Question

A compound that weighs 56.2 g contains 23.6 g Sb (antimony) and 32.6 g F (fluorine). What is the percent composition of fluorine

Answer 1

Answer: 42.0%

Explanation: %Sb = 23.6 g/ 56.2 g x 100% = 42.0% Sb

Answer 2

The percent composition of fluorine is 58.0%.

How to calculate the percent composition of fluorine by a compound's weight?

• Antimony is a chemical element with the symbol Sb (from Latin: stibium) and atomic number 51.
• Fluorine is a chemical element with the symbol F and atomic number 9.
• Percent composition indicates the relative amounts of each element in a compound.

Percent composition = mass of compound ÷ molecular mass × 100

Mass of Compound = 32.6g
Molecular Mass = 56.2g

∴ percent composition = 32.6 ÷ 56.2 × 100

= 0.580× 100

= 58.0%

The percent composition of fluorine is 58.0%.

To learn more about percent composition, refer:

https://brainly.com/question/14853467

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