Question

What is the percent composition of carbon in a glucose molecule C6H12O6?

Answer 1

1) Find the molar mass (MM) of glucose

C: 12.0107 g/mol

H: 1.00794 g/mol

O: 15.9994 g/mol

MM C6H12O6 = (6*12.0107) +( 12*1.00794) + (6*15.9994) = 180.15588 g/mol

2) Find the mass of carbon in glucose

12.0107*6 = 72.0642

3) Percent composition

`\%_C=\frac{\text{mass of carbon}}{\text{mass of compound}}\cdot100`

`\%_C=\frac{72.0642\text{ g/mol}}{180.15588\text{ g/mol}}\cdot100=40.00\%`

The percent composition of carbon in glucose is 40.00%.

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