Question

#3: A compound contains 39.7 g carbon, 1/65 g hydrogen and 58.6 g of chlorine. If the molar mass is 181.5 g/mole, what is its molecular formula?

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Answer 1

$\boxed{\begin{array}{c|c|c|c}\boxed{\bf Element}&\boxed{\bf No\:of\:moles}&\boxed{\bf Sample\:ratio}&\boxed{\bf Simplest\:ratio}\\ \sf C&\sf \dfrac{39.7}{12}=3.3&\sf \dfrac{3.3}{1.65}=2 &\sf 2\\ \sf H&\sf \dfrac{1.65}{1}=1.65&\sf \dfrac{1.65}{1.65}=1 &\sf 1\\ \sf O&\sf \dfrac{58.6}{35.5}=1.65 &\sf\dfrac{1.65}{1.65}=1&\sf 1\end{array}}$

• Empirical formula:-C3HCl

Empirical formula mass=

$\\ \sf\longmapsto 3(12)+1+35.5=36+1+35.5=37+35.5=72.5g$

• Molecular formula mass=181.5g

$\\ \sf\longmapsto n=\dfrac{Molecular\: formula\:mass}{Empirical\: formula\:mass}=\dfrac{181.5}{72.5}=2.503\approx 3$

molecular formula

$\\ \sf\longmapsto 3(C3HCl)=C_9H_3Cl_3$