Calculate the empirical formula? 0.6g of magnesium burns to give 0.95g magnesium oxide.

1 Answer
Feb 13, 2018

#"Empirical formula"=MgO#

Explanation:

To find the empirical formula we find the MOLAR quantities of the constituent elements....

#"Moles of metal"=(0.60*g)/(24.31*g*mol^-1)=0.0247*mol#

#"Moles of oxygen"=(0.95*g-0.60*g)/(16.00*g*mol^-1)=0.0219*mol#...

...your data are not very good....the molar quantities should be spot on.......we divide thru by the lowest molar quantity to get...

#Mg_((0.0247*mol)/(0.0219*mol))O_((0.0219*mol)/(0.0219*mol))=Mg_(1.13)O#...and this we must accept as close enuff to #MgO#...i.e. a Group 2 cation, #Mg^(2+)#, and a Group 16 dianion, #O^(2-)#...recheck your data....