Question

A compound on analysis gave the following results C=54.54%, H=9.09% and vapour density of the compound==88. Determine the molecular formula of the compound

Answer 1

Answer:

percentage of oxygen 100 - (54.5 4+ 9.09)=36.37%

Answer 2

AnswEr:

Molecular formula of the compound is $\sf{C_{8} H_{16} O_{4}}$

ExplanaTion:

$\begin{array}{| c | c | c | c | c |}\cline{1-5} Elements & Percentage & Mass & Ratio & Simple\:ratio \\ \cline{1-5} C & 54.54 & 12 & 4.53 & 2\\ \cline{1-5} H & 9.09 & 1 & 9.09 & 4 \\ \cline{1-5} O & 36.37 & 16 & 2.27 & 1 \\ \cline{1-5}\end{array}$

Here, we find ratio using atomic mass and percentage of elements.

$\large{\boxed{\sf{Ratio\:=\:\dfrac{\%}{Atomic\:mass}}}}$

How, we'd find oxygen?

$\implies$ Left thing = 54.54 + 9.09 ( C + H )

$\implies$ 63.63

Now, we have to subtract left thing from 100.

→ Oxygen = 100 - 63.63

→ Oxygen = 36.37%

Thus, empirical formula formed is $\bold{\sf{C_{2} H_{4} O}}$

$\rule{200}2$

Empirical formula mass:

→ 12(2) + 4(1) + 16

→ 44

Molecular formula mass:

→ 2 × Vapour density

→ 2 × 88

→ 176

Now, we know that,

$\large{\boxed{\sf{\red{Common\:factor\:=\:\dfrac{Molecular\:formula\:mass}{Emperical\:formula\:mass}}}}}$

→ Common factor = $\sf{\dfrac{176}{44}}$

→ Common factor = 4

$\therefore$ Molecular formula = 4 × empirical formula

→ Molecular formula = 4 × $\sf{C_{2} H_{4} O}$

→ Molecular formula = $\sf{C_{8} H_{16} O_{4}}$

Hence, Molecular formula of the compound is $\bold{\sf{C_{8} H_{16} O_{4}}}$