# Question b6324

Mar 20, 2014

The empirical formula is $\text{NCl"_3}$.

#### Explanation:

The empirical formula is the simplest whole number molar ratio of the elements in the compound.

We must figure out the masses of $\text{N}$ and $\text{Cl}$, convert to moles, and then find their ratio.

Your compound contains $\text{0.77 mg N}$ and forms $\text{6.61 mg of chloride}$.

$\text{Mass of chloride = mass of N + mass of Cl}$

$\text{Mass of Cl" = "mass of chloride - mass of N" = "6.61 mg - 0.77 mg" = "5.84 mg of Cl}$

$\text{Moles of N" = 0.77 × 10^"-3" color(red)(cancel(color(black)("g N"))) × "1 mol N"/(14.01 color(red)(cancel(color(black)("g N")))) = 5.5 × 10^"-5"color(white)(l) "mol N}$

$\text{Moles of Cl" = 5.84 × 10^"-5" color(red)(cancel(color(black)("g Cl"))) × "1 mol Cl"/(35.45 color(red)(cancel(color(black)("g Cl")))) = 1.65 × 10^"-4"color(white)(l)"mol Cl}$

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

$\text{Element"color(white)(Ag) "Mass/mg"color(white)(Xm) "Moles"color(white)(Xm) "Ratio"color(white)(m)color(white)(l)"Integers}$
stackrel(————————————————————)(color(white)(m)"N" color(white)(XXXmm)0.77 color(white)(Xml)5.5 color(white)(l)× 10^"-5" color(white)(Xll)1color(white)(mmmmll)1)#
$\textcolor{w h i t e}{m} \text{Cl" color(white)(XXXXll)5.84 color(white)(mml)1.65 × 10^"-4} \textcolor{w h i t e}{X l} 3.00 \textcolor{w h i t e}{m m m l} 3$

There is 1 mol of $\text{N}$ for 3 mol of $\text{Cl}$.

The empirical formula is ${\text{NCl}}_{3}$.

Here is a video that illustrates how to determine an empirical formula.