Problem 1.2
A compound contains 4.07 % hydrogen,
24.27 % carbon and 71.65 % chlorine.
Its molar mass is 98.96 g. What are its
empirical and molecular formulas ?
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3
Answer:
step 1
Divide the given percentages of atoms with their molecular masses
H---4.07/1 =4.07
C---24.27/12 =2.02
Cl--71.65/35.5=2.01
step 2
divide all values with the lowest value obtained.
H---4.07/2.01=2
C---2.02/2.01=1
Cl---2.01/2.01=1
therefore the empirical formula is
CH
2
Cl
WEIGHT OF EMPIRICAL FORMULA=12+2+35.5=49.5
Given molecular weight =98.96 which is double of empirical weight .therefore molecular formula is C
2
H
4
Cl
2
.
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