Question

What is the percent composition of carbon in glucose (C6H12O6)? (The molar mass of C = 12.01, H = 1.0079 and O = 16.00.)

Answer 1

The molar mass of C6H12O6 is 180.1548 g/mol. The molar mass of C6 is 72.06 g/mol. So you divide the molar mass of C6 by the molar mass of C6H12O6 and multiply it by 100: (72.06g C6/180.1548g C6H1206)x100= .3999 x 100= 39.99% Carbon in glucose (40% rounded) 

Answer 2

Answer : The percent composition of carbon in glucose $(C_6H_{12}O_6)$ is, 6.66%

Solution : Given,

Molar mass of carbon = 12.01 g/mole

Molar mass of hydrogen = 1.0079 g/mole

Molar mass of oxygen = 16.00 g/mole

Molar mass of glucose $(C_6H_{12}O_6)$ = $(6\times 12.01)+(12\times 1.0079)+(6\times 16)=180.1548g/mole$

Formula used :

$\% \text{composition of carbon}=\frac{\text{Mass of carbon}}{\text{Total mass of glucose}}\times 100$

Now put all the given values in this formula, we get

$\% \text{composition of carbon}=\frac{12.01g/mole}{180.1548g/mole}\times 100=6.66\%$

Therefore, the percent composition of carbon in glucose $(C_6H_{12}O_6)$ is, 6.66%