From the Empirical Formula for MgO calculate Empirical Mass percent oxygen in Mg

Question

From the Empirical Formula for MgO calculate Empirical Mass percent oxygen in MgO, determine percent error introduced by rounding off by comparing experimentally determined mass percent to the empirical mass percent obtained after rounding off,

DATA AND OBSERVATIONS Determining the Empirical Formula of a Binary Compound 27539mg Mass of the dry crucible 27859mg Mass of Mg plus crucible 27970mg Mass of product (added .3 g magnesium oxide and 10 drops of Deionized water) plus crucible First Weighing 27969mg Mass of product plus crucible Second Weighing

Explanation / Answer

You need the molar masses of Mg (24.3g/mol) and O (16.0g/mol).

Now there is a one to one ratio of Mg and O, so one mole of MgO would weigh:

(24.3g/mol)(1mol) + (16.0g/mol)(1 mol) = 40.3 g

So there is 16 g of O in one mole of MgO at 40.3g.

From here, it's just a simple percentage calculation:

16g/40.3g *100% = 39.7%

From the given data

experimentally determined mass percent = 25.58%

empirical mass percent = 39.7%

percent error = 25.58/39.7 x100

                      = 64.4

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