An organic compound having molecular formula $C_{2} H_{6} O$ undergoes oxidation with $K_{2} C r_{2} O_{7} / H_{2} S O_{4}$ to produce X which contains 40% carbon, 6.7% hydrogen and 53.3% oxygen. The molecular formula of the compound X is:

C H_{2} O

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C_{2} H_{4} O_{2}

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C_{2} H_{4} O

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C_{2} H_{6} O_{2}

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Solution

The correct option is B $C_{2} H_{4} O_{2}$
$C_{2} H_{5} O H K_{2} C r_{2} O_{7} / H_{2} S O_{4} - --------- \to C H_{3} C O O H$

$% C = \frac{24}{60} \times 100 = 40.00$

$% H = \frac{4}{60} \times 100 = 6.66$

$% O = \frac{32}{60} \times 100 = 53.33$

Hence, option B is correct.

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Similar questions

An organic compound having molecular formula $C_{2} H_{6} O$ under goes oxidation with $\frac{K_{2} C r_{2} O_{7}}{H_{2} S O_{4}}$ to produce X which contains $40 %$ carbon, $6.7 %$ hydrogen and $53.3 %$ oxygen. The molecular formula of the compound X is

Q. An Annie compound is found to contain

40 %

carbon

6.7 %

hydrogenand

53.3 %

oxygen with a molecular mass of

60 g

. What is the molecular formulae of molecular compound?

Q. Oxidation product of X with molecular formula

C_{2} H_{4} O

is Y with molecular formula

C_{2} H_{4} O_{2}

.The compound Y is :

What is the empirical formula of a compound that contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen?

Q. An organic compound contains 20% carbon, 6.7% hydrogen, and 46.67% nitrogen. Its molecular weight was found to be 60. Find the molecular formula of the compound.

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An organic compound having molecular formula C2H6O undergoes oxidation with K2Cr2O7/H2SO4 to produce X which contains 40% carbon, 6.7% hydrogen and 53.3% oxygen. The molecular formula of the compound X is:

The correct option is B C2H4O2C2H5OHK2Cr2O7/H2SO4−−−−−−−−−−→CH3COOH% C=2460×100=40.00% H=460×100=6.66% O=3260×100=53.33Hence, option B is correct.

An organic compound having molecular formula $C_{2} H_{6} O$ undergoes oxidation with $K_{2} C r_{2} O_{7} / H_{2} S O_{4}$ to produce X which contains 40% carbon, 6.7% hydrogen and 53.3% oxygen. The molecular formula of the compound X is:

The correct option is B $C_{2} H_{4} O_{2}$
$C_{2} H_{5} O H K_{2} C r_{2} O_{7} / H_{2} S O_{4} - --------- \to C H_{3} C O O H$

$% C = \frac{24}{60} \times 100 = 40.00$

$% H = \frac{4}{60} \times 100 = 6.66$

$% O = \frac{32}{60} \times 100 = 53.33$

Hence, option B is correct.