Question

Asample of a compound contains 0.150 mol C, 0.105 mole H, 0.0300 mol O, and 0.0450 mol N. The molecular mass of the compound is found to be 402 g/mol. What is the compound's chemical formula?

Answer 1

The chemical formula is `"C"_20"H"_14"O"_4"N"_6`.

Explanation:

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`