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Understanding the Concept of a Mole in Chemistry: Calculations and Conversions, Study notes of Chemistry

Missouri State University (MSU)Chemistry

An explanation of the concept of a mole in chemistry, with examples of calculations and conversions using avogadro's number. Topics include determining the number of atoms in a given mass of an element, and converting between moles, mass, and atoms.

Typology: Study notes

Pre 2010

Uploaded on 07/23/2009

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CHEM 105 & 106 UNIT ONE, LECTURE SIX
1
CHM 105/106 Program 6: Unit 1 Lecture 6
IN THE PREVIOUS LECTURE WE WERE TALKING ABOUT MEASUREMENT RELATIVE TO CHEMISTRY AND
THE BASIC UNIT THAT WE WERE USING AND DO USE IN CHEMISTRY IS THE MOLE AND THE MOLE SAID
WAS A QUANTITY OF MATERIAL AND IT IS EQUAL TO SIX POINT ZERO TWO THREE (6.023) TIMES TEN TO
THE TWENTY THIRD UNITS, WHATEVER THAT UNIT MIGHT BE. IT COULD BE ATOMS, IT COULD BE
MOLECULES, IT COULD BE AUTOMOBILES,IT DOESN'T MAKE ANY DIFFERENCE, THAT IS THE NUMERIC
VALUE ASSOCIATED WITH THE MOLE. ALSO AS I INDICATED IN THE PREVIOUS LECTURE, THE MOLE IS
ABBREVIATED M-O-L SO WHEREVER YOU SEE THAT IT IS NOT MISSPELLED THAT IS JUST THE
ABBREVIATED FORM. RELATIVE THEN TO THE ELEMENTS AND THE COMPOUNDS WE FURTHER STATED
THAT ONE MOLE OF A MONOATOMIC ELEMENT WAS EQUAL TO THE MASS OF THE ELEMENT
EXPRESSED IN GRAMS, IN OTHER WORDS AGAIN, REFERRING NOW TO OUR PERIODIC TABLE THE
NUMBER THAT APPEARS BELOW THE SYMBOL ON THE PERIODIC TABLE IS CALLED THE ELEMENTS
MASS. IF WE EXPRESS THAT IN GRAMS THAT IS EQUIVALENT TO ONE MOLE OF THAT PARTICULAR
SUBSTANCE, AND THAT NUMBER OF GRAMS WOULD THEN CONTAIN THAT NUMBER OF ATOMS OF
THAT ELEMENT. FOR EXAMPLE, ONCE AGAIN IF WE WERE TO TAKE MERCURY WE SEE UP THERE HAS A
MASS OF TWO HUNDRED POINT FIVE NINE (200.59) AND IF WE MAKE THAT GRAMS IT THEN THEREFORE
EQUAL TO ONE MOLE OF THE MONOATOMIC ELEMENT MERCURY. AGAIN THAT WOULD CONTAIN SIX
POINT ZERO TWO (6.02) TIMES TEN TO THE TWENTY THIRD MERCURY ATOMS, ALRIGHT? IF IT IS A
POLYATOMIC WHICH REALLY WE ARE TALKING ABOUT ANYTHING WITH TWO OR MORE, POLYATOMIC
ELEMENT OR A COMPOUND THEN WE SAY THAT ONE MOLE IS EQUAL TO THE FORMULA MASS IN
GRAMS. NOW THE FORMULA MASS OF COURSE IS THEN, THE WEIGHT CONTRIBUTED BY EACH OF THE
ELEMENTS PRESENT IN THEIR CORRECT AMOUNT. WE LOOKED AT THOSE ELEMENTS THAT EXIST AS DI
AND POLYATOMIC ELEMENTS AND TALKED ABOUT HOW MUCH A MOLE OF THOSE WOULD BE, AND
ALSO IN TERMS OF A MOLECULE, IF FOR INSTANCE WE WERE TO LOOK AT WATER H20 AS A COMPOUND
WE HAVE TWO HYDROGENS AND EACH HYDROGEN WEIGHS ONE POINT ZERO ONE (1.01), SO THAT IS
TWO POINT ZERO TWO (2.02) GRAMS CONTRIBUTED BY THE HYDROGEN, WE HAVE ONE OXYGEN AND ITS
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Download Understanding the Concept of a Mole in Chemistry: Calculations and Conversions and more Study notes Chemistry in PDF only on Docsity!

