Question

Let’s work through another type of problem:

                 Butane is a ‘hydrocarbon’ compound containing only carbon and hydrogen. Butane is used as fuel in lighters. It’s empirical formula is C2H5, and it’s molar mass is 58.12 g/mole. Based on this data, what is the molecular formula of butane?

                 Begin by determining the empirical formula mass.

                 Next, divide the molar mass by the empirical formula mass.

                 Finally, multiply the empirical formula by this number.

    This results in a molecular formula of :

Putting all the steps together!

1.    A compound with the following mass percent composition has a molar mass (formula mass) of 60.10 g/mol. Find both the empirical formula and the molecular formula.

                                   39.97% C              13.41% H             46.62% N

2.    A compound was analyzed and found to contain the following percentages of the elements by mass: cobalt, 71.06%; oxygen, 28.94%. Determine the empirical formula.

3.    When 4.01 grams of mercury is strongly heated in air, the resulting oxide weighs 4.33 grams. Calculate the empirical formula of the resulting mercury oxide.

4.    A compound with empirical formula C2H5OH was analyzed and found to have a molar mass of approximately 90 g. What is the molecular formula of this compound?

5.    A 1.2569 gram sample of a new compound has been analyzed and found to contain the following masses of elements: carbon, 0.7238 g; hydrogen, 0.07088 g; nitrogen, 0.1407 g; and oxygen, 0.3214 g. Calculate the empirical formula of the compound.

    *Further analysis shows that the formula mass (molecular mass) is approximately 376-377 grams. Determine the molecular formula.

Answer 1

(1) percentage percntage / molar mass ratio simplest ratio

39.97% C 39.97 / 12 3.33 3.33 / 3.33 = 1

13.41% H 13.41 / 1 13.41 13.41 / 3.33 = 4

46.62% N 46.62 / 14 3.33 3.33 / 3.33 = 1

So empirical formula = CH₄N Molecular weight = 60.10

empirical formula mass = 12 + 1.01 + 14 = 30.01

value of n 30.10 / 30.01 = 2

So molecular formula of compound = (CH₄N)_n = (CH₄N)₂ = C₂H₈N₂

(2) percentage percntage / molar mass ratio simplest ratio

71.06% Co 71.06 / 58.93 1.205 1.205 / 1.205 = 1 so sipml ratio 2

28.94% O 28.94 / 16 1.808 1.808 / 1.205 = 1.5 so simple ratio 1.5 * 2 = 3

So empirical formula = Co₂O₃

(4) empirical formula = C₂H₅OH

Empirical formula weight = 12*2 + 1 *5 + 16 + 1 = 46

molecular weight of compund = 90

value of n = 90 / 46 = 2

so molecular formula = (C₂H₅OH)₂ = C₄H₁₀O₂H₂ = C₄H₁₂O₂

(5) % of Carbon in compund = (0.7238 / 1.2569 ) * 100 = 57.58%

% of Hydrogen in compund = (0.07088 / 1.2569) * 100 = 5.63%

  % of Nitrogen in compund = ( 0.1407 / 1.2569) * 100 = 11.19%

  % of Hydrogen in compund = ( 0.3214 / 1.2569 ) * 100 = 25.57%

  percentage percntage / molar mass ratio simplest ratio

57.58% C 57.58 / 12 4.79 4.79 / 0.799 = 6

5.63% H 5.63 / 1 5.63 5.63 / 0.799 = 7

11.19% N 11.19 / 14 0.799 0.799 / 0.799 = 1

25.57% O 25.57 / 32 0.799 0.799 / 0.799 = 1

So empirical formula = C₆H₇NO

Empirical weight = 12*6 + 1 *7 + 14 + 16 = 109

Molecular weight = 376

so molecular formula = (C₆H₇NO)₃ = C₁₈H₂₁N₃O₃