Calculate the empirical forumla of a compound which contain 4.07% hydrogen, 24.27%carbon and 71.65% chlorine
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Explanation:
Hydrogen Carbon Chlorine
mass 4.07 24.27 71.65
RAM 1 12 35.5
no. of moles 4.07/1 24.27/12 71.65/35.5
4.07 2.02 2.02
divide by the 4.07/2.02 2.02/2.02 2.02
least no 2 1 1
Empirical formula therefore is H2CCl
the molecular formula is (EF)n=MF
(1+1+12+35.5)=49.5......98.96/49.5=2 n=2
(H2CCl)2=H4C2Cl2
therefore the molecular formula is C2H4Cl2 (dichloroethane)
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