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?
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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`
`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`
- 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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