# Gravimetrics Lab Essay

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Gravimetric Analysis of a Metal Carbonate Post Lab TABLE Mass of Crucible + M2CO3 14.8145g Mass of Crucible + M2CO3 (1st weighing) 14.7915g Mass of Crucible 12.9405g Mass of M2CO3 1.851g Mass of filter paper + CaCO3 3.8025g Mass of filter paper 1.8120g Mass of CaCO3 1.9905g Moles of CaCO3 0.019887 mol Molar mass of M2CO3 93.08 g/mol Identity of M2CO3 Na2CO3 Percent Error 12.18% error ANALYSIS QUESTIONS 1. Calculate the moles of precipitated calcium carbonate, CaCO3. Enter this value in the Data Table. 1.9905g CaCO3 x 1 mol CaCO3 = 0.019887 mol CaCO3 1 100.09g 2. Calculate the molar mass of the unknown carbonate. Enter this value in the Data Table. Mass of M2CO3 / mol CaCO3 1.851g / 0.019887 mol = 93.08 g/mol M2CO3 3. From the calculated molar, identify the unknown. Calculate the percent error in the molar mass value. Enter both values in the Data Table. Molar Mass Na2CO3 = 105.99 g/mol – this is the closest molar mass to what I calculated, so the unknown M2CO3 must be sodium carbonate. Percent Error: 93.08g/mol – 105.99 g/mol (100) = 12.18% error 105.99 g/mol DISCUSSION Review the procedure and list the possible sources of error that would cause either the molar mass of the unknown to be (a) too high or (b) too low. The goal of this lab was to discover the unknown group 1 metal (M) of the compound M2CO3 by dissolving the compound in water and adding a solution of calcium chloride, CaCl2 to the solution in order to precipitate the carbonate ions to reveal the molar mass of the unknown element, thus determining the identity of the unknown element. The initial hypothesis was that the metal carbonate would be potassium carbonate, or K2CO3, because potassium is a group 1 element that is very similar to calcium, but there are many other group 1 elements that