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rasperas
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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
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
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