Combustion of a 0.525g sample of an unknown organic compound containing only C, H, and O...

Combustion of a 0.525g sample of an unknown organic compound containing only C, H, and O yielded 1.113g CO2 and 0.1708g of H20. What is the emperical formula of the compound? If the molar mass of the compound is 166.13g/mol, what is the molecular formula?

Please show work!!!!

Solutions

Expert Solution

moles of CO2 = 1.113 / 44 = 0.0253

moles of C= 0.0253

mass of Carbon = 12 x 0.0253 = 0.3036 g

moles of H2O = 0.1708 / 18 = 0.00949

moles of H = 2 x 0.00949 = 0.01898

mass of hydroegen = 0.01898 g

Oxygen mass = 0.525 - (0.3036 + 0.01898 ) = 0.2024 g

moles of O = 0.2024 /16 = 0.0126

C O H

0.0253 0.0126 0.01898

2 1 1.5

4 2 2

C4O2H2 -----------------------> empirical formula

empirical formula mass = 48 + 32 + 2 = 82

n = molar mass / empirical formula mass

= 166.13/82

= 2 (nearly)

molecular formula = n x empirical formula

= C8H4O4

queen_honey_blossom answered 1 year ago

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