Step 1. Calculate the empirical formula.
The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles, so our job is to calculate the molar ratio of $"C"$ to $"H"$ to $"O"$ to $"N"$.
I like to summarize the calculations in a table.
$"Element"color(white)(m)"Moles"color(white)(Xll) "Ratio"color(white)(mll)×2 color(white)(m) "Integers"$
$stackrel(——————————————————)(color(white)(m)"C" color(white)(XXXl)0.150 color(white)(Xml)5.00 color(white)(Xml)10.0color(white)(mm) 10$
$color(white)(m)"H" color(white)(XXXl)0.105 color(white)(mml)"3.50 color(white)(Xml)7.00color(white)(mmll) 7$
$color(white)(m)"O" color(white)(XXXl)0.0300 color(white)(mm)"1 color(white)(Xmmll)2 color(white)(mmmm)2$
$color(white)(m)"N" color(white)(XXXl)0.0450 color(white)(mm)"1.50 color(white)(Xml)3.00 color(white)(mml)3$
The empirical formula of the compound is $"C"_10"H"_7"O"_2"N"_3$.
Step 2. Calculate the empirical formula mass
The molecular mass of $"C"_10"H"_7"O"_2"N"_3$ is 201.18 u.
Step 3. Calculate the molecular mass.
The molecular mass must be an integral multiple multiple of the empirical formula mass.
$"MM" = n × "EFM"$
$n = "MM"/"EFM" = (402 color(red)(cancel(color(black)("u"))))/(201.18 color(red)(cancel(color(black)("u")))) = 2.00 ≈ 2$
Step 4. Calculate the Molecular Formula
The molecular formula must be twice the empirical formula.
The molecular formula is $"C"_20"H"_14"O"_4"N"_6$
