Question

A compound is found to contain 25.24 % sulfur and 74.76 % fluorine by weight. The molecular weight for this compound is 254.14 g/mol. What is the molecular formula for this compound

Answer 1

Answer:

$S_2F_{10}$

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:

$n_S=25.24gS*\frac{1molS}{32.01gS}=0.789molS\\\\n_F=74.76gF*\frac{1molF}{19.00gF}=3.93molF$

Next, we compute the subscript of each element in the formula by dividing each moles by the fewest moles:

$S=\frac{0.789}{0.789}=1\\\\F=\frac{3.93}{0.789} =5$

Whose molar mass is 127.01 g/mol. Now, we can compute the ratio between the molecular and empirical formulas as follows:

$ratio=\frac{254.14}{127.01} =2$

Thus, the molecular formula is:

$S_2F_{10}$

Regards!