CHM 105 & 106 MO1 UNIT TWO, LECTURE THREE 1 CHM 105/106 Program 10: Unit 2 Lecture 3 IN OUR PREVIOUS LECTURE WE TALKED ABOUT USING CHEMICAL EQUATIONS TO SHOW THE LAW OF MASS AND THE LAW OF CONSERVATION OF ATOMS AND WE TALKED ABOUT BALANCING CHEMICAL EQUATIONS. WE LOOKED AT DOING SOME EXAMPLES OF THAT. BALANCING EQUATIONS IN AND OF ITSELF OF COURSE IS NOT THE PURPOSE OF BALANCING THE EQUATION BUT RATHER TO HAVE A BALANCED EQUATION TO USE SO ONE CAN DO WHAT WE CALL A QUANTITATIVE PART OF A CHEMICAL REACTION OR AS WE REFER TO IT AS STOICHIOMETRY. STOICHEMETRIC CALCULATIONS THEN ARE CALCULATIONS BASED ON A BALANCED CHEMICAL EQUATION. SO BEFORE WE CAN DO THIS QUANTITATIVE STUFF WE HAVE TO BE ABLE TO BALANCE THEN OUR CHEMICAL REACTION. WE STARTED WITH REACTANTS. WE ENDED UP WITH PRODUCTS, AND WE HA VE TO HAVE THOSE BALANCE IN TERMS OF ATOMS. NOW THE QUANTITATIVE PART IS VERY IMPORTANT TO THE CHEMIST BECAUSE WHAT WE WANT TO DO IS TO BE ABLE TO USE AN EQUATION AND PREDICT HOW MUCH PRODUCT SHOULD I BE ABLE TO GET IF I COMBINE 10 GRAMS OF THIS WITH 20 GRAMS OF THAT? I SHOULD BE ABLE TO CALCULATE IT BEFOREHAND. IF IN FACT MASS IS CONSERVED IN A CHEMICAL REACTION AND WE HAVE A BALANCED CHEMICAL EQUATION WE SHOULD BE ABLE TO MAKE THAT SORT OF THEORETICAL CALCULATION. AND THAT TYPE OF CALCULATION IS WHAT WE RE TALKING ABOUT HERE THEN AS STOICHIOMETRY. NOW THERE S SORT OF A LITTLE ROAD MAP WE CAN FOLLOW. WE RE GOING TO LOOK AT BASICALLY THREE DIFFERENT TYPES OF PROBLEMS OR I GUESS I COULD EVEN SAY FOUR DIFFERENT TYPES OF PROBLEMS IN THIS CHAPTER DEALING WITH STARTING WITH KNOWN AMOUNT OF SOMETHING AND ASKING FOR A KNOWN AMOUNT OF AN UNKNOWN IN OUR OVERALL PROBLEM. AND WE HAVE SORT OF THEN A LITTLE ROADMAP HERE THAT WE CAN FOLLOW AS WE GO ALONG AND EACH ONE OF THESE ARROWS, EACH ONE OF THESE ARROWS NOW IS GOING TO CORRESPOND TO A MATHEMATICAL CONVERSION THAT WE NEED TO MAKE. SO IN BETWEEN EACH STEP HERE AS WE GO FOR INSTANCE FROM UNITS OF KNOWN TO MOLES OF KNOWN, THAT IS GOING TO INVOLVE THEN A CONVERSION FACTOR. SO WE RE GOING TO HAVE A CONVERSION FACTOR THAT WE RE GOING TO HAVE BETWEEN THOSE
CHM 105 & 106 MO1 UNIT TWO, LECTURE THREE 2 TWO PARTS. THAT LL BE ONE STEP. THEN WE RE GOING TO GO FROM MOLES OF KNOWN SUBSTANCE TO MOLES OF THE UNKNOWN, THE UNKNOWN BEING WHAT WE RE ASKING FOR. WE OBVIOUSLY HAVE TO KNOW WHAT WE RE SOLVING FOR SO IT S THE MATERIAL IDENTIFIED IN THE QUESTION IS HOW MUCH OF X, AND THEN X BECOMES OUR UNKNOWN. WE RE GONNA HAVE A CONVERSION FACTOR RIGHT HERE, AND THEN IF IT SAYS INSTEAD OF JUST HOW MANY MOLES OF SOMETHING IT MIGHT SAY HOW MANY GRAMS OF SOMETHING OR LATER ON WE LL LOOK IN THE CHAPTER ON GASES AND WE LL TALK ABOUT WHAT VOLUME OF SOMETHING, OR IN THE CHAPTER ON SOLUTIONS WE MAY ASK HOW MANY MILLILITERS OF A SOLUTION. BUT WHATEVER THE CASE IS THEN WE MAY HAVE A THIRD STEP IN HERE SO WE WILL HAVE A THIRD CONVERSION FACTOR AS WE GO FROM MOLES OF X TO GRAMS OF X TO LITERS OF X TO WHATEVER IS UNIT WISE THAT WE WANT. NOW THIS PARTICULAR ONE RIGHT HERE IS THE ONE THAT IS DETERMINED BY, BASED UPON, THE BALANCED CHEMICAL EQUATION. SO THIS IS THE STEP THAT WE ALWAYS HAVE TO TURN TO HAVING A BALANCED CHEMICAL EQUATION TO MAKE, THAT PARTICULAR STEP OF OUR SOLUTION. THE FIRST ONE IS GOING TO GO BACK TO SOME OF OUR INFORMATION THAT WE KNOW. FOR INSTANCE IF WE START WITH GRAMS AND WE WANT TO GO TO MOLES OF SOMETHING WELL WE VE DONE THAT. WE VE MADE THOSE CONVERSIONS. IF IT S AN ELEMENT WE MULTIPLY IT BY ONE MOLE OVER THE NUMBER OF GRAMS, ITS ATOMIC MASS. IF IT S A COMPOUND WE USE ONE MOLE OVER THE NUMBER OF GRAMS OF THE FORMULA UNIT. SO WE ALREADY HAVE DETERMINED HOW WE CAN GO FROM UNITS TO MOLES BUT THIS HERE IS THE PART THAT S GOING TO BE DETERMINED NOW BY OUR BALANCED EQUATION. NOW AS I SAID WE RE GOING TO REALLY LOOK AT THEN FOUR TYPES OF STOICHIOMETRIC PROBLEMS HERE TODAY. WE RE GOING TO LOOK AT WHAT WE CALL MOLE TO MOLE PROBLEMS. THAT DOESN T FOCUS REAL SHARPLY. MOLE TO MASS PROBLEM, MASS TO MOLE PROBLEM, AND FINALLY A MASS TO MASS PROBLEM. NOW WHAT THAT MEANS IS THAT IN THIS PARTICULAR CASE OUR KNOWN IS GIVEN IN TERMS OF MOLES AND THE QUESTION ASKS HOW MANY MOLES OF THE UNKNOWN COMPOUND. IN THE SECOND ONE, THE FIRST ONE WOULD BE GIVEN TO US IN THE PROBLEM IN MOLES, AND IT WOULD ASK US HOW MANY GRAMS OF SOMETHING WE ARE GOING TO GET. THIRD TYPE OF PROBLEM WOULD GIVE US THE KNOWN
