Solve Q2. Pls

Dear Student,

Molecular mass of the compound = 2 * vapour density (45)
The molecular mass of the compound = 90 grams

Element	Percentage Composition	Molar mass	Number of moles	Molar ratio $\frac{number of moles}{minimum moles (2.21)}$
C	26.59	12	$\frac{26.59}{12}=2.21$	1        $\left(\frac{2.21}{2.21}\right)$
H	2.22	1	$\frac{2.22}{1}=2.22$	1        $\left(\frac{2.22}{2.21}\right)$
O	71.19	16	$\frac{71.19}{16}=4.44$	2        $\left(\frac{4.44}{2.21}\right)$

According to this, CHO₂ is the empirical formula.
Empirical Mass = 12 + 1 + 2*16 =45 grams
​​​​​$\frac{Molar mass}{Empirical mass} =\frac{90}{45} =2$

Molecular formula=(Empirical Formula)₂
Hence C₂H₂O₄ - dicarboxylic acid COOH-COOH (oxalic acid)

Regards