Question e2fb6

Jan 13, 2016

${\text{N"_2"O}}_{4}$

Explanation:

As you know, a compound's empirical formula tells you the smallest whole number ratio that exists between the elements that make up said compound.

By comparison, the molecular formula tells you the exact number of atoms of each element that makes up the compound.

In essence, the molecular formula is a multiple of the empirical formula.

Your goal when dealing with such problems will be to determine how many empirical formulas are needed to get to the molecular formula.

Notice that the compound is said to have a molar mass of $\text{92 g/mol}$. So, if the molecular formula is a multiple of the empirical formula, you can use the molar mass of the empirical formula to get the multiplication factor that exists between the two.

The empirical formula is said to be ${\text{NO}}_{2}$. To get its molar mass, use the molar masses of the individual atoms

overbrace(1 xx "14.007 g/mol")^(color(red)("1 atom of N")) + overbrace(2 xx "15.9994 g/mol")^(color(purple)("2 atoms of O")) = "46.006 g/mol"#

This means that you have

$46.006 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) xx color(blue)(n) = 92 color(red)(cancel(color(black)("g}}}}$

This will get you

$\textcolor{b l u e}{n} = \frac{92}{46.006} = 1.9997 \approx 2$

Therefore, the molecular formula for your compound is

$\left({\text{NO"_2)_color(blue)(2) = color(green)("N"_2"O}}_{4}\right) \to$ dinitrogen tetroxide