CHM 105 & 106 MO1 UNIT TWO, LECTURE THREE 3 SUBSTANCE IN GRAMS AND ASK FOR HOW MANY MOLES OF SOMETHING WE RE GOING TO GET. AND THE LAST ONE S GOING TO GIVE US THE NUMBER OF GRAMS OF SOMETHING AND ASK FOR GRAMS OF SOMETHING. SO THAT S HOW WE IDENTIFY THEM, NOT THAT THERE S ANY MAGICAL SIGNIFICANCE BY THOSE TERMS. BUT THAT S WHAT WE RE GIVEN IN SOME UNIT AND WE RE ASKED FOR IN SOME FINAL UNIT AND THAT S THE WAY WE CATEGORIZE THEN THE STOICHIOMETRIC PROBLEM. NOW IF WE RELAYED THESE BACK TO THIS ROADMAP, NOTICE THAT IF I STARTED WITH MOLES OF KNOWN I WOULD BE RIGHT HERE, AND I WANT AN A NSWER IN MOLES OF UNKNOWN MEANS I M ONLY GOING RIGHT THERE AND THAT TELLS ME THAT IN THAT TYPE OF PROBLEM I M ONLY GOING TO NEED ONE CONVERSION FACTOR. THAT S ALL I M GOING. I M JUST GOING ONE STEP ON THAT ROAD MAP BECAUSE WE RE ALREADY PARTWAY THERE WHEN WE STARTED AND WE RE ONLY GOING TO GO PARTWAY TO THE END AND WE ONLY NEED THEN ONE CONVERSION FACTOR. IF WE LOOK AT THE NEXT ONE WE RE GOING TO START WITH MOLES SO WE RE RIGHT HERE, AND IT WANTS UNITS IN MASS BUT WE SEE WE RE GOING TO HAVE TO GO FROM MOLES TO MOLES, THAT S ONE CONVERSION FACTOR AND THEN CONVERT FROM MOLES TO GRAMS AND SO IN THAT TYPE OF PROBLEM WE RE GOING TO HAVE TWO CONVERSION FACTORS IN OUR PROBLEM SETUP. IN THE THIRD TYPE WE RE STARTING WITH MASS MEANS THAT WE RE STARTING OVER HERE. WE HAVE TO CONVERT IT TO MOLES OF THE KNOWN SUBSTANCE BUT WE WANT TO GO ALL THE WAY TO MOLES OF UNKNOWN SO WE RE GOING OT HAVE TO GO TO HERE SO WE SEE THAT WE RE GOING TO INVOLVE TWO CONVERSION FACTORS IN THAT TYPE OF PROBLEM. AND IN THE THIRD ONE WE RE STARTING HERE AND WE RE GOING OVER HERE. SO WE RE GOING TO GO ONE, TWO, THREE, WE HAVE TO HAVE THREE CONVERSION FACTORS INVOLVED IN OUR PROBLEM TO GO FROM THE START TO THE FINISH. ALRIGHT, NOW WE RE GOING TO USE UNIT ANALYSIS AS WE HAVE IN PREVIOUS CHAPTERS. IN OTHER WORDS, WE RE GOING TO ASK THE QUESTION, HOW MANY MOLES? HOW MANY GRAMS? HOW MANY SOMETHINGS? AND THEN WE RE GOING TO SAY WHAT IS IT THAT WE RE STARTING WITH. WE RE GOING TO BE LOOKING TEN FOR CONVERSION FACTORS AND THOSE CONVERSION FACTORS MAY BE THE MASS TO MOLE RATIO. IT MAY BE THE MOLE TO MOLE RATIO FROM THE BALANCED EQUATION. BUT AGAIN, WE RE GOING TO PAY VERY
CHM 105 & 106 MO1 UNIT TWO, LECTURE THREE 4 CLOSE ATTENTION TO THE UNITS AS WE SET THE PROBLEM UP. WHEN WE END UP WITH THE SAME UNITS ON THE RIGHT THAT WE ASK FOR ON THE LEFT WE RE COMFORTABLE THAT WE HAVE THE PROBLEM SETUP CORRECTLY. SO THIS IS SORT OF JUST A QUICK GUIDELINE THAT WHEN WE LOOK AT A TYPE OF PROBLEM WE MENTALLY THEN CAN CLICK IN AND SAY, OH GOSH, THAT SHOULD HAVE AT LEAST TWO CONVERSION FACTORS IN IT. OR THAT ONE SHOULD ONLY NEED ONE CONVERSION FACTOR. I STILL HAVE TO DETERMINE WHAT THE CORRECT CONVERSION FACTOR IS, BUT AT LEAST THAT GIVES ME SOMETHING TO LOOK AT TO SAY HOW MANY STEPS I NEED TO GO. LET S TAKE A LOOK AT THIS CHEMICAL REACTION RIGHT HERE. WE HAVE THE REACTION OF C3H8, THAT IS A HYDROCARBON. AS A MATTER OF FACT, IT HAS THE CHEMICAL FORMULA OF CNH2N + 2. IN OTHER AS A MATTER OF FACT, IT S A SATURATED HYDROCARBON AND WHAT WOULD THAT HYDROCARBON S NAME BE? THIS IS ONE THAT WE SHOULD REMEMBER. ANYONE? HUH? (STUDENT RESPONSE NOT AUDIBLE) NO, NOT BUTANE. PROPANE. OKAY, METHANE WAS THE SINGLE CARBON. ETHANE THE TWO CARBON, PROPANE WAS THE THREE, AND THEN BUTANE WAS THE FOUR CARBON. SO THIS IS PROPANE. ALRIGHT SO LET S SUPPOSE THAT WE HAVE A PROPANE GRILL, AND WE START IT UP AND WE RE OPERATING IT THEN, WE RE REACTING IT WITH THE OXYGEN. WE RE BURNING THE PROPANE. THIS IS