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CHM 105 & 106 MO1 UNIT ONE, LECTURE EIGHT/ UNIT TWO, LECTURE ONE 1

CHM 105/106 Program 8: Unit 1 Lecture 8 / Unit 2 Lecture 1

IN THE END OF OUR PREVIOUS LECTURE WE WERE LOOKING AT THE AREA OF THAT

WE REFER TO AS , AND ORGANIC CHEMISTRY IS CHEMISTRY BASED

PRIMARILY ON CONTAINING COMPOUNDS. AND WITHIN THAT GROUP OF ORGANIC

CHEMISTRY THE SIMPLEST, I GUESS WE WOULD CALL IT SIMPLEST FROM THE STANDPOINT THAT

IT CONTAINS ONLY TWO ELEMENTS, ARE THE CHEMICAL COMPOUNDS CA LLED HYDRO

CARBONS. BY THEIR NAME, HYDROCARBONS ARE COMPOUNDS CONTAINING AND

CARBON ONLY, ONLY TWO ELEMENTS IN THERE. WE LOOKED AT ONE GROUP OF THOSE

YESTERDAY WHICH WE CALL SATURATED HYDROCARBONS, WHICH HAVE A GENERAL

CHEMICAL RELATING TO CARBON AND HYDROGEN, CNH2N+2, SO WHATEVER N IS FOR

CARBON, FOR INSTANCE IF WE HAD THEN A COMPOUND WHICH HAD, SAY, 10 HERE, SO

C10, THAT WOULD TELL US THEN THAT THE NUMBER OF WOULD BE TWO TIMES

THAT, 20 PLUS 2. SO THE FOR THE WOULD BE IN

THAT CASE. C10H22. WE CALL THIS GROUP THE , AS A FAMILY OF ORGANIC

COMPOUNDS. WE REFER TO THEM AS THE ALKANES, AND WE LOOKED AT THE FIRST 10 OF

THOSE ALKANES YESTERDAY, AND POINTED OUT THAT SOME OF THESE ARE FAIRLY COMMON

COMPOUNDS TO US, THINGS THAT WE EXPERIENCE OR UTILIZE ON A FAIRLY COMMON BASIS.

THE LATTER ONES ARE THOSE THAT ARE COMPONENTS OF GASOLINE THAT WE USE IN THE

AUTOMOBILE. THE FIRST ONE, THE SIMPLEST OF THE ALKANES, METHANE, WE INDICATED WAS

A COMPONENT OF NATURAL . PROPANE AND BUTANE OF COURSE ARE TWO OTHER FAIRLY

COMMON HYDROCARBONS. WE ALSO WENT ON AND TALKED A LITTLE BIT ABOUT THE

FACT THAT WHEN YOU HAVE THE FORMULA FOR A HYDROCARBON THAT IT IS POSSIBLE THAT

WE CAN PUT THESE TOGETHER, IT IS POSSIBLE THAT WE CAN PUT THESE TOGETHER IN MORE

THAN ONE WAY AND ACTUALLY PRODUCE THEN MORE THAN ONE CHEMICAL COMPOUND.

MEANING THAT THE COMPOUND SHOULD HAVE SOME DIFFERENT PHYSICA L PROPERTIES –

MELTING POINTS, BOILING POINTS, CHEMICAL , IF THEY IN FACT ARE UNIQUE. AND

JUST TO LOOK AT ONE HERE REAL QUICK A GAIN, IF FOR INSTANCE WE HAD A COMPOUND C4H10, CHM 105 & 106 MO1 UNIT ONE, LECTURE EIGHT/ UNIT TWO, LECTURE ONE 2

AND THAT IS AN ALKANE HYDROCARBON, MEETS THE FORMULA 2N, 2 TIMES N PLUS 2 FOR THE

HYDROGEN. WE COULD PUT THE 4 CARBONS TOGETHER LIKE SUCH, AND ATTACH THE 10

HYDROGENS, AND AS I INDICATED YESTERDAY, THE CHEMISTRY OF CARBON IS SUCH THAT IT

CAN HAVE 4 BONDS, 4 THINGS ATTACHED TO IT. IN OUR WE SHOW

THOSE AS DASHES. WE DON’T NEED TO WORRY ABOUT WHAT A IS AT THIS

POINT, BUT CARBON CAN HAVE 4 THING ESSEN TIALLY ATTACHED TO IT. HYDROGEN CAN HAVE

ONLY 1 BOND. AND SO THE ONLY WAY WE CAN HOOK HYDROGENS ON THERE IS WE CAN’T

HOOK TWO HYDROGENS TOGETHER AND THEN HAVE ANYTHING LEFT TO HOOK THE CARBON,

SO IF A HYDROGEN IS HOOKED TO A CARBON IT IS USED UP THEN ALL OF ITS ABILITY TO BOND.

WELL WE COULD DO IT THAT WAY, OR WE COULD A LSO HOOK 3 CARBONS LIKE THIS AGAIN

WITH THE HYDROGENS ATTACHED, AND THEN THE 4TH CARBON WE COULD PUT HERE AND

ATTACH HYDROGENS, AND IF WE LOOK WE SHOULD HAVE AGAIN 4 CARBONS AND 10

HYDROGENS SURROUNDING IT. AND THES E ARE TWO DIFFERENT CHEMICAL COMPOUNDS

WHICH WE REFER TO THEN AS , STRUCTURAL ISOMERS OF THE SAME CHEMICAL

FORMULA, BOTH OF THESE ARE C4H10, BOTH OF THESE ARE ALKANES, SATURATED

HYDROCARBONS, BUT THEY ARE TWO DISTINCT CHEMICAL COMPOUNDS. AND THE NAMING

RULES USED IN ORGANIC CHEMISTRY ARE CALLED THE IUPAC RULES . WHICH REFER TO IN THE

TEXT, BUT WE’RE NOT GOING TO SPEND ANY APPRECIABLE TIME AS WE GO THROUGH THE

COURSE WORRYING ABOUT APPLYING IUPAC RULES. WE’LL LEARN A FEW BASIC NAMES OF THE

ORGANICS AND USE THEM AS SUCH, BUT THE IUPAC RULE IS TO NAME THE LONGEST CARBON

CHAIN AS THE BASIC NAME AND THEN TO TELL WHAT OTHER THINGS MIGHT BE PRESENT IN THE

MOLECULE. IN THIS CASE, THE LONGEST CONTINUOUS CHAIN IS FOUR. THIS COMPOUND IS

MERELY CALLED BUTANE. THAT’S ALL WE HAVE TO SAY. THE ENDING, I DIDN’T MENTION THIS

IN THE PREVIOUS LECTURE, BUT THE ENDING OF ALL ALKANES IS A-N-E, THAT’S KIND OF A WAY

WE CAN GET AN IDEA WHEN WE SEE A NAME AS TO WHAT TYPE OF CHEMICAL COMPOUND,

WHAT TYPE OF ORGANIC COMPOUND IT IS. THIS ONE HAS 3 CARBONS IN A ROW. NO

WHICH WAY YOU GO, THE LONGEST CONTINUOUS CHAIN WOULD BE 3, AND SO THE BASE NAME

IS PROPANE. AGAIN NOTICE IT ENDS IN A-N-E. BUT THIS TIME WE HAVE THIS LITTLE GROUP CHM 105 & 106 MO1 UNIT ONE, LECTURE EIGHT/ UNIT TWO, LECTURE ONE 3

