CHEM 105 & 106 UNIT ONE, LECTURE SEVEN 1 IN OUR PREVIOUS LECTURE WE HAVE BEEN TALKING ABOUT IN CHAPTER TWO NOW THE

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1 CHEM 105 & 106 UNIT ONE, LECTURE SEVEN 1 CHM 105/106 Program 7: Unit 1 Lecture 7 IN OUR PREVIOUS LECTURE WE HAVE BEEN TALKING ABOUT IN CHAPTER TWO NOW THE MEASUREMENT IN CHEMISTRY, THE MOLE, MOLAR MASS, WE TALKED ABOUT HOW WE CAN DETERMINE PERCENT COMPOSITION. YESTERDAY WE EVEN LOOKED AT THEN, HOW WE CAN USE PERCENT COMPOSITION TO DETERMINE AN EMPIRICAL FORMULA. TODAY WE ARE GOING TO GO BACK AND LOOK AGAIN A LITTLE BIT AT HOW WE CAN APPLY THIS CONCEPT OF THE MOLE AND ITS RELATIONSHIP TO MASS, WE ARE ALSO GOING TO TAKE A LOOK AT A COUPLE MORE EMPIRICAL FORMULA CALCULATIONS AND WE WILL GO ON TO LOOK AT WHAT WE CALL THE MOLECULAR FORMULA, SO WE WILL START TODAY THEN BY TAKING A LOOK AT PROBLEM NUMBER NINE AT THE END OF THE CHAPTER, NINE PART C ASKS US TO CALCULATE THE NUMBER OF CHLORINE ATOMS IN TWELVE POINT SIX (12.6) GRAMS OF THE COMPOUND CCL2F2, THE CHEMICAL NAME OF THAT COMPOUND IN DICHLORAL DIFLORAL METHANE, YOU DON'T NEED TO REMEMBER THAT, BUT IT IS A CHEMICAL THAT WE CALL FREON, IT IS THE REFRIGERANT THAT IS USED, THAT HAS BEEN COMMERCIALLY USED FOR MANY YEARS, FREON. IT TELLS US THAT WE HAVE A MOLAR WEIGHT OR A FORMULA WEIGHT OF ONE HUNDRED AND TWENTY POINT NINE ONE (120.91) GRAMS, SO WE ALREADY HAVE THAT PIECE OF INFORMATION. NOW LETS ALSO REMEMBER WHAT ELSE WE KNOW ABOUT MATERIALS, WE KNOW THAT ONE MOLE OF CCL2F2 WOULD HAVE SIX POINT ZERO TWO (6.02) TIMES TEN TO THE TWENTY THIRD MOLECULES OF THAT SUBSTANCE, AVAGADRO'S NUMBER. IN ADDITION TO THAT WE ALSO KNOW FROM THE CHEMICAL FORMULA THAT THERE ARE TWO CHLORINE ATOMS IN EACH MOLECULE, SO WE HAVE THREE THINGS THEN THAT WE CAN STATE, WE KNOW ITS FORMULATED MASS, IT IS CALCULATED AND GIVEN TO US IN THE PROBLEM. WE KNOW THAT A MOLE OF THAT SUBSTANCE HAS THAT NUMBER OF MOLECULES AND WE KNOW HOW MANY ATOMS OF CHLORINE THERE ARE IN A MOLECULE, SO WE WILL HAVE THREE CONVERSION FACTORS THAT WE WILL BE USING TO CHANGE FORM GRAMS OF THIS COMPOUND TO ATOMS OF CHLORINE AS THE PROBLEM INDICATES. SETTING THE PROBLEM UP THEN, AND WHAT I AM GOING TO DO HERE BECAUSE WE WILL NEED A LITTLE BIT MORE SPACE, I AM GOING TO USE ANOTHER TRANSPARENCY HERE AND JUST SET IT SIDEWAYS SO I HAVE A LITTLE BIT MORE ROOM TO WRITE ACROSS. ALRIGHT, SO WHAT WE WANT

2 CHEM 105 & 106 UNIT ONE, LECTURE SEVEN 2 HERE ARE CHLORINE ATOMS, NOW LET ME POINT OUT THAT I AM GOING TO SET THE PROBLEM UP IN ONE CONTINUOUS SET UP, YOU CAN TAKE EACH INDIVIDUAL STEP AS A SEPARATE MATH CALCULATION AND THEN CARRY THE INFORMATION THAT YOU JUST CALCULATED INTO THE SECOND STEP AND INTO THE THIRD STEP OR THE FOURTH STEP WHATEVER NUMBER OF STEPS NECESSARY, SO YOU DON'T HAVE TO SET IT ALL UP IN ONE CONTINUOUS LONG EQUATION, IF YOU PREFER YOU CAN CERTAINLY DO IT PIECEMEAL INSTEAD. I WILL SET IT UP THIS WAY FIRST OF ALL CHLORINE ATOMS THEN ARE EQUAL TO TWELVE POINT SIX (12.6) GRAMS OF CCL2F2, MULTIPLIED BY ONE MOLE OF CCL2F2 PER ONE HUNDRED AND TWENTY POINT NINE ONE (120.91) GRAMS, NOW AT THIS POINT WE WILL HAVE ELIMINATED THEN GRAMS FROM OUR PROBLEM. WE HAVE THE NUMBER OF MOLES, THE NUMBER OF MOLES IS SORT OF GOING TO BE CENTRAL TO ALL OF OUR PROBLEM SOLVING OR CERTAINLY TO A MAJORITY OF OUR PROBLEM SOLVING IN CHEMISTRY, QUITE FREQUENTLY ONE OF THE FIRST THINGS WE ARE GOING TO ASK IN ANY PROBLEM RELATING HOW MANY MOLES? DO WE NEED TO FIND THE NUMBER OF MOLES? WILL FINDING THE NUMBER OF MOLES HELP US SOLVE THE PROBLEM? AND IN MOST CASES IT WILL, AND SO THAT IS SORT OF THE CENTRAL PART OF OUR PROBLEM. NOW, WE WANT TO CHANGE FROM MOLES TO MOLECULES, AND WE HAVE AVAGADRO'S NUMBER TO USE TO DO THAT, AND SO NOW WE CAN MULTIPLY IT BY SIX POINT ZERO TWO THREE (6.023) TIMES TEN TO THE TWENTY