THE REACTION THAT WE RE CARRYING OUT. PROPANE REACTING WITH THE OXYGEN IN THE AIR TO GIVE US CARBON DIOXIDE AND WATER. NOW I LL JUST ASK THE QUESTION AND I THINK YOU CAN SEE THAT THIS HAS NOTHING TO DO WITH THE MATHEMATIC PART, THE STOICHIOMETRIC PART. THIS JUST HAS TO DO WITH THE WAY THAT WE READ A CHEMICAL EQUATION. IF I WANT TO REACT ONE MOLE OF PROPANE HOW MANY MOLES OF OXYGEN DO I NEED? FIVE. THAT S WHAT THE EQUATION TELLS US RIGHT? THAT S THE WAY WE READ THAT EQUATION. ONE MOLE OF PROPANE REACTS WITH FIVE MOLES OF OXYGEN TO GIVE US THREE MOLES OF CARBON DIOXIDE AND FOUR MOLES OF WATER. ALRIGHT, SO IF I REACTED ONE MOLE OF PROPANE IT WOULD TAKE FIVE MOLES OF OXYGEN. IF I REACTED 10 MOLES OF PROPANE HOW MANY MOLES OF OXYGEN WOULD I NEED? 50. I MUST ALWAYS MAINTAIN THAT ONE TO FIVE MOLE RATIO. IF I START WITH A 10 TH OF A MOLE OF PROPANE I LL NEED 5/10 TH OF A MOLE OF OXYGEN. THAT S THAT CONVERSION FACTOR THAT WILL BE IN EVERY PROBLEM WE
CHM 105 & 106 MO1 UNIT TWO, LECTURE THREE 5 DO, EVERY STOICHIOMETRIC PROBLEM WE DO WE RE GOING TO HAVE THAT ONE CONVERSION FACTOR, THAT ONE THAT TAKES US FROM MOLES TO MOLES, WHICH IS THIS ONE RIGHT HERE, MOLES OF KNOWN TO MOLES OF UNKNOWN. THAT S THE CONVERSION FACTOR BASED ON THAT CHEMICAL EQUATION. THAT NUMBER RATIO NEVER CHANGES. THAT IS THE BALANCED MOLE RATIO FOR EVERY SPECIES IN THERE. ONE OF THESE GIVES ME FOUR WATERS. IF I GET FOUR WATERS I HAVE TO HAVE FORMED THREE MOLES OF CARBON DIOXIDE. THAT S WHAT THE EQUATION IS TELLING US. ALRIGHT, LET S LOOK THEN AT OUR FIRST SAMPLE PROBLEM FOR STOICHIOMETRY. AND THIS QUESTION SAYS HOW MANY MOLES OF CARBON DIOXIDE WOULD I BE ABLE TO PRODUCE FROM THE REACTION OF 2.85 MOLES OF OXYGEN? THE OXYGEN IS REACTING WITH THE PROPANE. SO WE HAVE THE PROPANE THERE BUT IF WE HAVE ONLY 2.85 MOLES OF OXYGEN HOW MANY MOLES OF CARBON DIOXIDE AM I GOING TO FORM? NOW NOTICE THAT THIS ONE, THE UNKNOWN IS ASKED FOR IN MOLES, THE KNOWN IS GIVEN TO US IN MOLES, THAT S A MOLE MOLE PROBLEM, AND WE RE ONLY GOING TO NEED ONE CONVERSION FACTOR IN THIS AND THAT CONVERSION FACTOR IS GOING TO BE BASED ON OUR CHEMICAL EQUATION. SO WE WRITE MOLES OF CARBON DIOXIDE IS EQUAL TO OUR MEASURED QUANTITY. THIS IS OUR UNKNOWN, 2.85 MOLES OF OXYGEN, AND NOW WE NEED A CONVERSION FACTOR. WE NEED TO CONVERT FROM MOLES OF THE KNOWN SUBSTANCE TO MOLES OF THE UNKNOWN. THAT IS DEPENDENT ON OUR CHEMICAL EQUATION. WELL GOING UP HERE WE SEE TH RATIO BETWEEN OXYGEN AND CARBON DIOXIDE ARE FIVE TO THREE OR THREE TO FIVE, WHICH EVER WAY WE NEED TO USE IT IN THE PROBLEM. IN THIS CASE WE WANT THE MOLES OF OXYGEN TO BE ON THE BOTTOM UNIT-WISE TO CANCEL. SO WE RE GOING TO HAVE FIVE MOLES OF OXYGEN ON THE BOTTOM AND WE RE GOING TO HAVE THREE MOLES OF CARBON DIOXIDE ON THE TOP. MOLES OF OXYGEN NOW WILL CANCEL OUT AND WE RE GOING TO BE LEFT THEN WITH THE UNITS OF MOLES OF CARBON DIOXIDE ON BOTH SIDES OF THE EQUATION. NOW ALL WE WOULD HAVE TO DO OF COURSE WOULD BE OUR NUMBER CRUNCHING TO DETERMINE THE ANSWER. NUMBER OF ALLOWED SIGNIFICANT FIGURES? THREE. THESE ARE EXACT NUMBERS. THAT S NOT A MEASURED QUANTITY THAT S A STATED QUANTITY BASED UPON OF COURSE THE LAW OF CONSERVATION OF ATOMS. SO THIS NUMBERS ARE EXACT. OKAY, SO THE CONTROLLING
CHM 105 & 106 MO1 UNIT TWO, LECTURE THREE 6 PART OF THE SIGNIFICANT FIGURES IS THIS ONE, NOT THIS. SO IT S NOT ONE AS IT MIGHT APPEAR, ITS RATHER THREE, THE MEASURED QUANTITY. AN EXAMPLE OF A MOLE TO MOLE PROBLEM. ALRIGHT LET S GO ON AND TAKE A LOOK AT ANOTHER ONE BASED ON THAT SAME EQUATION NOW. THIS TIME IT S ASKING US HOW MANY GRAMS OF WATER CAN WE GET IF WE REACT 3.44 MOLES OF OXYGEN AGAIN WITH THE PROPANE? WELL THIS TIME THEN WE RE LOOKING AT THE RELATIONSHIP OF OXYGEN/WATER BUT THIS TIME NOTICE THAT WE RE ASKED FOR MASS AND STARTING WITH MOLES. THAT WOULD BE THIS TYPE UP HERE. THAT WOULD BE A MOLE TO MASS, AND WE RE GOING TO NEED TWO CONVERSION FACTORS IN THERE, AND I THINK THESE WILL BECOME SOMEWHAT OBVIOUS AS WE SET THE PROBLEM