HANGING RIGHT HERE AND SO WE HAVE TO TELL IT IT’S THERE. IN THE DETAILED IUPAC

SYSTEM WE NUMBER THE CARBONS AND WE TELL WHAT CARBON THE GROUP IS ATTACHED. SO

IF WE WERE TO DO THAT AND CALL THIS CARBON 1 AND THIS ONE 2 AND THIS ONE 3, WE COULD

THEN CALL THIS 2, TO TELL THAT ON THE SECOND CARBON OF THE PROPANE THERE’S

SOMETHING ATTACHED, AND THEN WE WOULD SA Y WHAT’S ATTACHED? METHYL. THIS IS THE

NAME, METHYL IS THE NAME FOR THIS LITTLE CH3 GROUP HANGING ON THERE AND WE WOULD

CALL THAT THEN 2-METHYL PROPANE, AND THOSE ARE THE SAME RULES THAT WE WOULD USE

NO MATTER HOW COMPLI CATED THE M IGHT BECOME. ALRIGHT, SO THE MAIN

THING THAT WE’RE LOOKING AT HERE, THE MAIN THING THAT I WANT TO STRESS IS THE FACT

THAT WE HAVE SO MANY DIFFERENT WAYS OF PUTTING TOGETHER THESE CARBON HYDROGEN

STRUCTURES. AS I MENTIONED YESTERDAY, BY THE TIM E WE GET TO C10, C10H22 HAS ABOUT 70

DIFFERENT UNIQUE WAYS OF HOOKING IN TOGETHER. 70 DIFFERENT NAMES, 70 DIFFERENT

COMPOUNDS, ALL WITH THE SAME CHEMICAL FORMULA. SO CHEMICAL TELL US

WHAT AND HOW MUCH OF EACH ARE THERE, STRUCTURAL FORMULAS TELL US HOW THEY’RE

HOOKED TOGETHER, AND IF WE HAVE MORE THA N ONE WAY WE CAN HOOK THEM TOGETHER,

THOSE ARE REFERRED TO AS ISOMERS. NOW ONE OTHER TYPE OF HYDROCARBON THAT FALLS

INTO THIS SAME GENERAL CATEGORY ARE THE CYCLOALKANES. NOTICE THAT WE STILL HAVE

THIS NAME ALKANES, BUT THE CYCLO PART. THE NORMAL ALKANES A RE CHAINS, LONG

CHAINS, THINGS CAN BE BRANCHED OFF OF THOSE CHAINS, BUT AS W E SEE THEY’RE JUST

HOOKED TOGETHER, BUT CYCLOALKANES NOW BY THE TERM WOULD BE CYCLIC, OR IN OTHER

WORDS WE’RE GOING TO HOOK THINGS TOGETHER. LET’S TAKE BUTANE FOR A SECOND. THIS

IS THE WAY WE WOULD NORMALLY PICTURE BUTANE, 4 CARBONS INA ROW LIKE THIS. IF WE

WERE TO TAKE THE TWO END CARBONS, THIS WAS ONE CARBON AND AN END, THIS WAS THE

OTHER ONE. IF WE WERE TO BEND THEM AROUND WE CAN ACTUALLY HOOK THEM TOGETHER

LIKE SUCH AND WE WILL STILL HAVE HYDROGENS ON HERE. SO WE END UP WITH SORT OF A

SQUARE LOOKING SHAPE. NOW THE NAMING FOR THIS IS STILL THAT IT’S A BUTANE, BECAUSE

IT HAS 4 CARBONS, AND THAT WOULD BE THE RULE WE WOULD APPLY, BUT OBVIOUSLY THIS

ISN’T THE SAME BUTANE AS WHAT WE SAW JUST A MOMENT AGO WHERE THE 4 WERE INA ROW. CHM 105 & 106 MO1 UNIT ONE, LECTURE EIGHT/ UNIT TWO, LECTURE ONE 4

AS A MATTER OF FACT, IT DOESN’T HAVE THE SAME CHEMICAL FORMULA BECAUSE HERE THIS

HAS A CHEMICAL FORMULA OF C4H8. THIS IS NOT, THIS IS NOT AN ISOM ER OF THE PREVIOUS

CHEMICAL FORMULA. THE PREVIOUS CHEMICAL FORMULA WAS C4H10. THIS IS A NEW

COMPOUND, NEW CHEMICAL FORMULA. BUT HOW ARE WE GOING TO SHOW THIS NOW NOT

BEING THE SAME AS THE OTHER BUTANE? WE USE THE PREFIX “CYCLO.” SO THIS WOULD BE

CALLED CYCLOBUTANE, WHICH TELLS THE READER THAT 4 CARBONS, BUT THE END CARBONS

ARE HOOKED BACK TOGETHER TO MAKE IT INTO A CYCLING MATERIAL. AND THE TEXT SHOWS

A COUPLE OF CYCLIC COMPOUNDS. ONE THAT IS SOMEWHAT OF INTEREST, LET’S LOOK AT THIS

INE RIGHT HERE WHICH HAS 3 CARBONS ON IT AND EACH OF THESE CARBONS WOULD HAVE A

COUPLE OF HYDROGENS ATTACHED, LIKE SUCH. THIS IS CALLED CYCLO, AND THE NAME FOR 3

CARBONS WAS PROPANE, SO CYCLOPROPANE. CYCLOPROPANE IS AN ANESTHETIC. IT WAS

USED FOR MANY YEARS IN PLACE OF ETHER BECAUSE OF SOME OF THE ILL-EFFECTS THAT

ETHER HAS – AFTER-EFFECTS, NAUSEA, ETC, AND SO ETHER WAS REPLACED FOR SOME TIM E

WITH A CHEMICAL CALLED CYCLOPROPANE. THE ONLY PROBLEM IS CYCLOPROPANE IS VERY

REACTIVE AND SO NECESSARY PRECAUTIONS HAD TO BE TAKEN. YOU COULDN’T HAVE ANY

STATIC , THINGS HAD TO BE PROPERLY ELECTRICALLY GROUNDED, THE CLOTHING

WORN IN FOR AN OPERATING ROOM HAD TO BE CAREFUL THAT IT WASN’T CLOTHING…YOU

WOULDN’T WEAR INA WOOL SUIT, WHICH COULD PICK UP STATIC ELECTRICITY. SO

CYCLOPROPANE THEN WAS USED AS AN ANESTHETIC. OKAY? NOW, MOST OF THE, HERE WE GO,

MOST OF THE HYDROCARBONS OF COURSE ARE OBTAINED FROM OUR PET ROLEUM SUPPLY.

