Question

2) Determine the empirical formula for a compound that is 70.79% carbon, 8.91% hydrogen, 4.59% nitrogen, and 15.72 % oxygen.

Answer 1

The percent compositions of C, H, N, and O are as follows:

C = 70.79%

H = 8.91%

N = 4.59%

O =15.72%

Assume the percent compositions as masses in grams.

Thus, the masses of C, H, N, and O are 70.79 g, 8.91 g, 4.59 g, and 15.72 g respectively.

Determine the number of moles of each element from the given masses as follows:

For Carbon:

= 70.79 g C x ( 1 mol C /12.01 g C)

= 5.894 mol C

Similarly, for hydrogen:

= 8.91 g H x ( 1 mol H /1.008 g H)

= 8.84 mol H

Similarly, for nitrogen:

= 4.59 g N x ( 1 mol N /14.0067 g N)

= 0.3277 mol N

Similarly, for oxygen

= 15.72 g O x ( 1 mol O /15.9994 g O)

= 0.9825 mol O

Divide each mole by the least number of the mole as follows:

= 5.894 mol C / 0.3277 = 18 mol C

= 8.8392 mol H / 0.3277 = 27 mol H

= 0.3277 mol N / 0.3277 = 1 mol N

= 0.9825 mol O / 0.3277 = 3 mol O

Thus, the empirical formula for a compound is C₁₈H₂₇NO₃