UP. ALRIGHT SO LET S GO AHEAD, WE RE ASKED FOR GRAMS OF WATER, AND WE RE STARTING WITH 3.44 MOLES OF OXYGEN. NOW AGAIN IF WE WANT TO BACK UP HERE AND LOOK AT OUR ROAD MAP FOR JUST A SECOND, WE RE STARTING WITH MOLES OF OXYGEN, SO WE RE STARTING RIGHT HERE, MOLES OF KNOWN, AND WE RE GOING OVER HERE. SO OUR ROAD MAP TELLS US WE NEED TWO CONVERSION FACTORS, AND IT NOT ONLY TELLS US THAT, BUT IT TELLS US THAT OUR FIRST CONVERSION FACTOR IS THE ONE INVOLVING THE BALANCED CHEMICAL EQUATION, AND THAT WOULD PROBABLY MAKE SENSE WHEN WE LOOK AT UNITS BECAUSE SOMEHOW WE HAVE TO GET FROM MOLES OF OXYGEN TO MOLES OF WATER, THAT S OUR FIRST STEP IN THIS CASE. SO MULTIPLY BY FOUR MOLES OF WATER PER FIVE MOLES OF OXYGEN, THAT S FROM OUR STOICHIOMETRIC RELATIONSHIP IN OUR BALANCED CHEMICAL EQUATION BETWEEN THOSE TWO. NOW WE SEE THAT MOLES OF OXYGEN HAVE CANCELLED AND WE RE LEFT WITH MOLES OF WATER. BUT WE DIDN T ASK FOR MOLES OF WATER. WE ASKED FOR WHAT MASS, HOW MANY GRAMS/ OF COURSE WE VE DONE A LOT OF CALCULATIONS OF THAT TYPE IN THE PREVIOUS CHAPTER. SO NOW WE WANT TO MULTIPLY BY A RATIO OF GRAMS OF WATER TO MOLES OF WATER, AND SO HOW MUCH DOES ONE MOLE OF WATER WEIGHT? 18 GRAMS. 2 FOR THE HYDROGENS, 16 FOR THE OXYGEN, SO IT S JUST LIKE, YOU KNOW, ALL WE RE DOING IS CALCULATING THE FORMULA MASS. SO WE WOULD CALCULATE THE FORMULA MASS FOR WATER WHICH WOULD GIVE US 18.02 GRAMS USING 2 PLACES AFTER THE DECIMAL. THAT CANCELS, AND WE NOW NOTICE THAT WE HAVE GRAMS
CHM 105 & 106 MO1 UNIT TWO, LECTURE THREE 7 FOR UNITS ON THE RIGHT. WE HAVE GRAMS ASKED FOR, OUR UNITS MATCH, THE PROBLEM IS SET UP CORRECTLY. AND SO AT THIS POINT THEN IT WOULD ONLY BE A MATTER OF PLUGGING IN OUR NUMERIC VALUES HERE TO GET AN ANSWER. ALRIGHT, AGAIN AS I MENTIONED IN PROBLEM SOLVING IT S NOT A BAD IDEA TO MAKE A GUESSED ANSWER FIRST JUST SO THAT IF YOU PLUG INTO THE CALCULATOR AND SEE A NUMBER COMES OFF DRASTICALLY DIFFERENT THAN WHAT YOU PREDICTED YOU PROBABLY NEED TO GO BACK AND DOUBLE CHECK IT. BUT JUST LOOKING AT THIS HERE, 4/5, THAT WOULD BE A LITTLE BIT LESS THAN ONE TIMES THIS WOULD BE ABOUT THREE, AND THREE TIMES 18 WOULD BE ABOUT 54. SO I M JUST GUESSING THAT OUR ANSWER SHOULD BE SOMETHING AROUND 54 GRAMS, NOT 500, NOT 5, BUT 50, AND SO WHEN I PLUG IT INTO THE CALCULATOR THEN I AT LEAST HAVE SOME GENERAL IDEA OF WHAT KIND OF NUMBER I SHOULD COME OUT WITH. WELL LET S SEE WHAT DO WE GET HERE? WE HAVE 3.44 X 4, DIVIDED BY 5, X 18.02, AND OUR ACTUAL NUMBER TURNS OUT TO BE, WITH CORRECT NUMBER OF SIGNIFICANT FIGURES, 49.6 GRAMS IS OUR ACTUAL ANSWER OF THE PROBLEM. BUT YOU SEE THAT 54 AT LEAST GAVE US THE MAGNITUDE OF WHETHER IT S IN UNITS OF TENS OR HUNDREDS OR THOUSANDS OR TEN TO THE MINUS EIGHT, OR WHATEVER IT MAY BE. SO IT S A GOOD IDEA JUST TO KIND OF SCAN ACROSS AND GET A FEEL FOR WHAT YOUR ANSWER SHOULD BE. WELL THIS WAS AN EXAMPLE NOW OF OUR SECOND TYPE OF PROBLEM THERE, WHICH WE CALL MOLE-MASS STOICHIOMETRIC PROBLEM. LOOKING AT ANOTHER EQUATION HERE, THIS TIME WE HAVE A REACTION OF A COMPOUND CALLED CHLOROETHYLENE, THAT S THIS ONE HERE. REMEMBER IN A PREVIOUS CHAPTER WE TALKED ABOUT, WELL ACTUALLY I M INCORRECT ON THAT, IN THIS CHAPTER WE TALKED ABOUT THE REACTION OF HYDROCARBONS WITH THAT GROUP OF CHEMICALS CALLED THE HALOGENS CHLORINE, FLUORINE, BROMINE, AND IODINE. THIS IS THEN A HALOGENATED HYDROCARBON. I OTHER WORDS WE VE REACTED A COMPOUND THAT HAD ONLY CARBON AND HYDROGEN AND WE VE NOW REPLACED ONE OF THE HYDROGENS WITH A CHLORINE, WHICH IS ONE OF THE MEMBERS OF THEN THE HALOGEN FAMILY. OKAY, AND AS I SAY, THE NAME OF THIS WOULD BE CHLOROETHYLENE. THAT S A LITTLE DIFFERENT NAMING ENDING THAN WHAT WE WERE INTRODUCED TO SO FAR, BUT WE LL TALK ABOUT THAT TYPE OF NAME A LITTLE BIT LATER IN
