Chemistry
Glucose or blood sugar has the molecular formula C6H12O6. Determine % of each element.
Stoichiometry
5 Likes
Answer
Percentage of particular element = x 100
C6H12O6 = 6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 g per mole
Percentage of C = x 100 = 40%
Percentage of H = x 100 = 6.67%
Percentage of O = x 100 = 53.33%
Hence, Percentage of C = 40%, H = 6.67% and O = 53.33%
Answered By
3 Likes
Related Questions
Calcium, cobalt, nitrogen, magnesium, and silicon are some elements. Using this information complete the following.
(a) …………… is the most reactive metal.
(b) …………… is a metalloid.
(c) …………… is a non-metal.
Give reasons:
(a) Covalent compounds have low melting point.
(b) For covalent bond formation both atoms should have high ionisation.
(c) Covalent bond formation takes place when both atoms should have four or more electron in outermost shell.
Identify the functional group in the following organic compounds.
(a) Butene
(b) Propanoic acid
(a) State whether the following statements are true or false. Justify your answer.
(1) Electrolysis of molten lead bromide is considered to be a redox reaction.
(2) Copper is a good conductor of electricity but it is a non-electrolyte.
(b) State the reaction occurring at the anode during electrolysis of copper sulphate solution using a copper anode and platinum anode respectively.