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allsm [11]
3 years ago
7

8) what is the empirical formula for ethyl fluoride if the compound contains 49.97% carbon, 10.51% hydrogen, and 39.52% fluorine

by mass?
Chemistry
2 answers:
Novay_Z [31]3 years ago
7 0
43.97/12.01=3.661=2
10.51/1.01=10.406=5
39.52/19.00=2.075=1
C2H5F
Ilia_Sergeevich [38]3 years ago
4 0

Answer: C2H5F

Explanation:

Take each percentage of the element and divide it by that element's mass. (This is the biggest number for the element found at the bottom of its cell on the periodic table.) Round up or down to whole numbers, and then divide each answer by the <em>lowest</em> answer you got from doing this.

So, Carbon: (49.97/12.01)= 4.

Hydrogen: (10.51/1.01)= 10.

Fluorine: (39.52/19)=2. Fluorine has the lowest answer, therefore we will go:

4/2= 2.

10/2=5

2/2=1.

There you have your answer! C2H5F

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