CHM 105/106 Program 6: Unit 1 Lecture 6

IN THE PREVIOUS LECTURE WE WERE TALKING ABOUT MEASUREMENT RELATIVE TO CHEMISTRY AND

THE BASIC UNIT THAT WE WERE USING AND DO USE IN CHEMISTRY IS THE MOLE AND THE MOLE SAID

WAS A QUANTITY OF MATERIAL AND IT IS EQUAL TO SIX POINT ZERO TWO THREE (6.023) TIMES TEN TO

THE TWENTY THIRD UNITS, WHATEVER THAT UNIT MIGHT BE. IT COULD BE ATOMS, IT COULD BE

MOLECULES, IT COULD BE AUTOMOBILES,IT DOESN'T MAKE ANY DIFFERENCE, THAT IS THE NUMERIC

VALUE ASSOCIATED WITH THE MOLE. ALSO AS I INDICATED IN THE PREVIOUS LECTURE, THE MOLE IS

ABBREVIATED M-O-L SO WHEREVER YOU SEE THAT IT IS NOT MISSPELLED THAT IS JUST THE

ABBREVIATED FORM. RELATIVE THEN TO THE ELEMENTS AND THE COMPOUNDS WE FURTHER STATED

THAT ONE MOLE OF A MONOATOMIC ELEMENT WAS EQUAL TO THE MASS OF THE ELEMENT

EXPRESSED IN GRAMS, IN OTHER WORDS AGAIN, REFERRING NOW TO OUR PERIODIC TABLE THE

NUMBER THAT APPEARS BELOW THE SYMBOL ON THE PERIODIC TABLE IS CALLED THE ELEMENTS

MASS. IF WE EXPRESS THAT IN GRAMS THAT IS EQUIVALENT TO ONE MOLE OF THAT PARTICULAR

SUBSTANCE, AND THAT NUMBER OF GRAMS WOULD THEN CONTAIN THAT NUMBER OF ATOMS OF

THAT ELEMENT. FOR EXAMPLE, ONCE AGAIN IF WE WERE TO TAKE MERCURY WE SEE UP THERE HAS A

MASS OF TWO HUNDRED POINT FIVE NINE (200.59) AND IF WE MAKE THAT GRAMS IT THEN THEREFORE

EQUAL TO ONE MOLE OF THE MONOATOMIC ELEMENT MERCURY. AGAIN THAT WOULD CONTAIN SIX

POINT ZERO TWO (6.02) TIMES TEN TO THE TWENTY THIRD MERCURY ATOMS, ALRIGHT? IF IT IS A

POLYATOMIC WHICH REALLY WE ARE TALKING ABOUT ANYTHING WITH TWO OR MORE, POLYATOMIC

ELEMENT OR A COMPOUND THEN WE SAY THAT ONE MOLE IS EQUAL TO THE FORMULA MASS IN

GRAMS. NOW THE FORMULA MASS OF COURSE IS THEN, THE WEIGHT CONTRIBUTED BY EACH OF THE

ELEMENTS PRESENT IN THEIR CORRECT AMOUNT. WE LOOKED AT THOSE ELEMENTS THAT EXIST AS DI

AND POLYATOMIC ELEMENTS AND TALKED ABOUT HOW MUCH A MOLE OF THOSE WOULD BE, AND

ALSO IN TERMS OF A MOLECULE, IF FOR INSTANCE WE WERE TO LOOK AT WATER H20 AS A COMPOUND

WE HAVE TWO HYDROGENS AND EACH HYDROGEN WEIGHS ONE POINT ZERO ONE (1.01), SO THAT IS

TWO POINT ZERO TWO (2.02) GRAMS CONTRIBUTED BY THE HYDROGEN, WE HAVE ONE OXYGEN AND ITS

MASS IS SIXTEEN AND SO THEREFORE WE HAVE EIGHTEEN POINT ZERO TWO (18.02) AS THE FORMULA

MASS. IF WE SAY GRAMS EIGHTEEN POINT ZERO TWO (18.02) GRAMS THAT IS ONE MOLE OF WATER

THEN EQUALS EIGHTEEN POINT ZERO TWO (18.02) GRAMS, AND THAT AMOUNT OF WATER WOULD

CONTAIN SIX POINT ZERO TWO (6.02) TIMES TEN TO THE TWENTY THIRD WATER MOLECULES. WATER

MOLECULES IS OUR UNIT THIS TIME BECAUSE WE ARE TALKING ABOUT A COMPOUND. NOW, WITH

THOSE CONCEPTS IN MIND, A MOLE IS EQUAL TO THIS, AS A MATTER OF FACT WE IDENTIFIED THIS IN

THE PREVIOUS LECTURE AS AVOGADRO'S NUMBER (6.023) TIMES TEN TO THE TWENTY THIRD, WITH

THESE RELATIONSHIPS THEN A MOLE IS EQUAL TO THIS NUMBER, A MOLE IS EQUAL TO THE ATOMIC

MASS, A MOLE IS EQUAL TO THE FORMULA MASS, THEN WE CAN USE THOSE RELATIONSHIPS TO DO A

WHOLE VARIETY OF DIFFERENT TYPES OF CALCULATIONS, AND SO TODAY WE WILL CONTINUE ON

THEN AND LOOK AT SOME OTHER EXAMPLE PROBLEMS DEALING WITH THIS CONCEPT OF THE MOLE.

