An unknown hydrocarbon compound was analyzed for hydrogen by elemental
analysis and results show that it contains 5.611% H. What is the empirical
formula?

The answer to your question is    C₁.₄H₁

Explanation:

Data

Percent of H = 5.611%

Percent of C = 100 - 5.611

                     = 94.389 %

Empirical formula = ?

Process

1.- Express the percents as mass

Hydrogen = 5.611 g

Carbon = 94.389 g

2.- Convert the grams to moles

Hydrogen

                         1 g 1 mol

                      5.611 g x

                         x = (5.611 x 1) / 1

                             x = 5.611 moles

Carbon

                      12 g 1 mol

                     94.389 g  x

                         x = (94.389 x 1) / 12

                         x = 7.866 moles

3.- Divide by the lowest number of moles

Hydrogen    5.611 / 5.611 = 1

Carbon         7.866 / 5.611 = 1.4

4.- Write the empirical formula

                             C₁.₄H₁

< > according to newton's third law of motion, forces always act in equal but opposite pairs. another way of saying this is for every action, there is an equal but opposite reaction.< >

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