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BACKGROUND:

Magnesium is a very reactive metal in the alkaline earth metal group. When magnesium reacts with oxygen, when it is burned, it can either produce magnesium oxide or magnesium peroxide. In order to determine which is produced it is necessary to determine the empirical formula of the product. The empirical formula is the simplest ratio of elements in a compound and can be determined by calculating mole ratios of the products and reactants, which can be done by taking the masses of the elements in the product, dividing by their molar mass, and dividing the result by the smallest quotient. In this experiment, the mass of the magnesium and the product, whether magnesium oxide or magnesium peroxide, were known. The mole ratio of magnesium and oxygen were determined showing that the product was magnesium oxide. The determination of the empirical formula can be seen in Figure #1. In addition, the reactive nature of magnesium causes it to not only react with oxygen when it is burned, but also with nitrogen gas present in the air. In this experiment, when the magnesium was burned it produced magnesium nitride which interfered with the reaction between magnesium and oxygen because it used up some of the magnesium. In order to undo this reaction water was added creating magnesium hydroxide and ammonia through double displacement, then the magnesium hydroxide decomposed into magnesium oxide and water because when it is exposed to heat it decomposes. These reactions can be seen in Figure #2

Three basic types of chemical reactions are synthesis, double displacement, and decomposition. Synthesis occurs when two elements go through a chemical reaction, switching their ionic charges to form a compound. In this experiment, magnesium reacted with oxygen and magnesium reacted with nitrogen to create magnesium oxide and magnesium nitride. Diatomic elements, such as oxygen and nitrogen, receive a two subscript when they are alone. In order to complete the reaction, the equation must be balanced, so that there are an equal amount of elements on each side of the equation. In order to balance these reactions a two was placed in front of both magnesium, magnesium oxide, and magnesium nitride. This type of reaction can be seen in Figures #2 and #3. Double displacement is when two compounds go through a chemical reaction, switching their positive and negative elements to create two new compounds. When elements come together to form compounds, they switching their charges, creating subscripts. In this experiment, magnesium nitride reacted with water producing magnesium hydroxide and ammonia. Decomposition occurs when a compound breaks up into two or more elements or compounds. In this experiment, magnesium hydroxide decomposed into magnesium oxide and water. In order to complete both of these reactions, the equations must be balanced, so that there are an equal amount of elements on each side of the equation. The double displacement reaction required a two in front of the water because there were two moles of hydroxides and the decomposition reaction was already balanced. These reactions can be seen in Figure #3.

Stoichiometry is the quantitative study of chemical reactions. There is a specific process that is executed in order to complete stoichiometry. The goal is to solve for the mass of one substance when the mass is known of another substance. In this experiment, the mass of magnesium was known and the mass of magnesium oxide was solved for. The basic procedure for this process is taking the known mass, converting it to moles, then to the mole ratio, then to the unknown mass. In order to complete these steps, it is necessary to know how to calculate the molar mass of compounds. The molar mass is simply the sum of the atomic masses of the elements in the compound. One mole is 6.02*10²³ formula units or the molar mass of an element. The mole ratio that is calculated in this process is the ratio of the coefficients in the balanced equation of the compounds. The stoichiometry done is this experiment is shown in Figure #3.

pH paper is used in order to determine the basicity or acidity of a solution visually. It is a chemical detector of hydronium ions and uses various colors to represent the pH scale. If the solution is acidic then the pH paper should correlate with numbers between 0 and 7 and if the solution is basic then the pH paper should represent numbers from 7 to 14. Although reading pH paper colors is very subjective, therefore not very precise, it is helpful in determining whether a solution falls into the basic, acidic or neutral category. In this experiment, nitrogen was driven off as ammonia meaning that the pH paper should have turned a color correlating to the pH between 7 and 14 because ammonia is a weak base.

PURPOSE:

This experiment was designed to determine the mass of magnesium and its percent composition in magnesium oxide.

