# Show that its molecular formula of the organic compound in the following case is C2H4O2:A 2.203 g sample of an organic compound was extracted from a plant. When it was burned in oxygen, the...

Show that its molecular formula of the organic compound in the following case is C2H4O2:

A 2.203 g sample of an organic compound was extracted from a plant. When it was burned in oxygen, the hydrogen in the compound was converted to 1.32 g of water and the carbon was oxidized to 3.23 g of carbon dioxide.

Another sample was analyzed in a mass spectrometer. The mass spectrum produced showed that the molar mass of the compound was 60.0 g mol^-1.

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A 2.203 g sample of an organic compound was extracted from a plant. When it was burned in oxygen, the hydrogen in the compound was converted to 1.32 g of water and the carbon was oxidized to 3.23 g of carbon dioxide.

Assuming the organic compound consists of only carbon, hydrogen and oxygen, only carbon and hydrogen burn to form carbon dioxide and water.

The percentage mass of hydrogen in water is 2/18. 1.32 g of water contains 0.1466 g of hydrogen. The percentage of carbon in carbon dioxide is 12/44. 3.23 g of carbon dioxide contains 0.8809 g of carbon.

As the mass of the sample was 2.203 g, the mass of oxygen is 2.203 - 0.1466 - 0.8809 = 1.1755 g.

The number of moles of the three elements in the sample is 0.1466 of H, 0.0629 of carbon and 0.0734 of oxygen.

The best integral ratio of H, C and O is 2:1:1

This shows that the empirical formula is CH2O

The molecular mass of the compound is 60 g/mole. The sum of the mass of C, H and O from the empirical formula gives 30.

The molecular formula is obtained by multiplying the quantities of all the elements in the empirical formula by 2 which gives C2H4O2. The molecular formula of the organic compound is C2H4O2.

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