THIRD MOLECULES, I AM STILL GOING TO RUN OUT OF SPACE, PER ONE MOLE OF CCL2F2 AND NOW THE MOLES WILL CANCEL, BUT I AM STILL NOT AT ATOMS AND SO THE NEXT STEP WOULD BE TO PUT IN A CONVERSION FACTOR THAT WOULD CONVERT FROM MOLECULES OF CCL2 TO ATOMS OF CHLORINE, WELL I GUESS WHAT I WILL DO IS I WILL DO IT THIS WAY FOR THIS TIME. WE ARE GOING TO MULTIPLY NOW BY TWO CHLORINE ATOMS PER ONE CCL2F2 MOLECULE AND SO THE MOLECULE WILL CANCEL AND OUR FINAL UNIT LEFT IN THE PROBLEM IS CHLORINE ATOMS. NOW, OBVIOUSLY SOME OF IT IS ALREADY OFF THE PAGE OVER THERE SO WE CAN'T SEE IT ALL, BUT, WE DO HAVE THEN THE THREE STEPS AND OUR UNITS THAT WE FINALLY HAVE, CHLORINE ATOMS IS IN FACT WHAT WE ASKED FOR IN THE PROBLEM. NOW WE WOULD DO THE MATHEMATICAL STEPS THEN TO GET US TO A NUMERIC ANSWER AND I WILL SLIDE THIS BACK JUST A LITTLE BIT SO WE CAN AT LEAST SEE THE NUMBERS THAT WE ARE GOING TO INVOLVE IN OUR CALCULATION. SO, WE ARE GOING TO HAVE THEN TWELVE POINT SIX (12.6) GRAMS MULTIPLIED BY SIX POINT ZERO TWO THREE (6.023) TIMES TEN TO THE TWENTY

3 CHEM 105 & 106 UNIT ONE, LECTURE SEVEN 3 THIRD, THEN AS I SUGGESTED YESTERDAY SO THAT NO MATTER WHICH WAY YOU PROCEED ON THE CALCULATOR IN BETWEEN EACH MATH STEP IF YOU HIT THE EQUAL SIGN, YOU CAN'T MAKE AN ERROR IN TERMS OF THE LOGIC YOU PUT INTO THE CALCULATOR, SO I AM GOING TO HIT EQUAL NOW AND THEN I AM GOING TO HIT MULTIPLY BY TWO AND I AM GOING TO HIT THE EQUAL AND I AM NOW GOING TO HIT DIVIDED BY ONE HUNDRED AND TWENTY POINT NINE ONE (120.91) AND EQUAL AGAIN AND I HAVE NOW PUT IN ALL OF THE NUMERIC VALUES AND I HAVE A FINAL ANSWER THEN OF ONE POINT TWO (1.2) NOW LOOKING AT OUR NUMBERS HERE WE HAVE ONLY THREE SIGNIFICANT FIGURES HERE, FIVE, FOUR AND UNLIMITED, I MEAN THERE ARE EXACTLY TWO CHLORINE ATOMS PER MOLECULE, THERE CAN'T BE ANY FRACTIONAL AMOUNT SO THAT IS AN EXACT NUMBER, SO WE ARE LIMITED TO THREE SIGNIFICANT FIGURES WHICH GIVES US THEN A FINAL ANSWER OF ONE POINT TWO SIX (1.26) TIMES TEN TO THE TWENTY THIRD CHLORINE ATOMS. NOW, AS I SAID THAT IS ONE WAY YOU CAN PROCEED SET IT ALL UP IN A PROBLEM IF YOU HAVE ROOM ON THE PAPER AND FEEL COMFORTABLE TO DO THAT, THAT IS CERTAINLY A LOGICAL WAY TO PROCEED. HOWEVER, ALSO AS I SAID YOU COULD SET THE PROBLEM UP IN THREE INDIVIDUAL STEPS, LET ME JUST QUICKLY GO AHEAD, WE WON'T DO THE MATH PART, BUT, LETS LOOK AT HOW WE COULD PROCEED, THE FIRST QUESTION I WOULD ASK BASED ON THE INFORMATION WE HAVE IS HOW MANY MOLES OF CCL2F2 I HAVE, AND IF I DID THAT I WOULD SAY ALRIGHT WE HAVE TWELVE POINT SIX (12.6) GRAMS OF CCL2F2 MULTIPLIED BY ONE MOLE OF CCL2F2 PER ONE HUNDRED AND TWENTY POINT NINE ONE (120.91) GRAMS OF THE COMPOUND. NOW, AT THIS POINT THEN I WOULD MATHEMATICALLY CALCULATE THE NUMBER OF MOLES OF CCL2. MY NEXT STEP WOULD BE TO ASK THE QUESTION WELL NOW THAT I KNOW HOW MANY MOLES, HOW MANY MOLECULES DO I HAVE? AND SO I WOULD NOW SET UP THE NEXT PART OF THE PROBLEM, I WOULD SAY MOLECULES OF CCL2F2 WOULD BE EQUAL TO THE NUMBER OF MOLES, WHICH WE JUST CALCULATED, MULTIPLIED TIMES SIX POINT ZERO TWO THREE (6.023) TIMES TEN TO THE TWENTY THIRD MOLECULES AGAIN, I AM RUNNING OUT OF SPACE HERE, PER ONE MOLE AND SO OUR UNIT MOLES WOULD CANCEL, CALCULATED NOW WE WOULD HAVE THE NUMBER OF MOLECULES AND THEN I KNOW THAT THE QUESTION IS THE NUMBER OF ATOMS, I WOULD THEN ADD THE FINAL STEP CHLORINE ATOMS WOULD BE EQUAL TO THE NUMBER OF MOLECULES THAT WE JUST CALCULATED MULTIPLIED BY TWO CHLORINE ATOMS PER ONE MOLECULE, SO WE CAN THEN SET IT UP

