Acetic acid, HC2H3O2
First, calculate for the molar mass of acetic acid as shown below.
M = 1 + 2(12) + 3(1) + 2(16) = 60 g
Then, calculating for the percentages of each element.
Hydrogen:
P1 = ((4)(1)/60)(100%) = 6.67%
Carbon:
P2 = ((2)(12)/60)(100%) = 40%
Oxygen
P3 =((2)(16) / 60)(100%) = 53.33%
Glucose, C6H12O6
The molar mass of glucose is as calculated below,
6(12) + 12(1) + 6(16) = 180
The percentages of the elements are as follow,
Hydrogen:
P1 = (12/180)(100%) = 6.67%
Carbon:
P2 = ((6)(12) / 180)(100%) = 40%
Oxygen:
P3 = ((6)(16) / 180)(100%) = 53.33%
b. Since the empirical formula of the given substances are just the same and can be written as CH2O then, the percentages of each element composing them will just be equal.