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An analysis of a compound having a molar mass of 180 g/mol gives the following percentage composition: 40.00% C, 6.67% H, and 53.33% 0.​

Sagot :

90Angleidn

Answer:

For this question Atomic Mass of Carbon C = 12 g/mol, Hydrogen H = 1 g/mol and oxygen O = 16 g/mol.

The Molecular Mass of the compound is 60 g/mole so you have to find the Mass of each element from its percentage as follows:

Carbon 40% of 60 g/mol = 40/100 x 60 = 24 g we divide this by the Atomic Mass of Carbon which is 12 so 24/12 = 2 Carbon atoms in the compound.

Hydrogen 6.5% of 60 g/m = 6.5/100 x 60 = 3.9 g we divide this by the atomic Mass of Hydrogen which is 1 so 3.9/1 = 4 atoms of H in the Compound.

Finally Oxygen 53.5% of 60 g/mol = 53.5/100 x 60 = 32.1 g we divide by the atomic Mass of Oxygen which is 16 so 32.1/16 = 2 atoms of oxygen in the compound.

So we have 2 Carbons, 4 Hydrogens and 2 Oxygen so the Molecular formula of the Compound is C2H4O2

Answer: Molecular Formula of the Compound = C2H4O2

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