Chemistry
An organic compound on analysis gave H = 6.48% and O = 51.42%. Determine it's empirical formula if the compound contains 12 atoms of carbon. [C = 12, H = 1, O = 16]
Stoichiometry
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Answer
| Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
|---|---|---|---|---|
| Oxygen | 51.42 | 16 | = 3.213 | = 1 |
| Hydrogen | 6.48 | 1 | = 6.48 | = 2.016 = 2 |
| Carbon | 100 - (51.42 + 6.48) = 42.1 | 12 | = 3.508 | = 1.091 = 1 |
Simplest ratio of whole numbers = O : H : C = 1 : 2 : 1
As number of carbon atoms = 12
Therefore, O : H : C = 12 : 24 : 12
Hence, empirical formula is C12H24O12
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