THE FIRST ONE THEN THAT I WOULD LIKE TO LOOK AT WILL BE PROBLEM NUMBER THREE AT THE END

OF THE CHAPTER, WHICH STATE THAT WE HAVE POINT ONE FOUR SIX (.146) MOLES OF COPPER (Cu) HOW MANY GRAMS OF COPPER DO WE HAVE? SO, AS WE DISCUSSED IN CHAPTER ONE ON PROBLEM SOLVING WE WILL CONTINUE TO USE THE SAME METHOD FOR SOLVING PROBLEMS HERE IN OTHER WORDS, UNIT ANALYSIS. SO LETS SEE, WE ARE SAYING THAT WE HAVE THIS AMOUNT OF COPPER, WE WANT TO KNOW MASS IN GRAMS OKAY, SO WE ARE GOING TO SAY GRAMS ARE EQUAL TO ZERO POINT ONE FOUR SIX (0.146) MOLES OF COPPER. NOW, COPPER IS A MONOATOMIC ELEMENT AND WE SAID THAT A MOLE OF A MONOATOMIC ELEMENT IS EQUAL TO ITS ATOMIC MASS EXPRESSED IN GRAMS, SO WE HAVE A RELATIONSHIP BETWEEN MOLES AND MASS THAT WE CAN USE, AND WE WILL DO SO AGAIN IN THE TERMS OF A CONVERSION FACTOR, SO WE ARE GOING TO MULTIPLY BY, LOOKING UP AT COPPER ALSO, AS I MENTIONED IN THE PREVIOUS LECTURE AS WELL AS THE PERIODIC TABLE, SUCH AS WHAT WE HAVE UP THERE OR AT THE BACK AT THE BOOK, WE ALSO HAVE THE TABLE AT THE END OF THE TEXT WHICH ALREADY HAS THE ATOMIC WEIGHTS ROUNDED OFF TO THE NEAREST HUNDREDTH OF A GRAM, TWO PLACES AFTER THE DECIMAL, AND THOSE ARE THE NUMBERS THAT WE WILL USE IN PROBLEM SOLVING. YOU DON'T NEED TO USE ALL OF THOSE NUMBERS, YOU USE TWO PLACES AFTER THE DECIMAL. ALRIGHT, SO FOR COPPER THEN WE ARE LOOKING AT SIXTY THREE POINT FIVE FIVE

PREVIOUS ONE. WE ARE GOING TO SAY GRAMS OF H2SO4 EQUALS ONE POINT THREE FOUR (1.34) TIMES

TEN TO THE MINUS TWO MOLES OF H2SO4 MULTIPLIED BY NINETY EIGHT POINT ZERO NINE (98.09)

