlorine by mass. The molecular mass of tha compound is 98.98 amu

Answer 1

Masses:-(will be provided in the exams)

chlorine=35 amu

carbon=12amu

hydrogen=1amu

Atomic ratio:-(% of the compound÷atomic mass)

Cl= 71.65÷35=2.04 (always bring to 2 decimal place)

C=24.27÷12=2.02

H=4.07÷1=4.07

Simplest ratio:-(Smallest atomic ratio÷All atomic ratio)

Cl=2.04÷2.02= 1.009=1(approx)

C=2.02÷2.02=1

H=4.07÷2.02= 2.01=2(approx)

Empharical formula= ClCH2

Empharical formula mass

= Cl+C+2H

=35+12+2

=49 amu

Molecular mass=98.98 amu

Molecular mass÷Empharical formula mass= n

or, 98.98/49=n

or, n=2.02 =2(approx)

Molecular formula=n(ClCH2)

=2×(ClCH2)

=Cl2C2H4

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