Combustion of a 1.025 g sample of a compound containing only carbon, hydrogen and oxygen produced 2.265 g of CO2 and 1.236 g of H2O. What is the empirical and molecular formulas of the sample compound, if its molecular weight has been roughly determined to be 363 g/mol by mass spectrometer?
1) Moles of CO2 = Mass / Molar mass
= 2.265 g / 44
= 0.0514 moles of CO2
If our sample contains 0.0514 moles of CO2 then our sample must contains 0.0514 moles of C, becouse CO2 prepared from 1 moles of carbon & 2 moles of oxygen
Then mass of carbon = Molar mass x moles = 12.011 x 0.0514
= 0.617 g of C.
2) Moles of Water = Mass / Molar mass
= 1.236 g / 18
= 0.0686 moles
One mole water prepared from one mole of oxygen & Two moles hydrogen. Therefore moles of hydrogen
= 2 x 0.0686
= 0.1372 moles of H2
Now, Mass of H2 = Molar mass x Moles
= 1 x 0.1372
= 0.1372 g of H2
When we add mass of carbon and Hydrogen together then we get,
= 0.617 g C + 0.1372 g H2
= 0.7542 g
3) We cumbust 1.025 g of sample. Then we can find the mass of oxygen as,
= 1.025 - 0.7542
= 0.2708 g of Oxygen.
Moles of O2 = Mass / molar mass
= 0.2708 / 16
= 0.01692 moles
Now, we get,
a) Moles of Carbon = 0.0514 moles
b) Moles of Hydrogen = 0.1372 moles
c) Moles of Oxygen = 0.01692 moles
Dividing by smallest number to above values,
a) Carbon = 0.0514 / 0.01692 = 3
b) Hydrogen = 0.1372 / 0.01692 = 8.1 = 8
c) Oxygen = 0.01692 / 0.01692 = 1
Therefore the empirical formula = C3H8O
Empirical formula mass = ( 3 x 12 ) + ( 1 x 8 ) + ( 16 x 1 )
= 36 + 8 + 16
= 60
4) Molecular formula = Molecular wt / 60
= 363 g/mol / 60
= 6.05
Multiplying by 6 to empirical formula we get,
C18H48O6 . This is the Molecular formula.
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