So ns am do the efforts to find the percent composition of Magnesium Oxide and also my masses wereMagnesium: 0.2 gramsOxygen: 0.06 gramsMagnesium Oxide: 0.26 grams

So i did (0.2/0.26) x 100 and also got mine percentage and did (0.06/0.26) x 100 and got that percentage.. Magnesium: 76.9% Oxygen: 23.1%

HOWEVER, after ~ seeing online I"ve seen world get the atomic mass that Magnesium (24.31) and also divide it by Molar mass of Magnesium Oxide (40.31) to acquire a percentage that way (which is different than the before percentage). So currently I"m conflicted top top what need to I do.. Specifically after this I require to uncover the percent error as well..

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Chemistry The Mole concept Percent composition
2 answers

Stefan V.
Oct 1, 2017

Here"s exactly how you deserve to do that.


The first thing that you should do here is to use the molar masses of magnesium oxide, magnesium, and elemental oxygen to calculate the theoretical percent composition that the oxide.

This will act together the theoretical value, i.e. The accepted value in your percent error calculation.

So, you know that #1# mole that magnesium oxide, #"MgO"#, includes #1# mole of magnesium and also #1# mole the oxygen.

This means that if you pick a sample that consists of exactly #1# mole that magnesium oxide, this sample will have a fixed of #"44.3044 g"# because magnesium oxide has actually a molar massive of #"44.3044 g mol"^(-1)#.

Similarly, this sample will certainly contain #"24.305 g"# the magnesium since magnesium has actually a molar fixed of #"24.305 g mol"^(-1)#.

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This indicates that the percent composition of magnesium in magnesium oxide will certainly be

#(24.305 color(red)(cancel(color(black)("g"))))/(44.3044color(red)(cancel(color(black)("g")))) * 100% = 60.304%#

Consequently, the percent ingredient of oxygen will be

#100% - 60.304% = 39.696%#

So in theory, you should have a #60.304%# percent composition of magnesium in magnesium oxide.

Now, you performed an experiment and found that a #"0.26-g"# sample the magnesium oxide includes #"0.2 g"# the magnesium.

You can use your data to calculation an experimental value for the percent composition the magnesium in magnesium oxide.

#(0.2 color(red)(cancel(color(black)("g"))))/(0.26color(red)(cancel(color(black)("g")))) * 100% = 76.923%#

To discover the percent error, use

#"% error" = (|"accepted worth " - " speculative value"|)/"accepted value" * 100%#

For the percent composition that magnesium in magnesium oxide, you have actually a percent error the

#"% error" = (| 60.304 color(red)(cancel(color(black)(%))) - 76.923color(red)(cancel(color(black)(%)))|)/(60.304color(red)(cancel(color(black)(%)))) * 100%#

#"% error" = 27.56%#

I"ll leave the answer rounded to four sig figs, however keep in mind that your values only justify one far-reaching figure here.

That"s rather a hefty percent error, most most likely an indicator the some severe experimental/procedural errors.

You deserve to follow the same approach to uncover the percent error because that the percnet ingredient of oxygen in the oxide.