# Determining the Empirical Formula of Magnesium Oxide

1511 WordsNov 29, 20127 Pages
Chemistry Lab Report example: Chemistry Laboratory Report (Magnesium Oxide) INTRODUCTION: As we learned before on how to determine the empirical formula of a compound based on the test and also chemical analysis on it. Hence this experiment is mainly goes around with how to determine the empirical formula of Magnesium Oxide following various tight procedures in order to get the knowledge and apply it onto another compounds. We are investigating the empirical formula of Magnesium Oxide in this experiment. RESEARCH QUESTION: How empirical formula of Magnesium Oxide is obtained by heating Magnesium in the presence of air? HYPOTHESIS: When Magnesium and Oxygen are heated together, they readily undergo a chemical change Magnesium +…show more content…
The ribbon is not. The crucible is heated for another 10 minutes. Step 9 is repeated until the ribbon become whitish. Then the crucible is allowed to cool. 10. The crucible and the end product is weighed. DATA TABLE: Items Mass(+/- 0.001)g Mass of crucible + lid 21.880 Mass of crucible + lid + magnesium 22.190 Mass of crucible + lid + magnesium oxide 22.380 Mass of magnesium oxide 0.500 Mass of magnesium 0.310 Mass of oxygen 0.190 DATA PROCESSING: Mass of magnesium used = 22.190g – 21.880g = 0.310g Moles of magnesium =0.310g/24.310 g mol -1 =0.01275 mol Mass of oxygen that combines with magnesium = 22.380g – 22.190g = 0.190g Moles of oxygen = 0.190g/16.000g mol -1 = 0.011875 mol Empirical formula Magnesium (Mg) Oxygen (O) Mass 0.310g 0.190g Mole =0.310g/24.310g mol -1 =0.012752 mol =0.190g/16.000g mol -1 =0.011875 mol Ratio = 0.012752mol/0.011875mol =1.1 =1 =0.011875mol /0.011875mol = 1 Formula MgO MgO Experimental percent oxygen in Magnesium Oxide Mass of MgO = 0.190g =0.190g/0.500g x 100 =38% Experimental percent magnesium in Magnesium Oxide Mass of magnesium= 0.310g =0.310g/0.500g x 100 = 62% Theoretical percent oxygen in Magnesium Oxide Formula: 2Mg + O2 → 2MgO Moles of MgO = 0.500g/(24.310+16.000)g mol -1 = 0.01240387 mol From the equation: 2 mol of MgO→1 mol of O2 0.01240387 mol→0.006219 mol Mass = 0.062019g x 32 g mol -1 = 0.1984000g Percent =