Question

What is the empirical formula a for a compound that is 15.3% magnesium, 44.5% chlorine, and 40.2 % oxygen

Answer 1

Answer:

$MgCl_{2} O_{4}$

Explanation:

When calculating an empirical formula from percentages, assume you have a 100g sample. This allows you to convert the percentages directly to grams, because X % of 100g is X grams.

So:

15.3 % = 15.3g Mg, 44.5% = 44.5g Cl, 40.2% = 40.2g O

The next step is to divide each mass by their molar mass to convert your grams to moles.

15.3/24.3 = .630 mol

44.5/35.5 = 1.25 mol

40.2/16.0 = 2.5 mol

Then you will divide all of your mol values by the SMALLEST number of moles. This gives you whole numbers that are the mole ratio (subscripts) of the empirical formula.

.630 mol/.630mol = 1

1.25 mol/.630mol = 2

2.5 mol/.630mol = 4

So the empirical formula is

$MgCl_{2} O_{4}$