Question

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

Answer 1

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Answer 2

Step-by-step 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((1mol)/(23g))=1.88mol\\C=11.3g((1mol)/(12g) )=0.94mol\\O=45.3g((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=(1.88)/(0.94)=2\\ C=(0.94)/(0.94)=1\\ O=(2.83)/(0.94) =3`

So, your empirical formula would be:

`Na_2CO_3`