Question

Determine the molecular formula for the compound with a molar mass of 46.08 g/mol and the following percent compostion:

C:H:N:26.06%13.13%60.81%

Answer 1

Answer:

      $\large\boxed{\large\boxed{CH_6N_2}}$

Explanation:

1. First determine the empirical formula.

a) Base: 100 g of compound

             mass       atomic mass      number of moles

                g                g/mol                    mol

C           26.06            12.011             26.06/12.011   = 2.17

H           13.13              1.008              13.13/1.008    =  13.03

N           60.81             14.007            60.81/14.007 = 4.34

b) Divide every number of moles by the smallest number: 2.17

mass       number of moles        proportion

C                  2.17/2.17                         1

H               13.03/2.17                         6

N                4.34/2.17                          2

c) Empirical formula

      $CH_6N_2$

d) Mass of the empirical formula

      $1\times 12.011g/mol6+6\times 1.008g/mol+2\times 14.007g/mol=46.07g/mol$

2. Molecular formula

Since the mass of one unit of the empirical formula is equal to the molar mass of the compound, the molecular formula is the same as the empirical formula:

                   $CH_6N_2$