AND IF YOU TRAVEL ACROSS TEXAS, OKLAHOMA, EVEN PARTS OF MISSOURI, OHIO, YOU MAY

SEE IN THE FIELDS ALONG THE ROAD THESE PUMPS. THESE PUMPS THAT ARE MOVING UP AND

DOWN ARE PUMPING, LITERALLY PHYSICALLY PUMPING THE OIL OUT OF THE GROUND. THE OIL

OR PETROLEUM IS A , A MIXTURE OF ALL SORTS OF HYDROCA RBONS, INCLUDING ALL

OF THOSE ALKANES THAT WE’RE TALKING ABOUT EXCEPT FOR METHANE AND , WHICH

ARE USUALLY AT ROOM TEMPERATURE, SO THEY WOULD HAVE ESCAPED AS GASES, BUT

THE REST OF THEM WILL COME OUT IN THIS MATERIAL CALLED PETROLEUM. MOST OF THE

WELLS DRILLED EARLY ON WERE WHAT WE WOULD REFER TO AS VERTICAL DRILLING. IN CHM 105 & 106 MO1 UNIT ONE, LECTURE EIGHT/ UNIT TWO, LECTURE ONE 5

OTHER WORDS THEY WOULD PUT UP A DRILL SITE, THEY WOULD DRILL A HOLE STRAIGHT

DOWN INTO THE OIL POCKET, OR THE SUSPECTED OIL POCKET, AND THEN GET THE OIL OUT.

NOW THE MOVIES OF COURSE WE OFTEN SEE THE DRILL AND THEY HIT AND THE OIL COMES

BLOWING OUT AND SPRA YS ALL OVER EVERYBODY AND THERE’S A LOT OF CELEBRATION AND

THAT TYPE OF THING. THAT IS TRUE OCCASIONALLY WHERE WE HAVE OIL UNDER GROUND

THAT’S UNDER HIGH PRESSURE, BUT MOST OF THE OIL UNDER GROUND DOESN’T COME

SHOOTING OUT. IT’S NOT UNDER HIGH PRESSURE, AND SO THEREFORE, TO GET IT OUT WE HAVE

TO PUMP IT OUT. IT DOESN’T COME SPRAYING OUT, AND SO THIS IS WHAT WE SEE AS WE

TRAVEL ACROSS THEN, THIS. NOW ONE OF THE PROBLEMS WITH THE VERTICAL DRILLING

PROCESS IS THE FACT THAT YOU HAVE TO HAVE ONE OF THOSE LITTLE DRILLS, ONE OF THOSE

LITTLE PUMPS FOR EVERY LITTLE PUDDLE OF OIL THAT’S DOWN THERE. IF YOU’VE EVER

DRIVEN AROUND LOS ANGELES FOR INSTANCE, THERE ARE PLACES THAT THESE PUMPS ARE SO

CLOSE THEY’RE ALMOST SITTING ON TOP OF EACH OTHER. THEY’VE DRILLED ALL OF THESE

LITTLE DRILL HOLES DOWN INTO DIFFERENT POCKETS OF OIL TO PUM P IT OUT. WELL THERE

ARE SOME NEW TECHNIQUES BEING DEVELOPED AND THIS IS CALLED HORIZONTAL DRILLING,

AND SO THIS HERE IS TRYING TO DEMONSTRATE THE DIFFERENCE BETWEEN HORIZONTAL

DRILLING AND VERTICA L DRILLING. AS THEY SAY, MOST OF THE OIL THAT WE SEE TODAY HAS

STILL BEEN DRILLED VERTICALLY. BECAUSE WE’RE PUMPING OUT MOST OF THE BIG PUDDLES,

WE NOW HAVE TO START LOOKING AT RECOVERING OUR PETROLEUM FROM THE LITTLE

PUDDLES THAT ARE DOWN THERE, AND IT’S NOT ECONOMICALLY FEASIBLE TO DRILL DOWN

AND TAP IN TO EACH OF THOSE LITTLE PUDDLES DOING IT VERTICALLY, BUT A TECHNIQUE HAS

BEEN DESIGNED WHICH WE CALL HORIZONTAL DRILLING WHERE THEY ACTUALLY GO INTO AN

OIL FIELD AND THEY DRILL, FIRST OF COURSE VERTICALLY, BUT THEN THEY HAVE THE ABILITY

TO BEGIN TURNING THE DRILLING BIT SO THAT MECHANICALLY IT STARTS TO CURVE, AND

NOW INSTEAD OF DRILLING DOWN THEY’RE DRILLING HORIZONTALLY THROUGH THE OIL

FIELD. WELL WHAT THIS DOES IS THIS ALLOWS THEM NOW TO DRILL THROUGH MANY MANY

MANY LITTLE PUDDLES OF OIL INSTEAD OF JUST GOING INTO ONE. SO INSTEAD OF PUTTING AN

OIL DRILL HERE AND HERE AND HERE, THEY PUT ONE OVER THERE AND THEY DRILL AND THEN CHM 105 & 106 MO1 UNIT ONE, LECTURE EIGHT/ UNIT TWO, LECTURE ONE 6

GO HORIZONTALLY TO PICK UP THOSE OTHER OIL DEPOSITS. OKAY, AND IN THE PROCESS OF

COURSE WE CAN THEN ECONOMICALLY RECOVER SOME OIL THAT PREVIOUSLY WE COULD NOT,

AND OF COURSE IF WE CAN’T RECOVER THE OIL WE’RE GOING TO HAVE TO DEPEND MORE ON

EXPORTED OIL, AND IF WE DO THAT, I MEAN IMPORTED OIL, NOT EXPORT ED, IMPORTED OIL,

WHICH OF COURSE THEN PUTS OUR ECONOMY AT THE HANDS OF OTHER COUNTRIES, AND SO

WE WOULD RATHER HAVE OUR OWN. THIS GIVES ABOUT A 90% INCREA SE IN OIL RECOVERY,

THE HORIZONTAL VERSUS THE VERTICAL. ECONOMICALLY WE CAN DO IT FOR ABOUT 10% OF

THE COST, OR RECOVER ABOUT 90% MORE OF THE PETROLEUM. WELL ONCE THE PETROLEUM

REACHES THE SURFACE OF COURSE IT’S NOT VERY USEABLE. IF YOU’VE EVER SEEN CRUDE

PETROLEUM OR CRUDE OIL IT’S A NASTY LOOKING, SYRUPY, BLACK MATERIAL THAT DOES N’T

SEEM TO HAVE ANY PRACTICAL USE. AND SO FROM THAT POINT THEN WE HAVE TO PROCESS IT.

