Question

A compound contains 43.3% sodium,11.3% carbon,and 45.3% oxygen. Calculate its empirical formula

Answer 1

Explanation:

the percentage can be also written as grams.

$Sodium=43.3g\\Carbon=11.3g\\Oxygen=45.3g$

Using their atomic mass, convert from grams to mol.

$Sodium=23g/mol\\Carbon=12g/mol\\Oxygen=16g/mol$

Convert.

$Na=43.3g(\frac{1mol}{23g})=1.88mol\\C=11.3g(\frac{1mol}{12g} )=0.94mol\\O=45.3g(\frac{1mol}{16g})=2.83mol$

Now find the lowest number of moles and divide each element by it. In this case, the lowest number is 0.94mol

$Na=\frac{1.88}{0.94}=2\\ C=\frac{0.94}{0.94}=1\\ O=\frac{2.83}{0.94} =3$

So, your empirical formula would be:

$Na_2CO_3$

Answer 2

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