# Asample of a compound contains 0.150 mol C, 0.105 mole H, 0.0300 mol O, and 0.0450 mol N. The molecular mass of the compound is found to be 402 g/mol. What is the compound's chemical formula?

Aug 31, 2016

The chemical formula is ${\text{C"_20"H"_14"O"_4"N}}_{6}$.

#### Explanation:

Step 1. Calculate the empirical formula.

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

I like to summarize the calculations in a table.

$\text{Element"color(white)(m)"Moles"color(white)(Xll) "Ratio"color(white)(mll)×2 color(white)(m) "Integers}$
stackrel(——————————————————)(color(white)(m)"C" color(white)(XXXl)0.150 color(white)(Xml)5.00 color(white)(Xml)10.0color(white)(mm) 10
$\textcolor{w h i t e}{m} \text{H" color(white)(XXXl)0.105 color(white)(mml)} 3.50 \textcolor{w h i t e}{X m l} 7.00 \textcolor{w h i t e}{m m l l} 7$
$\textcolor{w h i t e}{m} \text{O" color(white)(XXXl)0.0300 color(white)(mm)} 1 \textcolor{w h i t e}{X m m l l} 2 \textcolor{w h i t e}{m m m m} 2$
$\textcolor{w h i t e}{m} \text{N" color(white)(XXXl)0.0450 color(white)(mm)} 1.50 \textcolor{w h i t e}{X m l} 3.00 \textcolor{w h i t e}{m m l} 3$

The empirical formula of the compound is ${\text{C"_10"H"_7"O"_2"N}}_{3}$.

Step 2. Calculate the empirical formula mass

The molecular mass of ${\text{C"_10"H"_7"O"_2"N}}_{3}$ is 201.18 u.

Step 3. Calculate the molecular mass.

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

$\text{MM" = n × "EFM}$

n = "MM"/"EFM" = (402 color(red)(cancel(color(black)("u"))))/(201.18 color(red)(cancel(color(black)("u")))) = 2.00 ≈ 2

Step 4. Calculate the Molecular Formula

The molecular formula must be twice the empirical formula.

The molecular formula is ${\text{C"_20"H"_14"O"_4"N}}_{6}$