- WELCOME TO THE VIDEO TUTORIAL COMPOUNDS PART THREE, THE MOLE.
THIS VIDEO TUTORIAL WILL GIVE US AN INTRODUCTION
TO BOTH THE QUALITIES AND THE QUANTITIES
OF THE CONCEPT OF A MOLE IN CHEMISTRY, VERY KEY CONCEPT.
LET'S GET STARTED. OKAY.
SO THE MOLE IS THE AMOUNT OF A SUBSTANCE
AND SOME PEOPLE EVEN THINK
THAT WE SHOULD CALL IT THE CHEMICAL AMOUNT
BECAUSE IT'S NEAR AND DEAR TO A CHEMIST'S HEART.
WHY IS THAT? WELL, HERE'S THE MOLE.
THE MOLE CAN BE A SIMPLE
AS SAYING 6.02 X 10 TO THE 23RD ITEMS,
ATOMS, ELEMENTS, COMPOUNDS, ANYTHING.
ATOMS OR COMPOUNDS. WE'LL JUST STAY THERE. OKAY.
NOW, THIS DEFINITION RIGHT HERE WAS CALLED AVOGADRO'S NUMBER
AND IT'S AS SIMPLE AS THAT, JUST THE DEFINITION
SO WE COULD HAVE A MOLE OF ANYTHING, A MOLE OF WATERMELONS,
A MOLE OF SHOES, A MOLE OF HATS,
BUT IT MAKES MORE SENSE TO USE IT IN CHEMISTRY
WITH ATOMS OR COMPOUNDS BECAUSE THEY'RE SO TINY.
BUT A MOLE CAN ALSO BE DEFINED AS A SPECIFIC NUMBER OF GRAMS
OF A SAMPLE.
AND WHEN WE DEFINE THE MOLE THIS WAY, WE CALL IT THE MOLAR MASS.
ALL RIGHT.
NOW, WHEN WE'RE THINKING ABOUT AVOGADRO'S NUMBER
AND THE ATOMS THAT THE NUMBER OF ATOMS
OR THE NUMBER OF COMPOUNDS,
THIS IS MORE OF A THEORETICAL PERCEPTION, RIGHT?
THIS IS WHERE FOR THINKING ABOUT A CHEMICAL FORMULA, RIGHT?
CHEMICAL FORMULAS OR WE'RE GOING TO WORK OUR WAY UP
TO CHEMICAL REACTIONS, YOU KNOW,
THERE'S GOING TO BE MILLIONS AND MILLIONS AND MILLIONS
OF ATOMS OR COMPOUNDS
BUT WE'RE GOING TO THINK ABOUT JUST ONE OR TWO AT A TIME.
THAT'S WHAT OUR BRAIN CAN DO OR SOLUTION COMPOSITION.
WHAT EXACTLY IS FLOATING AROUND IN THAT SOLUTION, RIGHT?
SO THESE ARE THEORETICAL IDEAS THAT WILL DRIVE OUR WORK
BUT WE CAN'T LOOK AT THESE INDIVIDUAL ATOMS OR COMPOUNDS.
WE DON'T HAVE ATOMIC TWEEZERS.
SO, IN THE LAB, WE NEED TO LOOK AT THE MOLAR MASS.
GRAMS PER MOLE, RIGHT?
AND SO THIS IS OUR EXPERIMENTAL, RIGHT?
WHEN WE GO INTO THE LAB, RIGHT?
WHEN WE GO INTO THE LAB WE CAN'T TAKE OUT OUR ATOMIC TWEEZERS
AND GRAB ONE ATOM AT A TIME
BUT WE CAN WEIGH OUT GRAMS OR MILLILITERS OF A SAMPLE, RIGHT?
WE CAN MAKE SOME MEASUREMENTS.
SO THE IMPORTANT THING IS THE POWER OF THE MOLE IS IT LINKS
WHAT WE'VE LEARNED ABOUT AT THE ATOMIC LEVEL, RIGHT?