SO HERE WE’RE SHOWING NOW THE PETROLEUM IS COMING INTO THEN A REFINERY AS WE

REFER TO IT. WE REF INE THINGS, WE SEPARATE THINGS. THE OIL IS COMING IN, THE FIRST

THING WE DO IS HEAT IT. ALRIGHT, SO WE BEGIN BOILING IT. WE HEAT IT TO HIGHER

TEMPERATURES. SOME OF IT TURNS TO A GASEOUS STATE AT THAT POINT. AND WE THEN

BEGIN PUMPING IT INTO WHAT IS CALLED A FRACTIONATING POWER. A BIG COLUMN PACKED

FULL OF MATERIAL, SOLID MATERIAL WHERE THIS GASEOUS CRUDE THAT WE’VE NOW HEATED

TO HIGH TEMPERATURE BEGINS MOVING THROUGH. THE BOTTOM IS QUITE WARM, AND IT

COOLS OFF AS IT GOES TOWARD THE TOP. SO ONLY THE MOST GASEOUS MATERIALS MAKE IT

ALL THROUGH THE FRACTIONATING COLUMN AND ARE TAPPED OFF OF THE TOP, AND WE SEE

HERE THEN IT SHOWS UP HERE THAT, NUMBER ONE, AND THIS HAS A TEMPERATURE OF ABOUT

100 DEGREES, FRACTION NUMBER ONE COMES OFF AT THE TOP. FRACTION NUMBER ONE

CONTAINS THOSE THINGS THAT WOULD BE NORMALLY GASES ANYWAY, METHANE, ETHANE,

AND PROPANE. SO WE THEN TAKE THOSE OFF, AND THEN THE SECOND LEVEL THERE, THE 100

TO THE 200 DEGREE PA RT OF THAT FRACTIONATING COLUMN WE TAP OFF ANOTHER GROUP OF

HYDROCARBONS WHICH CONTAIN THEN THE C4 THROUGH C12, WHICH IS PRIMARILY THE

AUTOMOBILE FUEL, AND SO ON DOWN THE LINE. THE NEXT IS THE DIESELS, THE KEROSENES.

FINALLY WE GET THE MATERIALS THAT ARE LI QUIDS EVEN AT THE HIGHER TEMPERATURES. CHM 105 & 106 MO1 UNIT ONE, LECTURE EIGHT/ UNIT TWO, LECTURE ONE 7

THOSE ARE LUBRICATING OILS, AND THERE ARE SOME THINGS THAT WE JUST CAN’T DO A

WHOLE LOT WITH BECAUSE THEY’RE REALLY BIG HYDROCARBONS AND THOSE ARE THE TAR

AND ASPHALT MATERIAL THAT WE END UP WITH FROM THE CRUDE OIL. NOW THAT

PARTICULAR PROCESS AS I SAID IS REFERRED TO AS FRACTIONAL DISTILLATION. FRACTIONAL

BECAUSE WE’RE TAKING OFF FRACTIONS OR PORTIONS OF THE COMPOUND AS IT GOES

THROUGH THE COLUMN. FRACTIONAL DISTILLA TION. IF YOU DRIVE PAST A REFINERY THIS IS

ONE OF THE MAJOR THINGS THAT’S GOING ON THERE IS WE’RE SEPARATING OUT INTO THE

VARIOUS FRACTIONS. NOW DIFFERENT PETROLEUM, DIFFERENT CRUDE OILS HAVE DIFFERENT

PERCENTAGES OF THESE, AND OF COURSE OUR DEMAND FOR CERTAIN THINGS ARE DIFFERENT.

WE HAVE A VERY HIGH DEMAND IN THIS PART OF THE WORLD FOR GASOLINE, AND SO

OBVIOUSLY THAT’S ONE FRACTION THAT WE NEED, WE NEED A LOT OF. BUT SOMETIMES W E’RE

GOING TO END UP WITH A WHOLE LOT OF KEROSENE, DIESEL FUEL, LUBRICATING OILS ON

HAND BECAUSE WE’RE PRODUCING AND TAKING THE GAS FRACTION OFF VERY RAPIDLY AND

USING IT. WELL THEN WE HAVE TO LOOK FOR SOME OTHER PROCESSING METHODS, AND ONE

THAT IS USED IS CALLED CRACKING. CRACKING IS A VERY IMPORTA NT PART OF THE

PETROLEUM INDUSTRY. BY THE NAME YOU MIGHT GUESS, CRACKING M EANS TO BREAK

MOLECULES INTO SMALLER PARTS. WHAT WE’RE GOING TO DO IS WE’ RE GOING TO TAKE SOM E

OF THIS MATERIAL THAT IS LESS USEFUL LIKE KEROSENE FOR INSTANCE AND WE’RE GOING TO

TAKE SOME OF THAT C10 AND C14’S AND WE’RE GOING TO ADD A CHEMICAL, A CRACKING

AGENT, AND MIX IT WITH THAT AND BREAK IT INTO SMALLER UNITS OF SAY C8’S AND C6’S SO

THAT THEY CAN BE USED NOW AS GASOLINE. SO WE CAN IMPROVE THE PRODUCTIVITY OF

CERTAIN OF THESE FRACTIONS BY USING A CHEMICAL CRACKING PROCESS. OKAY. WELL THE

PETROLEUM INDUSTRY AND ALL OF THE OTHER ORGANIC COMPOUNDS THAT ARE ASSOCIATED

WITH IT IS PROBABLY THE MAJOR SINGLE ONE CHEMICAL INDUSTRY IN THE UNITED STATES.

