0.078 g of hydrocarbon occupies 22.4 ml of volume at 1 atm and 0° C. The empirical formula of the hydrocarbon is CH. The molecular formula is:

Question

0.078 g of hydrocarbon occupies 22.4 ml of volume at 1 atm and 0° C. The empirical formula of the hydrocarbon is CH. The molecular formula is: (A) C 2 H 2 (B) C 4 H 4 (C) C 6 H 6 (D) C 8 H 8

Tardigrade Admin · Accepted Answer

22.4 mL of hydrocarbon weighs 0.078 g Then, 22400 mL of hydrocarbon weighs =(0.078/22.4) × 22400 g =78 g = molar mass Emprical mass = at. mass of C + at. mass of H =12+1=13 g n=(78/13)=6 ∴ M.F =( E ⋅ F ) n =( CH )6 or C 6 H 6

Q. $0.078 g$ of hydrocarbon occupies $22.4 m l$ of volume at $1$ atm and $0^{\circ} C$ . The empirical formula of the hydrocarbon is $C H$ . The molecular formula is:

$C_{2} H_{2}$

$C_{4} H_{4}$

$C_{6} H_{6}$

$C_{8} H_{8}$

$22.4 m L$ of hydrocarbon weighs $0.078 g$
Then, $22400 m L$ of hydrocarbon weighs $= \frac{0.078}{22.4} \times 22400 g$
$= 78 g =$ molar mass
Emprical mass $=$ at. mass of $C +$ at. mass of $H$
$= 12 + 1 = 13 g$
$n = \frac{78}{13} = 6$
$∴ M . F = (E \cdot F)_{n} = (C H)_{6}$
or $C_{6} H_{6}$

Q. 0.078g of hydrocarbon occupies 22.4ml of volume at 1 atm and 0∘C. The empirical formula of the hydrocarbon is CH. The molecular formula is:

C2​H2​

C4​H4​

C6​H6​

C8​H8​