An organic compound contains 80% (by wt.) carbon and the remaining percentage of hydrogen. The right option for the empirical formula of this compound is: [Atomic wt. of C is 12, H is 1] 1. CH3 2. CH4 3. CH 4. CH2

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An organic compound contains 80% (by wt.) carbon and the remaining percentage of hydrogen. The right option for the empirical formula of this compound is: [Atomic wt. of C is 12, H is 1] 

1. CH3

2. CH4

3. CH

4. CH2

Answer:1. CH3

Topic: 

Empirical Formula for Molecular Formula

👉An empirical formula represents the simplest
whole number ratio of various atoms present in
a compound, whereas, the molecular formula
shows the exact number of different types of
atoms present in a molecule of a compound.

👉If the mass per cent of various elements

present in a compound is known, its empirical

formula can be determined. Molecular formula

can further be obtained if the molar mass is

known.

👉

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A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it
in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A
volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g.
Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular
formula.

Solution: 

Since, 3.38 g of carbon dioxide are obtained, the mass of C present in the sample of welding fuel gas is
$\frac{12}{44}\times 3.38=0.92$  g.
Since, 0.690 g of water are obtained, the mass of hydrogen present in the welding fuel gas sample is
$\frac{2}{18}\times 0.690=0.077$g.
The percentage of C is $\frac{0.92}{0.92+0.077}\times 100=92.3$ %.
The percentage of H is $\frac{0.077}{0.92+0.077}\times 100=7.7$%.

(i) The number of moles of carbon $=\frac{92.2}{12}=7.7$.
The number of moles of hydrogen $=\frac{7.7}{1}=7.7$.
The mole ratio $C:H=\frac{7.7}{7.7}=1:1$.
Hence, the empirical formula of the welding fuel gas is CH.

(ii) The weight of 10.0 L of gas at N.T.P is 11.6 g.
22.4 L pf gas at N.T. P will weigh $\frac{11.6}{10.0}\times 22.4=26g/mol$
This is the molar mass.

(iii) Empirical formula mass is $12+1=13$.
Molecular mass is 26.
The ratio, n of the molecular mass to empirical formula mass is $\frac{26}{13}=2$.
The molecular formula is $2(CH)=C_2{H}_2$.

Hence,Empirical formula CH, molar mass 26.0 g mol–1, molecular formula C2H2

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Option (a) is correct

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A solution containing 1.8 g of a compound 

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B

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C

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D

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