Assume that you have #"195 g (1mol)"# of the compound.
Determine the mass of each element in the compound by multiplying its percentage times its molar mass #("195 g")#.
#0.495xx195 "g"=96.5 "g C"#
#0.0515xx195 "g"=10.0 "g H"#
#0.289xx195"g"=56.4 "g N"#
#0.165xx195"g"=32.2 "g O"#
Determine the Molecular Formula
Determine the number of moles of each element using its calculated mass and its molar mass (atomic weight on the periodic table in grams/mole).
For example, the molar mass of #"C"# is 12.0107 g/mol, which means that 1 mol of C has a mass of 12.0107 g. We can convert that into two conversion factors: #(12.0107"g C")/(1"mol C")# and #(1"mol C")/(12.0107"g C")#. We will use the second conversion factor, so we will essentially be dividing the mass of the element in the compound by its molar mass.
#96.5 cancel"g C"xx(1"mol C")/(12.0107 cancel"g C")="8.04 mol C"##~~"8 mol C"#
#10.0 cancel"g H"xx(1"mol H")/(1.00794 cancel"g H")="9.92 mol H"##~~"10 mol H"#
#56.4 cancel"g N"xx(1"mol N")/(14.0067 cancel"g N")="4.03 mol H"##~~"4 mol N"#
#32.2 cancel"g O"xx(1"mol O")/(15.999 cancel"g O")="2.01 mol O"##~~"2 mol O"#
The molecular formula is #"C"_8"H"_10"N"_4"O"_2"#.
Emperical Formula
The empirical formula for a compound represents the lowest whole number ratio of elements in the compound. If we look at the molecular formula, #"C"_8"H"_10"N"_4"O"_2"#, we see that the subscripts have a common factor of #2#, which we can factor out of the molecular formula.
Therefore, the empirical formula is #"C"_4"H"_5"N"_2"O"#.
