Answer:
[tex]SF_5[/tex]
Explanation:
Hello!
In this case, when determining empirical formulas by knowing the by-mass percent, we can first assume we have 25.24 g of sulfur and 74.76 g of fluorine, so we compute the moles:
[tex]n_S=25.24g*\frac{1mol}{32.07g} =0.787mol\\\\n_F=74.76g*\frac{1mol}{19.00g}= 3.935mol[/tex]
Now, we divide the moles of both S and F by the fewest moles, in this case, those of S in order to determine the subscripts in the empirical formula:
[tex]S=\frac{0.787}{0.797}=1 \\\\F=\frac{3.935}{0.797}=5[/tex]
Therefore, the empirical formula is:
[tex]SF_5[/tex]
Best regards!
