Question

14. A certain hydrocarbon has the following percentage composition C=83.72% , H = 16.27%. The vapour density of the compound was found to be 43. The molecular formula of the compound is-​

Answer 1

Answer:

Empirical formula is C₂H5

Molecular weight = 2 x Vapour density

Molecular weight = n x (Empirical formula

weight)

58 - nx (12 x 2 + 1 x 5)

n = 2

So, molecular formula = C4H10

Answer 2

ANSWER:

Find moles of each element

Assume 100g of HC

$(83.72/100) * 100g/ 12g/mole = 0.69moles of C$

$(16.27/100) * 100g/1g/mole = 0.162moles of H$

Those are subscripts.

We have $C0.69H0.162$

Divide each subscript by smallest subscript

$CH_{2}.4$

Multiply by factor to get whole numbers for subscripts must be whole number.

Multiply by 5

$C_{5}H_{12}$ = ?

It has a v.p. at STP. Liquid $76.992$kPa$(at 20^{0}C)$  isopentane

$0.076atm$ and $PV=nRT$

And$n/V = \frac{P}{RT} = 0.076atm/(0.082 * (20 + 273)) = 0.032mole/L$

$0.032Mole/L * 72.15 g/mol = 2.4g/L$

Not correct. So $36g/L/0.032Mole/L = 112.5g/Mole$ is the mw

$C_{8}H_[19}???$nothing with v.d. close.