##”NH”_2″Cl”##
As you know, the empirical formula tells you what the smallest whole number ratio that exists between the atoms that make up a compound is.
In your case, you know that the empirical formula is ##”NH”_2″Cl”##, which means that the regardles of how many atoms of each element you get in the actual compound, the ratio that exists between them will always be ##1 : 2 : 1##.
What you actually need to determine is how many empirical formulas are needed to get to the molecular formula.
Notice that the problem provides you with the molar mass of the compound. This means that you can use the molar mass of the empirical formula to determine exactly how many atoms you need to form the compound’s molecule.
##color(blue)(“molar mass empirical formula” xx n = “molar mass compound”)##
To get the molar mass of the empirical formula, use the molar masses of its constituent atoms
##”14.0067 g/mol” + 2 xx “1.00794 g/mol” + “35.453 g/mol” = “51.48 g/mol” ~~ “51.5 g/mol”##
This means that you have
##51.5 color(red)(cancel(color(black)(“g/mol”))) xx n = 51.5color(red)(cancel(color(black)(“g/mol”)))##
As you can see, you have ##n=1##.
This means that the empirical formula and the molecular formula are equivalent, ##”NH”_2″Cl”##.