CHM 105 & 106 MO1 UNIT TWO, LECTURE THREE 8 CHAPTER SIX OR SEVEN. ALRIGHT, AS A MATTER OF FACT THIS ALSO HAS A COMMON NAME OF VINYLCHLORIDE, AND IT IS USED TO MAKE A CHEMICAL CALLED POLYVINYLCHLORIDE, WHICH IS ONE OF OUR EVERY DAY PLASTIC MATERIALS. AGAIN WE LL LOOK AT THAT WHEN WE TALK MORE ABOUT CHEMICAL BONDING. BUT HERE WE HAVE THEN THIS HALOGENATED HYDROCARBON REACTING WITH OXYGEN AND IN THE PROCESS WE GET CARBON DIOXIDE, WATER, AND HYDROCHLORIC ACID, AND THEN I M GOING TO PAUSE A SECOND, WE HAVE A QUESTION. (STUDENT RESPONSE NOT AUDIBLE) YES. THE QUESTION IS, DO WE NEED TO WORRY ABOUT BALANCING THE EQUATION? THE ANSWER IS YES, YOU MAY NEED TO BALANCE THE EQUATION, AND THAT S WHY OF COURSE WE SPENT SOME TIME LOOKING AT HOW WE GO ABOUT BALANCING AN EQUATION. THESE CALCULATIONS ARE DEPENDENT UPON A PROPERLY BALANCED EQUATION. MOST OF THE ONES IN THE PROBLEM SETS HOWEVER, THE BALANCED EQUATION IS ALREADY GIVEN, SO UNLESS IT INDICATES THAT YOU MUST BALANCE THE EQUATION OR IT DOESN T GIVE YOU AN EQUATION AT ALL, THEN OF COURSE YOU WOULD HAVE TO WRITE IT AND BALANCE IT. BUT FOR MOST OF THE PROBLEMS YOU WILL FIND A BALANCED CHEMICAL EQUATION. OKAY, ALRIGHT LET S SEE THEN WHAT THE QUESTION IS THIS TIME. THIS TIME IT SAYS, HOW MANY MOLES OF OXYGEN ARE NEEDED TO REACT 10 GRAMS OF THE CHLOROETHYLENE? NOW THIS TIME NOTICE THAT WE RE ASKING FOR THE UNKNOWN IN MOLES AND WE RE STARTING WITH THE KNOWN IN MASS, AND IF WE GO BACK ONCE AGAIN TO OUR ROAD MAP HERE LOOKING AT, THIS TIME WE RE STARTING WITH MASS OVER HERE, AND WE RE ASKED FOR NUMBER OF MOLES, SO WE RE GOING HERE, SO WE SEE THAT WE WOULD HAVE TWO STEPS THAT WE RE INVOLVED WITH. WE WOULD NEED TWO CONVERSION FACTORS TO MAKE THAT CALCULATION, AND OF COURSE ONE OF THOSE CONVERSION FACTORS, AND I LL STRESS THIS OVER AND OVER, BUT ONE OF THOSE CONVERSION FACTORS IS GOING TO INVOLVE THE BALANCED CHEMICAL EQUATION. THAT ONE IS ALWAYS IN OUR STOICHIOMETRIC PROBLEM. THAT S THE ONE STEP THAT WE ALWAYS ARE GOING TO BE LOOKING FOR. ALRIGHT, SO WE START, WE RE LOOKING FOR MOLES OF OXYGEN, AND WE RE STARTING WITH TEN GRAMS OF THE COMPOUND C2H3CL. THE FIRST CONVERSION FACTOR THAT WE MUST MAKE IS TO CONVERT FROM GRAMS OF THE KNOWN MATERIAL TO
CHM 105 & 106 MO1 UNIT TWO, LECTURE THREE 9 MOLES OF THE KNOWN MATERIAL, AND OF COURSE THAT S GOING TO INVOLVE THEN THE FORMULA MASS. SO MULTIPLIED BY ONE MOLE OF C2H3CL OVER WHATEVER NUMBER OF GRAMS OF C2H3CL THAT WE HAVE, AND OF COURSE WE WOULD THEN HAVE TO CALCULATE THAT. SO CALCULATING IT WE WOULD HAVE THEN OF COURSE TWO CARBONS SO THAT WOULD BE 2 X 12.01, 24.02, AND WE D HAVE 3 HYDROGENS, 3 X 1.01, 3.03, AND 1 CHLORINE 35.45, AND GIVING US THEN A FORMULA MASS THEN OF 0, 5, 7, 12, 6. 62.50, AND NOW NOTICE OUR UNITS OF GRAMS HAVE CANCELLED, OKAY. NOW WE HAVE MOLES OF THE UNKNOWN SUBSTANCE, I MEAN, EXCUSE ME, MOLES OF THE KNOWN SUBSTANCE. NOW WE HAVE TO GO TO MOLES OF THE UNKNOWN, AND FOR THAT WE RE GOING TO TURN TO OUR BALANCED CHEMICAL EQUATION. THIS CASE WE RE RELATING NOW, OXYGEN TO THE COMPOUND AND WE SEE THAT 5 OXYGENS TO 2 OF THOSE, AND SO NOW WE LL PUT IN ANOTHER CONVERSION FACTOR HERE. WE LL NOW MULTIPLY BY 5 MOLES OF OXYGEN PER 2 MOLES OF C2H3CL. NOW THOSE UNITS CANCEL AND WE RE FINALLY LEFT THEN WITH MOLES OF OXYGEN WHICH WAS WHAT WE ASKED FOR. ALRIGHT NOW AGAIN WE CAN GET A ROUGH IDEA OF THE ANSWER BECAUSE WE SEE TWO INTO THAT ABOUT TWO AND A HALF. 2 ½ X 10 WOULD BE 25, AND 25/62.5 WOULD BE JUST A LITTLE BIT LESS THAN THE HALF, SO WE WOULD EXPECT OUR ANSWER IN SOMETHING ABOUT MAYBE.35. SO WE LL GO AHEAD AND PLUG THROUGH ON OUR CALCULATOR QUICKLY THEN. WE HAVE 50, I M TAKING THE 10 X 5, DIVIDED BY 2, AND THEN ALL OF THAT DIVIDED BY 62.5, AND WE END UP WITH AN ANSWER OF, ACTUAL ANSWER OF 0.400 MOLES OF OXYGEN, AND WE CAN HAVE THREE SIGNIFICANT FIGURES BECAUSE WE HAD 3 IN THE MASS OF THE COMPOUND THAT WE STARTED WITH. ANY QUESTION ON ANY STEP OF THAT ONE? THE LONGEST TYPE ARE OF COURSE THOSE THAT TAKE US FROM MASS TO MASS, AND THIS IS AN EXAMPLE HERE WHERE I ASK FOR HOW MAY GRAMS OF HYDROGEN CHLORIDE, SO HCL, ARE WE GOING TO PRODUCE FROM 10 GRAMS OF C2H3CL? SO WE HAVE GRAMS OF THIS AND WE WANT GRAMS OF THAT AND IN THAT CASE THEN WE RE GOING TO ONCE AGAIN, FOLLOWING OUR ROADMAP, WE RE GOING TO GO FROM GRAMS TO MOLES, MOLES TO MOLES OF THE UNKNOWN AND MOLES OF UNKNOWN TO GRAMS. SO WE NEED THREE CONVERSION FACTORS IN A MASS TO MASS STOICHIOMETRIC PROBLEM. SO LOOKING AT THIS ONE THEN AND SETTING IT UP,