WE DEPEND A LOT ON PETROLEUM AND PETROLEUM PRODUCTS, NOT ONLY TO DRIVE OUR CAR,

BUT ALL THE PLASTICS THAT WE USE ARE PETROLEUM BASED MATERIALS. THAT ALL COMES

FROM CHEMICALS THAT WE TAKE OUT OF PETROLEUM AND THEN CHEMICALLY TREAT THEM

TO FORM , TO FORM PLASTICS, ETC. MOST OF THE RUBBER WE USE IS NOT NATURAL CHM 105 & 106 MO1 UNIT ONE, LECTURE EIGHT/ UNIT TWO, LECTURE ONE 8

RUBBER IT’S SYNTHETIC RUBBER. AGAIN, IT’S A PETROLEUM BASED PRODUCT. SO THIS IS THE

STARTING POINT OF SOME OF THE MAJOR INDUSTRIES IN THE UNITED STATES, IN THE WORLD

FOR THAT MATTER. ALRIGHT, WELL THAT WRAPS UP WHAT I HAVE TO SAY AS FAR AS CHAPTER

TWO, BUT I THOUGHT THAT IT MIGHT BE JUST A GOOD IDEA JUST SORT OF AS A WRAP UP OF THE

MATH PART OF IT TO LOOK AT ONE LAST PROBLEM FROM CHAPTER TWO, AND THEN WE’LL GO

ON AND TALK A LITTLE BIT ABOUT CHAPTER THREE THIS MORNING. PROBLEM NUMBER 22

SAYS HOW MUCH, WHAT IS THE OF 8.46 X 10^21 OF THE CHEMICAL

COMPOUND CH3OH? THAT PARTICULAR CHEMICAL COMPOUND IN KNOWN AS METHANOLE,

ALSO CALLED METHYL ALCOHOL, ALSO HAS A COMMON NAME WOOD ALCOHOL, IT’S AN

ORGANIC COMPOUND. W HAT WE WANT TO KNOW IS HOW MUCH WOULD THIS NUMBER OF

MOLECULES OF THAT WEIGH? WELL ONCE AGAIN JUST AS A BRIEF REVIEW, KEEPING IN MIND

THAT WE KNOW THAT ONE OF A COMPOUND IS EQUAL TO IT’S FORMULA MASS. SO WE

HAVE THAT RELATIONSHIP, AND WE KNOW HOW TO FIND NOW FORMULA MASS. TO FIND

FORMULA MASS WE TAKE THE ATOMIC MASSES IN THEIR CORRECT RATIO AND ADD THEM

TOGETHER. SO IN THIS PARTICULAR CASE THE FORMULA MASS FOR W OOD ALCOHOL, WE HAVE

ONE CARBON, SO WE HA VE 1 X 12.01 FOR THE CARBON. WE HAVE FOUR HYDROGENS, 4 X 1.01,

AND WE HAVE ONE , 1 X 16, AND SO W E WOULD HAVE THEN A FORMULA MASS OF 32.05

GRAMS. SO WE KNOW THAT ONE MOLE OF WOOD ALCOHOL WEIGHS THAT AMOUNT. WE ALSO

KNOW THAT ONE MOLE OF ANYTHING CONTAINS 6.023 X 10^23 UNITS. IN THIS PARTICULAR CASE

THE UNITS BECAUSE WE TALKING ABOUT A COM POUND WOULD BE MOLECULES. SO WE HAVE

TWO RELATIONSHIPS, A RELATIONSHIP OF MOLECULES TO MOLES. WE HAVE A RELATIONSHIP

OF MOLES TO MASS, AND OF COURSE USING THOSE TWO CONVERSION FACTORS THEN WE

SHOULD BE ABLE TO ANSWER THE PROBLEM, AND I’LL GO AHEAD AND DO IT ON ANOTHER

TRANSPARENCY SO I CA N WORK SIDEWAYS SO WE CAN SHOW BOTH CONVERSION FACTORS. SO

THE QUESTION WAS NOW HOW MANY GRAMS OF THIS CH3OH DO WE HAVE IN 8.46 X 10^21

MOLECULES? NOW THE FIRST RELATIONSHIP WE’RE GOING OT USE THEN IS THE ONE THAT W E

SEE JUST ABOVE THAT, THE RELATIONSHIP OF MOLECULES TO MOLES. SO WE CAN SAY

MULTIPLIED BY ONE MOLE OF CH3OH PER 6.023 X 10^23 MOLECULES, I’LL JUST ABBREVIATE IT CHM 105 & 106 MO1 UNIT ONE, LECTURE EIGHT/ UNIT TWO, LECTURE ONE 9

AS “M”, AND SO MOLECULES ARE CANCELLED, AND WE’RE NOW AT MOLES OF CH3OH, AND

THEN WE KNOW THAT WE CAN NOW BRING IN THE RELATIONSHIP OF MASS. SO MULTIPLIED BY

32.05 GRAMS OF CH3OH PER 1 MOLE CH3OH AND MOLES CANCEL AND W E’RE LEFT THEN WITH

GRAMS WHICH IS OF COURSE THE UNIT THAT W E WERE SEEKING. ONCE AGAIN THEN WE’RE

READY TO PLUG IN TO OUR CALCULATOR AND OBTAIN THE FINAL NUMERIC ANSWER, AND

LET’S SEE WE WOULD HAVE 8.46 EXPONENTIAL 21, AND AGAIN I STRESS, BE SURE NOT TO GET

THE TIMES 10 IN THERE, THOSE THAT MAY BE USING THIS FOR THE FIRST TIME IN A COURSE,

DIVIDED BY 6.023 EXPONENTIAL 23. THEN I’LL HIT THE EQUAL SIGN AND FINALLY MULTIPLIED

BY 32.05, AND MY FINAL ANSWER IS THEN THAT THAT NUMBER OF MOLECULES WHICH SOUNDS

LIKE A WHOLE LOT ONLY WEI GHS 0.541 GRAMS. WE WOULD BE LIMITED TO THREE SIGNIFICANT

FIGURES BASED ON THIS NUMBER HERE. THIS ONE WOULD HAVE FOUR, THIS ONE HAS FOUR,

BUT THAT ONE HAS THREE. SO .45, OH EXCUSE ME, 45, I MISSED A SPOT THERE, ZERO, THE

ACTUAL NUMBER ON THE CALCULATOR IS 4501793, AND OF COURSE THREE PLACES THEN

WOULD BE ZERO. ALRIGHT ANY QUESTION ON ANYTHING FROM CHAPTER TWO? CHAPTER

THREE NOW WE’RE GOING TO GO THE NEXT STEP. CHAPTER ONE WE TALKED ABOUT THE

ALPHABET, THE CHEMICAL ALPHABET, THE SYM BOLS. CHAPTER TWO WE TALKED ABOUT

CHEMICAL FORMULAS. PUT SYMBOLS TOGETHER TO MAKE WORDS. CHAPTER THREE WE’RE

GOING TO PUT ALL OF THAT TOGETHER, SYMBOLS AND THEN THE WORDS, THE CHEMICAL

FORMULAS TO TALK ABOUT CHEMICAL SENTENCES DEALING WITH CHEMICAL CHANGE, WHICH

WE REFER TO AS A CHEMICAL EQUATION. NOW, ONE OF THE CHARACT ERISTICS OF CHEMICAL

CHANGE AS IT IS WITH ALSO PHYSICAL CHANGE IS THAT WHEN SOMET HING CHANGES EITHER

CHEMICALLY OR PHYSICALLY WE DO NOT CREATE NOR DO WE DESTROY ANY MASS. IN OTHER

WORDS, WHATEVER AMOUNT THAT WE STARTED WITH IN A PHYSICAL PROCESS AND CHANGE

IN SOME WAY, WE STILL HAVE. FOR INSTANCE IF I TAKE 10 GRAMS OF SOLID , ICE CUBES,

AND I MELT IT AND EVAPORATE IT, I STILL HAVE 10 GRAMS OF WATER, IT’S IN THE GASEOUS

STATE, IT MAY OCCUPY A WHOLE LOT MORE SPACE, BUT WE STILL HAVE 10 GRAMS OF WATER.