GRAMS PER ONE MOLE, NOW I DID NOT LEAVE MYSELF ENOUGH ROOM THERE REALLY TO WRITE THE

UNITS BUT I REALLY DO RECOMMEND THAT YOU PUT THE UNITS IN, EVEN THOUGH WE CAN SEE IN THIS

CASE THAT THEY ARE GOING TO CANCEL OUT PUT THEM IN AND THEN CANCEL THEM OUT AND THIS

WILL KEEP YOU OUT OF TROUBLE IN MANY CASES IN PROBLEM SOLVING. NOW, LETS TAKE JUST A

SECOND HERE AND GO THROUGH THE CALCULATOR PART OF THIS PROBLEM, AND THE REASON I SAY

THAT IS BECAUSE WE ARE GOING TO USE AN EXPONENTIAL NUMBER IN OUR CALCULATION AND I

FREQUENTLY FIND THAT STUDENTS COME UP WITH ANSWERS ONE POWER OF TEN DIFFERENT THEN

WHAT THEY SHOULD. I HAVE A PRETTY GOOD IDEA OF WHY THAT OCCURS AND SO JUST TO MAKE

SURE THAT WE ARE ALL FOLLOWING THE CORRECT PROCEDURE ON THE CALCULATOR THEN WE WILL

GO AHEAD AND STEP WISE LOOK AT THIS ONE. NOW, I AM GOING TO REVERSE THE NUMBERS SO THAT I

CAN SHOW WHAT I AM DOING HERE, SO IF YOU HAVE YOUR CALCULATOR AND YOU ARE NOT

COMFORTABLE HAVING USED THE EXPONENTIAL PART OF THIS, THEN YOU MIGHT WANT TO FOLLOW

THE STEPS WHILE I AM DOING THEM. SO WHAT I AM GOING TO DO HERE, I AM GOING TO START AND I

AM GOING TO PUT IN NINETY EIGHT POINT ZERO NINE (98.09) AND THEN OF COURSE I AM GOING TO HIT

THE TIMES SIGN, AND NOW I AM GOING TO PUT IN THE NUMBER THAT IS IN EXPONENTIAL NOTATION

SO I WILL PUT IN ONE POINT THREE FOUR (1.34) YOU DO NOT HIT THE TIMES SIGN, YOU PUT IN ONE POINT

THREE FOUR (1.34), YOUR NEXT STEP TO TELL THE COMPUTER THAT WE ARE GOING TO GIVE IT A

NUMBER PERTAINING TO A TEN, A POWER, IS YOU LOOK FOR THE BUTTON ON THE CALCULATOR THAT

SAYS EE OR EX OR EXP, YOU HIT THAT BUTTON THEN, AND THAT POINT IF YOU LOOK ON THE DISPLAY

YOU SHOULD SEE A COUPLE OF LITTLE ZEROS SHOW UP, SMALLER ZEROS USUALLY BUT SOME

CALCULATORS WILL SHOW THEM AS THE SAME SIZE BUT THERE WILL PROBABLY BE A SPACE

BETWEEN THE NUMBER YOU PUT IN AND THE TWO ZEROS OK? REMEMBER THE KEY THING IS THAT

YOU DO NOT PUT IN TIMES AND YOU DO NOT PUT IN THE TEN. WHEN I HIT THE EE BUTTON OR THE EXP

BUTTON THAT IS ALREADY TELLING THE COMPUTER TO MULTIPLY THIS BY TEN, IT HAS ALREADY PUT

A TEN IN AND THE CALCULATOR IS NOW SAYING WHAT POWER OF TEN DO YOU WANT? IN THIS

PARTICULAR CASE WE WANT A MINUS TWO (-2) NOW, IF YOU LOOK ON THERE YOU SHOULD FIND A

BUTTON THAT HAS BOTH A PLUS AND A MINUS SIGN ON IT THAT ALLOWS YOU TO CHANGE THEN

FROM A POSITIVE TO A NEGATIVE EXPONENTIAL SO IF I FIND THAT BUTTON AND I HIT IT, I WILL NOW

SEE THAT THE DISPLAY ON MY CALCULATOR WILL LOOK SOMETHING LIKE THIS ONE THREE FOUR DASH

ZERO TWO (134-02) AND THAT IS SAYING ONE POINT THREE FOUR (1.34) TIMES TEN TO THE NEGATIVE

TWO. NOW WE HAVE PUT THE NUMBER IN, WE HAVE PUT THE PREVIOUS NUMBER IN ALL WE ARE

GOING TO DO NOW IS HIT THEN THE NUMBER TWO AND YOUR DISPLAY SHOULD LOOK LIKE THIS, NOW

WE HIT THE EQUAL SIGN AND OF COURSE THEN WE WILL COME UP WITH THE FINAL ANSWER AND THE

FINAL ANSWER IS ONE POINT THREE ONE FOUR-FOUR ZERO SIX (1.314406) AND AGAIN ALWAYS KEEP IN

MIND SIGNIFICANT FIGURES, IN THIS PARTICULAR CASE WE SHOULD HAVE HOW MANY? THREE AND SO

WE COUNT OVER HERE THREE NOTICE THAT THE NUMBER AFTER THE LAST SIGNIFICANT FIGURE IS LESS

THEN FIVE SO THIS TIME WE ARE GOING TO DROP ALL THAT, AND OUR FINAL ANSWER IS MERELY ONE

POINT THREE ONE (1.31) GRAMS OF H2SO4, ANY QUESTIONS ON THAT? WELL LETS LOOK AT ANOTHER

EXAMPLE THEN, THIS ONE WILL BE A LITTLE DIFFERENT TWIST TO IT AND I AM GOING TO LOOK AT

PROBLEM NUMBER NINE IN THE TEXT THIS TIME. HOW MANY SILVER ATOMS ARE CONTAINED IN

THIRTEEN POINT EIGHT (13.8) GRAMS OF SILVER, HOW MANY SILVER ATOMS ARE THERE? NOW WE

KNOW THAT ONE MOLE OF SILVER OR ONE HUNDRED AND EIGHT POINT EIGHT SEVEN (108.87) GRAMS OF

SILVER WOULD CONTAIN AVAGADRO'S NUMBER, SIX POINT ZERO TWO THREE (6.023) TIMES TEN TO THE

TWENTY THIRD. WE CAN SEE HERE OF COURSE, QUITE OBVIOUSLY, THAT THIRTEEN POINT EIGHT IS LESS

THEN THE ONE HUNDRED AND EIGHT THAT IT TAKES FOR AN ENTIRE MOLE SO WE ALREADY KNOW

THAT OUR ANSWER SHOULD COME OUT SOMETHING LESS THEN AVAGADRO'S NUMBER. IF WE MAKE A

CALCULA TION AND WE GET AN ANSWER THAT IS BIGGER THAN AVAGADRO'S NUMBER WE KNOW

THAT WE HAVE DONE SOMETHING INCORRECTLY. NOW LETS CALL TO MIND ONCE AGAIN, WHAT

RELATIONSHIPS WE NOW KNOW, WE KNOW THAT ONE MOLE OF SILVER IS EQUAL TO ITS MOLAR MASS,

AND THAT FROM THE PERIODIC TABLE IS ONE HUNDRED SEVEN POINT EIGHT SEVEN (107.87) GRAMS. WE

ALSO KNOW THAT ONE MOLE OF SILVER IS EQUAL TO SIX POINT ZERO TWO (6.02) TIMES TEN TO THE

TWENTY THIRD SILVER ATOM. WE HAVE TWO CONVERSION FACTORS THIS TIME THAT WE ARE GOING

AGAIN I WILL STEP THROUGH THIS FOR THE EXPONENTIAL PART, SO I HAVE ALREADY SAID TIMES, SIX

POINT ZERO TWO THREE (6.023) AND THEN TO TELL THE COMPUTER TIMES TEN ALL WE DO IS HIT THE

EXPONENTIAL, WE SEE THE TWO ZEROS SHOW UP AND NOW WE PLUG IN THE POWER OF TEN, WHICH IN

THIS CASE IS TWENTY THREE, SO I SHOULD SEE ON MY CALCULATOR SIX POINT ZERO TWO THREE (6.023)