CHM 105 & 106 MO1 UNIT TWO, LECTURE THREE 10 ACTUALLY THE FIRST PART OF THIS IS THE SAME AS WE JUST DID, GRAMS OF HCL WILL BE EQUAL TO 10.0 GRAMS OF C2H3CL, MULTIPLIED Y, WE JUST CALCULATED THE MASS IN THE PREVIOUS PROBLEM SO WE ALREADY HAVE THAT RELATIONSHIP SET, MULTIPLIED BY 1 MOLE OF THE C2H3CL OVER 62.50 GRAMS AND I WON T BOTHER TO WRITE THE WHOLE FORMULA DOWN THIS TIME. NOW WE RE AT MOLES OF THE KNOWN SUBSTANCE. NOW WE RE GOING TO CONVERT TO MOLES OF THE UNKNOWN WHICH IS THE HCL, SO NOW WE NEED TO MULTIPLY BY THE NUMBER OF MOLES OF HCL WHICH IS 2 TO 2 MOLES OF THE OTHER ONE. SO WE HAVE 2 MOLES OF HCL PER 2 MOLES OF C2H3CL. NOW YOU MIGHT SAY WELL ISN T THAT KIND OF DUMB TO DO THAT? 2/2 IS OBVIOUSLY 1 AND 1 TIMES ANYTHING ISN T GOING TO CHANGE THE NUMBER. YES THAT S TRUE IT DOESN T CHANGE THE NUMBER BUT THE KEY THING IS THAT WE KEEP TRACK OF THE UNITS AND THAT IS HOW WE RE GOING TO CHANGE THEN FROM THE UNITS OF OUR KNOWN TO THE UNITS OF UNKNOWN, SO YES WE DO WANT TO INCLUDE IT EVEN THOUGH IT HAS A NUMBER OF ONE. SO FINALLY NOW HOWEVER WE DON T WANT MOLES OF HCL WE WANT GRAMS OF HCL AND SO WILL NEED TO PUT IN ONE MORE CONVERSION FACTOR. MULTIPLY BY THE NUMBER OF GRAMS OF HCL IN ONE MOLE OF HCL, AND THAT AGAIN WOULD BE ITS MOLAR MASS, ONE HYDROGEN, 1.01, ONE CHLORINE, 35.45, SO 36.46 GRAMS AND MOLES OF HCL NOW CANCEL. NOTICE AGAIN WE HAVE GRAMS ASKED FOR AND WE HAVE GRAMS DETERMINED. SO WE RE COMFORTABLE WE VE SET THE PROBLEM UP CORRECTLY AS LONG AS OUR UNITS MATCH WHICH THEY NOW DO AND SO THEREFORE WE WOULD BE READY TO SOLVE THE PROBLEM. WELL EYEBALLING A QUICK CHECK HERE THIS WOULD BE ABOUT 30/60 WHICH WOULD BE A HALF, AND A HALF OF 10 WOULD BE ABOUT 5 AND SO WE KNOW THAT OUR ANSWER IS APPROXIMATELY 5 GRAMS. SO AGAIN NOW WE WOULD PLUG IN OUR ACTUAL NUMBERS THERE SO WE WOULD HAVE 10 TIMES THAT SO I CAN JUST PUT IN 344.6 AND DIVIDE THAT BY 62.5 AND WE SEE THAT OUR ACTUAL FINAL A NSWER IS 5.51 GRAMS OF HCL. OKAY QUESTION? ( STUDENT RESPONSE NOT AUDIBLE) UH, YES IT SORTA LOOKS THAT WAY. GOOD QUESTION. THE QUESTION WAS, SHOULDN T THIS NUMBER BE 36.46, AND THAT IS RIGHT, WE JUST CALCULATED THAT HERE, 36.46, AND SO YES IN FACT THAT SHOULD HAVE BEEN 36.6, OUR ANSWER WOULD CHANGE SLIGHTLY, PROBABLY 6-0, AND I THREW THAT IN THERE AS A
CHM 105 & 106 MO1 UNIT TWO, LECTURE THREE 11 MISTAKE AND SOMEBODY WAS WIDE AWAKE ENOUGH TO CATCH THAT YOU SEE. OKAY SO YOU RE PAYING ATTENTION. ALRIGHT. ANY OTHER QUESTIONS ON THE STEP PROCESS OF THIS? ALRIGHT WELL LET S JUST LOOK AT 2 MORE EXAMPLES HERE. WE RE GOING TO DO MANY AS WE GO THROUGH IN NOT ONLY THIS CHAPTER WITH SOME OTHER ASPECTS OF THIS CHAPTER BUT ALSO FUTURE CHAPTERS, AND IT IS IMPORTANT, IT S IMPORTANT FOR YOUR SUCCESS IN THE COURSE ACTUALLY TO HAVE A GOOD FEELING FOR STOICHIOMETRIC PROBLEMS. WE RE GOING TO DO IT HERE, WE RE GOING TO DO IT IN THE NEXT CHAPTER, THE NEXT CHAPTER AFTER THAT, A CHAPTER ON GASES LATER ON. WE RE GOING TO RUN INTO STOICHIOMETRIC ABOUT HALF OF THE CHAPTERS WE WORK WITH SO NOW IS THE TIME TO GET DOWN EXACTLY THE STEPS THAT YOU WANT TO USE. ALRIGHT, LET S LOOK AT THIS PROBLEM HERE. THIS PARTICULAR PROBLEM REPRESENTS THE REACTION OF OXYGEN WITH A CHEMICAL NOTICE THIS IS A HYDROCARBON, BUT THAT PARTICULAR HYDROCARBON IS ACETYLENE, OKAY. THAT S NOT ONE OF THE ONES WE TALKED ABUT AS A SATURATED HYDROCARBON BUT IT S A HYDROCARBON. OXY-ACETYLENE TORCH. SO WE NEED FIVE MOLE OF OXYGEN TO REACT WITH EVERY TWO MOLES OF ACETYLENE, AND IN THE PROCESS WE SEE WE PRODUCE FOUR MOLES OF CARBON DIOXIDE AND TWO MOLES OF WATER AND OF COURSE HEAT ENERGY AND WE RE GOING TO LOOK AT ENERGY S CONTRIBUTION IN STOICHIOMETRY HERE JUST A LITTLE BIT LATER IN THE CHAPTER. NOW THE QUESTION THIS