Q44P - Chemistry: Molecular Nature Of Matter And Change

Question

Cortisol (m = 362.47 g/mol), one of the major steroid hormones, is a key factor in the synthesis of protein. Its profound effect on the reduction of inflammation explains its use in the treatment of rheumatoid arthritis. Cortisol is 69.6% C, 8.34% H, and 22.1% O by mass. What is its molecular formula?

Step-by-Step Solution

Answer

The obtained molecular formula is ${\mathbf{C}}_{\mathbf{21}}{\mathbf{H}}_{\mathbf{30}}{\mathbf{0}}_{\mathbf{5}}$ and the empirical formula is ${\mathbf{C}}_{\mathbf{21}}{\mathbf{H}}_{\mathbf{30}}{\mathbf{0}}_{\mathbf{5}}$

Step 1: Relation between mass and number of moles

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.

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

Step 2: Determine the empirical formula of the compound

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:${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{30}}{\mathbf{O}}_{\mathbf{5}}$

Step 3: Determine the concept of the whole-number multiple

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,

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

Step 4: Determine the molecular formula of the compound

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

The empirical formula mass is calculated as:

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

The molecular formula mass = 362.47 g/mol

The whole-number ratio is:

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

Multiplying subscripts of ${\mathbf{C}}_{\mathbf{21}}{\mathbf{H}}_{\mathbf{30}}{\mathbf{O}}_{\mathbf{5}}$ by 1, the molecular formula is given by

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