AND NOW I AM GOING TO HIT THE EQUAL SIGN, I DON'T NEED TO RECORD THIS, BUT THE CALCULATOR

HAS DISPLAYED AT THIS POINT WHAT THE ANSWER IS FOR THAT STEP NOW I AM GOING TO TELL TO

DIVIDE WHATEVER YOU HAD UP TO THAT POINT BY ONE ZERO SEVEN POINT EIGHT SEVEN (107.87) AND I

WILL HIT THE EQUAL AGAIN AND THE FINAL ANSWER THEN IS, SEVEN POINT SEVEN ZERO FIVE THREE-

THREE ZERO FIVE (7.7053305) AND THEN YOU SEE JUST A TWENTY TWO, REWRITING IT BACK THEN YOU

SAY TIMES TEN TO THE POSITIVE TWENTY TWO, IF IT IS A NEGATIVE NUMBER THERE WOULD BE A

LITTLE DASH IN FRONT OF IT. NOW LOOKING AT ALL OF THE NUMBERS WE HAVE USED IN OUR

CALCULATIONS, WHICH HAS ONLY INVOLVED MULTIPLYING AND DIVIDING, OUR ANSWER SHOULD NOT

HAVE MORE SIGNIFICANT FIGURES THEN THE POOREST NUMBER THAT WE HAVE USED, AND IN THAT

CASE THAT WOULD BE THE THIRTEEN POINT EIGHT (13.8) WHICH HAS ONLY THREE SIGNIFICANT

FIGURES. WE HAVE FOUR IN AVOGADRO'S NUMBER, WE HAVE FIVE IN THE MASS, BUT WE ARE LIMITED

TO THREE WHICH WAS OUR MEASURED QUANTITY. WE COME OVER HERE NOW, AND WE ARE GOING TO

THEN TAKE CARE OF THIS, NOW NOTICE THAT THIS IS A FIVE, BUT IT IS FOLLOWED BY SOMETHING

OTHER THAN ZERO'S WHICH MEANS THAT IT IS SLIGHTLY LARGER THEN FIVE, IT IS NOT AN EXACT FIVE

SO WE MUST ROUND THE PREVIOUS NUMBER UP, SO OUR FINAL ANSWER WOULD BE SEVEN POINT

SEVEN ONE (7.71) TIMES TEN TO THE TWENTY SECOND SILVER ATOMS. NOW WHAT IF IT HAD BEEN AN

EXACT FIVE AT THAT POINT? WE MENTIONED THIS IN CHAPTER ONE BUT LET ME JUST REFRESH YOUR

MEMORY ON IT, LETS SUPPOSE THAT THE NUMBER HAD IN FACT BEEN SEVEN ZERO FIVE AND THEN

ZERO-ZERO ZERO-ZERO TIMES TEN TO THE TWENTY SECOND, AND WE ARE LIMITING OUR SELVES TO

THREE SIGNIFICANT FIGURES, HOW DO WE HANDLE THAT? WELL THE RULE WAS, WE SAID THAT IF WE

HAVE AN EXACT FIVE, THE LAST SIGNIFICANT FIGURE SHOULD BE AN EVEN NUMBER IF IT IS ALREADY

AN EVEN NUMBER WE THROW THE FIVE AWAY, IF IT IS AN ODD NUMBER WE ROUND IT UP TO THE NEXT

EVEN NUMBER, WHICH MEANS THAT HALF OF THE TIME WE ARE THROWING AWAY AND THE OTHER

HALF OF THE TIME WE ARE ROUNDING UP, SO THAT WE HAVE A CONSISTENT WAY. IN THIS

PARTICULAR CASE THEN WE WOULD HAVE DROPPED ALL OF THIS BECAUSE THE NUMBER PROCEEDING

IT IS ALREADY AN EVEN NUMBER OKAY. ANY QUESTION ON ANY STEP OF THAT PARTICULAR

PROBLEM? ALRIGHT, ONE LAST ONE OF THIS TYPE. THIS ONE HOWEVER IS DEALING WITH A

COMPOUND, HOW MANY NITROGEN DIOXIDE MOLECULES, THAT IS THE WAY WE WOULD SAY THIS

CHEMICAL FORMULA, HOW MANY NITROGEN DIOXIDE MOLECULES ARE THERE IN ONE-HUNDRED AND

THIRTY-FIVE GRAMS OF NITROGEN DIOXIDE, NOW, I KNOW THAT ONE MOLE OF NITROGEN DIOXIDE IS

EQUAL TO ITS FORMULA MASS, WHICH OF COURSE WE CAN CALCULATE, THAT IS GOING TO BE EQUAL

TO ONE NITROGEN TIMES FOURTEEN POINT ZERO ONE (14.01) AND TWO OXYGENS TWO TIMES SIXTEEN

AND SO WE WILL HAVE THEN A TOTAL OF FORTY SIX POINT ZERO ONE (46.01) GRAMS, ONE MOLE OF NO

IS FORTY SIX POINT ZERO ONE (46.01) GRAMS. NOTE TAHT THE AMOUNT THAT WE ARE STARTING

WITH, JUST EYEBALLING IT, THE AMOUNT THAT WE ARE STARTING WITH IS ABOUT THREE TIMES

THAT MUCH AND SO THEREFORE I KNOW THAT I SHOULD END UP WITH THREE TIMES AS MANY

MOLECULES APPROXIMATELY AS I WOULD HAVE IN ONE MOLE OF NO2 AND HOW MANY MOLECULES

DO WE HAVE IN NO2? WE KNOW THAT ONE MOLE OF NO2 IA EQUAL TO SIX POINT ZERO TWO THREE

