Question

An organic compound containing carbon hydrogen and oxygen gave the following percentage composition C = 40.68% H = 5.08%. The vapour density of the compound is 59. Find the molecular formula of the compound

Answer 1

STEP 1 : RELATIVE NO OF ATOMS

in order to find the relative number of atoms

divide the % composition with the atomic mass of the element

$c \: = \frac{40.68}{12} = 3.39$

$h = \frac{5.08}{1} = 5.08$

STEP 2: SIMPLE RATIO

divide both the values from the previous step with the least no i.e 3.39

$c = \frac{3.39}{3.39} = 1$

$h = \frac{5.08}{3.39} = 1.5 = 2$

empirical formula = CH2

empirical mass = 12+2(1)=14 g

vapor density = 59

$vapour \: density = \frac{molar \: mass}{2}$

$molar \: mass = 59 \times 2 = 118g$

we know that ,

$molar \: mass \: = n(empirical \: mass)$

$n = \frac{118}{59}$

$n = 2$

molecular formula = n( empirical formula )

molecular formula = 2( CH2)

molecular formula = C2 H 4

Answer 2

Explanation:

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