A 0.0245 mg sample of nitrogen reacts with oxygen to form 0.0945 mg of the oxide. What is the empirical formula of this nitrogen oxide compound?

Aug 19, 2017

The empirical formula is ${\text{N"_2"O}}_{5}$.

Explanation:

The equation for the reaction is

${A}_{\text{r}} : \textcolor{w h i t e}{m m m l l} 14.01 \textcolor{w h i t e}{m m l} 16.00$
$\textcolor{w h i t e}{m m m m m l l} x \text{N"_2color(white)(m) + color(white)(ll) y"O"_2 → 2"N"_x"O"_y}$
$\text{Mass/µg":color(white)(m)24.5color(white)(mmm)70.0} \textcolor{w h i t e}{m m l} 94.5$

First, we must calculate the masses of $\text{N}$ and $\text{O}$ from the mass of the oxide.

$\text{Mass of N in oxide = 24.5 µg}$

$\text{Mass of O in oxide" = "mass of oxide - mass of N" = "(94.5 - 24.5) µg" = "70.0 µg}$

Now, we must convert these masses to moles and find their ratios.

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

$\underline{\boldsymbol{\text{Element"color(white)(X) "Mass/µg"color(white)(X) "Amt/µmol"color(white)(m) "Ratio"color(white)(ml)×2color(white)(mm)"Integers}}}$
$\textcolor{w h i t e}{m m} \text{N} \textcolor{w h i t e}{m m m m l} 24.5 \textcolor{w h i t e}{m m m m l l} 1.748 \textcolor{w h i t e}{X m m} 1 \textcolor{w h i t e}{m m m l} 2 \textcolor{w h i t e}{X m m m l} 2$
$\textcolor{w h i t e}{m m} \text{O} \textcolor{w h i t e}{X X X m l} 70.0 \textcolor{w h i t e}{m m m m l l} 4.375 \textcolor{w h i t e}{m m m} 2.502 \textcolor{w h i t e}{m l} 5.004 \textcolor{w h i t e}{X m l} 5$

The empirical formula is ${\text{N"_2"O}}_{5}$.