Question

Find the percentage composition of each element in Glucose (C6 H12 O6).

Answer 1

Given: Given compound is glucose $C_6H_{12}O_6$.

To find: We have to find the percentage composition of each element.

Solution:

The elements present in $C_6H_{12}O_6$ are C, Oxygen and H.

The atomic mass of Carbon, Hydrogen and Oxygen are 12,1 and 16.

The molecular weight of $C_6H_{12}O_6$ will be-

$6 \times 12 +12 \times 1 + 6 \times 16 \\ = 72 + 12 + 96 \\ = 180$

The percentage of composition of C is-

$\frac{12 \times 6}{180} \times 100 \\ = 40\%$

The percentage of composition of O is-

$\frac{16 \times 6}{180} \times 100 \\ = 53.33\%$

The percentage of composition of H is-

$\frac{1 \times 12}{180} \times 100 \\ = 6.66\%$

The percentage of composition of C, H and O in glucose is 40%, 6.66% and 53.33%.

Answer 2

Answer:

Given: Given compound is glucose C_6H_{12}O_6C

6

H

12

O

6

.

To find: We have to find the percentage composition of each element.

Solution:

The elements present in C_6H_{12}O_6C

6

H

12

O

6

are C, Oxygen and H.

The atomic mass of Carbon, Hydrogen and Oxygen are 12,1 and 16.

The molecular weight of C_6H_{12}O_6C

6

H

12

O

6

will be-

\begin{gathered}6 \times 12 +12 \times 1 + 6 \times 16 \\ = 72 + 12 + 96 \\ = 180\end{gathered}

6×12+12×1+6×16

=72+12+96

=180

The percentage of composition of C is-

\begin{gathered} \frac{12 \times 6}{180} \times 100 \\ = 40\%\end{gathered}

180

12×6

×100

=40%

The percentage of composition of O is-

\begin{gathered} \frac{16 \times 6}{180} \times 100 \\ = 53.33\%\end{gathered}

180

16×6

×100

=53.33%

The percentage of composition of H is-

\begin{gathered} \frac{1 \times 12}{180} \times 100 \\ = 6.66\%\end{gathered}

180

1×12

×100

=6.66%

The percentage of composition of C, H and O in glucose is 40%, 6.66% and 53.33%.