Question

A 170.00 g sample of an unidentified compound contains 29.84 g sodium, 67.49 chromium, and 72.67 g oxygen. What is the compound's empirical formula?

Answer 1

`Na_2Cr_2O_7`

Explanation:

So you have 29.84g of Sodium, 67.49g of chromium and 72.67g of oxygen

You would first find the number of mole of each substance in each given mass using:

`mols = (Mass)/(Molar Mass)`

`n(Na) =( 29.84g)/(22.9897g) = 1.298 mols`

`n(Cr) = (67.49g)/(51.996g) = 1.298 mols`

`n(O) = (72.67g)/(16g) = 4.54 mols`

Then you find the molar ratios between each element by dividing the moles by the lowest moles calculated

Molar ratio (Na) = `1.298/1.298 = 1`

Molar ratio (Cr) `1.298/1.298 = 1`

Molar ratio (O) `4.54/1.298 = 3.5`

However 3.5 is not a whole number, to make a whole number we multiply it by 2 however if we do that we have to multiply each mol by 2 as well.

Molar ratio (Na) = `1*2 = 2`

Molar ratio (Cr) = `1*2 = 2`

Molar ratio (O) = `3.5*2 = 7`

Then we can write the empirical formula

`Na_2Cr_2O_7`