THINKING ABOUT THE STRUCTURE OF THESE ATOMS, RIGHT?
WITH CARBON OR SILVER
THAT THEY HAVE DIFFERENT NUMBERS OF PROTONS AND NEUTRONS
AND SO THEY HAVE DIFFERENT MASSES, RIGHT?
SO WHEN WE LOOK AT CARBON,
WE KNOW THAT THE AVERAGE MASS WOULD BE 12.01 AMU/ATOM
IF WE AVERAGED OUT ALL THE ISOTOPES.
THE BEAUTY OF THE MOLE IS THAT THE SAME VALUE,
THE NUMERICAL VALUE THAT WORKS IN THE THEORY WORLD
WITH INDIVIDUAL ATOMS,
THE SAME NUMERICAL VALUE 12.01 IS ALSO THE NUMBER OF GRAMS
IN A MOLE OF CARBON.
SO WHEN WE HAVE EXACTLY 6.02 X 10 TO THE 23RD ATOMS
WE GET THIS BEAUTIFUL CONNECTION HERE
WHERE WE GET TO KEEP THE NUMBER THE SAME
AND WE JUST GET TO SWAP OUT THE UNITS
AND IT WORKS FOR EVERYBODY.
SO FOR SILVER FOR EXAMPLE
WHERE WE HAVE A DIFFERENT RATIO OF ISOTOPES,
THE AVERAGE ATOMIC MASS OF ONE SILVER ATOM IS 107.87 AMU
BECAUSE IT'S A MUCH LARGER ATOM.
SO IF WE WANTED TO HAVE THE SAME NUMBER OF SILVER ATOMS
AS COPPER ATOM--OH, EXCUSE ME, CARBON ATOMS, RIGHT?
IF WE WANTED TO HAVE ONE MOLE OF SILVER ATOMS,
IF WE WANTED TO HAVE 6.02 X 10 TO THE 23RD SILVER ATOMS,
WE WOULD NEED TO WEIGH OUT 107.87 GRAMS TO HAVE A MOLE.
SO HERE, SO I'M GOING TO SHOW YOU AND--
SO LET'S LOOK AT THIS BRIEFLY, PUT SOME PAPER DOWN, RIGHT?
SO HERE IS--HERE'S OUR CARBON, RIGHT?
AND THEN HERE'S THE SILVER, RIGHT?
SO AS WE WOULD EXPECT, SO THERE'S ONE MOLE OF CARBON HERE
AND THEN THERE'S ONE MOLE OF SILVER.
SO BECAUSE THE SILVER ATOMS HAVE MORE PROTONS AND NEUTRONS,
THEY HAVE MUCH GREATER MASS,
SO BOTH OF THESE SAMPLES, WHAT DO THEY HAVE THAT'S THE SAME?
THEY BOTH HAVE 6.02 X 10 TO THE 23RD ATOMS.
THEY HAVE THE SAME NUMBER OF ATOMS
BUT THERE'S A DIFFERENT MASS PRESENT
BASED ON THE ATOMIC STRUCTURE.
ALL RIGHTY.
NOW, LET'S TAKE THIS--WELL, LET'S--WE'LL FOCUS ON ELEMENTS
FOR A LITTLE BIT LONGER
AND THEN WE'LL WORK OUR WAY UP TO COMPOUNDS.
ALL RIGHTY.
SO HERE'S ALL OF OUR DEFINITIONS AT THE TOP OF THE PAGE, RIGHT?
SO HERE'S OUR AVOGADRO'S NUMBER OR MOLAR MASS, RIGHT?
SO THIS IS--AVOGADRO'S NUMBER IS A DEFINED TERM.
NOW, THIS MOLAR MASS, RIGHT NOW IT'S REALLY SIMPLE.
WE JUST LOOK ON THE PERIODIC TABLE FOR THE ATOMIC MASS
OF THE INDIVIDUAL ELEMENT, RIGHT?
WHEN WE GET THE COMPOUNDS,
WE'LL HAVE TO COMBINE THE RATIO OF ATOMS WITH THE INFORMATION
ON THE PERIODIC TABLE.