4 CHEM 105 & 106 UNIT ONE, LECTURE SEVEN 4 IN INDIVIDUAL STEPS AND IF THIS SEEMS EASIER TO YOU THEN TRYING TO PUT CONTINUOUS CANCELLATION CONVERSION UNITS CERTAINLY FEEL FREE TO DO SO. THERE IS NO ONE CORRECT WAY TO DO A PROBLEM, YOU HAVE TO DECIDE WHAT MAKES SENSE TO YOU, THE EASIEST APPROACH FOR YOU AND THAT THEN IS THE WAY YOU PROCEED. THE NEXT PROBLEM THAT I WANT TO LOOK AT TAKES US BACK TO WHAT WE WERE TALKING ABOUT IN THE PREVIOUS LECTURE ON CALCULATING EMPIRICAL FORMULAS REMEMBER THAT AN EMPIRICAL FORMULA IS THE SMALLEST WHOLE NUMBER AND THAT IS CRITICAL, WHOLE NUMBER RATIO OF ATOMS IN A CHEMICAL COMPOUND. IF WE HAVE EITHER MASS INFORMATION OR PERCENT INFORMATION GIVEN TO US WE CAN PROCEED, IT DOESN'T HAVE TO BE IN PERCENT, IT DOESN'T HAVE TO MASS, IT CAN BE IN EITHER, OF COURSE, ALL OF THE ELEMENTS HAVE TO BE EXPRESSED IN THE SAME MANNER. THE EXAMPLE THAT WE DID IN THE PREVIOUS LECTURE, THE INFORMATION WAS GIVEN TO US IN THE NUMBER OF GRAMS IN EACH OF THE ELEMENTS, THIS EXAMPLE GIVES US THE INFORMATION FOR THE ELEMENTS IN TERMS OF PERCENT, BUT YOU SEE IF I WERE TO TAKE A ONE HUNDRED GRAM SAMPLE OF THIS COMPOUND I WOULD HAVE THEN FORTY TWO POINT NINE GRAMS OF ZINC. I WOULD HAVE TWENTY POINT THREE GRAMS OF PHOSPHOROUS AND I WOULD HAVE THIRTY SIX POINT EIGHT GRAMS OF OXYGEN. I COULD TAKE ANY SIZE SAMPLE THAT I WANTED TO AND SO WHETHER IT IS IN PERCENT OR MASS WE PROCEED EXACTLY THE SAME WAY. THE FIRST STEP IS TO CHANGE FROM A MASS RATIO, WHICH IS WHAT THIS IS, THAT IS PERCENT IN MASS TO A MOLE RATIO AND SO WHAT WE ARE GOING TO DO IS CALCULATE THE NUMBER OF MOLES OF EACH OF THESE ELEMENTS, SO MOLES OF ZINC WILL BE EQUAL TO FORTY TWO POINT NINE AND WE WILL MAKE IT GRAMS NOW, WE WILL TAKE A HUNDRED GRAM SAMPLE TIMES ONE MOLE OF ZINC PER, ALRIGHT REMEMBER THAT A MOLE OF AN ELEMENT IS EQUAL TO ITS ATOMIC MASS, EXPRESSED IN GRAMS, SO LOOKING UP THERE AT ZINC, SIXTY FIVE POINT THREE EIGHT (65.38) GRAMS. WE NOW HAVE MOLES OF ZINC. WE DO THE SAME THING NOW FOR PHOSPHOROUS SO FOR PHOSPHOROUS WE WILL HAVE TWENTY POINT THREE GRAMS MULTIPLIED BY ONE MOLE OF PHOSPHOROUS AND LOOKING AT THE PERIODIC TABLE AT THE END OF THE BOOK, WE FIND THIRTY POINT NINE SEVEN (30.97) GRAMS FOR THAT AND OXYGEN THIRTY SIX POINT EIGHT (36.8) MULTIPLIED BY ONE MOLE OF OXYGEN AGAIN REFERRING TO ITS ATOMIC MASS, SIXTEEN POINT ZERO-ZERO (16.00) GRAMS AND SO WE NOW HAVE EVERYTHING IN TERMS OF A MOLE RATIO. WELL LETS GO AHEAD AND

5 CHEM 105 & 106 UNIT ONE, LECTURE SEVEN 5 CALCULATE THIS SET OF NUMBERS SO WE HAVE FORTY TWO POINT NINE (42.9) DIVIDED BY SIXTY FIVE POINT THREE EIGHT (65.38) AND THE FIRST ONE THEN GIVES US A VALUE OF ZERO POINT SIX FIVE SIX (0.656) TWO POINT THREE DIVIDED BY THREE ZERO POINT NINE SEVEN (30.97) WE HAVE A VALUE OF ZERO POINT SIX FIVE-FIVE (0.655) AND THIRTY SIX POINT EIGHT (36.8) DIVIDED BY SIXTEEN GIVES US THE VALUE OF TWO POINT THREE ZERO (2.30). NOW ONCE WE REACH THIS POINT WE NOW HAVE A RATIO EXPRESSED IN TERMS OF MOLES OF ATOMS BUT OBVIOUSLY IT IS NOT THE SMALLEST WHOLE NUMBER, NONE OF THOSE ARE WHOLE NUMBERS. SO THE NEXT STEP WE TAKE TO GUARANTEE THAT WE HAVE AT LEAST ONE IN A WHOLE NUMBER FORM IS WE ARE NOW GOING TO DIVIDE EACH OF THE NUMBER WE JUST CALCULATED BY THE SMALLEST NUMBER THAT WE HAD, AND THE NUMBERS THAT WE HAVE HERE THAT MEANS WE ARE GOING TO DIVIDE BY ZERO POINT SIX FIVE-FIVE (0.655), DIVIDE BY ZERO POINT SIX FIVE-FIVE (0.655), DIVIDE BY ZERO POINT SIX FIVE-FIVE (0.655). WHEN WE DO SO OBVIOUSLY WE ARE GOING TO GET ONE AND ONE FOR THE FIRST TWO AND THEN DIVIDING TWO POINT THREE (2.3) BY POINT SIX FIVE-FIVE (.655) WE GET THREE POINT FIVE (3.5) ALRIGHT, NOW, WE HAVE TWO OF THE THREE ELEMENTS EXPRESSED AS WHOLE NUMBERS. OBVIOUSLY THE OXYGEN IS NOT, AND IF WE HAVE A FRACTIONAL AMOUNT WE HAVE TO BEGIN LOOKING FOR A MULTIPLIER. I SAID THAT IF WE ARE WITHIN PLUS OR MINUS A TENTH, WE CAN OFTEN ROUND OFF ONLY AFTER WE HAVE MADE THIS DIVISION, WE CANT ROUND OFF BACK HERE, WE CAN ONLY ROUND OFF AFTER WE HAVE DONE THE DIVISION. WELL IN THIS PARTICULAR CASE I THINK THAT IT IS SOMEWHAT OBVIOUS THAT IF WE WERE TO NOW MULTIPLY SO WE ARE GOING TO ADD ONE FINAL STEP