Calculate the percent composition by mass of C6H12O6.?

Question

please explain how you got the answer. THANKS!

Answer 1

6 mol C x (12 g C/1 mol C) = 72 g C
12 mol H x (1 g H/1 mol H) = 12 g H
6 mol O x (16 g O/1 mol O) = 96 g O

mol of element x (molar mass/ 1 mol element)
molar mass / 1 mol element is a conversion factor

72 + 12 + 96 = 180 = total molar mass

then take each of the numbers (mass of each element of the compound) and divide it by the total molar mass so...
72/180 = .40 move the decimal 2 places to the right to get a percent = 40%

12/180 = .07 = 7%

96/180 = .53 = 53%

therefore
the perecent composition = 40% Carbon, 7% Hydrogen, 53% Oxygen

Answer 2

take one molecule.contribution of C6 is 6*12=72.since one atom of carbon equals 12 units of mass.similarly H12 is 12*1=12.O6 is 6*16=96. total mass is 72+12+96=180.
now percentage of C is (72/180)*100=40%.
similarly for H is 6.66666...%and for O is 53.333....%
here we used masses of atoms so these are percent composition by mass.

Answer 3

C6 ==>72
H12 ==> 12.096
O6 ==> 95.996
MW = 180.092
C ==> 72/180.092 = 39.98%
H ==> 12.096/180.092 = 6.72%
O ==> 95.996/180.092 = 53.30%