BUT DON'T WORRY, WE'LL GET THERE.
LET'S JUST STAY WITH SINGLE ATOMS FOR THE TIME BEING, RIGHT?
SO RIGHT, WITH CARBON 12.01 GRAMS IS A MOLE, RIGHT,
VERSUS SILVER, WHICH HAS MORE PROTONS AND NEUTRONS
AND SO IT HAS A MUCH LARGER MASS, RIGHT,
A 107.87 GRAMS PER MOLE.
ALL RIGHTY. NOW, WHICH SAMPLE CONTAINS MORE ATOMS?
SO IF WE HAVE 21 GRAMS OF CARBON, RIGHT,
VERSUS 21 GRAMS OF SILVER.
OKAY. SO JUST INTUITIVELY WHICH SAMPLES CAN HAVE MORE ATOMS?
GO AHEAD AND MAKE A PREDICTION. ALL RIGHTY.
NOW, LET'S DO SOME CALCULATIONS TO SEE HOW YOU DID. ALL RIGHT.
SO WE KNOW THAT FOR EVERY 12 GRAMS
WE WILL HAVE 1 MOLE OF CARBON.
SO IF WE PUNCH--WE TAKE 21/12.01 WE HAVE 1.7 MOLES OF CARBON.
IF WE HAVE 21 GRAMS OF SILVER, RIGHT?
THAT BECAUSE OF THE LARGER MASS OF EACH SILVER ATOM,
IT TAKES A 107.87 GRAMS TO GET A MOLE A SILVER.
SO KNOW WE CAN SEE THAT IF WE HAVE 21 GRAMS OF SILVER,
WE ONLY HAVE 0.19 MOLES OF SILVER, RIGHT?
SO DEFINITELY, RIGHT, 1.7 MOLES, THERE WE GO.
THIS IS THE LARGER NUMBER, RIGHT,
1.7 MOLES OF CARBON MEANS WE HAVE MORE ATOMS.
BUT LET'S SAY YOU'RE NOT CONVINCED.
LET'S KEEP GOING, RIGHT?
SO FIRST WE USE THE MOLAR MASS.
WE WENT FROM GRAMS TO MOLES
AND WE COULD COMPARE THE MOLES DIRECTLY.
BUT LET'S SAY I'M REALLY CURIOUS.
I WANT TO KNOW EXACTLY HOW MANY ATOMS THAT WOULD BE, RIGHT?
1.7 MOLES OF CARBON, EACH ONE OF THE MOLES WOULD HAVE 6.02 X 10
TO THE 23RD ATOMS.
SO THAT 1.7 MOLES WOULD GIVE US 1.0 X 10
TO THE 24TH CARBON ATOMS, RIGHT,
OR WE'LL JUST SAY 10 X 10 TO THE 23RD CARBON ATOMS.
AND YOU MIGHT WONDER WHY I'M DOING THIS
BUT I'LL SHOW YOU JUST A SEC, RIGHT?
SO WE ONLY HAD 0.19 MOLES OF SILVER
AND THE MOLE DEFINITION NEVER CHANGES
WHEN WE'RE LOOKING FOR ATOMS OR COMPOUNDS, ONE MOLE OF SILVER,
ONE OF ANYTHING IS 6.02 X 10 TO THE 23RD.
ALL RIGHT.
SO WE PUNCH THAT THROUGH OUR CALCULATION AND WE GET 1.1 X 10
TO THE 23RD SILVER ATOMS, RIGHT?
SO THAT'S WHY I DID THIS, ALL RIGHT,
SO WE CAN SEE THAT 10, RIGHT.
THE EXPONENTS ARE THE SAME, SO 10 IS MUCH BIGGER.
SO PROBABLY BELABOR THE POINT A LITTLE BIT BUT THE--
I REALLY WANT TO EMPHASIZE THIS.