HERE. IF WE WERE TO MULTIPLY EACH OF THESE BY TWO WE WILL END UP WITH WHOLE NUMBERS OF TWO, TWO, AND SEVEN AND SO THE EMPIRICAL FORMULA FOR THIS COMPOUND WOULD BE ZN2P2O7 AND THAT IS THE EMPIRICAL FORMULA OF THIS COMPOUND CONTAINING ZINC, PHOSPHOROUS AND OXYGEN. AN EXAMPLE WE DID IN THE PREVIOUS LECTURE AFTER WE MADE THIS DIVISION WE ENDED UP WITH ALL EVEN NUMBERS, I BELIEVE THE EXAMPLE WE HAD IT CAME OUT ONE-ONE AND TWO OR ONE POINT NINE EIGHT (1.98) WHICH WE COULD ROUND OFF JUSTIFIABLY TO TWO, BUT IF IT COMES OUT THEN SOMETHING GREATER THAN A TENTH THEN WE WILL HAVE TO LOOK FOR A MULTIPLIER. NOW, THE THINGS THAT WE ARE GOING TO MOST TYPICALLY SEE AS DECIMALS OUT HERE WOULD BE SOMETHING LIKE POINT THREE-

6 CHEM 105 & 106 UNIT ONE, LECTURE SEVEN 6 THREE (.33) IF WE SEE THAT WHAT ARE WE GOING TO MULTIPLY IT BY? WE ARE GOING TO MULTIPLY BY A THREE TO GET TO A WHOLE NUMBER ONE THIRD TIMES THREE OR WE MIGHT SEE POINT TWO FIVE (.25) IF WE HAVE A POINT TWO FIVE (.25) WHAT WOULD WE PROBABLY USE AS OUR MULTIPLIER? TO GET IT TO A WHOLE NUMBER? FOUR OK, IF WE HAD POINT FIVE (.5) OBVIOUSLY WE HAVE ALREADY DONE THAT IT IS TWO, IF WE HAVE PINT SIX-SIX (.66) WHAT ARE WE GOING TO DO? MULTIPLY IT BY THREE, ONE THIRD TIMES THREE WOULD GIVE US A TWO. AND LETS SEE, THE OTHER COMMON ONE MIGHT BE POINT SEVEN FIVE (.75) WHAT COULD WE MULTIPLY TO GET TO A WHOLE NUMBER THERE? BACK TO FOUR, RIGHT, TIMES FOUR. SO, THOSE ARE THE ONES THAT WE ARE LIKELY TO ENCOUNTER IN EMPIRICAL FORMULA CALCULATIONS. ALRIGHT, ANY QUESTION ON ANY STEP OF THAT PARTICULAR ONE? NOW, IF WE HAVE CALCULATED AN EMPIRICAL FORMULA AND IF WE KNOW THE MOLAR MASS OF THE COMPOUND WE CAN CALCULATE, BUT WE CAN DETERMINE THE MOLECULAR FORMULA, WE CAN DETERMINE ACTUAL CHEMICAL FORMULA. REMEMBER THE EMPIRICAL FORMULA MAY NOT BE HOW THE MOLECULE EXISTS IT IS JUST THE SIMPLEST WHOLE NUMBER RATIO OF THE ELEMENTS. NOW, HOW DO WE GO ABOUT DETERMINING THE MOLECULAR FORMULA THEN IF WE HAVE THIS INFORMATION GIVEN? WELL THE MOLECULAR FORMULA IS EQUAL TO THE EMPIRICAL FORMULA MULTIPLIED BY MOLAR MASS DIVIDED BY EMPIRICAL FORMULA MASS. AND THIS WILL COME OUT TO BE A WHOLE NUMBER OR IT HAS TO COME OUT TO BE A WHOLE NUMBER AND AGAIN WE MIGHT DO SOME SMALL ROUNDING OFF BECAUSE OF SCIENTIFIC MEASUREMENTS SO IN THIS PARTICULAR EXAMPLE PROBLEM THAT WE ARE LOOKING AT HERE THEN, IT ASKS US TO CALCULATE THE MOLECULAR FORMULA PROBLEM, SEVENTEEN AT THE END OF THE CHAPTER AND SO WE WOULD SAY ALRIGHT THE MOLECULAR FORMULA, THEN WHAT THE MOLECULE ACTUALLY IS THEN AND I AM JUST GOING TO ABBREVIATE THAT M.F. MOLECULAR FORMULA WOULD BE EQUAL TO THE EMPIRICAL FORMULA CH2 MULTIPLIED BY THE MOLAR MASS SEVENTY POINT ONE FIVE (70.15) DIVIDED BY THE EMPIRICAL FORMULA MASS SO WE HAVE TO CALCULATE HOW MUCH THIS WEIGHS. WELL A CARBON IS TWELVE POINT ZERO ONE FOR A CARBON AND TWO HYDROGENS WOULD BE TWO POINT ZERO TWO (2.02) TWO TIMES ONE POINT ZERO ONE (1.01) AND SO WE HAVE FOURTEEN POINT ZERO THREE (14.03) FOR THE EMPIRICAL MASS. WE COME BACK UP HERE FOURTEEN POINT ZERO (14.03) NOW SEVENTY POINT ONE FIVE DIVIDED BY FOURTEEN POINT ZERO THREE IS EQUAL TO WHAT? IT IS EQUAL TO FIVE

7 CHEM 105 & 106 UNIT ONE, LECTURE SEVEN 7 AND SO OUR MOLECULAR FORMULA IS THEN CH2 MULTIPLIED BY FIVE. NOW THE FIVE IS USED TO MULTIPLY THE EXPONENTS SO OUR FINAL MOLECULAR FORMULA IS C5H10 SO THE ACTUAL MOLECULE CONTAINS FIVE CARBONS AND TEN HYDROGENS. THE EMPIRICAL FORMULA SHOWS US A RATIO OF ONE TO TWO CARBON TO HYDROGEN ATOMS. OBVIOUSLY THAT IS FIVE TO TEN BUT THE ACTUAL MOLECULE, THE ACTUAL CHEMICAL WE ARE LOOKING AT THEN ACTUALLY HAS A FORMULA OF C5H10 NOT CH2. TO GET TO THIS POINT HOWEVER ONE HAS TO BE ABLE TO EXPERIMENTALLY DETERMINE THE MOLAR MASS AND AS WE GO ON WE ARE GOING TO LOOK AT A COUPLE OF CHAPTERS ON HOW WE CAN DO THIS EXPERIMENTALLY. WE ARE GOING TO