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Question: Glucose is a physiological sugar. If the mass percentage of \(C = x\% \), mass percent of \(H = y\% ...

Glucose is a physiological sugar. If the mass percentage of C=x%C = x\% , mass percent of H=y%H = y\% , and mass percent of O=z%O = z\% in glucose (C6H12O6{C_6}{H_{12}}{O_6}), then the value of x,yx,y and zz are, respectively:
A. 40,40, 6.67,6.67, 53.3353.33
B. 30,30, 6.67,6.67, 43.3343.33
C. 40,40, 7.67,7.67, 63.3363.33
D.None of the above

Explanation

Solution

We have to remember that the percentage composition of elements in a compound expresses the ratio of each element of the amount to the total amount of individual elements present in the compound multiplied by 100. The sum of percentage of all elements must add up to 100%\% .

Complete step by step answer:
We can calculate the percent composition of component in a compound by:
Each atom has a mass (atomic weight) and the masses are added to get the molar mass.
Divide the mass due to the component by the total molar mass of the component on the compound and multiply by 100.
Percentage composition =(Mass of the componentTotal molar mass of compound)×100 = \left( {\dfrac{{{\text{Mass of the component}}}}{{{\text{Total molar mass of compound}}}}} \right) \times 100
We have to remember that the formula of glucose is C6H12O6{C_6}{H_{12}}{O_6}. From the molecular formula, glucose molecules contain 66 carbon atoms,1212 hydrogen atoms and 66oxygen atoms.
Mass of C,HC,H and OO are 12,12, 1,1, and 1616 respectively.
Total molecular mass of glucose =(6×12)+(12×1)+(6×16)=180g = \left( {6 \times 12} \right) + \left( {12 \times 1} \right) + \left( {6 \times 16} \right) = 180g
In glucose there are 66 carbon atom, atomic weight carbon is 1212, so the molar mass is 7272 and the percentage composition of carbon in C6H12O6=(72180)×100%=40%{C_6}{H_{12}}{O_6} = \left( {\dfrac{{72}}{{180}}} \right) \times 100\% = 40\% percentage by the mass.
For hydrogen, there are 1212 hydrogen atom in glucose, atomic weight is 11, so the molar mass is1212 and percent composition of hydrogen in C6H12O6=(12180)×100%=6.7%{C_6}{H_{12}}{O_6} = \left( {\dfrac{{12}}{{180}}} \right) \times 100\% = 6.7\% percentage by the mass.
Similarly, for oxygen, there are 66 hydrogen atom, atomic weight is 1616, so the molar mass is 9696 and percent composition of oxygen in C6H12O6=(96180)×100%=53.3%{C_6}{H_{12}}{O_6} = \left( {\dfrac{{96}}{{180}}} \right) \times 100\% = 53.3\% percentage by the mass.
We know the sum of the percentage of all elements must add up to 100100. So, we add the percentage composition of glucose by mass the mass of carbon, hydrogen and oxygen is
40%+6.7%+53.33%=100.03%40\% + 6.7\% + 53.33\% = 100.03\% . By considering the whole number, then the percentage is 100100.
The mass percentage of C=40%C = 40\% , mass percent of H=6.67%H = 6.67\% , and mass percent of O=53.33%O = 53.33\% in glucose and the values of x,x, yy and zz are 40,40, 6.67,6.67, and 53.3353.33 respectively.

So, the correct answer is Option A.

Note: In chemical analysis, the percentage composition plays an important role. It is found in honey, fruits, vegetables and sugars. It is mainly stored in muscle as glycogen and liver. Glucose is the major energy source in cell function. Without glucose, the brain could not work well and is of great importance in the regulation of metabolism.