WE DON’T LOSE ANY MASS IN THE PHYSICAL CHANGE. IN A WE DO NOT

CREATE AND WE DO NOT DESTROY ANY MASS. IT MAY BE HOOKED TOGETHER DIFFERENTLY, CHM 105 & 106 MO1 UNIT ONE, LECTURE EIGHT/ UNIT TWO, LECTURE ONE 10

BUT WHATEVER AMOUNT OF CHEMICALS WE START WITH IS EQUAL TO THE AMOUNT OF

CHEMICALS THAT WE END UP WITH. THE LAW OF CONSERVATION OF MASS THEN PERTAINS TO

CHEMICAL CHANGE. TO REPRESENT A CHEMICA L CHANGE OF COURSE W E USE WHAT WE CALL

A CHEMICAL EQUATION. AND A CHEMICAL EQUATION CONSISTS OF TWO SIDES. THE FIRST, THE

CHEMICALS WE’RE STARTING WITH, WHICH WE CALL THE REACTANTS, AND WHAT WE END UP

WITH WHICH WE CALL THE PRODUCTS, AND WHAT WE’RE SAYING HERE THEN IN THE LAW OF

CONSERVATION OF MASS, WHATEVER MASS OF REACTANTS WE START WITH IS GOING TO BE

EQUAL TO THE MASS OF PRODUCTS THAT WE END UP WITH. WE ‘RE NOT GOING TO LOSE ANY,

WE’RE NOT GOING TO GENERATE ANY. WE’RE NOT GOING TO PRODUCE ANY. NOW LET’S LOOK

AT A PARTICULAR CHEMICAL EQUATION AND SEE WHAT KIND OF THINGS A CHEMICAL

EQUATION CAN TELL US. NOW ONE THING IS THAT ON THIS SIDE AS I SAY WE HAVE THE

REACTANTS, AND WE HA VE TO HAVE CORRECT CHEMICAL FORMULAS FOR EACH OF THE

REACTANTS INVOLVED. WE MUST HAVE CORRECT CHEMICAL FORMULAS FOR EACH OF THE

PRODUCTS THAT IS FORMED. WE ALSO IN SOME CASES IF WE KNOW THE INFORMATION WE

INCLUDE THE PHYSICAL STATE FOR THE REACTANTS AND THE PRODUCTS. SO WHAT WE’RE

GOING TO DO HERE IS WE’RE GOING TO PUT A LITTLE SUBSCRIPT “S” IS A , IS A

SOLID UNDER NORMAL CONDITIONS, AND THAT’S PROBABLY THE WAY WE WOULD BE USING IT

IN A CHEMICAL REACTION. ALRIGHT SO I PUT A LITTLE SUPERSCRIPT “S” TO REPRESENT SOLID.

WATER NORMALLY WE KNOW IS A LIQUID, AND SO WE PUT A LITTLE SUPERSCRIPT “L” TO GIVE

THE PHYSICAL STATE OF WATER. SODIUM , THAT’S THE CHEMICAL NAME FOR THIS

CHEMICAL COMPOUND OVER HERE, NAOH, THIS IS ACTUALLY GOING TO BE A . IT’S A

WATER SOLUTION, AND WHAT WE USE A “AQ” IN PARENTHESIS, WHICH STANDS FOR AQUEOUS.

WATER IS AQUA, AND SO AQUEOUS IS A WATER SOLUTION OF THE SODIUM HYDROXIDE, AND

FINALLY OVER HERE WE HAVE HYDROGEN. HYDROGEN REMEMBER WE TA LKED ABOUT SOME

OF THE GASES – HYDROGEN, OXYGEN, FLUORINE, ETC. HYDROGEN IS A GAS AT ROOM

TEMPERATURE, AND WE PUT A LITTLE THEN SUBSCRIPT “G”. SO WE CAN INCLUDE THAT KIND

OF INFORMATION. A LITTLE LATER IN CHAPTER THREE WE’LL ALSO INCLUDE SOME

INFORMATION PERTAINING TO ENERGY. WHETHER ENERGY IS PUT IN OR ENERGY IS TAKEN OF CHM 105 & 106 MO1 UNIT ONE, LECTURE EIGHT/ UNIT TWO, LECTURE ONE 11