(6.023) TIME TEN TO THE TWENTY THIRD NO2 MOLECULES, I ALREADY KNOW ABOUT WHAT MY

ANSWER IS GOING TO BE, MY ANSWER IS GOING TO BE APPROXIMATELY EQUAL TO EIGHTEEN TIMES

TEN TO THE TWENTY THIRD OR WE COULD WRITE THAT BETTER IN SCIENTIFIC NOTATION AS ONE

POINT EIGHT (1.8) TIMES TEN TO THE TWENTY FOURTH. NOW, I JUST POINT THIS OUT, THIS IS PROBABLY

USEFUL WAY TO LOOK AT PROBLEMS THAT HAVE SEVERAL NUMBERS IN THEM, TRY TO GET AN

APPROXIMATE ANSWER IT IS PRETTY EASY TO PUNCH IN WRONG NUMBERS ON A CALCULATOR, BUT IF

I CAN MENTALLY JUST LOOK AT IT I SAID THIS GOES INTO THERE ABOUT THREE TIMES AND THEREFORE

I HAVE ABOUT THREE TIMES THAT AMOUNT, THREE TIMES SIX WOULD BE EIGHTEEN TIMES TEN TO THE

TWENTY THIRD MOLECULES SO JUST AN EYEBALL ESTIMATE OF THE ANSWER. NOW WE WILL GO

AHEAD AND ACTUALLY CALCULATE THE TRUE VALUE, SO, MOLECULES OF NO2 EQUAL TO ONE

HUNDRED AND THIRTY FIVE GRAMS AND NO2, I REALLY SHOULD BE WORKING ON THESE WITH THE

TRANSPARENCY SIDEWAYS AND I WILL CHANGE THAT IN THE FUTURE, ONE HUNDRED THIRTY FIVE

CHAPTER ITSELF. THE QUESTION THERE IS, WHAT IS THE PERCENT CARBON IN THE COMPOUND C2H4O2?

NOW, IF WE ARE TALKING ABOUT PERCENT THEN BY DEFINITION, PERCENT IS EQUAL TO X, IN THIS

CASE WE ARE GOING TO BE WORKING IN MASS PERCENTS SO WE WILL SAY THE GRAMS OF X OVER

GRAMS TOTAL TIMES TEN TO THE SECOND, TIMES ONE HUNDRED TO PUT IT INTO PERCENT, THAT IS

THE WAY WE CALCULATE THE PERCENT OF ANYTHING. IN THIS CASE HOWEVER WE ARE DEALING

WITH PERCENTS BASED ON MASS, SO IN ORDER TO CALCULATE THE PERCENT CARBON IN THIS

COMPOUND WE NEED TWO PIECES OF INFORMATION. WE NEED TO KNOW HOW MANY GRAMS OF

CARBON WE HAVE AND WE NEED TO KNOW WHAT THE FORMULA WEIGHT IS, BECAUSE THE FORMULA

WEIGHT IS GOING TO BE GRAMS TOTALED THAT WE NEED, SO WE NEED TWO PIECES OF INFORMATION.

LETS GO AHEAD HERE AND DETERMINE THE FORMULA WEIGHT FIRST, IF WE DO THAT THEN WE SEE

THAT WE HAVE TWO CARBONS TWELVE POINT ZERO ONE (12.01) FOR EACH CARBON TWENTY FOUR

POINT ZERO TWO (24.02) AND WE HAVE FOUR HYDROGENS, EACH HYDROGEN IS ONE POINT ZERO ONE

(1.01) AND WE HAVE THEN FOUR POINT ZERO FOUR (4.04) AND WE HAD TWO OXYGENS IN THE FORMULA

TIMES SIXTEEN IS THIRTY TWO POINT ZERO-ZERO (32.00) AND WE HAVE THEN SIX EIGHT TEN AND SIXTY

AND SO THE TOTAL WEIGHT, THE FORMULA WEIGHT IS SIXTY POINT ZERO SIX (60.06). SO TO ANSWER

OUR QUESTION WHICH WAS DETERMINE THE PERCENT CARBON WE ARE GOING TO TAKE THE NUMBER

OF GRAMS OF CARBON THAT WERE IN THE COMPOUND, WE JUST CALCULATED THAT RIGHT HERE SO

WE HAVE TWENTY FOUR POINT ZERO TWO (24.02) GRAMS ALL DIVIDED BY THE GRAMS TOTAL SIXTY

POINT ZERO SIX (60.06) GRAMS, OUR UNITS OF GRAMS WILL HAVE CANCELED, PERCENT IS A UNITLESS

NUMBER, TIMES TEN TO THE SECOND AND SO NOW IF WE PLUG THIS INTO OUR CALCULATOR THEN WE

WILL TAKE TWENTY FOUR POINT ZERO TWO (24.02) DIVIDED BY SIXTY POINT ZERO SIX (60.06) AND WE

HAVE A PERCENTAGE THEN OF CARBON IN THE COMPOUND EQUAL TO THIRTY NINE POINT NINE-NINE

THREE-THREE FOUR (39.99334) PERCENT. THIS TIME HOW MANY SIGNIFICANT FIGURES ARE WE

ALLOWED? FOUR AREN'T WE? WE KNEW THE GRAMS OF CARBON BEFORE WE KNEW THE GRAMS OF

TOTAL BEFORE SO WE ARE JUSTIFIED WITH FOUR, AND SO WE WOULD WRITE OUR ANSWERS THIRTY

NINE POINT NINE-NINE PERCENT (39.99%) ALRIGHT? ANY QUESTIONS ON ANY STEP OF THAT ONE?

