Question

A GASEOUS HYDROCARBON CONTAINS 82.76% OF CARBON .FIND ITS EMPIRICAL FORMULA. (ATOMIC WEIGHT OF C=12,H=1)

Answer 1

Answer:

HERE IS YOUR ANSWER

Explanation:

ANSWER

Element %composition Mol. mass No. of C atoms Simplest Rounding

ratio off

C 82.76 12 6.89 1× 2 2

H 17.24 1 17.24 2.5× 2 5

Number of carbon atoms present =

12

82.76

=6.89

Number of hydrogen atoms present =

1

17.24

=17.24

simplest ratio =

6.89

6.89

=1

simplest ratio =

6.89

17.24

=2.5

∴ Empirical fromula is C

2

H

5

Empirical formula mass= 2×12+5 = 29

∴ V.D. = 29

Mol. mass = 2 × V.D.

=58

n=

Empiricalformulamass

Molecularmass

=

29

58

=2

∴ Molecular formula = C

4

H

10

.

Answer 2

Answer:

$Hey \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] make brainliest please$

Explanation:

Number of carbon atoms present = 82.76/12​=6.89

Number of hydrogen atoms present = 17.24/1​=17.24

simplest ratio = 6.89/6.89​=1

simplest ratio = 17.24​/6.89=2.5

∴ Empirical fromula is C2​H5​