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

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

HOWEVER, ~ seeing virtual I"ve seen people get the atomic mass of Magnesium (24.31) and divide that by Molar fixed of Magnesium Oxide (40.31) to acquire a percent that means (which is different than the prior to percentage). So now I"m conflicted on what have to I do.. Especially after this I require to uncover the percent error as well..

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Chemistry The Mole principle Percent composition

Stefan V.
Oct 1, 2017

Here"s just how you can do that.

Explanation:

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

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

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

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

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

See more: Which Bond Is Most Polar? ? A Which Molecule Contains The Most Polar Bonds

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

#100% - 60.304% = 39.696%#

So in theory, friend should have actually a #60.304%# percent ingredient of magnesium in magnesium oxide.

Now, you performed one experiment and also found that a #"0.26-g"# sample that magnesium oxide has #"0.2 g"# that magnesium.

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

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

To find the percent error, use

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

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

#"% 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 leaving the prize rounded to 4 sig figs, however keep in mind that your values just justify one far-ranging figure here.

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

You can follow the same approach to discover the percent error for the percnet composition of oxygen in the oxide.