What is the percent composition of C in C6H12O6
The percent composition of C in C6H12O6 can be found by dividing the number of carbon atoms in one molecule of C6H12O6 by the molar mass of C6H12O6 and multiplying by 100.
The molar mass of C6H12O6 can be calculated using the atomic masses of each element:
C = 12.01 g/mol
H = 1.01 g/mol
O = 16.00 g/mol
Molar mass of C6H12O6 = (6 * 12.01) + (12 * 1.01) + (6 * 16.00) = 72.06 + 12.12 + 96.00 = 180.18 g/mol
In one molecule of C6H12O6, there are 6 carbon atoms.
Percent composition of C = (6 * 12.01 g/mol) / 180.18 g/mol * 100 = 40.01%
Therefore, the percent composition of C in C6H12O6 is 40.01%.