Analysis of a gas gave $:$ $C = 85.7 %$ and H $= 14.3 %$ . If the formula mass of this gas is $42$ a.m.u, what are the empirical formula and the true formula of the compound?

C H; C_{4} H_{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C H_{2}; C_{3} H_{6}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C H_{3}; C_{3} H_{9}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C_{2} H_{2}; C_{3} H_{6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C_{2} H_{4}; C_{3} H_{6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $C H_{2}; C_{3} H_{6}$
100 a.m.u of compound will contain 85.7 a.m.u C and 14.3 a.m.u H.

42 a.m.u (1 mole) of the compound will contain $\frac{42 \times 85.7}{100} = 36$ a.m.u C and $\frac{42 \times 14.3}{100} = 6$ a.m.u H.

The atomic masses of C and H are 12 a.m.u and 1 a.m.u respectively.

The number of moles of C present in 1 mole of compound $= \frac{36}{12} = 3$ moles.

The number of moles of H present in 1 mole of compound $= \frac{6}{1} = 6$ moles.

Hence, the molecular formula of the compound is $C_{3} H_{6}$ .

The empirical formula of the compound is obtained by dividing molecular formula with 3. It is $C H_{2}$ .

Option B is correct.

Suggest Corrections

Similar questions

The empirical formula of an organic gas containing carbon and hydrogen is CH₂. The mass of 1 litre of this organic gas is exactly equal to that of 1 litre of N₂. Therefore, the molecular formula of organic compound is:

Q. (a) What is the general formula of alkenes? Identify the alkenes from the following:
C₂H₆, C₂H₄, C₃H₄, C₂H₂, C₃H₆
(b) Which of the following can give addition reactions?
C₂H₆, C₃H₈, C₃H₆, C₂H₂, CH₄
Give reason for your choice.

Q. An organic compound containing

C

and

H

gave the following analysis:

C = 40 %, H = 6.7 %

and the remaining, oxygen. Its empirical formula would be:

Q. An organic compound contains carbon and hydrogen. Its elemental analysis gave

C - 75 %

and

H - 25 %

. The empirical formula of the compound would be:

Q. An organic compound on analysis gave

C = 54.2

H =

9.2

by mass. Its empirical formula is :

Join BYJU'S Learning Program

Analysis of a gas gave: C=85.7% and H=14.3%. If the formula mass of this gas is 42 a.m.u, what are the empirical formula and the true formula of the compound?

Analysis of a gas gave $:$ $C = 85.7 %$ and H $= 14.3 %$ . If the formula mass of this gas is $42$ a.m.u, what are the empirical formula and the true formula of the compound?