CH 101 1st Edition Lecture 2 Outline of Last Lecture I Basic Chemistry Definitions A Converting from Grams to Moles and Grams to Molecules II Balancing Chemical Equations III Finding the Molar Mass of a Molecule IV Introduction to Stoichiometry A Converting from Grams to Moles Outline of Current Lecture I Converting from Grams to Moles to Grams A The Difference between O and O2 II How to Determine the Limiting Reagent in a Chemical Reaction A The Drawing it Out Method B The Math Method III The Scientific Method A Dalton s Atomic Theory IV Defining Matter and Energy Current Lecture I Converting from Grams to Moles to Grams You will use stoichiometry to convert from grams to moles to grams when you are dealing with a compound such as Ca NO3 2 and you have to find the mass of one of one of the elements that make up the compound This form of stoichiometry is also used if you are given the mass of one of the elements or molecules that make up the compound and are asked to find the mass of the compound When given a question like this there are certain steps to follow Step 1 Take the amount of grams given and convert it to moles by multiplying it by the ratio of moles to grams 1 mole the atomic mass Step two Multiply by the ratio of moles of the substance that you know the mass of to the moles of the substance that you are trying to find the mass of Step 3 multiply by the atomic mass of the substance that you are trying to find the mass of over I mole atomic mass 1 mole Ex A sample of Ca NO3 2 contains 2 04 grams of Nitrogen What is the mass of the Ca NO 3 2 sample 2 04 g N 1 mole N 14 007 g N Step 1 This step gives you the amount of moles in a 2 04 g sample of Nitrogen These notes represent a detailed interpretation of the professor s lecture GradeBuddy is best used as a supplement to your own notes not as a substitute 2 04 g N 1 mole N 14 007 g N 1 mole Ca NO3 2 2 moles N Step 2 This step give you the moles of Ca NO3 2 in a 2 04 grams of Nitrogen 2 04 g N 1 mole N 14 007 g N 1 mole Ca NO3 2 2 moles N 164 086 g Ca NO3 2 1 mole Ca NO3 2 Step 3 This step tells you the mass of Ca NO3 2 that is produced from a 2 04 g sample of Nitrogen 11 9 grams Ca NO3 2 Answer A The Difference between O and O2 Knowing the difference between when to use O and O2 is important because that s the only way you can determine the correct molar ratios and properly balance equations O has 1 mole and O2 has 2 moles O O is an atom not an element or a molecule Has a mass of 16 found on the Periodic Table Only occurs after a chemical reaction that separates the two Oxygen atoms in O2 O2 O2 is an element and a molecule Has a mass of 32 Is Oxygen s natural state Oxygen in the air is O2 Ex How many grams of oxygen does a 2 303 sample of Al2O3 contain In this equation the content tells us that oxygen means O not O2 because in the molecule Al2O3 contains 3 oxygen atoms which means O2 is slip apart to fulfill the 3 atoms needed to create Al2O3 Ex How many grams of oxygen are required to react with 2 303 grams of Al to produce Al 2O3 In this equation the content tells us that oxygen means O2 because they are asking about oxygen reacting with Al As a reactant oxygen is in its natural state and therefore is O2 II How to Determine the Limiting Reagent in a Chemical Reaction The limiting reagent is the reactant in a chemical reaction that produces the least product You can find the limiting reagent by drawing out the amount of moles of each reactant or by using an equation A The drawing it out method starts by finding the moles of each reactant and then drawing a circle to represent 1 mole for each reactant To find the limiting reagent you draw a circle around the atoms that form 1 mole of the product until you are left with not enough atoms to form the product The atoms that are left are in excess and the atoms you have no more of is the limiting reactant Ex You are given 223 4 g of Fe and 354 5 g of Cl 2 which react to produce the molecule Fe2Cl3 Find the limiting reagent and determine how many moles of the product are produced 4Fe s 3Cl2 g 2Fe2Cl3 First you ALWAYS have to balance your equation 223 4 g Fe 1 mole Fe 55 85 g Fe 4 moles Fe Find the number of moles of each reactant by converting from grams to moles 354 5 g Cl2 1 mole Cl2 70 906 g Fe 5 moles Cl2 The large circles represent moles of Fe and the small red circles represent moles Cl 2 each mole of Cl2 has 2 Cl atoms therefore you draw a total of 8 Cl circles Group together the atoms the make up the product in this case it s Fe2Cl3 which contains 2 Fe atoms and 3 Cl atoms Answer now you can see that you can produce 2 moles of Fe2Cl3 and that the limiting reagent is Cl2 which as 1 mole left over B The Math Method starts the same way as the drawing it out method but instead of drawing out the moles of each reactant you multiply the moles of each reactant by the ratio of moles of product over moles of reactant The reactant with the smallest amount of product produced is the limiting reactant Ex You are given 223 4 g of Fe and 354 5 g of Cl2 which react to produce the molecule Fe2Cl3 Find the limiting reagent and determine how many moles of the product are produced 4Fe s 3Cl2 g 2Fe2Cl3 Balance equation 223 4 g Fe 1 mole Fe 55 85 g Fe 4 moles Fe Find the number of moles of one of Fe 4 moles Fe 2 moles Fe2Cl3 4 moles Fe 2 moles Fe2Cl3 Multiply by the ratio of moles of Fe2Cl3 to moles of Fe in order to find the amount of product produced 354 5 g Cl2 1 mole Cl2 70 906 g Fe 5 moles Cl2 Now find the moles of the other reactant Cl 5 moles Cl2 2 moles Fe2Cl3 3 moles Cl2 3 33 moles Fe2Cl3 Compare the moles of Fe2Cl3 produced by Cl2 to the moles produced my Fe to determine that Fe is the limiting reagent Since we know Fe is the limiting reagent we also know that only 2 moles of Fe2Cl3 can …