# Question 138ca

Sep 9, 2017

WARNING! Long answer! The molecular formula is ${\text{C"_6"H"_5"NO}}_{2}$.

#### Explanation:

I believe the correct vapour density is 61.5. See below.

Our first task is to calculate the empirical formula of the compound.

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

The compound contains 58.55 % $\text{C}$, 4.05 % $\text{H}$, and 11.36 % $\text{N}$.

That adds up to 73.96 %.

The remainder, 26.04 % must be $\text{O}$.

So, our job is to calculate the molar ratio of $\text{C}$ to $\text{H}$ to $\text{N}$ to $\text{O}$.

Assume that you have 100 g of sample.

Then it contains 58.55 g of $\text{C}$, 4.05 g of $\text{H}$, 11.36 g of $\text{N}$, and 26.04 g of $\text{O}$.

$\text{Moles of C" = 58.55 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)( "g C")))) = "4.8751 mol C}$

$\text{Moles of H" = 4.05 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "4.018 mol H}$

$\text{Moles of N" = 11.36 color(red)(cancel(color(black)("g N"))) × "1 mol N"/(14.01 color(red)(cancel(color(black)( "g O")))) = "0.810 85 mol N}$

$\text{Moles of O" = 26.04 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)( "g O")))) = "1.6275 mol O}$

From this point on, I like to summarize the calculations in a table.

$\boldsymbol{\underline{\text{Element"color(white)(m) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio"color(white)(m)"Integers}}}$
color(white)(m)"C" color(white)(XXXmm)58.55 color(white)(Xm)4.8751 color(white)(mm)6.0120color(white)(mmm)6
$\textcolor{w h i t e}{m} \text{H} \textcolor{w h i t e}{X X X X m l l} 4.05 \textcolor{w h i t e}{m m} 4.018 \textcolor{w h i t e}{X m l} 4.955 \textcolor{w h i t e}{m m m l l} 5$
$\textcolor{w h i t e}{m} \text{N"color(white)(XXXmm)11.36 color(white)(Xm)"0.810 85} \textcolor{w h i t e}{m} 1 \textcolor{w h i t e}{m m m m m l l} 1$
$\textcolor{w h i t e}{m} \text{O} \textcolor{w h i t e}{X X X m m} 26.04 \textcolor{w h i t e}{X m} 1.6275 \textcolor{w h i t e}{m l l} 2.0072 \textcolor{w h i t e}{m m m l} 2$

The empirical formula is ${\text{C"_6"H"_5"NO}}_{2}$.

Now, we find the molecular formula.

$\text{Empirical formula mass" = "(72.06 + 5.04 + 14.01 + 32.00) u" = "123.10 u}$

"Vapour density" = 6.15 = M_text(r gas)/M_text(r H₂) = M_text(r gas)/2.016

M_text(r gas) = 6.15 × 2.016 = 12.40

This is impossible!

However, if the vapour density is 61.5, we get

M_text(r gas) = 61.5 × 2.016 = 124.0

The molecular mass must be an integral multiple of the empirical formula mass.

"MM"/"EFM" = 124.0/123.10 = 1.007 ≈ 1#

∴ The molecular formula must be the same as the empirical formula: ${\text{C"_6"H"_5"NO}}_{2}$.