PROBLEM NUMBER TWELVE AT THE END OF THE CHAPTER ASKS US TO CALCULATE THE PERCENTAGE

OF EACH ELEMENT IN THE COMPOUND CALLED AMMONIUM PHOSPHATE, BUT WHAT I AM GOING TO

DO HERE IS I AM JUST GOING TO DO ONE WE DON'T NEED TO DO ALL, THE ONLY REASON I AM GOING TO

DO THIS IS SO THAT WE HAVE A REVIEW OF HOW WE CALCULATE THE FORMULA WEIGHT FOR A

COMPOUND WHICH HAS A SET OF PARENTHESES WITH A SUBSCRIPT NUMBER. LETS CALCULATE THE

PERCENT PHOSPHOROUS, THAT IS WHAT WE WANT TO CALCULATE SO WE KNOW THAT WE NEED TO

KNOW THE GRAMS OF PHOSPHOROUS AND WE NEED TO KNOW THE FORMULA WEIGHT OR FORMULA

MASS ONCE AGAIN. SO, WE GO AHEAD AND THEN AND WE ARE GOING TO HAVE TO CALCULATE THE

FORMULA MASS, REMEMBER THAT WHEN WE ARE DEALING WITH A FORMULA WHICH HAS A

PARENTHESES WITH A SUBSCRIPT FOLLOWING IT THEN EVERYTHING INSIDE THE PARENTHESES IS

MULTIPLIED BY THAT SUBSCRIPT NUMBER, SO, WE HAVE THREE TIMES ONE, WE HAVE THREE

NITROGENS AND EACH OF THOSE WEIGHS FOURTEEN POINT ZERO ONE (14.01). WE HAVE THREE TIMES

FOUR WHICH IS TWELVE HYDROGENS MULTIPLIED BY ONE POINT ZERO ONE (1.01). WE HAVE ONE

PHOSPHOROUS SO WE ARE GOING TO MULTIPLY THAT, LOOKING AT THE CHART UP THERE, BY THIRTY

POINT NINE SEVEN (30.97) AND WE HAVE FOUR OXYGENS MULTIPLIED BY SIXTEEN SO THAT GIVES US

THEN, ALL OF THE ELEMENTS AND THEIR PROPER AMOUNT IN THE FORMULA AND SO THE FIRST ONE

WOULD BE FORTY TWO POINT ZERO THREE (32.03), TWELVE POINT ONE TWO (12.12), THIRTY POINT NINE

SEVEN (30.97) AND SIXTY FOUR POINT ZERO-ZERO (64.00), ADDING THEM UP THEN WE HAVE ONE

HUNDRED AND FORTY NINE POINT ONE TWO (149.12). THIS WOULD BE THE FORMULA MASS, THAT

WOULD BE OF COURSE, HOW MUCH ONE MOLE OF THE COMPOUND WOULD WEIGH, SO FINALLY THE

PERCENT PHOSPHOROUS WOULD BE EQUAL TO THE GRAMS CONTRIBUTED BY THE PHOSPHOROUS

THIRTY POINT NINE SEVEN (30.97) GRAMS DIVIDED BY ONE HUNDRED FORTY NINE POINT ONE TWO

(149.12) GRAMS, GRAMS CANCEL, MULTIPLY IT BY TEN TO THE SECOND TO GIVE US AN ANSWER IN

PERCENT, SO THIRTY POINT NINE SEVEN (30.97) DIVIDED BY ONE HUNDRED FORTY NINE POINT ONE TWO

(149.12) AND WE KNOW IN THIS CASE THAT OUR ANSWER SHOULD HAVE FOUR SIGNIFICANT FIGURES,

AND SO OUR ANSWER WOULD BE TWENTY POINT SEVEN-SEVEN PERCENT (20.77%). ANY QUESTION ON

ANY PART OF THAT ONE? NOW AS I SAID THE REAL IMPORTANCE OF USING PERCENT IS TO BE ABLE TO

CALCULATE THE CHEMICAL FORMULA, IN OTHER WORDS WE HAVE AN UNKNOWN COMPOUND, WE

AGAIN THAT GETS RID OF THAT AND WE ARE AT MOLES, AND MOLES OF CHLORINE WOULD BE EQUAL

TO ZERO POINT SEVEN ONE SEVEN (0.717) GRAMS OF CHLORINE MULTIPLIED BY ONE MOLE OF Cl PER THIRTY FIVE POINT FOUR FIVE (35.45) GRAMS OF Cl, AGAIN, IT IS ATOMIC WEIGHT. NOW WE HAVE THE NUMBER OF MOLES OF EACH OF THE ELEMENTS PRESENT, SO LETS CALCULATE THAT TO SEE WHAT WE GET FOR A NUMBER HERE AND SO WE HAVE POINT TWO FOUR THREE (.243) DIVIDED BY TWELVE POINT ZERO ONE (12.01) AND MY FIRST NUMBER HERE IS ZERO POINT ZERO TWO ZERO TWO (0.0202). THE NEXT NUMBER POINT ZERO FOUR ZERO FOUR (.0404) DIVIDED BY ONE POINT ZERO ONE (1.01) WE HAVE A NUMBER OF ZERO POINT ZERO FOUR