how can we form magnesium oxide at a lab? explain with suitable diagram.
Answers
Answer:when we burn magnesium in air, it forms burns with white dazzling flame and forms magnesium oxide
Explanation:
Mg+O2 gives 2MgO which resembles as ash
Answer:
MARK BRAINLIEST
Explanation:
Lab 2 - Determination of the Empirical Formula of Magnesium Oxide
GOAL AND OVERVIEW
The quantitative stoichiometric relationships governing mass and amount will be studied using the combustion reaction of magnesium metal. Magnesium is reacted with oxygen from the air in a crucible, and the masses before and after the oxidation are measured. The resulting masses are used to calculate the experimental empirical formula of magnesium oxide, which is then compared to the theoretical empirical formula. A crucible and Bunsen burner will be used to heat magnesium metal to burning.
Objectives and Science Skills
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Determine the empirical formula and percent yield of the ionic oxide produced by the reaction of Mg with O2 based on experimental data.
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Quantitatively and qualitatively evaluate experimental results relative to those theoretically predicted based on known chemical principles and stoichiometric calculations.
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Identify and discuss factors or effects that may contribute to deviations between theoretical and experimental results and formulate optimization strategies.
SUGGESTED REVIEW AND EXTERNAL READING
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Data analysis introduction (online), reference materials
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Textbook information on ionic compounds and empirical formulas
BACKGROUND
A great deal of chemical knowledge has been amassed by using simple combustion experiments conducted with crucibles, burners, and balances. In this experiment, you are using this technique to experimentally determine the empirical formula of magnesium oxide.
This lab illustrates (1) the law of conservation of mass and (2) the law of constant composition.
1
The total mass of the products of a reaction must equal the total mass of the reactants.
2
Any portion of a compound will have the same ratio of masses as the elements in the compound.
Molecular composition can be expressed three ways:
1
In terms of the number of each type of atom per molecule or per formula unit (the formula).
2
In terms of the mass of each element per mole of compound.
3
In terms of the mass of each element present to the total mass of the compound (mass percent).
The empirical formula of a compound gives the lowest whole-number ratio of the constituent atoms that is consistent with the mass ratios measured by experiment.
In this lab, magnesium metal (an element) is oxidized by oxygen gas to magnesium oxide (a compound). Magnesium reacts vigorously when heated in the presence of air. The Mg-O2 reaction is energetic enough to allow some Mg to react with gaseous N2. Although there is a higher percentage of N2 gas in the atmosphere than O2, O2 is more reactive and the magnesium oxide forms in a greater amount than the nitride. The small amount of nitride that forms can be removed with the addition of water, which converts the nitride to magnesium hydroxide and ammonia gas. Heating the product again causes the loss of water and conversion of the hydroxide to the oxide.
The unbalanced equations are:
( 1 )
Mg(s) + N2(g) + O2(g) → MgO(s) + Mg3N2(s)
( 2 )
MgO(s) + Mg3N2(s) + H2O(l) → MgO(s) + Mg(OH)2(s) + NH3(g)
( 3 )
MgO(s) + Mg(OH)2(s) → MgxOy(s) + H2O(g)
Balancing the reactions is not necessary because the theoretical reaction product and yield is based on the amount of Mg available to react. The expected product is MgO, so the 1-to-1 mole ratio Mg to O in the product is all that is required.
Based on the masses of the solid reactant (Mg) and product (MgxOy), the mass in grams and the amount in moles of Mg and O in the product can be determined. Recall that the conversion factor relating grams to moles is molar mass.
( 4a )
mass of Mg + mass of O = mass of MgxOy
or
( 4b )
mass of O = mass of MgxOy − mass of Mg
( 5a )
mol Mg = w grams Mg ×
1 mol Mg
24.31 g Mg
( 5b )
mol O = z grams O ×
1 mol O
16.00 g O
,
where w is the grams of Mg used and z is the grams of O incorporated.
The empirical formula of magnesium oxide, MgxOy, is written as the lowest whole-number ratio between the moles of Mg used and moles of O consumed. This is found by determining the moles of Mg and O in the product; divide each value by the smaller number; and, multiply the resulting values by small whole numbers (up to five) until you get whole number values (with 0.1 of a whole number).
For example, if 0.0109 moles of Mg are combined with 0.0103 moles of O:
( 6 )
Mg0.0109/0.0103O0.0103/0.0103 = Mg1.06O1.00 ⇒ no need to multiply ⇒ MgO
For example, if 0.0129 moles of Mg are combined with 0.0103 moles of O:
( 7 )
Mg0.0129/0.0103O0.0103/0.0103 = Mg1.25O1.00 ⇒ multiply by 4 ⇒ (Mg1.25O1.00)4 ⇒ Mg5O4
EXPERIMENTAL NOTES
MATERIAL REQUIRED:-
Safety goggles
Magnesium ribbon, Mg
Balance (to 0.0001g)
Ring stand
Bunsen burner
Ring support/ clay triangle
Crucible/ lid
Tongs
Clay tile
Crucible Use
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