A carbon compound on analysis gives the following percentage composition. Carbon $14.5$ , hydrogen $1.8$ chlorine $64.6$ , oxygen $19.24$ . The empirical formula of the compound is:

C_{2} H_{3} C l_{3} O_{2}

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C_{2} H C l_{3} O_{4}

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C_{2} H_{2} C l_{3} O_{3}

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C_{2} H_{4} C l_{3} O_{2}

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Solution

The correct option is A $C_{2} H_{3} C l_{3} O_{2}$
Percentage compostion of the elements present in the compound,
$C : H : C l : O = 14.5 : 1.8 : 64.46 : 19.24$

Dividing by molar mass we get the mole ratio as,
$C : H : C l : O = 1.21 : 1.8 : 1.81 : 1.2$

Dividing by the smallest number and converting the same to whole numbers, we get the simplest whole number ratio.
$C : H : C l : O = 2 : 3 : 3 : 2$

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A carbon compound on analysis gives the following percentage composition. Carbon 14.5 , hydrogen 1.8 chlorine 64.6 , oxygen 19.24 . The empirical formula of the compound is:

A carbon compound on analysis gives the following percentage composition. Carbon $14.5$ , hydrogen $1.8$ chlorine $64.6$ , oxygen $19.24$ . The empirical formula of the compound is: