An organisc compound, whose vapour density is 45, has the following percentage composition,
H = 2.22 % O = 71 . 19% and remaining carbon.

Calculate:
(a) its empirical formula, (b) its molecular formula.

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Solution

Element % at. mass atomic ratio simple ratio

C 26.59 12 $\frac{26.59}{12} = 2.21$ 1

H 2.22 1 $\frac{2.22}{1} = 2.22$ 1

O 71.19 16 $\frac{71.19}{16} = 4.44$ 2

(a) its empirical formula =
(b)empirical formula mass = 45
vapour density = 45
so molecular mass=VD $\times$ 2 = 90
so molecular formula

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. The molecular formula of the compound is:

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vapor density of that compound is 90. Find out the molecular formula of the compound. Calculate the percentage composition of carbon, hydrogen, and oxygen respectively of that compound.

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Element	%	at. mass	atomic ratio	simple ratio
C	26.59	12	$\frac{26.59}{12} = 2.21$	1
H	2.22	1	$\frac{2.22}{1} = 2.22$	1
O	71.19	16	$\frac{71.19}{16} = 4.44$	2

An organisc compound, whose vapour density is 45, has the following percentage composition, H = 2.22 % O = 71 . 19% and remaining carbon. Calculate: (a) its empirical formula, (b) its molecular formula.

Element % at. mass atomic ratio simple ratio C 26.59 12 26.5912=2.21 1 H 2.22 1 2.221=2.22 1 O 71.19 16 71.1916=4.44 2 (a) its empirical formula = (b)empirical formula mass = 45 vapour density = 45 so molecular mass=VD × 2 = 90 so molecular formula

An organisc compound, whose vapour density is 45, has the following percentage composition,
H = 2.22 % O = 71 . 19% and remaining carbon.

Calculate:
(a) its empirical formula, (b) its molecular formula.

Element % at. mass atomic ratio simple ratio

C 26.59 12 $\frac{26.59}{12} = 2.21$ 1

H 2.22 1 $\frac{2.22}{1} = 2.22$ 1

O 71.19 16 $\frac{71.19}{16} = 4.44$ 2

(a) its empirical formula =
(b)empirical formula mass = 45
vapour density = 45
so molecular mass=VD $\times$ 2 = 90
so molecular formula