LOOK AT ONE METHOD THAT WE CAN USE IN THE CHAPTER ON SOLUTIONS WE ARE GOING TO LOOK AT ANOTHER METHOD WE CAN USE IN LABORATORY WORK WE ARE GOING TO LOOK AT A THIRD METHOD WE CAN USE FROM THE STANDPOINT OF ACID BASED REACTIONS SO WE ARE GOING TO LEARN HOW WE CAN DO THIS PART BUT FOR NOW ALL WE CAN SAT IS IF WE HAVE THIS GIVEN AND WE HAVE EITHER THE EMPIRICAL FORMULA OR THE INFORMATION ON THE PERCENT COMPOSITION WE CAN IN FACT DETERMINE THEN THE MOLECULAR FORMULA. PROBLEM NUMBER NINETEEN THEN COMBINES THESE TWO SO WE WILL TAKE A QUICK LOOK AT IT, GIVES US THE CHEMICAL COMPOSITION OF THE COMPOUND, GIVES US THE MOLAR MASS OF THE COMPOUND, AND ASKS US TO DETERMINE THE MOLECULAR FORMULA. TO DO SO WE MUST FIRST DETERMINE THE EMPIRICAL FORMULA. SO WE ARE GOING TO DO THIS JUST LIKE WE DID IN THE PREVIOUS ONE WE ARE GOING TO START WITH CARBON, WE HAVE EIGHTY FIVE POINT SIX ZERO (85.60) MULTIPLIED BY ONE MOLE OF CARBON OVER TWELVE POINT ZERO ONE (12.01) GRAMS SO THE GRAMS CANCEL THEN HYDROGEN FOURTEEN POINT FOUR ZERO (14.40) GRAMS TIMES ONE MOLE OF HYDROGEN OVER ONE POINT ZERO ONE (1.01) GRAMS AND WE MATHEMATICALLY THEN NEED TO DETERMINE THIS SO WE HAVE EIGHTY FIVE POINT SIX (85.6) DIVIDED BY TWELVE POINT ZERO ONE (12.01) AND THE FIRST ONE I GET THERE IS SEVEN POINT ONE TWO SEVEN (7.127)AND WE HAVE FOURTEEN POINT FOUR (14.4) DIVIDED BY ONE POINT ZERO ONE (1.01) AND WE GET FOURTEEN POINT TWO FIVE (14.25) IF WE DIVIDE EACH BY THE SMALLEST THAT WE OBTAINED, SO THIS IS OUR SECOND STEP WE GET A ONE AND A TWO AND SO OUR EMPIRICAL FORMULA FOR THIS KIND OF COMPOUND IS THE SAME AS WHAT WE SAW IN SEVENTEEN A, THE EMPIRICAL FORMULA IS EQUAL TO CH2 NOW, TO DETERMINE THE MOLECULAR FORMULA THEN WE ARE GOING TO HAVE TO USE THE MOLAR MASS AND THE EMPIRICAL FORMULA

8 CHEM 105 & 106 UNIT ONE, LECTURE SEVEN 8 MASS. IN THIS CASE IT JUST HAPPENS THAT WE ALREADY HAVE IT CALCULATED BECAUSE IT IS THE SAME EMPIRICAL FORMULA AS THE ONE IN SEVENTEEN A. SO, THE MOLECULAR FORMULA IS EQUAL TO CH2 MULTIPLIED BY OUR MOLAR MASS UP THERE, FORTY TWO POINT ONE (42.1) DIVIDED BY THE EMPIRICAL MASS, FOURTEEN POINT ZERO THREE (14.03) THAT IS EQUAL TO CH2 TIMES THREE SO THE MOLECULAR FORMULA IS IN FACT C3H6. NOW NOTICE PROBLEM SEVENTEEN A WE HAD AN EMPIRICAL FORMULA OF CH2 PROBLEM NUMBER NINETEEN WE HAVE AN EMPIRICAL FORMULA OF CH2 BUT THEY ARE NOT THE SAME CHEMICAL COMPOUND, EVEN THOUGH THEY BOTH HAD ONE TO TWO CARBON TO HYDROGEN RATIO. THE ACTUAL COMPOUND THAT WE HAVE TO KNOW THE MOLAR MASS TO GET TO THAT POINT. THESE TWO COMPOUNDS C5H10 AND C3H6 ARE TWO TOTALLY DIFFERENT COMPOUNDS, TWO DIFFERENT ORGA NIC COMPOUNDS THAT WE REFER TO AS HYDROCARBONS, ORGANIC COMPOUNDS THAT CONTAIN HYDROGEN AND CARBON ONLY ARE CALLED HYDROCARBONS. THESE ARE TWO DIFFERENT HYDROCARBONS, TOW DIFFERENT ORGANIC COMPOUNDS. ALRIGHT, ANY QUESTIONS ON ANY STEP OF THAT? WELL THEN I WOULD LIKE TO SPEND THE REMAINDER OF THIS LECTURE GOING BACK AND PICKING UP A LITTLE BIT AT THE EARLIER PART OF THE CHAPTER AS I MENTIONED WHEN WE WERE TALKING ABOUT TWO TYPES OF FORMULAS WE TALKED ABOUT, FIRST OF ALL THE CHEMICAL FORMULA AND THEN WE TALKED ABOUT THE STRUCTURAL FORMULA, AND THE STRUCTURAL FORMULA TELLS US HOW THE ATOMS ARE HOOKED TOGETHER. THE CHEMICAL FORMULA TELLS US HOW MANY OF EACH TYPE OF ATOM ARE PRESENT AND THAT IS ALL IT TELLS US. IF YOU LOOK BACK IN THE FIRST LECTURE IN THIS CHAPTER YOU WILL SEE THAT WE LOOKED AT A COUPLE OF COMPOUNDS, SULFURIC ACID, WE SHOWED ITS CHEMICAL FORMULA AND ITS STRUCTURAL FORMULA. WHERE WE ARE GOING TO APPLY THIS TODAY IS IN THE AREA OF CHEMISTRY THAT WE REFER TO AS ORGANIC CHEMISTRY. ORGANIC CHEMISTRY IS THE CHEMISTRY THAT IS BASED ESSENTIALLY ON CARBON, CARBON IS THE BACKBONE OF ALL OF THE COMPOUNDS THAT FALL IN THIS CATEGORY CALLED ORGANIC CHEMISTRY. ONE OF THE FIRST GROUPS OF ORGANIC CHEMICAL COMPOUNDS THAT WE ARE GOING TO LOOK AT ARE THE ONES THAT I JUST MENTIONED, THE HYDROCARBONS, AND THE HYDROCARBONS CONTAIN ONLY TWO ELEMENTS HYDROGEN AND CARBON. THERE ARE MANY DIFFERENT HYDROCARBONS THAT WE HAVE AND WE ARE GOING TO LOOK AT IN THIS PARTICULAR CHAPTER, FIST OF ALL WE ARE GOING TO LOOK AT THE SIMPLEST GROUP OF

