Indicate whether the following data for Part B, Part C, or Part D. _____ Identif
ID: 531069 • Letter: I
Question
Indicate whether the following data for Part B, Part C, or Part D. _____ Identify the original sample _____ 1. Mass of crucible and lid (g) 2. Mass of crucible, lid, and sample (g) 3. Mass of sample (g) 4. Instructor's approval of flame and apparatus 5. Mass of crucible, lid, and product 1st mass measurement (g) 2nd mass measurement (g) 3rd mass measurement (g) 6. Final mass of crucible, lid, and compound (g) 7. Mass of compound (g) 8. Part B. Combination Reaction of Magnesium and Oxygen a. Mass of O (g) b. Mass ratio of Mg to O c. Moles of Mg (mol) d. Moles of O (mol) e. Mole ratio of Mg to O f. Consensus empirical formula of magnesium oxide g. Percent by mass (%): _____ % Mg; _____ % O 9. Part C. Decomposition Reaction of a Pure Compound a. Mass of product = mass CaO (g) b. Mass of CO_2 (g)Explanation / Answer
The above experiment is to study the combination reaction of Mg and O. The first compound in trial 1 (0.18 g) is of Mg. After combustion Mg burns to give MgO as per the reaction:
Mg + 1/2 O2 -------> MgO
Clearly massof MgO must be more than Mg alone. But in the experiment it is observed that MgO has mass 0.025 g which is less than 0.18 g. This can be due to experimental error.
Hence we will solve the problem theoretically. Suppose mass of MgO comes to 0.30 g instead of 0.025 gm.
In that case mass of O= mass of MgO - Mass of Mg = 0.30 - 0.18 = 0.12 g
Mass ratio of Mg to O = 0.18/ 0.12 = 1.5
Moles of Mg in sample = 0.18/24 = 0.0075 g where 24 g is the molecular weight of Mg
Moles of O in the compound : 0.12/16 = 0.0075 where 16 g is the molecular weight of O.
Mole ratio of Mg to O = 0.0075 : 0.0075 = 1:1
Empirical formula for MgO is : Mg1O1 = MgO
Percent by mass : (0.18/ 0.30) *100= 60% of Mg and (0.12/0.30) * 100 = 40% of O