Question

A gaseous hydrogen and carbon containing compound is decomposed and formed to contain 82.66% carbon and 17.34% hydrogen by mass. The mass of 158 mL of the gas, measured at 556 mmHg and 25C, is 275g. What is the molecular formula of the compound?

Answer 1

Step-by-step explanation:

Mole percentage of carbon =
`(percentage given)/(molar mass of carbon)`

=
`(82.66)/(12)`

= 6.89

Mole percentage of hydrogen =
`(percentage given)/(molar mass of hydrogen)`

=
`(17.34)/(1)`

= 17.34

Now, dividing mole percentage of both the atoms by 6.89.

Then, C = 1 and H =
`(17.34)/(6.89)` = 2.5

Hence, empirical formula is
`C_(2)H_(5)`.

As, it is given that P = 556 mm Hg. Convert mm Hg into atm as follows.

`(556 mm Hg * atm)/(760 mm Hg)`

= 0.7316 atm

Volume is given as 158 mL. So, in liter volume is
`(158)/(1000)` equals 0.158 L.

According to ideal gas equation, PV = nRT

`0.7316 atm * 0.158 L = (mass)/(molar mass) * 0.082 atm L/mol K * 298 K`

`0.7316 atm * 0.158 L = (0.275 g)/(molar mass) * 0.082 atm L/mol K * 298 K`

molar mass = 58.2 g

Hence, molecular weight of
`C_(2)H_(5)` is
`12 * 2 + 5` = 29.

Therefore,
`(C_(2)H_(5))_(n)` = 58

29 × n = 58

n = 2

Thus, molecular formula of the compound is
`C_(4)H_(10)`.