9 CHEM 105 & 106 UNIT ONE, LECTURE SEVEN 9 HYDROCARBONS WHICH ARE REFERRED TO AS SATURATED HYDROCARBONS. SATURATED HYDROCARBONS MEANS THAT WE HAVE THE MAXIMUM CARBON HYDROGEN RATIO THAT THERE IS IN ANY ORGANIC COMPOUND, THAT IS WHY IT IS CALLED THE SATURATED HYDROCARBON. THEY WILL ALL HAVE A CHEMICAL FORMULA OF CN WHERE N IS ANY NUMBER THAT WE WANT TO PLUG IN H2N THAT MEANS TWO TIMES WHATEVER NUMBER WE USED FOR THE CARBON PLUS TWO. THEY WILL ALL HAVE THAT CARBON TO HYDROGEN RATIO, IN OTHER WORDS, IF I HAVE A SATURATED HYDROCARBON WHICH HAS SIX CARBONS THEN IT WOULD TELL ME THAT TO BE A SATURATED HYDROCARBON IT WOULD NEED TO HAVE TWICE THAT MANY, TWO TIMES SIX WOULD BE TWELVE PLUS THE TWO SO THE CHEMICAL FORMULA FOR A SATURATED HYDROCARBON WITH SIX CARBONS WOULD BE CH14 O.K. ALL SATURATED HYDROCARBONS FIT THIS CHEMICAL FORMULA RIGHT HERE THIS GENERAL FORMULA C6H14. THESE ALSO ARE SOMETIMES REFERRED TO AS THE ALKANE (ANE) AND I WILL COME BACK AND MENTION THIS ANE ASPECT HERE IN A MOMENT, THESE ARE CALLED THE ALKANE AS A GROUP IN ORGANIC CHEMISTRY AND WHAT WE ARE GOING TO TRY TO DO AS WE GO ALONG IN THE DIFFERENT CHAPTERS IS THAT WE ARE GOING TO TRY TO LOOK AT SOME DIFFERENT ORGANIC CHEMICAL GROUPS OF COMPOUNDS THAT WILL FALL INTO THIS AREA OF ORGANIC CHEMISTRY. ORGANIC CHEMISTRY IS PROBABLY THE BIGGEST INDIVIDUAL SUBGROUP OF CHEMISTRY THAT WE HAVE. WE ARE PROBABLY MOST INVOLVED AND IMPACTED BY ORGANIC TYPE CHEMICAL COMPOUNDS. THE FUEL IN YOUR AUTOMOBILE THE RUBBER MAKING UP THE TIRES, THE PLASTIC, THE CLOTHING THAT YOU ARE WEARING, MUCH OF THE FOOD THAT WE EAT ARE IN THIS CATEGORY OF ORGANIC CHEMICALS AND SO WE ARE GOING TO LOOK AT SOME KEY CATEGORIES, SOME KEY GROUPS OF ORGANIC COMPOUNDS AS WE GO ALONG. NOT SO MUCH AS TO TRY AND MEMORIZE ALL OF THEM IN TERMS OF CHEMICAL FORMULAS, NAMES, AND ALL OF THAT, BUT TO HAVE A BASIC UNDERSTANDING OF THIS MAJOR, MAJOR AREA OF CHEMISTRY CALLED ORGANIC CHEMISTRY. NOW, TABLE TWO POINT ONE IN THE TEXT LISTS TEN OF THE ALKANE HYDROCARBONS ALL OF THESE HAVE THAT CNH TWO N PLUS TWO RELATIONSHIP, AND WE WANT TO TAKE A LOOK AT A FEW OF THESE BECAUSE THESE ARE COMPOUNDS THAT WE DO ENCOUNTER ON A RATHER OR REGULAR ROUTINE BASIS AND SO LETS TAKE A LOOK AND I WOULD LIKE YOU TO AND AT LEAST IN THIS CASE KNOW THE NAMES OF THESE AND THERE CHEMICAL FORMULA, NOW THAT MIGHT SEEM QUITE DIFFICULT AT FIRST BUT WHEN WE POINT

10 CHEM 105 & 106 UNIT ONE, LECTURE SEVEN 10 OUT A FEW THINGS HERE I THINK YOU WILL SE THAT IT IS NOT THAT MUCH OF A PROBLEM. THE FIRST FOUR THAT WE HAVE UP HERE IS METHANE CH4 THAT IS THE SIMPLEST HYDROCARBON, METHANE IS THE MAJOR CONSTITUENT OF NATURAL GAS. IF YOU HAVE A NATURAL GAS FURNACE IN YOUR HOME, IT IS HEATED BY THAT, THE MAJOR COMPONENT OF THE GAS THAT IS BEING BURNED BY THAT FURNACE IS METHANE CH4. SO NATURAL GAS OR ACTUALLY THE NAME METHANE PROBABLY ARRIVED FROM ITS EARLY NAME WHICH WAS CALLED MARSH GAS, WHEN MARSH'S UNDERGO DECAY THE BIO MASS BEGINS TO DECAYING ONE OF THE PRODUCTS THAT WE HAVE FROM THAT DECAY IS METHANE OR NATURAL GAS. ANOTHER COMPONENT OF NATURAL GAS BUT IN A MUCH LESSER AMOUNT THIS USUALLY, PROBABLY IN THE RANGE OF ABOUT NINETY PERCENT OF NATURAL GAS, ABOUT TEN PERCENT IS THE SECOND ONE CALLED ETHANE. WE WON'T WORRY TO MUCH ABOUT ETHANE RIGHT NOW, BUT WE WILL COME BACK AND TALK ABOUT ITS CHEMICAL IMPORTANCE IN FUTURE CHAPTERS BECAUSE AS IT SAYS IT IS A STARTING MATERIAL FOR A CHEMICAL COMPOUND CALLED ETHENE, AND WHEN WE TALK ABOUT ETHENE WE