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What is the empirical formula of a compound that contains 31.42% sulfur, 31.35% oxygen, and 37.23% fluorine by mass, with a molecular weight of 102.2 g/mol? Also, determine its molecular formula.

User Sdexp
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1 Answer

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Firstly, the assumption in these problems is that you have a 100g sample.

Sulfur = 31.42 g
Oxygen = 31.35 g
Fluorine = 37.23 g


Then, these values are converted to moles by multiplying by the ratio of 1 mole divided by the molar mass of the element.


Sulfur:
31.42 g * 1mol/32.06 u = 0.98 mol Sulfur

Oxygen:
31.35 g * 1mol/15.999u = 1.96 mol Oxygen

Fluorine:
37.23 g * 1mol/18.998u= 1.96 mol Fluorine



Rounding these numbers, you have 1 mole of Sulfur, 2 moles of Oxygen, and 2 moles of Fluorine, and so the empirical formula is

SO2F2







User Arabinelli
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