Chemistry Problem- Burning Magnesium

[rasperas]

Short Question- What is the product resulted from burning magnesium?
More Detail- For a chemistry lab I had to react magnesium oxygen- and thereby find out the empirical formula. Below is the lab procedure and my work.
Procedure-
1) Heat a clean dry crucible to redness. (to prevent extraneous r/x's)
2) Allow crucible to cool and mass and record.
3) Cut a piece of Mg ribbon about 5 cm. long.
4) Measure the length of Mg and record (+_ .0001 m) mass 1.00m MG = 1.2819
5) Break the Mg ribbon into small pieces into the bottom of the crucible.
6) Set crucible on ring stand/ clay triangle.
7) Heat 7-10 minutes.
8) Check for and unreacted Mg.
9) Allow crucible contents to cool.
10) Add just enough distilled water to cover the contents in the crucible.
11) Gently heat the crucible again to dry the product.

Data
a. Mass crucible 19.29
b. length Mg .0461 M (46.1 mm)
c. mass crucible and product 19.41 g.

Analysis

Mass Mg 1.2819 x .0491 = .0591
Mass O 19.41-19.29-.0591
Mole Mg .0591 g / 24.305 = .00243 mol Mg
Mole O .06 g / 15.999 = .004 mol O
Mol Ratio .004/.00243 = 1.6 mol 0 per 1 mol Mg x 2 (b/c empirical formulas contain integers only)
Allowing for room of error I get Mg2O3. However here's the rub. Mg= charge 2+
O = charge 2-. 2 x 2 = Total charge of 4+ on Mg. 2 x 3 = total charge of 6- on O. Unablanced.

Is this answer (Mg2O3) correct or is it wrong -(too many innacurate measurments in the data).?

Out of curiosity- What is the purpose of adding distilled water and then evaporating it? The cover to the crucible was on most of the lab- but never-the-less, is this sufficient to prevent nitrogen from skewing the results by reacting with Mg?

Thanks,
Rod Aspera

 

[dextercioby]

Short Question- What is the product resulted from burning magnesium?

Magnesium oxyde.$MgO$

More Detail- For a chemistry lab I had to react magnesium oxygen- and thereby find out the empirical formula. Below is the lab procedure and my work.
Procedure-
1) Heat a clean dry crucible to redness. (to prevent extraneous r/x's)
2) Allow crucible to cool and mass and record.
3) Cut a piece of Mg ribbon about 5 cm. long.
4) Measure the length of Mg and record (+_ .0001 m) mass 1.00m MG = 1.2819
5) Break the Mg ribbon into small pieces into the bottom of the crucible.
6) Set crucible on ring stand/ clay triangle.
7) Heat 7-10 minutes.
8) Check for and unreacted Mg.
9) Allow crucible contents to cool.
10) Add just enough distilled water to cover the contents in the crucible.
11) Gently heat the crucible again to dry the product.

Data
a. Mass crucible 19.29
b. length Mg .0461 m (46.1 mm)
c. mass crucible and product 19.41 g.

Analysis

Mass Mg 1.2819 x .0491 = .0591
Mass O 19.41-19.29-.0591
Mole Mg .0591 g / 24.305 = .00243 mol Mg
Mole O .06 g / 15.999 = .004 mol O
Mol Ratio .004/.00243 = 1.6 mol 0 per 1 mol Mg x 2 (b/c empirical formulas contain integers only)
Allowing for room of error I get Mg2O3. However here's the rub. Mg= charge 2+
O = charge 2-. 2 x 2 = Total charge of 4+ on Mg. 2 x 3 = total charge of 6- on O. Unablanced.

Is this answer (Mg2O3) correct or is it wrong -(too many innacurate measurments in the data).?

Out of curiosity- What is the purpose of adding distilled water and then evaporating it? The cover to the crucible was on most of the lab- but never-the-less, is this sufficient to prevent nitrogen from skewing the results by reacting with Mg?

Thanks,
Rod Aspera

Check your numbers,again.Give units in SI:m,Kg,s.What is the linear density of Mg??Give the figure in Kgm^{-1}.
There's no such thing as
$$Mg_{2}O_{3}$$
The mole ratio should be
2moles of Mg+1mole 0_{2}=2moles MgO.
One mole of oxigen has 32g=0.032Kg,not 16g=0.016Kg

Daniel.

 

[rasperas]

One mole of oxigen has 32g=0.032Kg,not 16g=0.016Kg

For personal simplicity, I was just expressing oxygen as non-diatomic; everything in the problem was adjusted to allow this.

Thankyou for the help.

After thinking about the problem, I realized it is an issue of accuracy. The rulers provided by school are standard "walmart" rulers. The scales also aren't scientific quality.

 

[dextercioby]

For personal simplicity, I was just expressing oxygen as non-diatomic; everything in the problem was adjusted to allow this.

Thankyou for the help.

After thinking about the problem, I realized it is an issue of accuracy. The rulers provided by school are standard "walmart" rulers. The scales also aren't scientific quality.

For "personal simplicity" means entering conflict with science.One mole of Oxigen has N_{A} molecules and that's exactly 32g,not 16g.Oxigen is diatomic,and in chemical reactions it will always enter as a diatomic molecule;only under exceptional conditions it appears monoatomic,e.g.the redox reactions involving $H_{2}SO_{4},H_{2}O_{2},HNO_{3},H_{2}SO_{5}$ and so on.but in this case,it is a product of reaction,not a reactant.

Daniel.

Daniel.

 

[ArrowHeart]

My question is alwo related to Mg burning. We did the experiment at school & after MgO was produced, we massed it. It turned out that the amount of oxygen in MgO was negative--MgO weighed less than the Mg I started out with.
I'm assuming it's because smoke was given off. But why was smoke produced & what exactly is in that smoke?