TIME IS HOW MANY MOLES OF OXYGEN DO WE NEED TO REACT 86.4 GRAMS OF THE ACETYLENE. NOTICE THAT OUR KNOWN IS GIVEN TO US IN MASS AND OUR UNKNOWN IS GIVEN TO US IN MOLES. THIS IS A MASS TO MOLE TYPE PROBLEM SO JUST MENTALLY STARTING TO THINK OF THE STEPS BEFORE WE EVER WRITE ANYTHING DOWN I M GOING TO THINK. I NEED TO CHANGE FROM GRAMS OF THIS KNOWN SUBSTANCE TO MOLES, STEP NUMBER ONE, AND THEN I NEED TO CHANGE FROM MOLES OF THIS TO MOLES OF THE UNKNOWN, STEP NUMBER TWO, AND IN THAT CASE THAT S AS FAR AS I NEED TO GO. SO I NEED TWO STEPS IN THE PROBLEM SO MOLES OF OXYGEN ARE EQUAL TO 86.4 GRAMS OF C2H2, AND THE FIRST THING WE NEED TO DO IS CHANGE FROM GRAMS TO MOLES. SO MULTIPLIED BY ONE MOLE OF ACETYLENE PER THE NUMBER OF GRAMS WHICH WOULD BE 26.04 GRAMS. THAT S MOLAR MASS, SO I JUST CALCULA TED THAT QUICKLY IN MY HEAD THERE SO
CHM 105 & 106 MO1 UNIT TWO, LECTURE THREE 12 THAT WE COULD USES IT. NOW WE HAVE THE MOLES OF THE ACETYLENE, NOW WE NEED TO USE OUR STOICHIOMETRIC RELATIONSHIP TO GET TO THE NEXT STEP AND WE SEE THAT FROM OUR EQUATION UP HERE WE HAVE A FIVE TO TWO MOLE RATION, AND WE USE IT IN WHATEVER FASHION WE NEED TO CANCEL THE UNITS OUT. IN THIS CASE WE WANT THE FIVE MOLES OF OXYGEN ON TOP AND WE WANT THE TWO MOLES OF ACETYLENE, HC2H2, ON THE BOTTOM. OUR UNITS CANCEL AND NOTICE AT THIS POINT WE HAVE MOLES OF OXYGEN ON BOTH SIDES OF THE EXPRESSION UNITWISE, AND WE RE COMFORTABLE WE HAVE THE PROBLEM SET UP CORRECTLY. AGAIN, RUNNING THROUGH A QUICK CALCULATION HERE WE WOULD HAVE ABOUT, THIS IS 2 ½ X 100, IT D BE 250/25, SO OUR ANSWER IS GOING TO BE PROBABLY SOMETHING A LITTLE LESS THAN 10, SAY ABOUT 9 MOLES, AND WE LL GO AHEAD AND PLUG IN THEN AND SEE WHAT WE ACTUALLY GET. 86.4 DIVIDED BY 26.04 AND MULTIPLIED BY 2.5 AND WE GET AN ANSWER OF 8.29 MOLES OF OXYGEN. OKAY? ALRIGHT, NOW. THE LAST ONE OF THIS TYPE THAT WE LL DO THEN IS GRAMS OF CARBON DIOXIDE PRODUCED ALONG WITH 6.85 GRAMS OF WATER. SO IN THIS PARTICULAR ONE WE RE SAYING IF IN THE REACTION I GOT 6.85 GRAMS OF WATER PRODUCED, HOW MANY GRAMS OF THE CARBON DIOXIDE WOULD I HAVE ALSO PRODUCED? NOW OUR RATIO DOESN T HAVE TO BE REACTANT TO REACTANT, IT DOESN T HAVE TO BE REACTANT TO PRODUCT. IT CAN BE PRODUCT TO PRODUCT OR PRODUCT TO REACTANT OR REACTANT TO REACTANT. IT CAN BE ANY RELATIONSHIP AS LONG AS WE HAVE THE BALANCED EQUATION WE CAN RELATE THIS TO THAT OR THIS TO THIS OR THIS TO THAT, DOESN T MAKE ANY DIFFERENCE. ALRIGHT, IN THIS CASE THEN WE RE ASKING FOR HOW MANY GRAMS OF CARBON DIOXIDE WOULD I HAVE PRODUCED IF I HAD PRODUCED 4.85 GRAMS OF WATER. NOW NOTICE THIS TIME WE RE STARTING WITH MASS, WE RE ASKING FOR MASS. SO THINKING OF OUR ROADMAP WE RE GOING TO GO FROM MASS TO MOLES, ONE CONVERSION STEP, MOLES TO MOLES USING OUR BALANCED EQUATION TWO, AND MOLES TO MASS THE THREE CONVERSION FACTORS IN OUR ROADMAP. SO NOW WE RE READY TO SET UP THE PROBLEM KEEPING IN MIND WE NEED THREE STEPS IN THERE. SO GRAMS OF CO2 ARE EQUAL TO 6.85 GRAMS OF WATER, AND THE FIRST STEP THEN WILL BE TO CHANGE THAT TO MOLES, SO WE RE GOING TO MULTIPLY BY ONE MOLE OF WATER OVER 18.02 GRAMS OF WATER. WE
CHM 105 & 106 MO1 UNIT TWO, LECTURE THREE 13 CALCULATED THAT JUST A MOMENT AGO, SO WE VE DONE THE FIRST STEP. WE NOW NEED TO USE OUR BALANCED EQUATION TO RELATE THEN THE MOLES OF THE CO2 TO THE MOLES OF WATER, AND WE LOOK UP AT OUR BALANCED EQUATION, 2 TO 4, AND SO WE SEE THAT WE HAVE 4 MOLES OF CARBON DIOXIDE FOR EACH 2 MOLES OF WATER THAT WERE FORMED AND WE NOW NOTICE THAT OUR WATERS CANCEL OUT. AND FINALLY THEN WE NOW NEED TO CHANGE FROM MOLES OF CARBON DIOXIDE TO GRAMS OF CARBON DIOXIDE, WHOOPS, NOT AN EQUAL SIGN, A TIMES, AND HOW MANY GRAMS OF CO2 ARE THERE IN ONE MOLE OF CO2? WELL WE WOULD HAVE TO ADD IT UP, 12 FOR THE CARBON AND 2 X 16, 32 FOR THE OXYGENS. WE WOULD HAVE 44.01 GRAMS OF CARBON DIOXIDE PER ONE MOLE. AGAIN WE NOTE NOW THAT WE HAVE GRAMS OF CARBON DIOXIDE ON THE RIGHT, WE ASK FOR CARBON DIOXIDE, OUR PROBLEM SHOULD BESET UP CORRECTLY. AND