So,
We're going to assume a 100g mass of substance, and then interrogate the molar quantities of each element. The steps that we're going to follow, are:
Step 1. Divide the percentage of each element by its molecular mass: (These are the moles)
[tex]\begin{gathered} C\colon \\ \frac{63.3}{12}=5.275 \\ H\colon \\ \frac{7.2}{1.00784}=7.1439 \\ N\colon \\ \frac{2.2}{14}=0.15714286 \\ O\colon \\ \frac{27.3}{16}=1.70625 \end{gathered}[/tex]
Step 2. We're going to divide each result by the smallest, in this case, it is 0.157.
[tex]\begin{gathered} C\to\frac{5.275}{0.15714286}=33.56 \\ H\to\frac{7.1439}{0.15714286}\approx47 \\ N\to\frac{0.15714286}{0.15714286}=1 \\ O\to\frac{1.706265}{0.15714286}=10.85 \end{gathered}[/tex]
Now, approximate to the nearest number.
Finally, we obtain:
[tex]C_{34}H_{47}NO_{11}[/tex]
The mass of this compound is 645.315g/mol. So, the empirical and molecular formulas are the same.
