# How to find Percent Composition of Magnesium Oxide with Masses of Magnesium and Oxygen? After that how do I find percent error/

## Hi all, So I am trying to find the percent composition of Magnesium Oxide and my masses were Magnesium: 0.2 grams Oxygen: 0.06 grams Magnesium Oxide: 0.26 grams So I did (0.2/0.26) x 100 and got my 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 people get the Atomic mass of Magnesium (24.31) and divide it by Molar mass of Magnesium Oxide (40.31) to get a percentage that way (which is different than the before percentage). So now I'm conflicted on what should I do.. especially after this I need to find the percent error as well..

Oct 1, 2017

Here's how you can do that.

#### Explanation:

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

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

So, you know that $1$ mole of magnesium oxide, $\text{MgO}$, contains $1$ mole of magnesium and $1$ mole of oxygen.

This means that if you pick a sample that contains exactly $1$ mole of magnesium oxide, this sample will have a mass of $\text{44.3044 g}$ because magnesium oxide has a molar mass of ${\text{44.3044 g mol}}^{- 1}$.

Similarly, this sample will contain $\text{24.305 g}$ of magnesium because magnesium has a molar mass of ${\text{24.305 g mol}}^{- 1}$.

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 composition 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 $\text{0.26-g}$ sample of magnesium oxide contains $\text{0.2 g}$ of magnesium.

You can use your data to calculate 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 value " - " experimental value"|)/"accepted value" * 100%

For the percent composition of magnesium in magnesium oxide, you have 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 leave the answer rounded to four sig figs, but keep in mind that your values only justify one significant figure here.

That's quite a hefty percent error, most likely an indicator of some serious experimental/procedural errors.

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

Oct 1, 2017

$\text{For the empirical formula.....}$

$\text{....you divide thru by the atomic mass of each element.....}$

#### Explanation:

$\text{Moles of magnesium} = \frac{0.20 \cdot g}{24.31 \cdot g \cdot m o {l}^{-} 1} = 8.23 \times {10}^{-} 3 \cdot m o l$

$\text{Moles of oxygen} = \frac{0.060 \cdot g}{16.00 \cdot g \cdot m o {l}^{-} 1} = 3.75 \times {10}^{-} 3 \cdot m o l$

We divide thru by the LOWEST molar quantity, that of oxygen to get a trial empirical formula of $M {g}_{\frac{8.23 \times {10}^{-} 3 \cdot m o l}{3.74 \times {10}^{-} 3 \cdot m o l}} {O}_{\frac{3.74 \times {10}^{-} 3 \cdot m o l}{3.74 \times {10}^{-} 3 \cdot m o l}} \equiv M {g}_{2} O$.

This result makes no sense at all, given that we expect a formula of $M g O$.

And this would give a percentage with respect to mass of metal versus mass of oxygen of Mg%=60.3;O%=39.7%.

See if there is an error in your data... If no, blag someone else's data, and use that for your calculation, while acknowledging your data are unsound.