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Analysis of a compound known to contain only Mg, P,and O gives this analysis.21.8% ...
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`Mg = 24`
`O = 16`
`P = 31`
Let us say we have 100g of compound.
Then we have 21.8g of Mg,27.7 of P and 50.5g of O.
Moles of Mg `= 21.8/24 = 0.908`
Moles of P `= 27.7/31 = 0.894`
Moles of O `= 50.5/16 = 3.156`
`M:P:O = 0.908:0.894:3.156 ~= 1:1:3.5 = 2:2:7`
So the empirical formulae would be `Mg_2P_2O_7`
Posted by jeew-m on June 8, 2013 at 5:39 AM (Answer #1)
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