Question

What is the percentage of oxygen in C6H1206?​

Answer 1

number of oxygen : 6

number of whole atoms in molchol :

6 + 12 + 6 = 24

Percentage of oxygen = n o o / n o a i m × 100

P o o = 6 / 24 × 100

P o o = 1/4 × 100

P o o = % 25

Answer 2

$\huge{ \mathrm{  \underline{ Answer} \:  \:  ✓ }}$

Mass of elements in $\mathrm{C_6H_{12}O_6}$ are :

• O = $16 u$

• C = $12 u$

• H =$1 u$

Total mass of $\mathrm{C_6H_{12}O_6}$ is :

• $(12u \times 6) + (1u \times 12) +( 16u \times 6)$

• $72u + 12u + 96u$

• $180u$

Total Mass of Oxygen in $\mathrm{C_6H_{12}O_6}$ is :

• $16u \times 6$

• $96u$

Percentage of Oxygen by mass in $\mathrm{C_6H_{12}O_6}$ is :

$\boxed{ \frac{total \: \: mass \: \: of \: \: oxygen}{total \: \: mass \: of \: glucose } \times 100 }$

• $\dfrac{96u}{180u} \times 100$

• $\dfrac{32u}{3u} \times 5$

• $\dfrac{160u}{3u}$

• $53.33 \: \%$

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