# A compound has 69.8% carbon, 18.6% oxygen and 11.6% hydrogen. When 5.3 g of this compound is vapourised at..........125 degree celcius and 102 KPa, it occupies a volume of 1.0L. Detremine the...

A compound has 69.8% carbon, 18.6% oxygen and 11.6% hydrogen. When 5.3 g of this compound is vapourised at......

....125 degree celcius and 102 KPa, it occupies a volume of 1.0L. Detremine the molecular formula of this compound?

jeew-m | College Teacher | (Level 1) Educator Emeritus

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Molecular mass

C = 12g/mol

H = 1g/mol

O = 16g/mol

Let us say we have 100g of the compound.

Then we have 69.8g of C, 18.6g of O and 11.6g of H in the compound.

Number of C moles `= 69.8/12 = 5.8167`

Number of O moles `= 18.6/16 =1.1625`

Number of H moles `= 11.6/1 =11.6`

Mole ratio

`C:H:O = 5.8167:11.6:1.1625`

To simplify the ratios we have to divide the whole by the smallest number among the three which is 1.1625.

`C:H:O = 5.8167/1.1625:11.6/1.1625:1.1625/1.1625`

`C:H:O = 5:10:1`

So the empirical formula of the compound will be `C_5H_10O`

It is given that when 5.3g of this compound is vaporized we will get 1L volume at 125C temperature and 102kpa pressure.

Using ideal gas law;

`PV = nRT`

`P = 102kpa = 1.00atm`

`V = 1.00L`

`R = 0.08206(Latm)/(molK)`

`T = 125+273 K = 398K`

`n = PV/(RT)`

`n = (1xx1)/(0.08206xx398)`

`n = 0.0306`

So in the compound we have 0.0306 moles in 5.3g.

Molecular weight of compound `= 5.3/0.0306 = 173.2g/(mol)`

Let us say the formula of the compound is `(C_5H_10O)_n`

Then we can write;

`(12xx5+1xx10+16xx1)xxn = 173.2`

`86xxn = 173.2`

`n = 2.01`

usually n is a whole number. So we can say n = 2

So the molecular formula of the compound is;

`(C_5H_10O)_2 = C_10H_20O_2`

Assumptions

• The vapour of the compound act as a ideal gas
• The compound convert to a gas completely when it is heated

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