WILL SEE ITS COMMERCIAL IMPORTANCE. IT IS A VERY, VERY IMPORTANT INDUSTRIAL CHEMICAL. THE NEXT TWO ARE TWO THAT YOU ARE FAMILIAR WITH C3H8 PROPANE IF YOU GO CAMPING YOU PROBABLY TAKE ALONG SOME SORT OF A COOK STOVE THAT OPERATES ON PROPANE, THIS LPG LIQUID PETROLEUM GAS, LPG IS LIQUIFIED PROPANE. IF YOU HAVE A GRILL IN THE BACK OF YOU HOME OR SOMETHING AND YOU HAVE A TANK ON IT THAT IS PROPANE. IF YOU LIVE IN A RURAL AREA AND DON'T HAVE ACCESS TO NATURAL GAS YOU ARE PROBABLY HEATING YOUR HOME WITH PROPANE, THE LARGE TANK THEY BRING IT OUT AS A LIQUID AND PUT IT IN THE TANK. THE NEXT ONE BUTANE IS THE GAS, IS THE HYDROCARBON WE FIND IN CIGARETTE LIGHTERS, BUTANE CIGARETTE LIGHTER IS THEN USING BUTANE FOR ITS FUEL AND BUTANE AT ROOM TEMPERATURE AND A SMALL AMOUNT OF PRESSURE IS A LIQUID. IT CAN BE LIQUIFIED RELATIVELY EASILY. NOW THE NEXT SIX IF YOU LOOK A THEIR NAMES IS WHAT MAKES IT EASY TO REMEMBER THEM. LOOK AT THEIR NAMES, PENTANE, WHAT IS THE PREFIX FOR FIVE? PENTA. HEXANE, WHAT IS THE PREFIX FOR SIX? HEXA. EACH OF THESE COMPOUNDS THEN FROM C4 ON STARTING WITH C5 ARE NAMED USING THE PREFIX SYSTEM. PENTANE, HEXANE FOR SIX, HEPTANE FOR SEVEN, OCTANE, AND I AM SURE YOU HAVE ALL HEARD OF THAT ONE IT MEANS EIGHT CARBONS, SO WHEN YOU SEE AN OCTANE READING ON A FUEL PUMP WHAT THEY ARE TALKING ABOUT IS THAT THIS

11 CHEM 105 & 106 UNIT ONE, LECTURE SEVEN 11 FUEL IS COMPARED TO A PARTICULAR SAMPLE OF GASOLINE THAT IS RATED AT ONE HUNDRED AND SO IF YOU THEN BUY GASOLINE THAT SAYS EIGHTY-SEVEN OCTANE IT MEANS THAT IT HAS THE BURNING RATE, THE POWER RATE OF ABOUT EIGHTY-SEVEN PERCENT OF WHAT THE PURE IS OCTANE USED AS A REFERENCE WOULD BE, ALRIGHT? NONANE AND THEN FINALLY DECANE. YOU CAN SEE THAT A LOT OF THESE ARE COMPONENTS OF GASOLINE, OKAY, SO GASOLINE IS A MIXTURE AS WE POINTED OUT IN CHAPTER ONE, IT IS NOT A COMPOUND, IT IS A MIXTURE OF MANY DIFFERENT CHEMICAL COMPOUNDS. NOW ONE OF THE THINGS ABOUT ORGANIC COMPOUNDS AND STRUCTURAL FORMULAS, LETS TAKE A LOOK AT C5H12 FOR A SECOND, THAT IS THE CHEMICAL FORMULA C5H12. NOW, STRUCTURAL FORMULA WISE I COULD HOOK THE FIVE CARBONS TOGETHER AND I COULD PUT HYDROGENS ON, WE WILL FIND IN A LATER CHAPTER ON CHEMICAL BONDING THAT CARBON HAS THE ABILITY TO FORM FOUR CHEMICAL BONDS BUT WE WILL SEE HERE IS THAT EACH CARBON IN THIS STRUCTURAL FORMULA HAS FOUR DASHES, FOUR CHEMICAL BONDS ASSOCIATED WITH IT. WE WILL LATER FIND OUT THAT HYDROGEN CAN HAVE ONLY ONE CHEMICAL BOND, SO WE SEE THAT EACH HYDROGEN IS HOOKED TO A CARBON BUT WE COULD NOT HAVE A HYDROGEN HOOKED TO A CARBON HOOKED TO ANOTHER HYDROGEN THAT IS NOT A POSSIBILITY FOR HYDROGEN TO DO, IT CAN ONLY FORM ONE CHEMICAL BOND, AND IF IT IS USED IT TO HOOK TO THE CARBON THEN OBVIOUSLY IT CAN'T HOOK TO SOMETHING ELSE. NOW THAT IS ONE WAY THAT I CAN HOOK THESE TOGETHER, BUT IN ORGANIC CHEMISTRY IT TURNS OUT THAT MANY OF THE CHEMICAL FORMULAS CAN BE SHOWN IN MANY DIFFERENT STRUCTURAL FORMS, FOR INSTANCE, I CAN TAKE THIS SAME ONE, I WILL PUT FOUR IN A ROW LIKE THIS AND I WILL BRANCH ONE OFF THIS WAY, IF I HOOK THAT HYDROGENS, REMEMBER THAT I CAN ONLY HAVE FOUR DASHES ON EACH CARBON HERE AS I JUST SAID SO IF WE DO THIS AND WE PUT THE HYDROGENS ON THERE AND WE COUNT UP THE HYDROGENS WE WILL FIND THAT WE HAVE TWELVE HYDROGENS. THAT IS A UNIQUE MOLECULE THAT IS THE SAME CHEMCIAL FORMULA BUT IT IS A TOTALLY NEW STRUCTURAL FORMULA. IT HAS DIFFERENT BOILING POINTS, IT HAS DIFFERENT CHEMICAL REACTIVITY, IT HAS A DIFFERENT CHEMICAL COMPOUND. EVEN THOUGH IT HAS THE SAME CHEMICAL FORMULA IT IN FACT IS NOT THE SAME CHEMICAL COMPOUND. WELL AS A MATTER OF FACT LETS SEE, WE COULD MAYBE DO SOMETHING LIKE THIS, WE COULD PUT THREE IN A ROW AND WE COULD PUT A COUPLE OF CARBONS OFF THE MIDDLE LIKE THAT AND THEN WE COULD PUT HYDROGENS

