Question

Experiment 7 Empirical Formula

Objectives:

• Determine the expected formula for the ionic oxide expected when Mg reacts with O2
• Find the theoretical and actual yields of magnesium oxide
• Evaluate results using stoichiometry and error analysis

Introduction:

The goal of this experiment is to determine the Empirical Formula of a Compound. (The Empirical Formula of a Compound is the simplest whole number ratio between the elements of a compound) If one can synthesize a compound from elements, then it is possible to determine an experimental empirical formula for the compound, from its molar and stoichiometric ratios.

I hypothesize that by reacting Mg with pure O, (or air, and then using water and heat to remove any magnesium nitride formed upon exposure to the air), and then calculating the ratio between the moles of Mg and O, I should be able to determine experimentally the empirical formula of MgO.

Calculations:

Mg(s) + N2(g) + O2(g) → MgO(s) + Mg3N2(s)

MgO(s) + Mg3N2(s) + H2O(l) → MgO(s) + Mg(OH)2(s) + NH3(g)

MgO(s) + Mg(OH)2(s) → MgxOy(s) + H2O(g)

mass of Mg + mass of O = mass of MgxOy

mass of O = mass of MgxOy − mass of Mg

mol Mg = w grams Mg × 1 mol Mg/ 24.31 g Mg

Mass of crucible (g):	15.961g	Mass of product (g):	0.027 g
Mass of crucible and sample (g):	15.982g	Mass of oxygen (g):	0.006g
Mass of Sample (g);	0.021g	Mole ratio of Mg to O:	3:2
1st mass instrument (g):	15.991g	Moles of Mg:	3
2nd mass instrument (g):	15.988g	Moles of Oxygen:	2
Percent by mass (%):	77.78 % Mg	22.22 % O

I need help writing the conclusion

Answer 1

Step 2: Write the masses of Mg and O

Sample = 0.021 g

O = 0.006 g

Hence Mg = 0.021 - 0.006 = 0.015 g

Step 3: Calculate the number of moles:

n_Mg = 0.015 / 24 = 0.000625

n_O = 0.006 / 16 = 0.000375

Step 4: Divide the number of moles by the lowest number ( here it is 0.000375) to find the ratio of the number of moles between Mg and O

Mg = 0.000625 / 0.000375 = 1

O = 0.000375 / 0.000375 = 1

Step 5: Determine the empirical formula:

MgO

But there is something wrong with your observations.

Considering the mole ratio of Mg:O which is 3:2 the empirical formula should be Mg3O2 which is incorrect. Kindly recheck if you did any experimental error whether all the Mg was converted to MgO, whether the lid was kept on etc.