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?

1 Answer
Aug 31, 2016

Answer:

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#