Question

Glucose is a physiological sugar. If the mass percentage of $C = x\% $, mass percent of $H = y\% $, and mass percent of $O = z\% $ in glucose (${C_6}{H_{12}}{O_6}$), then the value of $x,y$ and $z$ are, respectively:
A. $40,$$6.67,$$53.33$
B. $30,$$6.67,$$43.33$
C. $40,$$7.67,$$63.33$
D.None of the above

Answer 1

Hint: 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 $ = \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 ${C_6}{H_{12}}{O_6}$. From the molecular formula, glucose molecules contain $6$ carbon atoms,$12$ hydrogen atoms and $6$oxygen atoms.
Mass of $C,H$ and $O$ are $12,$ $1,$ and $16$ respectively.
Total molecular mass of glucose $ = \left( {6 \times 12} \right) + \left( {12 \times 1} \right) + \left( {6 \times 16} \right) = 180g$
In glucose there are $6$ carbon atom, atomic weight carbon is $12$, so the molar mass is $72$ and the percentage composition of carbon in ${C_6}{H_{12}}{O_6} = \left( {\dfrac{{72}}{{180}}} \right) \times 100\% = 40\% $ percentage by the mass.
For hydrogen, there are $12$ hydrogen atom in glucose, atomic weight is $1$, so the molar mass is$12$ and percent composition of hydrogen in ${C_6}{H_{12}}{O_6} = \left( {\dfrac{{12}}{{180}}} \right) \times 100\% = 6.7\% $ percentage by the mass.
Similarly, for oxygen, there are $6$ hydrogen atom, atomic weight is $16$, so the molar mass is $96$ and percent composition of oxygen in ${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 \[100\]. 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\% $. By considering the whole number, then the percentage is \[100\].
The mass percentage of $C = 40\% $, mass percent of $H = 6.67\% $, and mass percent of $O = 53.33\% $ in glucose and the values of $x,$ $y$ and $z$ are $40,$ $6.67,$ and $53.33$ 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.