A REACTION WE WILL SHOW THAT AS WELL, BUT FOR THE TIME BEING WE’LL LIMIT OURSELVES

TO THIS RIGHT HERE. NOW LET’S SEE WHAT THE EQUATION SAYS TO US, HOW WE COULD READ

THIS EQUATION. WELL ONE WAY WE COULD, OBVIOUSLY WE COULD SA Y THAT TWO SODIUM

ATOMS, IN OTHER WORDS A CHEMICAL EQUATION IS AN ABBREVIATED FORM FOR PUTTING

DOWN WHAT WE WOULD SAY IN NORMAL WORDS. WE CAN SAY TWO SODIUM REACT

WITH TWO WATER MOLECULES TO PRODUCE TWO MOLECULES OF SODIUM HYDROXIDE AND

ONE MOLECULE OF HYDROGEN. SO WHAT OUR CHEMICAL EQUATION DID WAS TOOK THAT

GROUP OF WORDS THEN AND PUT IT INTO SHORTHAND FASHION THAT W E CALL THE CHEMICAL

EQUATION. NOW WE KNOW THAT TYPICALLY WE DON’T DEAL IN ATOMS AND MOLECULES. OUR

BASIC UNIT OF MEASUREMENT IN CHEMISTRY AS WE MENTIONED IN CHAPTER TWO AND WILL

USE THROUGHOUT HE COURSE IS WHAT? THE MOLE. AND SO WE’RE GOING TO SAY THE SAME

THING NOW, BUT THIS TIME WE’RE GOING TO PUT IT IN THE UNITS THAT WE WOULD NORMALLY

TALK ABOUT A REACTION, WHICH WOULD BE MOLES, AND SO WE COULD THEN SAY TWO MOLES

– AND WE CAN ABBREVIATE IT, TWO MOLES OF NA, OR SODIUM, REACT WITH, WOOPS, REACT

WITH TWO MOLES OF WATER TO PRODUCE TWO MOLES OF SODIUM HYDROXIDE AND ONE MOLE

OF HYDROGEN. SO NOW WE’VE EXPRESSED IT IN TERMS OF MOLES RATHER THAN ON TERMS OF

ATOMS AND MOLECULES, WHICH OF COURSE IS A MORE REALISTIC WAY. BUT WE ALSO SAID AT

THE BEGINNING OF THIS THAT A CHEMICAL REACTION SHOULD SHOW THE LAW OF

CONSERVATION OF MASS. SO IN OTHER WORDS, WHATEVER I STARTED WITH MASS-WISE ON

THIS SIDE WE SHOULD END UP WITH MASS-WISE ON THAT SIDE. WELL LET’S TAKE A LO OK ONCE

HERE. THEN WHAT WE STARTED UP WITH AND WHAT WE ENDED UP WITH MASS-WISE ON THIS

SIDE WE HAD TWO SODIUMS, SO WE HAVE 2 X 22.99 GRAMS, SODIUM’S ATOMIC MASS PLUS

WATER. OKAY WE HAVE TWO UP THERE AND WATER WEIGHS, THE HYDROGEN WEIGHS

1.01 AND OXYGEN WEIGHS 16 SO THE WATER WOULD WEIGH 18.02 GRAMS, AND THERE ARE TWO

OF THEM, THAT’S WHY WE’RE MULTIPLYING BY TWO HERE FOR EACH. THIS SHOULD EQUAL

NOW TWO TIMES THE FORMULA MASS OF SODIUM HYDROXIDE, AND LET’S SEE WE WOULD

HAVE TO CALCULATE THAT. THAT WOULD BE 22.99 + 1.01 FOR THE HYDROGEN + 16 FOR THE

OXYGEN, THAT’D BE 24, WOULD BE 40.00 GRAMS + THE HYDROGEN, AND THERE’S ONLY ONE OF CHM 105 & 106 MO1 UNIT ONE, LECTURE EIGHT/ UNIT TWO, LECTURE ONE 12

THOSE AND IT WEIGHS 2.02 GRAMS. WELL IF WE MULTIPLY THIS OUT LET’S SEE WE WOULD END

UP WITH 45.98 GRAMS + 36.04 GRAMS = 80 GRAMS + 2.02 GRAMS, AND WE ADD THIS TOGETHER,

WE SEE THAT WE HAVE 82.02 GRAMS IS EQUAL TO 82.02 GRAMS. THE FACT IS THAT YES, MASS

HAS BEEN CONSERVED. WE HAVE SHOWN THAT WE HAVE THE SAME AMOUNT OF PRODUCT

FORMED ON A MASS STANDPOINT, AS WE HAD THEN REACTANTS TO START WITH. NOW, KEEP IN

MIND THAT IN A CHEMICAL REACTION THAT ATOMS HAVE A SPECIFIC MASS, AND IF ATOMS

HAVE A SPECIFIC MASS THEN IT WOULD SEEM LOGICAL THAT IF IN FACT MASS IS CONSERVED IN

A CHEMICAL REACTION THEN ATOMS MUST ALSO BE CONSERVED INA CHEMICAL REACTION,

AND SO WE HAVE AN ADDITIONAL LAW HERE WHICH IS THE LAW OF CONSERVATION OF ATOMS.

NOW AN INTERESTING CHEMICAL REACTION THAT IS INDICATED IN THE TEXT IS ONE THAT I

MENTIONED THE OTHER DAY. HERE IS THE ELEMENT SODIUM. SODIUM IS A METAL, IT’S A SOFT

METAL, LOOKS LIKE A BLOCK OF BUTTER HERE, YOU CAN CUT IT WITH A BUTTER KNIFE

ACTUALLY IT’S THAT SOFT. IT’S A SHINY, SOFT METAL, AND WE HAVE CHLORINE GAS HERE

AND CHLORINE GAS IS A YELLOWISH LOOKING GAS, YELLOWISH-GREEN GAS, AND AS I

MENTIONED BEFORE, BOTH OF THESE ARE VERY TOXIC TO US, BUT IF WE PUT THE TWO

TOGETHER WE GET A SPECTACULAR CHEMICAL REACTION WITH THE RELEASE OF A LOT OF

HEAT, LIGHT AND SPARKING, AND THAT OCCURRING, AND WE END UP WITH A PRODUCT OF

SODIUM WHICH AS I INDICATED IS ESSENTIAL FOR OUR DIET. SO A CHEMICAL

REACTION, THIS PARTICULAR CHEMICAL REACTION REALLY DOES EMPHASIZE WHAT HAPPENS

IN CHEMICAL CHANGE. WE HAVE A MAJOR CHANGE IN THE PROPERTIES OF THE MATTER

INVOLVED, UNLIKE WE DO IN A PHYSICAL CHANGE. NOW IF, IN FACT, ATOMS ARE GOING TO BE

THEN CONSERVED IN A CHEMICAL REACTION THEN WE CAN CARRY OUT A PROCESS WHICH WE

CALL BALANCING THE CHEMICAL EQUATION. W E STARTED IN THE PREVIOUS ONE, THE

EQUATION WAS ALL BALANCED. WE HAD TWO M OLES OF SODIUM PLUS TWO MOLES OF WATER

TO GET TWO MOLES OF SODIUM HYDROXIDE AND ONE MOLE OF HYDROGEN. THAT EQUATION

WAS ALREADY BALANCED, BUT IF WE START WITH REACTANTS AND PUT THEM TOGETHER AND

MAKE PRODUCTS AND KNOW WHAT THEY ARE, TO WRITE THE CHEMICAL EQUATION WE HAVE

TO BALANCE IT. WE HAVE TO BALANCE THE ATOMS, AND SO LET’S TAKE A LOOK AT HOW WE CHM 105 & 106 MO1 UNIT ONE, LECTURE EIGHT/ UNIT TWO, LECTURE ONE 13

GO ABOUT THE PROCESS. ANYONE HERE THAT HAPPENS TO BE AN ACCOUNTING MAJOR THIS

SHOULD BE RIGHT UP YOUR ALLEY BECAUSE ALL WE’RE DOING IS A BOOKKEEPING PROCESS.

