What is the percentage composition of #CF_4#?

1 Answer
Dr. Hayek
Feb 8, 2016

#13.6% C and 86.4%F#

Explanation:

The percent composition of an element in a compound is given by:

#(("Atomic mass")xx("subscript"))/("molar mass of the compound")xx100%#

Therefore, for the compound #CF_4#:

#%C=(12.0xx1)/(88.0)xx100%=13.6%#

#%C=(19.0xx4)/(88.0)xx100%=86.4%#

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