WE CAN COMPARE MOLES OF A SUBSTANCE
OR WE CAN COMPARE ATOMS OF A SUBSTANCE CHEMICALLY.
THAT IS CHEMICALLY SOUND.
NOTICE WE CANNOT--BOTH SAMPLES WERE 21 GRAMS.
SO THE MASSES WERE THE SAME
BUT CHEMICALLY THEY HAD DIFFERENT AMOUNTS OF SUBSTANCE.
SO IF WE WANTED TO DO REACTIONS,
WE HAVE A LOT MORE ATOMS OF CARBON AVAILABLE
THAN WE DO SILVER, RIGHT?
SO THE BRIDGE--THIS IS VERY-- THE MOLE--
THIS IS VERY IMPORTANT BRIDGE
THAT GETS US FROM THE INDIVIDUAL ATOMS--OOPS, AND COMPOUNDS,
RIGHT, TO THE LAB WHERE WE CAN WEIGH WITH GRAMS.
OKAY.
AND THE COOL PART IS IT'LL-- THIS--IT--THIS BRIDGE HAPPENS
BECAUSE IT ACTUALLY TAKES INTO ACCOUNT THE DIFFERENCES
IN THE ATOMIC STRUCTURE OF THE ELEMENTS
FORMING THE COMPOUNDS OF INTEREST.
SO THAT IS WHAT MAKES THE MOLE SUCH AN IMPORTANT
AND POWERFUL CONCEPT.
SO WE'VE SPENT SOME TIME GETTING TO KNOW THE MOLE
FOR INDIVIDUAL ELEMENTS.
LET'S GO AHEAD NOW AND GET A LITTLE MORE CHALLENGING.
LET'S LOOK AT COMPOUNDS. ALL RIGHT.
SO THIS JUST ADDS ANOTHER LAYER TO IT
BECAUSE NOW WE HAVE A CHEMICAL FORMULA.
SO LET'S LOOK HERE.
REMEMBER THAT THE CHEMICAL FORMULA IS THE FIXED RATIO
OF ATOMS
AND NOW WE CAN TAKE THAT ONE STEP FURTHER.
WE CAN TALK--THINK ABOUT IT IN TERMS OF THE MOLE RATIO, RIGHT,
OF ATOMS THAT FORM ONE MOLE OF THE COMPOUND.
ALL RIGHT.
SO LET'S COMPARE CALCIUM CHLORIDE WITH GLUCOSE, RIGHT.
SO WE'RE THINKING--BECAUSE--
OH, THERE'S ONE CALCIUM ION AND TWO CHLORIDE IONS
BUT NOW THAT WE KNOW ABOUT THE MOLE WE COULD SAY,
OH, THERE'S A MOLE OF CALCIUM AND TWO MOLES OF CHLORIDE,
ALL RIGHT.
SEE HOW WE CAN INTERCHANGE INDIVIDUAL IONS WITH MOLES.
ALL RIGHT.
SO LET'S COMPARE THESE.
WHAT TYPE OF COMPOUNDS ARE THESE, RIGHT?
WELL, WITH CALCIUM CHLORIDE, RIGHT, WE RECOGNIZE THAT WE SEE,
WE HAVE A METAL, RIGHT,
AND THEN WE HAVE TWO CHLORIDE IONS ARE NONMETAL, RIGHT?
SO YOU HAVE METALS AND NONMETALS,
SO WE KNOW THAT THIS AN IONIC COMPOUND
AND THEN WE LOOK AT GLUCOSE, RIGHT.
WE SEE ALL NONMETALS. OKAY.
AND MAYBE WE SHOULD DRAW SOME NOTES HERE, RIGHT?
SO WITH CALCIUM, WE RECOGNIZE CALCIUM AS A METAL
AND CHLORIDE AS A NONMETAL.
SO WHENEVER WE HAVE A MIX OF METAL AND NONMETAL,
WE KNOW THAT WE'RE GOING TO BE FORMING AN IONIC COMPOUND,
RIGHT, WITH THE LOSS AND GAIN OF ELECTRONS
AND HERE WITH ALL NONMETALS,
THIS IS GOING TO BE A COVALENT COMPOUND, OKAY.
