A compound is found to contain 15.94 percent boron and 84.06 percent fluorine by mass.?

A compound is found to contain 15.94 percent boron and 84.06 percent fluorine by mass.

Mole ratio B : F
= 15.94/10.81 : 84.06/19.00
= 1.475 : 4.424
= 1.475/1.475 : 4.424/1.475
= 1 : 3

Empirical formula = BF₃


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A compound is found to contain 39.12 percent carbon, 8.772 percent hydrogen, and 52.11 percent oxygen by mass.

Mole ratio C : H : O
= 39.12/12.01 : 8.772/1.008 : 52.11/16.00
= 3.257 : 8.702 : 3.257
= 3.257/3.257 : 8.702/3.257 : 3.257/3.257
= 1 : 2.67 : 1
= 1×3 : 2.67×3 : 1×3
= 3 : 8 : 3

Empirical formula = C₃H₈O₃

...
B (15.94 / 10.81) F (84.06 / 19.00) [convert g to mol]
B 1.4745 F 4.4242
[divide by smaller]
B 1.00 F 3.00
so
BF3

using same steps as above
C (39.12 / 12.01) H (8.772 / 1.01) O (52.11 / 16.00)
C 3.257 H 8.685 O 3.257
C 1.00 H 2.6667 O 1.00
[round to closest fraction]
C 1/1 H 8/3 O 1/1
multiply by 3 to get rid of frations
C3H8O3

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