a compound has a percent compostion of carbon equal to 48.8383%, hydrogen equal to 8.1636%, and oxygen equal to 43.1981%. what is the mepirical formula

C₂H₃O

Explanation:

From the question given above, the following data were obtained:

Carbon (C) = 48.8383%

Hydrogen (H) = 8.1636%

Oxygen (O) = 43.1981%

Empirical formula =?

The empirical formula of the compound can be obtained as follow:

C = 48.8383%

H = 8.1636%

O = 43.1981%

Divide by their molar mass

C = 48.8383 / 12 = 4.07

H = 8.1636 / 1 = 8.1636

O = 43.1981 / 16 = 2.7

Divide by the smallest

C = 4.07 / 2.7 = 2

H = 8.1636 / 2.7 = 3

O = 2.7 /2.7 = 1

Thus, the empirical formula of the compound is C₂H₃O

< > 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.< >

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).