SO THERE--HERE, ALL OF THE ATOMS WILL SHARE.
NOW, THIS HAS A HUGE EFFECT
WHEN WE LOOK AT THE STRUCTURE OF THESE COMPOUNDS
WHEN WE DISSOLVE THEM IN WATER, RIGHT?
AND WE TALKED ABOUT--THAT WAS ANOTHER IMPORTANT WAY TO THINK
ABOUT THE MOLE, RIGHT, SOLUTION COMPOSITION.
SO, HERE WE HAVE OUR SALT CRYSTAL OF CALCIUM CHLORIDE
AND WE PUT IT INTO WATER, RIGHT?
WE'RE GOING TO GET CALCIUM IONS
AND THEN WE'RE GOING TO GET TWO CHLORIDE IONS
OR FOR EACH MOLE OF CALCIUM CHLORIDE,
WE'LL GET ONE MOLE OF CALCIUM IONS
AND TWO MOLES OF CHLORIDE IONS, RIGHT?
SO WE WILL GET THAT ONE TO TWO RATIO
WHICH WE SEE IN THE CHEMICAL FORMULA.
ON THE OTHER HAND, WHEN WE LOOK AT GLUCOSE,
NOW THE STRUCTURE'S A LITTLE TRICKY.
SO YOU'RE JUST GOING TO HAVE TO TAKE--JUST TRUST ME ON THIS,
I'M GOING TO SHOW IT THIS WAY.
PRETTY SOON THIS WILL BE AN OLD FRIEND, RIGHT?
BUT NOTICE THE STRUCTURE OF GLUCOSE.
LOOK AT ALL OF THESE HYDROXYL GROUPS. OKAY.
SO I DID THIS IN A SKELETAL LINE STRUCTURE.
EACH CORNER IS A CARBON
AND WE'LL TALK MORE ABOUT THAT VERY SOON
BUT THE IMPORTANT THING TO NOTICE
IS LOOK AT ALL OF THESE HYDROXYL GROUPS,
THEY LOVE TO INTERACT WITH WATER
AND SO GLUCOSE LIKES TO DISSOLVE IN WATER, RIGHT?
BUT THIS IS AN IMPORTANT THING TO KEEP IN MIND
THAT FOR THE IONIC COMPOUND,
THESE BECOME INDEPENDENT IONS IN SOLUTION, RIGHT?
WHEREWITH COVALENT COMPOUNDS, THE MOLECULES STAY IN TACKED,
RIGHT?
SO, FORMULA UNITS SEPARATE OUT, MOLECULES STAY TOGETHER.
SO, A VERY IMPORTANT DISTINCTION
AND A GOOD EXAMPLE OF HOW WE CAN LOOK AT THE MOLE RATIO.
NOW, LET'S LOOK AT SOME QUESTIONS SIMILAR
TO WHAT WE DID WITH SILVER AND CARBON.
LET'S SAY WE HAVE 21 GRAMS SAMPLE OF CALCIUM CHLORIDE,
RIGHT?
SO IF WE HAVE 21 GRAMS OF CALCIUM CHLORIDE,
HOW MANY MOLES WOULD WE HAVE?
WELL, WE'RE GOING TO NEED TO DETERMINE THE MOLAR MASS, RIGHT?
SO THE MOLAR MASS COMES FROM THE CHEMICAL FORMULA
IN THE PERIODIC TABLE.
SO FOR CALCIUM, RIGHT, WE HAVE ONE CALCIUM
AND IF WE LOOK ON THE PERIODIC, TABLE
THAT CALCIUM WEIGHS 40.08 GRAMS.
AND THEN THE CHLORIDES, WE HAVE TWO OF THOSE, RIGHT?
SO THE ONE-- YEAH, I'LL PUT THE ONE THERE.
MAYBE WE SHOULD PUT THOSE IN RED. OKAY.
