# How do you find the empirical formula for this scenario? A compound is composed of 38.8% chlorine and 61.2% oxygen. The molecular weight is 182.9 grams.

The empirical formula of a chemical species can be thought of as the formula that shows the ratio of the atoms of different elements in the given molecule. Also, note that the actual number of atoms of various elements in the molecule is not shown by the empirical formula. The...

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The empirical formula of a chemical species can be thought of as the formula that shows the ratio of the atoms of different elements in the given molecule. Also, note that the actual number of atoms of various elements in the molecule is not shown by the empirical formula. The molecular formula of a given species shows those numbers.

In the given case, it is given that the molecule is composed of 38.8% chlorine and 61.2% oxygen.

Since we are dealing with percentages, it is easier if we assume that we have 100 g of the compound. In that case, 38.8% of the compound, or 38.8 g of it, is chlorine, while 61.2%, or 61.2 g, are composed of oxygen.

Using the atomic mass of each of the elements, we can figure out how many moles of each is in the compound.

Since the atomic mass of chlorine is (about) 35.5 g/mol, 38.8 g of chlorine is equal to 38.8/35.45 ~ 1.09 moles.

Similarly, the atomic mass of oxygen is 16 g/mol, and hence 61.2 g of oxygen is equal to 61.2/16 = 3.8 moles.

Taking the ratio of the moles of each of the species, 3.8/1.09 ~ 3.5.

That is, for every atom of chlorine, there are 3.5 atoms of oxygen.

Since we cannot have decimals in the molecular formula, let's multiply each with 2. This would mean that for every 2 atoms of chlorine, there will be 7 atoms of oxygen.

Using these numbers, the empirical formula of the given compound is Cl2O7.

Let's calculate the mass of 1 mole of this compound by using the empirical formula.

Molar mass = 2 x 35.5 + 7 x 16 = 183 g, which is almost the same as the given molecular weight.

Hence, we can see that Cl2O7 is both the empirical and molecular formula.

Hope this helps.

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