ZERO-ZERO (0.0400) AND WE HAVE POINT SEVEN ONE SEVEN (.717) DIVIDED BY THIRTY FIVE POINT FOUR FIVE (35.45) AND WE HAVE ZERO POINT ZERO TWO ZERO TWO (0.0202) NOW, WE KNOW THAT ONE MOLE OF ANYTHING CONTAINS THE SAME NUMBER OF ATOMS, SO ACTUALLY IF WE WERE TO TAKE THOSE MOLES AND MULTIPLY BY AVAGADRO'S NUMBER IT WOULD TELL US THE NUMBER OF ATOMS OF EACH OF THOSE PRESENT. IN OUR FORMULA OF COURSE WE ARE USUALLY TALKING ABOUT ONE, NOW WE COULD WRITE THE FORMULA AS C0.0202 H0.0400 AND Cl0. WELL WE KNOW THAT IN CHEMICAL FORMULAS THAT WE DON'T HAVE FRACTIONAL ATOMS, WE HAVE WHOLE ATOMS IN CHEMICAL FORMULAS. AND SO AT THIS POINT OUR MANIPULATION IS MERELY HOW CAN WE CONVERT THIS SET OF NUMBERS, THIS RATIO WHICH IS CORRECT, HOW CAN WE CONVERT THAT NOW TO WHOLE NUMBERS, AND MAKE THEM AS SMALL AS POSSIBLE, THE EMPIRICAL FORMULA IS THE SMALLEST WHOLE NUMBER RATIO. WELL THE WAY WE CAN DO THIS IS TO DIVIDE EACH BY THE SMALLEST NUMBER SO IF WE NOW TAKE CARBON, TAKE THE NUMBER THAT WE HAD FOR IT ZERO POINT TWO ZERO TWO (0.0202) AND THE SMALLEST NUMBER OF THE MOLES THAT WE HAD FOR EVERYTHING WAS, WHOOPS WE NEED ANOTHER ZERO IN HERE, ZERO POINT ZERO TWO ZERO TWO (0.0202) AND THEN HYDROGEN WAS ZERO POINT FOUR ZERO-ZERO (0.400) DIVIDED BY ZERO POINT ZERO TWO ZERO TWO AND CHLORINE ZERO POINT ZERO TWO ZERO TWO (0.0202) DIVIDED BY ZERO POINT ZERO TWO ZERO TWO (0.0202) WELL QUITE OBVIOUSLY THE CARBON AND THE CHLORINE ARE NOW ONES, WHEN WE DIVIDE THEM BY THE SMALLEST NUMBER WHICH IN FACT IS THEIR SAME NUMBER THEY ARE GOING TO GIVE US ONE, THE OTHER NUMBER WE GET THEN WOULD BE POINT ZERO FOUR (.04) DIVIDED BY POINT ZERO TWO ZERO TWO (.0202) AND THAT GIVES US A NUMBER HERE EQUAL TO ONE

POINT NINE EIGHT (1.98) NOW AT THIS POINT WHEN WE HAVE DIVIDED THE FIRST TIME BY THE

SMALLEST WHOLE NUMBER OF MOLES FROM OUR FIRST CALCULATION WE CAN NOW DO A LITTLE BIT

OF ROUNDING OFF AND WE CAN ROUND OFF UP TO A MAXIMUM OF ABOUT ONE TENTH (1/10) SO IN

THIS PARTICULAR CASE ONE POINT NINE EIGHT (1.98) JUSTIFIABLY BECAUSE WE ARE WORKING WITH

EXPERIMENTAL DATA COULD BE CALLED TWO AND OUR EMPIRICAL FORMULA THEN WOULD BE C ONE

WE DON'T WRITE THE ONE, H TWO Cl AND THAT WOULD BE CALLED THE EMPIRICAL FORMULA FOR THIS CHEMICAL COMPOUND ALRIGHT? NOW, WE MAY NOT ALWAYS GET WHOLE NUMBERS IN THIS STEP, WHEN WE DIVIDE THROUGH WE MAY NOT GET WHOLE NUMBERS, WE MIGHT GET SOMETHING LIKE ONE POINT FIVE (1.5), OR ONE POINT THREE-THREE (1.33) WE CAN'T ROUND OFF THAT AMOUNT AND IN THAT CASE WE NEED TO LOOK FOR A SECOND MATHEMATICAL STEP TO CHANGE EVERYTHING TO A WHOLE NUMBER, IF IT WERE ONE POINT FIVE (1.5) IF THIS WERE ONE, ONE POINT FIVE AND ONE WE COULD MULTIPLY EVERYTHING THROUGH BY TWO AND THAT WOULD OF COURSE GIVE US A RATIO OF TWO TO THREE TO TWO. SO WE MUST LOOK FOR A MULTIPLIER SOMETIMES TO GET A WHOLE NUMBERS IN THE EMPIRICAL FORMULA, WELL IN THE NEXT LECTURE WE WILL DO A COUPLE MORE EXAMPLES OF THIS AND CONTINUE ON THEN AND TALK SOME ABOUT MOLECULAR FORMULAS AND LOOK AT STRUCTURAL FORMULAS ONCE AGAIN THAT WE TALKED ABOUT AT THE BEGINNING OF CHAPTER TWO.