SO THE ONE AND THE TWO, THOSE COME FROM THE RATIO.
SO ONE CALCIUM, TWO CHLORIDES.
THE MASK COMES FROM THE PERIODIC TABLE.
SO WHEN WE ADD THOSE ALTOGETHER, 110.98 GRAMS PER MOLE.
IT'S VERY IMPORTANT TO PUT THE UNITS.
THEY'RE GOING TO HELP YOU, RIGHT?
SO HERE, WE'RE BEING ASKED TO CALCULATE THE MOLES
AND WE'RE GIVEN THE GRAMS
SO IT TAKES 110.98 GRAMS OF CALCIUM CHLORIDE
TO HAVE ONE MOLE OF CALCIUM CHLORIDE.
SO, WE CAN CANCEL OUT THE GRAMS AND WE'RE LEFT WITH MOLES,
PUNCH THIS IN TO OUR CALCULATOR AND WE SEE THAT WE GET--
LET'S ANSWER IT, 0.19 MOLES OF CALCIUM CHLORIDE.
OKAY. NOW, LET'S TRY GLUCOSE, RIGHT?
AS WE COULD START TO SUSPECT, RIGHT?
ASK YOURSELF RIGHT NOW,
DO WE EXPECT THERE TO BE MORE MOLES OF GLUCOSE OR LESS?
IT'S A MUCH BIGGER MOLECULE, HAS A LOT MORE ATOMS.
THEY DON'T NECESSARILY WEIGHT AS MUCH
BUT THERE'S A LOT MORE OF THEM.
WELL, LET'S DOUBLE CHECK, RIGHT?
CARBON, CARBON, WE'LL KEEP GOING WITH THE COLOR CODING, RIGHT?
SO THERE'S SIX CARBON ATOMS,
THERE'S TWELVE HYDROGENS AND THERE'S SIX OXYGENS.
SO THESE NUMBERS IN RED CAME FROM OUR MOLE RATIO.
THEN WE LOOK ON THE PERIODIC TABLE AND WE GET THE MOLAR MASS
FOR EACH ELEMENT, RIGHT?
AND WE ADD THOSE ALTOGETHER.
GLUCOSE WEIGHS 180.18 GRAMS PER MOLE.
NOW THAT WE HAVE THE MOLAR MASS, WE CAN SAY,
IF WE HAVE 21 GRAMS IN GLUCOSE, WE'LL JUST GIVE THAT A BIG G,
RIGHT?
SO IT TAKES 180 GRAMS OF GLUCOSE TO MAKE ONE MOLE OF GLUCOSE.
SO WE CAN SEE HERE THAT 21 GRAMS
IS ONLY GOING TO BE 0.12 MOLES OF GLUCOSE.
ALL RIGHT. SO IT SEEMS A BIT COUNTERINTUITIVE, RIGHT?
BUT IT--IF WE THINK ABOUT IT A LITTLE DEEPER,
IT STARTS TO MAKE SENSE.
IF THE MOLAR MASS IS SMALLER, IT TAKES LESS GRAMS TO MAKE A MOLE
SO WE END UP WITH MORE SUBSTANCE.
IF OUR SAMPLE IS HEAVIER, IF IT HAS A LARGER MOLAR MASS,
IT TAKES A BIGGER SAMPLE TO HAVE THE SAME NUMBER OF COMPOUNDS
OF MOLECULES.
ALL RIGHTY.
SO LET'S JUST DO A COUPLE MORE QUESTIONS TO WRAP THIS UP
FOR THIS PAGE, RIGHT?
HOW MANY MOLES OF CHLORIDE IONS ARE PRESENT IN THAT 21 GRAMS?
OKAY. SO WE'LL KEEP GOING, RIGHT?
SO THE 21 GRAMS GAVE US 0.19 MOLES OF CALCIUM CHLORIDE
SO HOW MANY MOLES OF CHLORIDE IONS, RIGHT?
SO FOR EVERY MOLE OF CALCIUM CHLORIDE WE WOULD HAVE TWO MOLES
OF CHLORIDE IONS, RIGHT?
