Week 5 notes; molarity
Week 5 notes; molarity CHEM 120
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This 2 page Class Notes was uploaded by Leslie Pike on Friday September 25, 2015. The Class Notes belongs to CHEM 120 at Western Kentucky University taught by Dr. Darwin Dahl in Summer 2015. Since its upload, it has received 53 views. For similar materials see College Chemistry I in Chemistry at Western Kentucky University.
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Date Created: 09/25/15
Chem 120 notes Week 5 Important Exam 2 covers chapters 3 4 and 5 and will be given on October 12 Combustion analysis burns an organic compound contains only carbon hydrogen possibly oxygen in excess of oxygen The resulting water and carbon dioxide are weighed To calculate the molecular formula of the compound rst take the weight of carbon dioxide and water and convert this into moles Water weighs 18 grams per mole and carbon dioxide weighs 46 grams per mole Then do molestomoles conversion two moles of hydrogen in one mole of water one mole of carbon in one mole of carbon dioxide You now have your ratio of carbon to hydrogen IMPORTANT NOTE YOU CANNOT USE THIS SAME PROCEDURE TO CALCULATE THE PERCENTAGE OF OXYGEN THE COMPOUND IS BURNED IN AN ATMOSPHERE OF OXYGEN THIS MEANS THE PRODUCTS WILL HAVE MORE OXYGEN THAN THE ORIGINAL COMPOUND DID AND YOU HAVE NO MEANS OF TELLING HOW MUCH MORE JUST BY LOOKING AT THE NUMBERS To calculate the amount of oxygen convert your carbon and hydrogen back to grams If this total weight is less than the weight of the original sample the remainder of the weight will be oxygen Convert this weight to moles atomic oxygen weighs 16 grams per mole Stoichiometry problems It is VERY IMPORTANT that you learn how to work these You WILL have these types of problems in future exams Stoichiometry starts with a balanced equation Often the equation you will be given will not be balanced and you will be required to balance it For example the decomposition of potassium chlorate Just an interesting side note WKU Chem club fries gummy bears with this reaction for entertainment purposes They heat a test tube containing about a teaspoon of potassium chlorate When it boils meaning that 02 is being released they poke in a gummy bear The bear catches re and the ame is bright bluepurple and there is a whistling noise KCIO3 I KCI 02 This equation is unbalanced there are three oxygen moles on the left and two on the right Dr Dahl did not explain this in class but there is a shortcut to balancing equa ons KCIO3 I KCI 3202 Tada there are now three oxygen moles on both sides Some professors do not like fractions some are ne with them To be safe multiply everything by the denominator to remove the fraction Stoichiometry problems give you the mass of one reactant and ask you how many grams of product can be produced from this reactant Alternatively they can give you the mass of the product and ask you how many grams of reactant are needed As you progress through chemistry you will also be asked to work with volumes of gas given in liters and joules of energy but the basic process is the same First convert whatever you are given to moles For the time being it is grams later on in the course it might be something else but the concept is the same convert it to moles Then nd the molestomoles ratio of product and reactant Then if the problem requests it convert your new number of moles back to grams Molarity Again very important concept here Expect to see molarity problems on the next exam One molar symbolized M is equivalent to one mole of solute over one liter of solution If you have one mole of sucrose and you put it in a oneliter volumetric ask and you ll the ask to the line you have a 1 M solution of sucrose If you have two moles of sucrose and you put it in a oneliter volumetric ask and you ll the ask to the line you have a 2 M solution of sucrose Molarity is a measurement of concentration The greater the molarity the greater the concentration IMPORTANT The molarity is the moles of solute divided by the liters of solution NOT the moles of solute divided by the amount of water added to it We DO NOT CARE how much water was added to the solution we ONLY care that it was enough to bring the TOTAL volume to however many liters Stoichiometry problems may give you amounts in molars instead of moles Converting between the two is very easy multiply molars by liters of solution to get moles Continue as you would for any other stoichiometry problem