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The empirical formula is the simplest formula for a compound, which is defined as the ratio of the indices of the smallest possible whole number of elements present in the formula.

The molar mass of a substance is the weight of one mole of a compound in grams.

Given that in the manufacture of certain resins, analysis of urea reveals that, by mass $20.1\%$carbon, $6.7\%$hydrogen, $46.5\%$nitrogen, and the balance oxygen.

Assume the mass is $100\text{g}$. So, there is $20.1\text{g}$carbon, $\text{6.7g}$hydrogen, $\text{46.5g}$nitrogen, and oxygen can be calculated as follow.

$\begin{array}{rcl}\text{O}& =& 100\text{g}−(20.1+6.7+46.5)\text{g}\\ & =& 100\text{g}−73.3\text{g}\\ & =& 26.7\text{g}\end{array}$

The molar mass of carbon is $12\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$.

The molar mass of hydrogen is role="math" localid="1663680274716" $1\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$.

The molar mass of nitrogen is $14\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$.

The molar mass of oxygen is $16\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$.

Hence, the moles of carbon, hydrogen, nitrogen and oxygen can be calculated as follow.

$\begin{array}{rcl}\text{³¾´Ç±ô±ð²õ o´Ú​â¶Ä‰C}& =& \text{³¾²¹²õ²õ o´Ú c²¹°ù²ú´Ç²Ô}×\frac{\text{1³¾´Ç±ô C}}{\text{³¾´Ç±ô²¹°ù″¾²¹²õ²õ}}\\ & =& 20.1\text{g}×\frac{\text{1³¾´Ç±ô }}{\text{12²µâ€‰}}\\ & =& 1.675\text{mol}\end{array}$

$\begin{array}{rcl}\text{³¾´Ç±ô±ð²õ o´Ú​â¶Ä‰H}& =& \text{³¾²¹²õ²õ o´Ú H}×\frac{\text{1molofH}}{\text{³¾´Ç±ô²¹°ù″¾²¹²õ²õ}}\\ & =& 6.7\text{g}×\frac{1\text{″¾´Ç±ô}}{1\text{ g}}\\ & =& 6.7\text{mol}\end{array}$

$\begin{array}{rcl}\text{³¾´Ç±ô±ð²õ o´Ú​â¶Ä‰N}& =& \text{³¾²¹²õ²õ o´Ú N}×\frac{\text{1molN}}{\text{³¾´Ç±ô²¹°ù″¾²¹²õ²õ}}\\ & =& 46.5\text{g}×\frac{1\text{³¾´Ç±ô }}{14\text{g}}\\ & =& 3.32\text{mol}\end{array}$

$\begin{array}{rcl}\text{³¾´Ç±ô±ð²õ o´Ú​â¶Ä‰O}& =& \text{²µ¾±±¹±ð²Ô​â¶Ä³¾²¹²õ²õ o´Ú O}×\frac{1\text{molofO}}{\text{³¾´Ç±ô²¹°ù″¾²¹²õ²õofO}}\\ & =& 26.7\text{g}×\frac{1\text{″¾´Ç±ô }}{16\text{g}}\\ & =& 1.67\text{mol}\\ & & \end{array}$

Moles of oxygen has the smallest value. Now divided moles of $\text{C,N,}$and $\text{H}$by moles of oxygen to get the empirical formula.

$\begin{array}{rcl}\text{C}& =& \frac{1.675\text{mol}}{1.67\text{mol}}\\ & =& 1\\ \text{H}& =& \frac{6.7\text{mol}}{1.67\text{mol}}\\ & =& 4\end{array}$

$\begin{array}{rcl}\text{N}& =& \frac{3.32\text{mol}}{1.67\text{mol}}\\ & =& 2\\ \text{O}& =& \frac{1.67\text{mol}}{1.67\text{mol}}\\ & =& 1\end{array}$

Hence, the empirical formula of urea is ${\text{CH}}_{\text{4}}{\text{N}}_{\text{2}}\text{O}$.

Given the osmotic pressure of solution of urea in water is $2.04\text{atm}$, concentration is $5\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$and temperature is $25°\text{C}$.

Hence, molarity is equal to the given concentration of solution divided by molar mass. So, it can calculate the moles of urea by using following formula.

$\mathrm{Î }=MRT$

Where, $\mathrm{Î }$is osmotic pressure, $M$is molarity of solution, $R$is gas constant $\left(0.082\raisebox{1ex}{$\text{L}â‹\dots \text{atm}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}â‹\dots \text{K}$}\right.\right)$, and $T$is temperature.

$2.04\text{atm}=\frac{\text{5g}/\text{L}}{\text{³¾´Ç±ô²¹°ù″¾²¹²õ²õ}}×\left(0.082\raisebox{1ex}{${\displaystyle \text{L}â‹\dots \text{atm}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}â‹\dots \text{K}$}\right.\right)×298\text{K}$

$\begin{array}{rcl}\text{³¾´Ç±ô²¹°ù″¾²¹²õ²õ}& =& \frac{\text{5}{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.}}{2.04\text{atm}}×\left(0.082\raisebox{1ex}{${\displaystyle \text{L}â‹\dots \text{atm}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}â‹\dots \text{K}$}\right.\right)×298\text{K}\\ & =& 59.94\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\end{array}$

Empirical formula for the mass is,

$\begin{array}{rcl}{\text{CH}}_{\text{4}}{\text{N}}_{\text{2}}\text{O}& =& 12\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.+\text{4}×\left(\text{1}\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\right)+2×\left(14\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\right)+16\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\\ & =& 60\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\end{array}$

From the above calculation it is clear that the value of calculated molar mass and empirical formula mass are approximately same.

Hence, there is no difference between empirical formula and molecular formula. It will be ${\text{CH}}_{\text{4}}{\text{N}}_{\text{2}}\text{O}$.