A QUICK SPAN ACROSS WE HAVE 6 OVER 18 WOULD BE ABOUT 1/3 X 2 WOULD BE 2/3, AND 2/3 OF THIS WOULD BE ABOUT 30, AND SO OUR ANSWER SHOULD BE APPROXIMATELY 30 GRAMS WHEN WE MAKE OUR CALCULATION. AGAIN WE LL CALCULATE IT QUICKLY, 6.85 DIVIDED BY 18.02, MULTIPLIED BY 2 AND MULTIPLIED BY 44.01, AND WE GET AN ANSWER OF 33.5 GRAMS OF CARBON DIOXIDE FOR OUR ANSWER. ANY QUESTION ON THAT PARTICULAR TYPE OF CALCULATION? NOW, ALL OF THESE PROBLEMS WE VE CALCULATED ARE BASED ON THE CONCEPT THAT WHEN WE PUT REACTANTS TOGETHER THAT THEY TOTALLY REACT TO FORM PRODUCTS. IN OTHER WORDS WHAT WE HAVE CALCULATED IN EACH CASE IS A THEORETICAL VALUE. THEORETICALLY THIS IS WHAT WE SHOULD GET, PROVIDED THAT EVERYTHING IN THE REACTION ACTUALLY DOES UNDERGO A CHEMICAL CHANGE. IN MANY REACTIONS IN NATURE THIS DOESN T OCCUR. WE ONLY GET PART OF THE REACTANTS THAT REALLY BECOME PRODUCTS, AND SO WHAT WE RE GOING TO LOOK AT IN THE NEXT PART OF STOICHIOMETRY IS WHAT WE REFER TO AS PERCENT YIELD. IN OTHER WORDS, HOW MUCH OF OUR PRODUCT DO WE REALLY GET IN THE CHEMICAL REACTION. PERCENT YIELD IS DEFINED AS THE GRAMS THAT WE ACTUALLY GET DIVIDED BY THE GRAMS THEORETICAL MULTIPLIED TIMES 10 TO THE SECOND TO PUT IT IN PERCENT. THIS RIGHT HERE, I LL PUT A LITTLE ASTERISK HERE, THAT IS WHAT WE GET, THAT S THE NUMBER WE OBTAIN WHEN WE DO A STOICHIOMETRIC PROBLEM. WHEN WE DO A STOICHIOMETRIC PROBLEM WE RE CALCULATING
CHM 105 & 106 MO1 UNIT TWO, LECTURE THREE 14 THE THEORETICAL VALUE. IN OTHER WORDS IN THAT LAST PROBLEM WHERE WE HAD THE 33 GRAMS OF CARBON DIOXIDE, THAT IS BASED ON THEORETICAL THAT 100% OF THE COMPOUND REACTED. ALRIGHT, LET S LOOK AT AN EXAMPLE HERE IN THE BOOK, AND THIS EXAMPLE GIVES US A REACTION OF FE2O3, REACTING WITH CARBON TO PRODUCE IRON AND CARBON MONOXIDE, AND THE QUESTION IS WHAT IS THE PERCENT YIELD IF A 438 GRAM SAMPLE OF THIS PRODUCE 275 GRAMS OF THIS. SO THIS IS WHAT WE ACTUALLY OBTAIN. THE QUESTION IS WHAT DID WE THEORETICALLY, WHAT SHOULD WE THEORETICALLY OBTAIN AND IF SO ONCE WE CALCULATED THE THEORETICAL WE CAN OF COURSE CALCULATE THE PERCENT YIELD. NOW THIS IS A MASS TO MASS PROBLEM, WHICH MOST OF THE PROBLEMS ARE GOING TO BE GIVEN AS, AND SO WE RE NEEDING TO GO FROM GRAMS OF THIS TO GRAMS OF THAT IN OUR CALCULATION. SO WE WOULD ASK THE QUESTION HOW MANY GRAMS OF IRON WOULD W E GET FROM 438 GRAMS OF FE2O3. WE WOULD HAVE TO CHANGE FROM GRAMS TO MOLES. SO ONE MOLE OF THE COMPOUND PER NUMBER OF GRAMS OF THE COMPOUND, AND WE WOULD LOOK UP THERE FOR IRON, 55.85, AND WE HAVE TWO OF THOSE AND WE HAVE 3 X 16, OR 48 FOR THE OXYGEN. A ND SO WE HAVE A MASS OF 0, 7, 9, 10, 159.70. THAT TAKES CARE OF THE GRAMS OF THAT. MULTIPLIED BY, OKAY, NOW WE RE AT MOLES, NOW WE NEED TO GO MOLES OF UNKNOWN TO MOLES OF IRON PER ONE MOLE OF FE2O3. SO THAT UNIT CANCELS AND THEN WE NEED TO CONVERT FROM MOLES OF IRON TO GRAMS OF IRON. MULTIPLIED BY 55.85 GRAMS OF IRON PER ONE MOLE OF IRON AND SO THAT UNIT CANCELS. NOW THIS IS GRAMS OF IRON THEORETICAL THAT WE RE CALCULATING. WELL LET S SEE I LL PLUG THAT IN REAL QUICK HERE BECAUSE WE NEED THAT NUMBER TO DO OUR FINAL STEP SO WE HAVE 438 DIVIDED BY 159.7, MULTIPLIED BY 2 AND MULTIPLIED BY 55.85. THEORETICALLY, IF ALL OF THIS REACTED WE SHOULD BE ABLE TO PRODUCE 306 GRAMS OF IRON. NOW WE SEE THAT WE DIDN T PRODUCE THAT AMOUNT. THE ACTUAL REACTION GAVE US ONLY 275 GRAMS AND SO THEREFORE NOW WE CAN CALCULATE WHAT PERCENT YIELD, WHAT PERCENT OF THE THEORETICAL AMOUNT DID WE ACTUALLY OBTAIN. SO PERCENT YIELD IS EQUAL TO 275 GRAMS DIVIDED BY 306 GRAMS TIMES 10 TO THE SECOND. OF COURSE OUR GRAMS CANCEL, PERCENT IS A UNITLESS NUMBER, AND WE HAVE A PERCENT YIELD OF 89.8% YIELD. ALRIGHT, IN
CHM 105 & 106 MO1 UNIT TWO, LECTURE THREE 15 OUR NEXT LECTURE WE WILL COME BACK AND EXPLORE PERCENT YIELD A LITTLE BIT FURTHER. WE LL ALSO GO ON AND TALK ABOUT ONE MORE ASPECT OF STOICHIOMETRY REFERRED TO AS THE LIMITING REACTANT PART OF STOICHIOMETRY AND WE LL TAKE THAT UP THEN IN OUR NEXT LECTURE.