Determine the empirical and molecular formulas for vanillin given that its molar mass is 152 g/mole and its

Question

composition is: 63.15% carbon, 5.30 % hydrogen and the rest oxygen

Answer 1

63.15 % Carbon
5.30 % Hydrogen
100 - (63.15+5.30) = 31.55 % Oxygen
To do this, divide each mass percentages with the corresponding weights of the elements to get the moles of each:
63.15 / 12 = 5.26
5.30 / 1 = 5.30
31.55 / 16 = 1.97
Now, divide all by the smallest, in this case, by 1.97,
5.26 / 1.97 = 2.67
5.30 / 1.97 = 2.69
1.97 / 1.97 = 1
Then multiply each by their atomic weights, and add:
(2.67 * 12) + (2.69 * 1) + (1 * 16) = 50.73
Then divide the given mass by the sum:
152 / 50.73 = 2.996
Finally, we multiply this by the empirical number of moles of each:
C: 2.67*2.996 = 7.99932, which is approximately 8
H: 2.69*2.996 = 8.059, which is approximately 8
O: 1*2.996 = 2.996, which is approximately 3

Therefore, vanillin is C8H8O3.

Hope this helps~ :)

Answer 2

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