Match the following:

Set – I	Set - II
A) %of 'S' in  ${\text{SO}}_2$	1) 27.27
B) %  of 'C' in  ${\text{CO}}_2$	2) 80
C) % of 'C' in Ethane	3) 40
D) %  of 'C' in Glucose	4) 50

For A: The gram molecular weight of   ${\text{SO}}_2$ is 64g

The gram atomic weight of S is 32g

${\text{S\hspace{0.17em}=\hspace{0.17em}32;\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}O}}_{\text{2}}\text{=\hspace{0.17em}32};{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}SO}}_{\text{2}}=64$

${\text{\%\hspace{0.17em}\hspace{0.17em}of S in\hspace{0.17em}SO}}_2=\frac{\text{Weight\hspace{0.17em}of\hspace{0.17em}sulphur}}{{\text{Gram\hspace{0.17em}molecular\hspace{0.17em}weight\hspace{0.17em}of\hspace{0.17em}SO}}_{\text{2}}}\times 100$

Therefore, $\%S=\frac{32}{64}\times 100=50\%$

For B:  The gram molecular weight of ${\text{CO}}_{\text{2}}=44$

Molar mass of carbon, C = 12

$\%\text{C}=\frac{12}{44}\times 100=27.27$

For C: The molecular formula of ethane is ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$

Then total mass of C in ${\text{C}}_2{\text{H}}_6\text{,\hspace{0.17em}\hspace{0.17em}2\hspace{0.17em}C = 2}\times \text{12 = 24}$   C = 24

Molar mass of  ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}=30$

Now, $\%\text{C}=\frac{24}{30}\times 100=80\%$

For D: The chemical formula of glucose is ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$

Total mass of carbon, $\text{C}=12\times 6=72$

Molar mass of glucose, ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}=180$

Then, $\%\text{C}=\frac{72}{180}\times 100=40\%$