Assume that you have "195 g (1mol)"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")(195 g).
0.495xx195 "g"=96.5 "g C"0.495×195g=96.5g C
0.0515xx195 "g"=10.0 "g H"0.0515×195g=10.0g H
0.289xx195"g"=56.4 "g N"0.289×195g=56.4g N
0.165xx195"g"=32.2 "g O"0.165×195g=32.2g 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"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")12.0107g C1mol C and (1"mol C")/(12.0107"g C")1mol C12.0107g 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".
