What is the empirical formula of a compound consisting of 29.6% oxygen and 70.4% fluorine by mass?

Answer : The empirical formula of the compound is,

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 O = 29.6 g

Mass of F = 70.4 g

Molar mass of F = 19 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of O =

Moles of F =

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

For O =

For F =

The ratio of O : F = 1 : 2

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

The Empirical formula =

Therefore, the empirical formula of the compound is,

n has four pairs of electrons in the valence shell, electrons repel one another ,n has 7 protons in its nucleus, and since n has 7 protons, it must fill the second shell, giving it 4 pairs of electrons. the electrons form 3 bonds and 1 lone pair of electrons. each pair of electrons repels the other pairs, so they are equally far apart.

so all of the above

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