Answer:
[tex]S_2F_{10}[/tex]
Explanation:
Hello!
In this case, since the empirical formula of a chemical compound is determined by assuming the by-mass percentages are masses, we can firstly compute the moles of sulfur and fluorine based on their atomic masses:
[tex]n_S=25.24gS*\frac{1molS}{32.01gS}=0.789molS\\\\n_F=74.76gF*\frac{1molF}{19.00gF}=3.93molF[/tex]
Next, we compute the subscript of each element in the formula by dividing each moles by the fewest moles:
[tex]S=\frac{0.789}{0.789}=1\\\\F=\frac{3.93}{0.789} =5[/tex]
Whose molar mass is 127.01 g/mol. Now, we can compute the ratio between the molecular and empirical formulas as follows:
[tex]ratio=\frac{254.14}{127.01} =2[/tex]
Thus, the molecular formula is:
[tex]S_2F_{10}[/tex]
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