The small amount of magnesium nitride that forms can be removed with the additio

Question

The small amount of magnesium 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 (egn 1) (eqn 2) (eqn 3) are: Mg(s) + N2(g)-02(g) MgO(s) + Mg;N2(s) MgO(s) + Mg,N2(s) + H2O(l) MgO(s) + Mg(OH)2(s) + NH3(g) MgO(s) Mg(OH)2(s) MgxOy(s) + H2O(g) A. In the lab you measure the masses of all objects on the balance to of a gram Notebook Page reference (1 Pt) B. In the lab you measure a clean dry crucible & cover to be 24.36 grams. You obtain a 2 cm piece of pure magnesium metal. After sand papering the magnesium you place it into your crucible and the mass of crucible, cover and magnesium masses out to be 24.66 grams. You observe that the magnesium is gray and shiny. You heat the crucible gently at first then vigorous for 10 minutes until the magnesium ignites. After the crucible cools you add a few drops of distilled water and heat again. The mass of the magnesium compound, crucible, and cover masses out to be 24.85 grams. What is the empirical formula of the compound? (4 Pts) Show all work & Units 1. Mass of the Mg 2. Mass of the MgO 3. Mass of O 4. Moles of Mg & Moles of O 5. Show mole to mole MgxOy 6. Empirical formula of compound C. A 0.2500 g sample of a compound known to contain carbon, hydrogen and oxygen undergoes complete combustion to produce 0.3664 g of CO2 and 0.1500 g of H2O. What is the empirical formula of this compound? (5 Pt) 1. Write an unbalanced combustion reaction 2. Grams and moles of C 3. Grams and moles of H 4. Grams and moles of O 5. Show mole:mole:mole of C:H:O & Empirical formula of hydrocarbon compound

Explanation / Answer

Experiment shown above,

B. From the experiment we get,

1. mass Mg = 24.66 - 24.36 = 0.30 g

2. mass MgO = 24.85 - 24.36 = 0.49 g

3. Mass of O = 0.49 - 0.30 = 0.19 h

4. moles Mg = 0.30 g/24 g/mol = 0.0125 mol

moles O = 0.19 g/16 g/mol = 0.012 mol

5. mole Mg/mol O = 0.0125/0.012 = 1

6. Empirical formula = MgO

---

C. compound containing C, H, and O

1. unbalanced equation CxHyOz ---> xCO2 + H2O

2. moles CO2 = moles C = 0.3664 g/44 g/mol = 0.00833 mol

grams C = 0.00833 mol x 12 g/mol = 0.1 g

3. moles H2O = 0.1500 g/18 g/mol = 0.00833 mol

moles H = 2 x 0.00833 mol = 0.02 mol

mass H = 0.0167 mol x 1 g/mol = 0.02 g

4. mass O = 0.25 g - 0.1 - 0.02 = 0.13 g

moles O = 0.13 g/16 g/mol = 0.00813 mol

5. moles : moles : moles of C : H : O = 1 : 2 : 1

Empirical formula of the hydrocarbon = CH2O

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.