Question

A sample of a pure compound that weighs 59.8 g contains 27.6 g Sb (antimony) and 32.2 g F (fluorine). What is the percent composition of fluorine

Answer 1

53.85%

Step-by-step explanation:

Data obtained from the question include:

Mass of antimony (Sb) = 27.6g

Mass of Fluorine (F) = 32.2g

Mass of compound = 59.8g

Percentage composition of fluorine (F) =..?

The percentage composition of fluorine can be obtained as follow:

Percentage composition of fluorine = mass of fluorine/mass of compound x 100

Percentage composition of fluorine = 32.2/59.8 x 100

= 53.85%

Therefore, the percentage composition of fluorine in the compound is 53.85%