SINGAPORE POLYTECHNIC SCHOOL OF CHEMICAL & LIFE SCIENCES CP 4001: ANALYTICAL & PHYSICAL CHEMISTRY Experiment 4: Gravimetric Analysis Prepared for: Mr Goh Tong Hng Submitted by: Ng Hui Shan (0900931) DBS/FT/1A/02 26th May 2009 CONTENTS 1. Synopsis 3 2. Objectives 4 3. Theory 1. Experimental Procedure 4 2. Stoichiometric Calculation 7 4. Procedure 7 5. Results & calculations 1. Amount of
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the following‚ f. Concentration‚ weight per unit weight‚ Ans: 0.1 kg solute/kg solvent g. Concentration‚ weight per unit volume‚ Ans: 104 kg solute/m3 solvent h. Molarity‚ Ans: 0.3 mole solute/L solution i. Mole Fraction‚ Ans: 0.0058 j. Molality‚ Ans: 0.325 mole solute/kg solvent k. Recalculate (a) – (e) if the‚ i. Sucrose solution contains 20 kg of sucrose in 80 kg of water‚ and density of the solution is 1083 kg/m3 ii. Sucrose solution
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____________________ 1. What does molarity mean? Number of moles of solute 1 liter solution 2. What is the molarity of a solution that contains 4.53 moles of lithium nitrate in 2.85 liters of solution? 4.53 mol LiNO3 = 1.59 M LiN03 2.85 L soln 3. What is the molarity of a solution that contains 0.00372 moles hydrochloric acid in 2.39 x 10-2 liters of solution? 0.00372 mol HCL =
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According to Figure 4 and 5‚ the moles of Cu initially obtained and at completion‚ differentiated. It was evident that the initial moles of Cu (0.0254)‚ did not regenerate all the amount‚ as 0.0124 moles of Cu was attained. In regards to this‚ the no. moles that was eliminated was approximately‚ 0.013. Respectively‚ in Figure 3‚ a large deviation amongst the initial and final quantity of copper‚ this implies that the rest of the mass that had diminished‚ was greater than the final product. These
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In order to find out the concentration of sodium carbonate used‚ the number of moles used need to be worked out first. 1. The equation for finding the number of moles is as follows: Amount of Na2CO3 (mol) = mass (g) / molar mass (g mol-1) The mass is 2.69g and the molar mass of Na2CO3 is [(26 x 2) + 12 + (16 x 3)] = 106 g mol-1 Therefore‚ 2.69 / 106 = 0.0253mol We can now use this number of moles to find the concentration of sodium carbonate used in solution. 2. The equation
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is uniform throughout f. Henry’s Law An expression for calculating the solubility of a gas in a fluid based on temperature and partial pressure g. solution mixture of two or more substances h. Molarity concentration measured by the number of moles of solute per liter of solution. i. Osmosis The tendency of molecules of a solvent to pass through a semipermeable membrane from a less concentrated solution into a more concentrated solution j. Solubility the quantity of a particular substance
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Emma Raggio Ms. Bednarska Chemistry Period B 30 September‚ 2014 Amedeo Avogadro Amedeo Avogadro was born in Turin‚ Italy‚ on 9th August‚ 1776. He was born to a family of lawyers. His father‚ Count Filippo Avogadro‚ was a well-known lawyer and public servant. Amedeo became a lawyer in 1796. Soon after he got his doctorate of law‚ he became interested in natural philosophy and mathematics. He left his successful legal career to teach mathematics and physics at the high school level in Vercelli‚
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Mole Conversion Web Quest EO 103 What the heck is a mole? We have Avagadro ( and others) to thank for this…….. 6.02 x 1023 is the number of “things” per mole of the substance. Here is a tutorial for an over view and for a reference as you work through these problems http://www.wiley.com/college/chem/spencer053872/tutorial/gramsmoles/gramsmoles1.html We can use this to do mathematical conversions to determine mass‚ volume‚ and number of atoms or molecules in a given substance. For example let’s
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The initial moles of copper used is .1574 moles. I calculated the mass of the oxygen by subtracting the mass of the crucible before the reaction and the afterwards. Then‚ in order to determine the moles of oxygen used during the reaction‚ I divided the mass of oxygen by its molar mass of 16.00 g/mol. The moles of oxygen was determined to be .1574. Afterwards‚ by dividing moles of copper over moles of oxygen‚ I determined that the molar relationship between
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100 mL gas (g) 0.274g 0.361g 0.100g c Molecular weight of the gas (g/mole) 44.10g/mol 58.12g/mol 16.04g/mol d Number of moles in the 100 mL sample 0.0062mol 0.0062mol 0.0062mol Average of all 3 gases: (0.0062+0.0062+0.0062) / 3 = 0.0062 2. To verify Avogadro’s Law‚ calculate the average number of moles for the three gases along with the percent deviation for each gas‚ according to the formula: % deviation = |(moles of gas) - (average for all gases)| / (average for all gases) * 100%
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