An unknown compound of carbon, hydrogen and oxygen contains 69.77% carbon and 11.63% hydrogen and has a molecular weight of 86. It does not reduce Fehling solution, but forms a bisulphite addition compound and gives a positive iodoform test. What are the possible structures for the unknown compound?
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Calculation of emperical formula:
We are given that the percentage of carbon and hydrogen in the given compound is 69.77 % 11.63 % respectively.
Therefore, the percentage of oxygen = (100 - 69.77 - 11.63) % = 18.60 %.
So, 100g of the sample will contain 69.77g of carbon, 11.63g of hydrogen and 18.6g of oxygen.
Number of moles in 69.77g of carbon = mass of carbon / molar mass of carbon = 69.77 / 12 = 5.814
11.63g of hydrogen = 11.63 / 1 = 11.63
18.6g of oxygen = 18.6 / 16 = 1.163
To obtain the simplest ratio of element, we will divide the number of moles of each element by the smallest value that is 1.163. So, we get
C : H : O = C5H10O
We are given that the percentage of carbon and hydrogen in the given compound is 69.77 % 11.63 % respectively.
Therefore, the percentage of oxygen = (100 - 69.77 - 11.63) % = 18.60 %.
So, 100g of the sample will contain 69.77g of carbon, 11.63g of hydrogen and 18.6g of oxygen.
Number of moles in 69.77g of carbon = mass of carbon / molar mass of carbon = 69.77 / 12 = 5.814
11.63g of hydrogen = 11.63 / 1 = 11.63
18.6g of oxygen = 18.6 / 16 = 1.163
To obtain the simplest ratio of element, we will divide the number of moles of each element by the smallest value that is 1.163. So, we get
C : H : O = C5H10O
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