91Ó°ÊÓ

The number of moles is calculated by the mass and Molar mass. The relationship between the number of moles, mass, and molar mass is given below.

$\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{moles}\mathbf{}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{molar}\mathbf{}\mathbf{mass}}$

The ratio of the number of moles of the C, H and O is:

$$\begin{gathered} \mathbf{n}\left(C\right)\mathbf{:}\mathbf{n}\left(H\right)\mathbf{:}\mathbf{n}\left(O\right)\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{696}}{\mathbf{12}\mathbf{.}\mathbf{01}}\mathbf{:}\frac{\mathbf{0}\mathbf{.}\mathbf{0843}}{\mathbf{1}}\mathbf{:}\frac{\mathbf{0}\mathbf{.}\mathbf{221}}{\mathbf{16}} \\ \mathbf{n}\left(C\right)\mathbf{:}\mathbf{n}\left(H\right)\mathbf{:}\mathbf{n}\left(O\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{058}\mathbf{:}\mathbf{0}\mathbf{.}\mathbf{0843}\mathbf{:}\mathbf{0}\mathbf{.}\mathbf{0138} \end{gathered}$$

Dividing the above value by the smallest n(0.0138)

$\mathbf{n}\left(C\right)\mathbf{:}\mathbf{n}\left(H\right)\mathbf{:}\mathbf{n}\left(O\right)\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{2}\mathbf{:}\mathbf{6}\mathbf{:}\mathbf{1}$

As we have whole numbers so it requires multiplication by 5. Then the empirical formula:${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{30}}{\mathbf{O}}_{\mathbf{5}}$

The whole number multiple given as the ratio of molar mass to the empirical formula mass of each element deduces molecular formula.

The whole number multiple is given as,

$\mathbf{Whole}\mathbf{-}\mathbf{number}\mathbf{}\mathbf{multiple}\mathbf{=}\frac{\mathbf{molar}\mathbf{}\mathbf{mass}\left(g/\mathrm{mol}\right)}{\mathbf{empirtical}\mathbf{}\mathbf{formulamass}\left(g/\mathrm{mol}\right)}$

The empirical formula is${\mathbf{C}}_{\mathbf{21}}{\mathbf{H}}_{\mathbf{30}}{\mathbf{O}}_{\mathbf{5}}$

The empirical formula mass is calculated as:

$$\begin{gathered} {\mathbf{M}}_{\mathbf{w}}\mathbf{}\mathbf{of}\mathbf{}\mathbf{empirical}\mathbf{}\mathbf{formula}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{21}\mathbf{×}\left(M_w\mathrm{ofC}\right)\mathbf{+}\mathbf{30}\mathbf{×}\left(M_w\mathrm{ofH}\right)\mathbf{+}\mathbf{5}\left(M_w\mathrm{of}O\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{21}\mathbf{×}\left(12.01g/\mathrm{mol}\right)\mathbf{+}\mathbf{30}\mathbf{×}\left(1.01g/\mathrm{mol}\right)\mathbf{+}\mathbf{5}\mathbf{×}\left(16.00g/\mathrm{mol}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{362}\mathbf{.}\mathbf{21}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{mol} \end{gathered}$$

The molecular formula mass = 362.47 g/mol

The whole-number ratio is:

$\mathbf{Whole}\mathbf{-}\mathbf{number}\mathbf{}\mathbf{multiple}\mathbf{=}\frac{\mathbf{362}\mathbf{.}\mathbf{47}\left(g/\mathrm{mol}\right)}{\mathbf{362}\mathbf{.}\mathbf{21}\left(g/\mathrm{mol}\right)}\mathbf{≈}\mathbf{1}$

Multiplying subscripts of role="math" localid="1657091949601" ${\mathbf{C}}_{\mathbf{21}}{\mathbf{H}}_{\mathbf{30}}{\mathbf{O}}_{\mathbf{5}}$by 1, the molecular formula is given by

Thus, the molecular formula is${\mathbf{C}}_{\mathbf{21}}{\mathbf{H}}_{\mathbf{30}}{\mathbf{O}}_{\mathbf{5}}$