An organic compound have oxygen 26.24% and hydrogen 4.92% . Find the empirical formula?
Give me answer with explaination.
Answers
ANSWER
Let amount of organic = 100 gm
amount of carbon = 54.55 g
amount of hydrogen = 9.09 g
Amount of oxygen = 36.26 g
number of mole of carbon =
12
34.55
=4.54583
number of mole of hydrogen =
1.018
9.09
=9
number of mole of oxygen =
16
36.26
=2.26
4.54833:9:2.26
2:4:1
organic compounds Empirical formula
=C
2
H
4
O
Molecular weight = 24+4+16=44
(for empirical)
Molecular weight = 2× vapour density
=2×44=88 g
To molecular formula be = C
4
H
8
O
2
H₃O
Mass percent of oxygen atom = 26.24%
Relative moles of atom of oxygen = 26.24/16 = 1.64
Mass percent of hydrogen atom = 4.92% .
Relative moles of atom of hydrogen atom 4.92 / 1 = 4.92
To find the empirical formula one need to find the simple molar ratio of both atoms for which Relative moles of both atom is divided by the smallest value of Relative moles of atom of oxygen atom
Simple molar ratio of oxygen atom = 1.64/1.64 = 1
Simple molar ratio of hydrogen atom = 4.92/1.64 = 3
Hence, empirical formula for given compound according to the question H₃O
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