MATERIALS:

• 1 electronic balance (3 decimal places)
• 1 clay triangle
• 1 crucible and lid
• 1 Bunsen burner
• 1 striker
• 1 ring stand clamp
• 1 ring stand
• 0.504 g magnesium turnings
• approximately 10 mL deionized water
• 10 mL beaker
• approximately 5 mL of 6 M hydrochloric acid
• 1 forceps
• 1 sheet of pH paper
• 1 stirring rod
• 1 dropper

PROCEDURE:

A ring stand was placed on the table and the ring stand clamp was secured onto the ring stand. A clay triangle was placed on the ring of the ring stand clamp. The crucible was massed on the three decimal electronic balance, the balance was zeroed and 0.504g of magnesium were weighed out. The crucible with the magnesium was placed on the clay triangle with the lid slightly ajar.  A flame was started with a striker and Bunsen burner and positioned under the crucible. The crucible was heated for 5 minutes, cooled for 30 seconds, and the contents were examined for any shiny parts signaling that the magnesium was not completely burned. Since the contents appeared to be converted to the dull gray oxide, the cover was returned to ajar and the crucible was heated in full heat for 5 minutes. The flame was removed from underneath the crucible and the crucible was cooled to room temperature. Once the crucible was removed form the clay triangle, using the stirring rod, any solids in the crucible were crushed. Any contents left on the stirring rod were washed into the crucible with deionized water and 10 drops of deionized water were added evenly throughout the crucible. The crucible was placed on the clay triangle with the lid slightly ajar and the heated once again to drive off the water added, as the water evaporated a moistened sheet of pH paper was placed in the stream of steam. The pH paper showed a basic response because any nitrogen that reacted with the magnesium was released as ammonia. Once all of the water was gone, the crucible was heated at full heat for 5 minutes. The crucible was cooled to room temperature and weighed, reheated for another 5 minutes, cooled, and reweighed. When the masses were within 0.005 grams the crucible was cleaned out with 6 M hydrochloric acid.

DATA:

Table #1: Mass of Magnesium Oxide Data

1. Mass of Crucible  21.289g
2. Mass of Crucible and Magnesium Turnings  21.793g
3. Mass of Magnesium Turnings  0.504g
4. Mass of Magnesium Oxide (Theoretical)  0.840g
5. Mass of Magnesium Oxide (Exp.) #1   0.844g
6. Mass of Magnesium Oxide (Exp.) #2  0.841g
7. Average Magnesium Oxide (Exp.)  0.843g
8. Percent Magnesium (Theoretical)   60.0%
9. Percent Magnesium (Experimental)  59.8%
10. Percent Error  0.357%
11. Percent Yield  100%

DISCUSSION/CALCULATIONS:

First, the empirical formula of the product was determined as shown in Figure #1.

Figure #1: Determining the Empirical Formula

0.843g Product - 0.504g Mg= 0.339g O in Product

$\frac{0.339\:g\:O}{16\:g\:O} = 0.0212$	$\frac{0.504\:g\:M\:g}{24\:g\:M\:g} = 0.0210$
$\frac{0.212}{0.210} = 1.01\:moles\:of\:O$	$\frac{0.210}{0.210} = 1.00\:moles\:of\:O$

The mass of oxygen in the product was determined by subtracting the mass of the magnesium in the product from the mass of the product obtained. Each mass of the each element in the product was divided by the molar mass. The masses were then divided by the smaller of the two quotients. This resulted in nearly one mole of each element in the product meaning that the product was magnesium oxide.

The chemical reaction was modeled with a balanced equation as show in Figure #2

Figure #2: Balancing Equations (Synthesis)

2Mg + O₂ → 2MgO

The balanced equation was written out stating which compounds would be reactants and which would be products. It was observed that there were two elements, signaling that a synthesis would occur. The outcome then showed a magnesium oxide compound. The equation was then balanced: there were two oxygen elements, therefore a two was placed before the magnesium oxide making two elements of magnesium, so a two was placed before the magnesium.

The reactivity of magnesium with nitrogen in the air to create ammonia is modeled in Figure #3.

Figure #3:Balancing Equations (Synthesis, Double Displacement, Decomposition)

2Mg + N2 → 2MgN	Synthesis
MgN + 2HOH → Mg(OH)2 + NH3	Double Displacement
Mg(OH)2 → MgO + HOH	Decomposition

Each balanced equation was written out stating which compounds would be reactants and which would be products. The first reaction was done the same way as in Figure #2. The second reaction had two compounds as reactants, signaling that a double displacement would occur. These reactants were magnesium and water, to undo the production of magnesium nitride. The outcome then showed a magnesium hydroxide compound, which is known to decompose in heat, therefore the magnesium hydroxide compound, was broken up, while the ammonia was omitted because it would be released as a gas. The equation was then balanced, but the number of elements in each side was equal therefore no balancing was necessary.

