KnowledgeBoat Logo
|

Chemistry

Glucose or blood sugar has the molecular formula C6H12O6. Determine % of each element.

Stoichiometry

5 Likes

Answer

Percentage of particular element = MassTotal mass\dfrac{\text{Mass}}{\text{Total mass}} x 100

C6H12O6 = 6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 g per mole

Percentage of C = 72180\dfrac{72}{180} x 100 = 40%

Percentage of H = 12180\dfrac{12}{180} x 100 = 6.67%

Percentage of O = 96180\dfrac{96}{180} x 100 = 53.33%

Hence, Percentage of C = 40%, H = 6.67% and O = 53.33%

Answered By

3 Likes


Related Questions