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.)

3.36%
6.71%
15.20%
40%
53.29%

Answer 1

Answer : The correct option is, 40 %

Explanation: Given,

Molar mass of C = 12.01 g/mole

Molar mass of H = 1.0079 g/mole

Molar mass of O = 16.00 g/mole

First we have to calculate the molar mass of glucose.

Molar mass of glucose
`(C_6H_(12)O_6)` =
`6(12.01)+12(1.0079)+6(16.00)=180.155g/mole`

Now we have to calculate the percent composition of carbon in glucose.

As we now that there are 6 number of carbon atoms, 12 number of hydrogen atoms and 6 number of oxygen atoms.

The mass of carbon =
`6* 12.01=72.06g`

Formula used :

`\%\text{ Composition of carbon}=\frac{\text{Mass of carbon}}{\text{Molar mass of glucose}}* 100`

Now put all the given values in this formula, we get the percent composition of carbon in glucose.

`\%\text{ Composition of carbon}=(72.06)/(180.155)* 100=39.9\%=40\%`

Therefore, the percent composition of carbon in glucose is, 40 %

Answer 2

Percent composition would be the amount of a substance per amount of the whole sample. In this case, we are asked for the grams carbon per gram of glucose. We do as follows:

1 g C6H12O6 ( 1 mol C6H12O6 / 180.16 g C6H12O6) ( 6 mol C / 1 mol C6H12O6 ) ( 12 g C / 1 mol C) = 0.3996 g C

Percent composition C = 0.3996 / 1 x 100 = 39.96% C