Figure #4: Stoichiometry

2Mg + O₂ → 2MgO
0.504g           ? g

$\left(0.504g\:Mg\right)\left(\frac{1\:mole\:Mg}{24g\:Mg}\right)\:\left(\frac{2\:moles\:MgO}{2\:moles\:Mg}\right)\left(\frac{40g\:MgO}{1\:mole\:MgO}\right)=0.840g\:MgO$

The second part of this experiment was conducting the stoichiometry. The calculations are shown in Figure #4. By calculating the stoichiometry, it was shown that the theoretical mass of the magnesium oxide is 0.840 g. This was obtained by taking the mass of the magnesium and converting to moles, then mole ratio, then finally grams.
The calculations of the percent of magnesium in magnesium oxide was determined by taking the mass of the magnesium turnings that were used, dividing that by the mass of the magnesium oxide, theoretical or experimental, and multiplying the result by 100. The calculation of percent of magnesium in magnesium oxide is show in Figure #5.

Figure #5: Percent of Magnesium in Magnesium Oxide

Theoretical: $\frac{2g\;Mg}{40g\;MgO}\cdot100\%=60.\%$

Experimental: $\frac{0.504g\;Mg}{0.843g\;MgO}\cdot100\%=59.8\%$

The calculations of percent error designated that some error occurred. Percent error is calculated by dividing the difference between the experimental and theoretical masses of magnesium oxide by the theoretical mass of magnesium oxide and multiplying the result by 100. The calculation of percent error for is shown in Figure #6.

Figure #6: Percent Error of Magnesium Oxide

$\frac{E-T}T\cdot100\%=\frac{0.843-0.840}{0.840}\cdot100\%=0.357\%$

The calculations of percent yield showed that all of the mass was “obtained”.  Percent yield is calculated by dividing the experimental mass of magnesium oxide by the theoretical mass of magnesium oxide and multiplying the result by 100. The calculation of percent error is show in Figure #7.

Figure #7:Percent Yield of the Mass of Magnesium Oxide (MgO)

$\frac ET\cdot100\%=\frac{0.843}{0.840}\cdot100\%=100.\%$

The percent error was very low, less than even a whole percent. This slight error may have occurred due to technique. When the magnesium oxide was crushed, the stirring rod may have had pieces left on it that were not washed away with the deionized water. This could have caused the mass to be slightly off when the crucible was massed. Also, when the water was added to wash off the stirring rod it was not measured, therefore if too much was added then it might have not completely evaporated, adding to the mass of the magnesium oxide. When the magnesium was first heated, it was necessary to make sure there were no shiny parts left before proceeding. There could have been pieces of magnesium that were shiny but were not noticed because they were so minor meaning that not all of the magnesium had burned. This would mean that the maximum amount of magnesium oxide would not be left, making the mass smaller than it could have potentially been. The percent yield was very high, meaning that a great amount of magnesium oxide was preserved throughout the experiment. Both the percent yield and percent error are beneficial as analysis of data. Percent error is better to see relatively how accurate the experiment was and highlights the faults of the data more than the accomplishments. The percent yield on the other hand, highlights the accuracy of the experiment and make the faults less dramatic. Percent yield is the better to calculate because with percent error only the error is seen and not the relationship of the error to the experiment. The percent yield may be greater than or less than 100%, but the error would still stay the same. This shows that with percent yield the results are more specific and error can be determined easier because a percent yield of less than 100% signifies that not all of the substance was gathered, while with a percent yield greater than 100% other substances were obtained along with the substance that is being gathered.

The methods and equipment were appropriate for conducting this experiment. This method would not be ideal for all substances.  If the substance created by products when reacting with water, than there would need to be a different way to wash away the pieces left on the stirring rod. Some compounds burn completely meaning that there would be no product to mass, so a different approach would be needed to calculate the products they create. If the reactant were too large or extremely small than a new instrument would need to be used in order to determine its mass.