Lab 3 Determination of an Empirical Formula Diana Bakos Sabrina Panetta SCH3U. 02 Wednesday, November 27, 2013 Introduction Stoichiometry is the study of quantitative relationships among the amounts of reactants used and amount of products produced. One of the main parts of stoichiometry is that it gives you a balanced equation which gives you the mole ratios from the coefficients, which also represents relative number of particles involved in the chemical reaction. For the following experiment used magnesium, this reacted with oxygen from the air in a crucible, while it was heated over a Bunsen burner. The masses before and after the reactants were oxidized were measured. With the results, the experimental empirical formula of magnesium oxide could be determined and compared with the theoretical empirical formula. Purpose 1. To prepare a binary compound 2. To find the percent composition by mass of the compound 3. To determine the formula of the compound Materials Please refer to page 1 of 2 of the lab handout Procedure Please refer to page 1 of 2 and 2 of 2 of the lab handout. Observations/Data Mass of magnesium, crucible and lid Mass of crucible and lid Mass of magnesium Mass of product crucible and lid Mass of product Increase in mass Mass of element that combined with magnesium 25.88g 25.57g 0.31g 26.03g 0.46g 0.19g 0.15g Calculations Calculate the percentage composition by mass of the compound Experimental percentage composition of Mg and O Experimental percentage composition of Mg and O you calculated in #1 determine the EF of the compound. Mg = Mass of magnesium/ Mass of product x 100 Mg = 0.31/0.46 x 100% = 67% O=Mass of element that combined with magnesium/ Mass of product x 100 O = 0.15/0.46x 100% = 33% Assume 100 g Mg = 67gMg x 1 mol/24.32 MgO =2.75mols O = 33g O x 1mol/15.99gMgO = 2 Mg:O 2.75:2 1.38:1 MgO Theoretical percentage composition of Mg and O you calculated in #1 determine the EF of the compound. Mg = 24.32gMg/(24.32+15.99MgO) x 100% = 60% O = 15.99gMg/(24.32 +15.99MgO) x 100% = 40% EF Assume 100g Mg = 60gMg x 1 mol/24.32 MgO =2.47mols O = 40g O x 1mol/15.99gMgO = 2.50 Mg:O 2. 47:2.50 1:1 Calculate the % error: Actual Yield Theoretical Yield 2Mg + O2 → 2MgO MgO Mg = (67-60)/60 x 100% = 7% O = (33-40) /40 x 100% = 7% Discussion 1. The observation that suggests that the formation of the compound was a chemical change rather than a physical one is oxidation of magnesium into magnesium oxide and a new substance that formed because heat burned the magnesium. 2. As you know, when magnesium is burnt it reacts with oxygen to produce magnesium oxide; thus the additional mass of the oxygen is being added to the original mass of magnesium to form magnesium oxide, which composes a compound with an increased mass. Magnesium + Oxygen → Magnesium Oxide 2Mg + O2 → 2MgO 3. One would want to heat the crucible strongly after the reaction had apparently been completed just in case there was some leftover residue that didn't react, and by heating it further one is increasing the likelihood of eliminating all that was left unreacted, thereby getting a more accurate weight of the yield. 4. The reason the lid must be left slightly ajar is so the combustion reaction can take place by allowing the oxygen in the air to enter in the crucible and react with the magnesium, but at the same time to stop to many fumes from escaping because that would reduce the mass of the magnesium. 5. The results of the experimental composition compared to the theoretical composition only have a 7% worth of error so Source of experimental error While the Mg ribbon was being heated over the Bunsen burner, to check The atmosphere is composed of 78.08% nitrogen, 20.95% oxygen, 0.93% argon, 0.039% carbon dioxide and about 1% water vapor depending on the temperature in the room. It’s unlikely that the Mg reacted with the 1% of H2O vapor, Argon or the CO2 that was in the air, but there is a likely chance. However, some of the nitrogen, because there it is present is a large quantity of it, could have reacted with the Mg, especially when the lid was lifted to check if the oxidization was occurring, which could have allowed for some of the MgO fumes to escape, and nitrogen to react. It’s difficult to indicate, which reaction took place because both Mg3N2 and MgO produce a powder. In order to have avoided this, water should have been added to form Mg(OH)2 and NH3 then one would again heat the product and produce MgO. Conclusions In conclusion, the empirical formula of Magnesium oxide (MgO) was determined. To find the empirical formula, first the mass of the O2, which was 0.15g, and mass of the MgO, which was 0.46g, had to be calculated. Then those digits, along with the original mass of the Mg, which was 0.31g, were used to calculate the experimental composition. For Mg the experimental composition was 67% and for O2 the experimental composition was 33%. Compared to the theoretical composition of Mg (60%) and O2 (40%) there was a 7% error for both, which is not accurate because it’s know, by mass, that the experimental composition should be close to 60%. However, dispite the 7% error the experiment was still a success as the empirical formula showed to be MgO, when Magnesium reacts with oxygen from the air. This also proves the law of definite proportions because the chemical compound contained exactly the same proportion of elements by mass.