THAT'S WHAT THE CHEMICAL FORMULA TELLS US.
IT GIVES US THE MOLE RATIO.
SO WE WOULD HAVE 0.38 MOLES OF CHLORIDE IONS AND SIMILARLY,
RIGHT?
NOW WE STILL HAVE 21 GRAMS OF GLUCOSE SO--RIGHT?
GLUCOSE HAS A HIGHER MOLAR MASS SO WE ENDED UP WITH LESS MOLES.
SO IF WE HAVE 0.12 MOLES OF GLUCOSE,
IF WE LOOK AT THE CHEMICAL FORMULA FOR GLUCOSE,
WE WOULD SAY THAT FOR EVERY MOLE OF GLUCOSE, RIGHT?
HOW MANY MOLES OF CARBON, RIGHT?
VERY STRAIGHT FORWARD, RIGHT THERE, RIGHT?
WE WOULD HAVE SIX MOLES OF CARBON ATOMS.
OKAY. SO ONCE AGAIN, RIGHT?
SEE LIKE MOLES ARE CANCELLING AND THEN WE'RE LEFT HERE. OKAY.
SO WE PLUG THAT IN OUR CALCULATOR,
0.72 MOLES OF CARBON ATOMS.
ALL RIGHTY.
SO WE WANT TO BE COMFORTABLE USING THE MOLAR MASS
TO GO FROM GRAMS OF A SUBSTANCE TO MOLES OF A SUBSTANCE
AND WE WANT TO RECOGNIZE THAT THE CHEMICAL FORMULAS
ARE ALSO MOLE RATIOS
WHEN WE THINK ABOUT HAVING A MOLE OF THE SUBSTANCE.
OKAY. AND NOW LET'S TAKE THIS ONE STEP FURTHER.
LET'S LOOK AT CHEMICAL REACTIONS, OKAY?
SO WE'VE TALKED ABOUT ALL THE DIFFERENT WAYS
THAT THE MOLE IS IMPORTANT.
SO LET'S DO A GRAND FINALE HERE. ALL RIGHT.
MOLECULES AND MOLES CAN BE USE INTERCHANGEABLY
BECAUSE THEY'RE LINKED BY AVOGADRO'S NUMBER.
SO I THINK THAT'S WHY CHEMISTS LOVE THEM SO MUCH. ALL RIGHT.
ATOMS OR MOLECULES. OKAY.
SO AS WE CAN SEE FROM THOSE PREVIOUS EXAMPLES, RIGHT?
WE CAN'T COMPARE GRAMS OF SAMPLES DIRECTLY.
THEY--WE HAVE TO ADJUST FOR THE DIFFERENT ATOMIC MASSES
OF THE ELEMENTS INVOLVED, RIGHT?
BECAUSE DIFFERENT ELEMENTS HAVE DIFFERENT NUMBERS OF PROTONS
AND NEUTRONS.
ALL RIGHT.
SO EVERYTHING LINKS BACK TO THAT BASIC ATOMIC STRUCTURE.
PRETTY COOL. ALL RIGHT. SO HERE'S OUR BUDDY GLUCOSE, RIGHT?
GLUCOSE CAN REACT WITH OXYGEN TO PRODUCE CARBON DIOXIDE AND WATER
AND THIS IS A BALANCED REACTION
SO LET'S LOOK AT WHAT THAT MEANS.
WELL, IN TERMS OF THE NUMBER OF MOLECULES, IS IT BALANCED?
WELL, WE HAVE ONE MOLECULE OF GLUCOSE
REACTING WITH SIX MOLECULES OF OXYGEN, RIGHT,
TO MAKE SIX MOLECULES OF CO2 AND WATER.
WELL, THAT DOESN'T LOOK BALANCED. HMM.
LET'S TRY MOLES, RIGHT?
WE COULD SAY ONE MOLE OF GLUCOSE REACTS WITH SIX MOLES OF OXYGEN
TO PRODUCE SIX MOLES OF CO2 AND SIX MOLES OF WATER.