HOW MANY ATOMS DO WE HAVE IN, HOW MANY DO WE GET OUT, AND WE HAVE TO MAKE SURE

THAT THEY BALANCE. SO THE FIRST ONE HERE, THERE ARE SOME RULES IN CHAPTER THREE,

EARLY IN CHAPTER THREE, THAT INDICATE THE ORDER THAT ONE SHOULD BALANCE THE

ELEMENTS IN THE EQUATION, AND THESE ARE FOUND ON PAGE 81. WE START FIRST OF ALL

WITH ATOMS OTHER THAN OXYGEN OR HYDROGEN, AND WE USUALLY START WITH THE

ELEMENTS FURTHEST TO THE LEFT IN THE PERIODIC CHART IF WE HA VE SOME CHOICE - LIKE

THOSE THAT WE CALL THE . THEN WE GO ON TOT HE OTHER ELEMENTS AND WE FINISH

BY BALANCING HYDROGEN AND THEN OXYGEN IF WE NEED TO DO SO. I THINK WHAT ‘M GOING

TO DO IS I’M GOING TO SKIP THIS ONE A SECOND. THIS IS PROBLEM NUMBER 2A AT THE END OF

THE CHAPTER. I’M GOING TO GO AHEAD AND DO 2C FIRST. WE HAVE CARBON, HYDROGEN, AND

OXYGEN TO BALANCE, A ND WE HAVE TO HAVE THE SAME NUMBER OF ATOMS OF THOSE OVER

ON THIS SIDE AS WE HAVE ON THIS SIDE. SO THE FIRST ELEMENT THAT WE’RE GOING TO LOOK

TO BALANCE THEN IS THE CARBON, BECAUSE IT’S NOT ONE OF THE HYDROGEN OR

WHICH WE’RE GOING TO KEEP FOR THE END. SO WE LOOK ON THIS SIDED AND WE SAY ALRIGHT

HOW MANY ATOMS OF CARBON DO WE HAVE ON THE. HOW MANY ATOMS OF CARBON DO WE

HAVE ON THE REACTANT SIDE? THREE RIGHT? CHEMICAL FORMULA C3 TELLS US WE HAVE

THREE CARBON ATOMS. NOW WE HAVE TO HAVE THREE CARBON ATOMS ON THE RIGHT-HAND

SIDE THEN, ON THE PRODUCT SIDE. WE SEE WE ONLY HAVE ONE, SO SOMEH OW WE HAVE TO

GET THREE OVER HERE. ALRIGHT NOW WHAT W E’RE GOING TO DO THEN IS WE’RE GOING TO

PUT A CO-EFFICIENT THREE IN FRONT OF THE CARBON DIOXIDE, IN FRONT OF THE CO2. YOU

CANNOT BALANCE BY CHANGING SUBSCRIPTS, THOSE SUBSCRIPTS ARE THE CHEMICAL

FORMULA OF THE SUBSTANCE. IF YOU CHANGE THE SUBSCRIPT YOU’RE MAKING IT A

DIFFERENT CHEMICAL, AND THAT’S NOT WHAT YOU’RE ASKING FOR. WE NOW HAVE THE

CARBONS BALANCED. NOW WE TURN TO THE HYDROGENS. ALRIGHT, AND SO WE SEE THAT WE

HAVE SIX HYDROGENS ON THIS SIDE AND WE’RE ONLY SHOWING TWO OVER HERE. WHAT

WOULD BE THE LEAST COMMON NUMBER FOR SIX AND TWO? WELL LEAST COMMON NUMBER CHM 105 & 106 MO1 UNIT ONE, LECTURE EIGHT/ UNIT TWO, LECTURE ONE 14

WOULD BE SIX, AND SO WE ALREADY HAVE THE SIX, WE’RE DOING THE SIX ONCE, BUT TWO

WOULD GO INTO SIX THREE TIMES OKAY. SO WHAT YOU WERE SAYING IS RIGHT, WE NEED TO

MULTIPLY THIS ONE BY THREE SO THAT WE HA VE SIX HYDROGENS ON EACH SIDE. ALRIGHT, SO

FAR WE HAVE EVERYTHING BALANCED THAT WAY. NOW WE TAKE A LOOK AT THE OXYGENS.

ALRIGHT, OVER HERE WE HAVE TWO OXYGENS. OVER ON THIS SIDE WE HAVE THREE TIMES

TWO, SO WE HAVE SIX OXYGENS HERE, AND THEN WE HAVE THREE TIM ES ONE, SO WE HAVE

THREE MORE OXYGENS HERE, SO WE HAVE A TOTAL OF NINE OXYGEN ATOMS ON THIS SIDE.

NOW WE MUST HAVE NINE OXYGEN ATOMS ON THE OTHER SIDE AS WELL. W ELL HOW ARE WE

GOING TO GET NINE OXYGEN ATOMS ON THIS SIDE OVER HERE? WELL EACH OXYGEN

MOLECULE CONTAINS HOW MANY OXYGEN ATOMS? TWO. SO IF WE WERE TO TAKE NINE

HALVES, OR IN OTHER WORDS IF WE TOOK FOUR AND A HALF OXYGEN MOLECULES - WE CAN

TAKE FRACTIONAL MOLECULES, WE CANNOT TAKE FRACTIONAL ATOMS. BUT SEEING THAT

OXYGEN HAS TWO ATOMS PER MOLECULE I CAN CUT IT IN HALF AND I CAN TAKE ONE OXYGEN

ATOM. REMEMBER ALL WE’RE DOING IS SHOWING A BALANCE IN TERMS OF ATOMS. SO WE

COULD SAY NINE HALVES OXYGEN MOLECULES, AND AT THIS POINT WE HAVE A BALANCE FOR

THE CARBON, WE HAVE A BALANCE FOR THE HYDROGEN, WE HAVE A BALANCE FOR THE

OXYGEN. HOWEVER, THE RULE THAT WE’RE GOING TO USE IS WE DO NOT INCLUDE FRACTIONS

IN BALANCED CHEMICAL EQUATIONS. SO I WANT TO GET RID OF THIS FRACTION. REMEMBER,

IT IS NOW BALANCED, BUT I CAN DO ANYTHING TO THAT EQUATION I WANT AS LONG AS I WERE

TO DO IT TO EVERYTHING IN THERE. IT WOULDN’T CHANGE THE BALANCE. SO IF I WANT TO

GET RID OF THIS TWO, IF I WANT TO GET RID OF THAT HALF OUTTA HERE WHAT COULD I DO? I

COULD MULTIPLY THIS WHOLE THING THROUGH BY TWO. SO IF I DO THAT MY FINAL BALANCE

IS GOING TO LOOK LIKE THIS: 2-CH3H6 + 9-O2 = 6-CO2 + 6-H2O. SO NORMALLY WE SAY THAT A

BALANCED CHEMICAL EQUATION SHOULD SHOW THE SMALLEST WHOLE NUMBER BALANCE

THAT WE CAN HAVE. WE SHOULD NOT INCLUDE FRACTIONS IN THE BALANCED EQUATION.

WELL I THINK WE’RE GOING TO GO AHEAD AND WRAP THIS UP TODAY AND WE WILL CONTINUE

ON WITH SOME MORE BA LANCING ON OUR NEXT LECT URE AND WE WILL CONTINUE ON IN OUR

DISCUSSION OF CHAPTER THREE.