Question

A compound contains 30. percent sulfur and 70. percent fluorine by mass. the empirical formula of the compound is

Answer 1

Answer : The empirical formula of a compound is, $SF_4$

Solution : Given,

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of S = 30 g

Mass of F = 70 g

Molar mass of S = 32 g/mole

Molar mass of F = 19 g/mole

Step 1 : convert given masses into moles.

Moles of S = $\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{30g}{32g/mole}=0.9375moles$

Moles of F = $\frac{\text{ given mass of F}}{\text{ molar mass of F}}= \frac{70g}{19g/mole}=3.684moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For S = $\frac{0.9375}{0.9375}=1$

For F = $\frac{3.684}{0.9375}=3.92\approx 4$

The ratio of S : F = 1 : 4

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $S_1F_4$  = $SF_4$

Therefore, the empirical formula of a compound is, $SF_4$

Answer 2

The empirical formula of the compound is: SF₄

The empirical formula of a compound is an expression of the formula of the compound in its simplest form showing the ratio of the atoms of the different elements present in the compound.

The empirical formula of the compound given in the question above can be obtained as follow:

Sulphur (S) = 30%

Fluorine (F) = 70%

Empirical formula =?

S = 30%

F = 70%

Divide by their molar mass

S = 30 / 32 = 0.9375

F = 70 / 19 = 3.6842

Divide by the smallest

S = 0.9375 / 0.9375 = 1

F = 3.6842 / 0.9375 = 4

Therefore, the empirical formula of the compound is SF₄

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