HMM. ALL RIGHT. LET'S DIG IN A LITTLE DEEPER.
LET'S LOOK AT THE MOLAR MASS. OKAY.
SO WE'VE ALREADY DONE THE MOLAR MASS FOR GLUCOSE
ON THE PREVIOUS PAGE.
IT'S 180.18 GRAMS PER MOLE. OXYGEN'S PRETTY SIMPLE, RIGHT?
2 X 16, SO 32 GRAMS PER MOLE. CARBON DIOXIDE, RIGHT?
ONE CARBON AND TWO OXYGENS, ALL RIGHT?
SO 44.01 GRAMS PER MOLE AND WATER, RIGHT?
TWO HYDROGENS AND ONE OXYGEN. OKAY. ALL RIGHTY.
SO NOW, LET'S SEE.
WHY DO WE SAY THAT THIS IS A BALANCED REACTION?
WELL, LET'S LOOK AT THE GRAMS FROM THE MOLE RATIOS, RIGHT?
SO IF WE HAVE ONE MOLE OF GLUCOSE AND THAT--
SO THAT WOULD BE THE 180.18 GRAMS
BUT WE HAVE SIX MOLES OF OXYGEN
AND THEY EACH WEIGH THIRTY-TWO, RIGHT?
SO IF WE HAVE SIX MOLES, THIRTY- TWO GRAMS PER MOLE, RIGHT?
SO THEN THAT GIVES US A HUNDRED AND NINETY-TWO GRAMS. OKAY.
CO2, RIGHT? ONCE AGAIN, WE'LL DO THE SAME THING.
WE HAVE SIX MOLES AND THEY EACH WEIGH FORTY-FOUR.
SO THAT'S GOING TO GIVE US--
I'M NOT GOING TO SHOW YOU HOW TO DO THE CALCULATION
BECAUSE WE JUST DID IT, RIGHT?
AND THEN WE HAVE SIX MOLES OF WATER, RIGHT?
SO EACH MOLE WEIGHS 18.02.
SO IF WE HAVE SIX MOLES, WE WOULD HAVE 108.12 GRAMS.
NOW LET'S LOOK, RIGHT?
IF WE ADD THIS ALL UP, 372.18 GRAMS IS THE REACTANTS
AND 372.18 GRAMS IS THE PRODUCT.
SO WE CAN SEE THAT OVER ALL,
THE BALANCED REACTION DOES OBEY THE CONSERVATION OF MASS,
WE HAVE THE SAME AMOUNT OF MASS AND WE COULD ALSO LOOK AT THIS
JUST IN TERMS OF ATOMS, RIGHT?
CARBON, HYDROGEN, AND OXYGEN, RIGHT?
IF WE LOOK ON THE LEFT, WE SEE ONLY SIX CARBON ATOMS
AND WHEN WE LOOK ON THE RIGHT,
OUR PRODUCT HAS SIX CARBON ATOMS.
HYDROGEN, THERE ARE 12 HYDROGEN ATOMS ON THE RIGHT AS A REACTANT
AND 12 HYDROGEN ATOMS-- NO, THAT WAS ON THE LEFT,
THERE'S THE RIGHT, 12 AND 12 AND THEN OXYGENS, RIGHT?
SIX AND TWELVE, SO 6 + 12 IS EIGHTEEN
AND HERE WE HAVE 12 + 6 IS 18.
SO WE SEE THAT THE MOLE IS THE ASPECT
THAT HELPS US CREATE BALANCE AND FOLLOW THE CONSERVATION OF MASS.
ALL RIGHTY.
SO THAT WRAPS UP THIS TUTORIAL INTRODUCING YOU
TO THE IMPORTANT CONCEPT OF THE MOLE.
PEASE SPEND SOME TIME AND WORK A FEW CALCULATIONS
TO REINFORCE YOUR UNDERSTANDING OF THIS IMPORTANT IDEA.