Question #c3e63

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Answer 1 · 2014-10-06T22:49:03

The empirical formula is $C_{3} F_{6} O_{2}$ $C_3F_6O_2$

Write down the ratio by mass C : F :O in grams:

163.8 : 518.1 : 145.4

Now work out the ratio by moles by dividing by the mass of 1 mole of each element. This is the $A_{r}$ $A_r$ value in grams.

$A_{r} [C] = 12$ $A_r[C]=12$ $A_{r} [F] = 19$ $A_r[F]=19$ $A_{r} [O] = 16$ $A_r[O] = 16$

Ratio moles:

$\frac{163.8}{12}$ $163.8/12$ : $\frac{518.1}{19}$ $518.1/19$ : $\frac{145.4}{16}$ $145.4/16$

$\Rightarrow$ $rArr$ 13.65 : 27.27 : 9.1

Divide through by 9.1 gives:

1.5 : 3 : 1

So x by 2 gives:

3 : 6 : 2

So the empirical formula is:

$C_{3} F_{6} O_{2}$ $C_3F_6O_2$