Dr. Moore has discovered a new vaccine composed of carbon, hydrogen, and oxygen. When the product was purified and sent off for elemental analysis it gave the following mass percents: 68.85% C, 4.95% H. Determine the empirical formula of this vaccine.

C4H3O

Explanation:

Data obtained from the question include:

Carbon (C) = 68.85%

Hydrogen (H) = 4.95%

Since the vaccine contains carbon, hydrogen and oxygen, the remaining percentage will be for oxygen. The percentage of oxygen will be:

Oxygen (O) = 100 - (68.85 + 4.95)

= 26.2%

The empirical formula for the vaccine can be obtained as follow:

Carbon (C) = 68.85%

Hydrogen (H) = 4.95%

Oxygen (O) = 26.2%

Divide the above by their individual molar mass as shown below:

C = 68.85/12 = 5.7375

H = 4.95/1 = 4.95

O = 26.2/16 = 1.6375

Next, divide by the smallest number as shown below:

C = 5.7375/1.6375 = 4

H = 4.95/1.6375 = 3

O = 1.6375/1.6375 = 1

Therefore, the empirical formula for the vaccine is C4H3O

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