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# How would you convert the empirical formula NO2 into the molecular formula, given the molar mass of this molecule is 92 g/mol?

Here’s how you’d do that.
A compound’s empirical formula tells you what the smallest whole number ratio between the atoms that make up that compound is.
I think of the empirical formula as a building block for molecules. A compound’s molecular formula will depend on how many building blocks are needed to build a molecule of a given substance.
In your example, you know that the empirical formula of the compound is ##”NO”_2##.
This tells you that the minimum ratio between nitrogen atoms and oxygen atoms is ##1:2##. In other words, the molecular formula will be a multiple of this building block.
##”molecular formula” = “empirical formula” xx color(blue)(n)” “##, where
##color(blue)(n)## – the number of building blocks needed to build the molecular formula.
You also know that the molar mass of your compound is ##”92 g/mol”##.
This means that the molar mass of all the atoms that make up that molecule must add up to give ##”92 g/mol”##.
So, how would you determine how many building blocks you need? Well, start by figuring out the molar mass of one building block, i.e. the molar mass of the empirical formula.
Since it contains one nitrogen atom and two oxygen atoms, you will get
##1 xx “14.0067 g/mol” + 2 xx “15.9994 g/mol” = “46.0055 g/mol”##
So, if one building block has a molar mass of ##”46.0055 g/mol”##, how many would you need to get the molecule?
##”46.0055 g/mol” * color(blue)(n) = “92 g/mol”##
##color(blue)(n) = (92color(red)(cancel(color(black)(“g/mol”))))/(46.0055color(red)(cancel(color(black)(“g/mol”)))) = 1.99976 ~~ 2##
This means that the compound’s molecular formula, which lists all the atoms that make up a molecule, will be
##(“NO”_2) xx 2 = “N”_2″O”_4 ->## dinitrogen tetroxide

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