12 CHEM 105 & 106 UNIT ONE, LECTURE SEVEN 12 AROUND HERE, I WON'T BOTHER TO DRAW EACH DASH THIS TIME, KEEPING IN MIND THAT EACH HYDROGEN HAS ONE DASH TO THE CARBONS AND THAT IS A THIRD WAY THAT WE CAN HOOK FIVE CARBONS AND TWELVE HYDROGENS TOGETHER TO MAKE THE SAME CHEMICAL FORMULA C5H12. AGAIN, EACH OF THESE IS A UNIQUE CHEMICAL COMPOUND, EACH OF THESE HAS A DIFFERENT CHEMICAL NAME OK? THIS ONE AS THE CHARTS SHOWED US WOULD BE CALLED PENTANE, THIS ONE HOWEVER, AND YOU DON'T NEED TO WRITE THESE DOWN, THIS ONE IS CALLED METHYL BUTANE AND THIS ONE IS CALLED TWO-TWO DIMETHYL PROPANE THEY HAVE DIFFERENT CHEMICAL NAMES FOR THE THREE DIFFERENT STRUCTURAL FORMULAS EVEN THOUGH THEY ARE THE SAME CHEMICAL FORMULAS. NOW, IN NATURE WHEN WE HAVE MORE THAN ONE WAY TO HOOK THE SAME NUMBER OF ATOMS TOGETHER WE REFER TO THESE DIFFERENT TYPES OF STRUCTURES AS ISOMERS, A TERM THAT YOU NEED TO REMEMBER. AND WHEN THESE ISOMERS ARE, BECAUSE OF THE DIFFERENT STRUCTURES WE HAVE DRAWN, WE SPECIFICALLY CALL THEM THEN STRUCTURAL ISOMERS. NOW, WHAT ARE STRUCTURAL ISOMER IS ONCE AGAIN, IT IS A CHEMICAL OR TWO DIFFERENT FORMS OF THE SAME CHEMICAL FORMULA HOOKED TOGETHER IN A DIFFERENT FASHION. WE HAVE TO OBEY THE BONDING ABILITY OF THE ELEMENTS WHEN WE DO THIS. WE CAN'T JUST HOOK THINGS TOGETHER AND MAKE ARBITRARY NEW STRUCTURAL FORMULAS, THEY DO HAVE TO FIT SOME SORT OF BONDING CRITERIA. YES, A QUESTION, SHOULD THE ONE IN THE MIDDLE BE NAMED TWO METHYL BUTANE? THAT IS A VERY GOOD QUESTION AND THE ANSWER IS, IT COULD BE BUT IT IS NOT NECESSARY, THE ONLY PLACE THAT I COULD PUT A METHYL GROUP IS THIS LITTLE CH3 THAT WE HAVE HANGING HERE, THE ONLY PLACE THAT IT COULD BE PUT WOULD BE ON THE SECOND CARBON, IF I PUT IT ON THE END I WOULD HAVE FIVE AND I WOULD BE BACK TO PENTHANE. IN THIS CASE I CAN ONLY PUT IT THERE, IF I PUT IT ON THIS ONE, REMEMBER I DREW THE MOLECULE, THE MOLECULE DOES NOT KNOW LEFT FROM RIGHT, IF I STOOD BEHIND THE BOARD AND LOOKED AT IT, IT WOULD LOOK THE REVERSE WAY, BUT THAT IS NOT AN UNIQUE MOLECULE. NOW IF THERE IS ANY QUESTION IN OUR NAMING WE COULD CERTAINLY USE THE TERM TWO METHYL BUTANE AND THAT IS CERTAINLY A GOOD ANSWER OR A GOOD NAME FOR IT. IN THIS PARTICULAR CASE AS I SAY, BECAUSE WE ARE NOT REALLY CONCERNED AT THIS POINT OF TRYING TO LEARN THE ORGANIC NOMENCLATURE BUT TO LOOK AT DIFFERENT GROUPS OF ORGANIC COMPOUNDS WE WON'T WORRY ABOUT EACH LITTLE INDIVIDUAL CRITERIA THAT WE NEED IN THE

13 CHEM 105 & 106 UNIT ONE, LECTURE SEVEN 13 NAMING. YES IT COULD BE CALLED TWO METHYL BUTANE WE COULD CALL IT JUST METHYL BUTANE, WE WOULD HAVE TO DRAW EXACTLY THE SAME STRUCTURE REGARDLESS OF WHAT THE NAME WAS, SO, THE MORE INFORMATION THAT WE PUT OF COURSE THE CLEARER IT IS TO SOMEONE BUT ALSO IN CHEMISTRY WE OFTEN TRY TO WRITE THINGS IN THE SIMPLEST FASHION THAT WE CAN TO STILL GET ACROSS THE INFORMATION THAT WE NEED. ALRIGHT? SO STRUCTURAL ISOMERS, SO WE CHEMICAL FORMULAS, WE HAVE A STRUCTURAL FORMULA THAT SHOWS HOW THINGS CAN BE HOOKED TOGETHER AND THEN WE SEE THAT IN SOME CASES WE CAN HOOK THINGS TOGETHER WITH THE SAME CHEMICAL FORMULA IN MORE THAN ONE WAY WHICH GIVES US THEN WHAT WE CALL STRUCTURAL ISOMERS. NOW, JUST TO GIVE YOU SOME IDEA OF THE MAGNITUDE OF THIS IF WE GO TO THE C8 COMPOUND, I BELIEVE THAT WE WILL HAVE EIGHTEEN DIFFERENT ISOMERS THAT WE CAN HAVE, IF WE GO TO C9 WITH THE CORRECT NUMBER OF HYDROGENS WE WILL HAVE SOMETHING LIKE THIRTY FIVE AND WHEN WE GO TO C10 WE HAVE SOMETHING APPROXIMATELY SEVENTY SO YOU SEE THE NUMBER OF DIFFERENT ISOMERS THAT ONE CAN HAVE FOR THE SAME CHEMICAL FORMULA BECOMES VERY, VERY LARGE AND WE HAVE MANY COMPOUNDS IN HYDROCARBONS THAT ARE TWELVE TO TWENTY FIVE CARBONS IN THEM, THERE ARE MANY DIFFERENT WAYS THAT WE CAN WRITE THEM. IN OUR NEXT LECTURE WE WILL TAKE A LITTLE BIT OF A QUICK LOOK AT HOW HYDROCARBONS ARE OBTAINED AND PROCESSED THEN FOR OUR JUICE.