## Empirical Formula-Part TwoPractice Problem Answers

Problem 1

1) Percent to mass.

carbon: 48.38 grams
hydrogen: 8.12 grams
oxygen: 43.50 grams

Note that the oxygen percentage came from 100% - (48.38 + 8.12) = 43.5%

(2) Mass to moles.

carbon: 48.38 / 12.011 = 4.028 mol
hydrogen: 8.12 / 1.008 = 8.056 mol
oxygen: 43.50 / 16.00 = 2.719 mol

(3) Divide by small:

carbon: 4.028 ÷ 2.719 = 1.48
hydrogen: 8.056 ÷ 2.719 = 2.96
oxygen: 2.719 ÷ 2.719 = 1.00

Note that the carbon value (4.028) is half the hydrogen value (8.056). That means there is one carbon for every two hydrogens in the answer. (4) Multiply 'til whole:

carbon: 1.48 x 2 = 3 hydrogen: 2.96 x 2 = 6 oxygen: 1 x 2 = 2

The empirical formula is C3H6O2

Problem 2

1) Percent to mass: N = 46.67 g; H = 6.70 g; C = 19.98 g; O = 26.65 g
2) Mass to moles: N = 3.33; H = 6.65; C = 1.66; O = 1.66
3) Divide by small: N = 2.01; H = 3.99; C = 1.00; O = 1.00
4) Multiply 'til whole: not required in this problem, the empirical formula is CH4N2O

Problem 3

The "EFW" of CH = 13.019.
78.11 ÷ 13.019 = 6.
The molecular formula is C6H6

Problem 4

The "EFW" of CH = 13.019.
26.04 ÷ 13.019 = 2.
The molecular formula is C2H2

Problem 5

1) Percent to mass: not required since the masses are given.
2) Mass to moles: N = 0.800; H = 0.156; C = 0.0400; O = 0.0400
3) Divide by small: N = 2.0; H = 3.9; C = 1.00; O = 1.00
4) Multiply 'til whole: not required in this problem, the empirical formula is CH4N2O
Note: examine the answers to problems one and five and you will see that these are the same empirical formula, but notice that the information is presented differently. You can generate the percentages by adding up the 4 masses given in problem five, then dividing each element's mass by the total to get the percentage. For example, nitrogen is 1.121 ÷ 2.402 = 0.46669 = 46.67%.