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THE MOLE
THE MOLETHE MOLE
• One way to measure how much substance available is to count the # of particles in that sample–However, atoms & molecules are
extremely small–Also, the # of individual particles in
even a small sample is very large–Therefore, counting the # of particles
is not a practical measure of amount• To solve this problem, scientists
developed the concept of the mole– It’s the “chemical counting unit”
• One way to measure how much substance available is to count the # of particles in that sample–However, atoms & molecules are
extremely small–Also, the # of individual particles in
even a small sample is very large–Therefore, counting the # of particles
is not a practical measure of amount• To solve this problem, scientists
developed the concept of the mole– It’s the “chemical counting unit”
How Scientists Keep Track of Atoms
How Scientists Keep Track of Atoms
Counting by weighing
• 1 Bean 5 grams• 5 beans 50 grams• HOW?
Average Mass
• Mass out 50 beans and find the average
• Just as a dozen eggs equals 12 eggs, a mole = 602,000,000,000,000,000,000,000– It is equal to that number no matter what
kind of particles you’re talking about– It could be represent marbles, pencils, or
bikes– usually deals with atoms and molecules
• The word “mole” was introduced about 1896 by wilhelm oswald, who derived the term from the latin word moles meaning a “heap” or “pile.”
• The mole, whose abbreviation is “mol”, is the SI base unit for measuring amount of a pure substance.
• Just as a dozen eggs equals 12 eggs, a mole = 602,000,000,000,000,000,000,000– It is equal to that number no matter what
kind of particles you’re talking about– It could be represent marbles, pencils, or
bikes– usually deals with atoms and molecules
• The word “mole” was introduced about 1896 by wilhelm oswald, who derived the term from the latin word moles meaning a “heap” or “pile.”
• The mole, whose abbreviation is “mol”, is the SI base unit for measuring amount of a pure substance.
• The mole is the chemist’s six-pack or dozen. Many objects in our everyday lives come in similar counting units.
• The mole is the chemist’s six-pack or dozen. Many objects in our everyday lives come in similar counting units.
1 dozen = 121 dozen = 121 mole = 6.02x10231 mole = 6.02x1023
1 dozen eggs = 12 eggs
1 dozen eggs = 12 eggs1 mole eggs = 6.02x1023 eggs
1 mole eggs = 6.02x1023 eggs2 doz of atoms = 24
atoms 2 doz of atoms = 24
atoms 2 mols of atoms=1. 20x1024
atoms
2 mols of atoms=1. 20x1024
atoms
or 6.02x1023
REPRESENTATIVE PARTICLESor 6.02x1023
REPRESENTATIVE PARTICLES
AVOGADRO’S #= AVOGADRO’S #=
•THE MOLE, AS A UNIT, IS ONLY USED TO COUNT VERY SMALL ITEMS–REPRESENTS A # OF ITEMS, SO, WE CAN KNOW EXACTLY HOW MANY ITEMS ARE IN 1 MOLE
•THE EXPERIMENTALLY DETERMINED NUMBER A MOLE IS THE EQUIVALENT OF IS CALLED:
•THE MOLE, AS A UNIT, IS ONLY USED TO COUNT VERY SMALL ITEMS–REPRESENTS A # OF ITEMS, SO, WE CAN KNOW EXACTLY HOW MANY ITEMS ARE IN 1 MOLE
•THE EXPERIMENTALLY DETERMINED NUMBER A MOLE IS THE EQUIVALENT OF IS CALLED:
• THE TERM REPRESENTATIVE PARTICLE REFERS TO THE SPECIES PRESENT IN A SUBSTANCE– USUALLY ATOMS– MOLECULES– OR FORMULA UNITS (IONS)
• IT’S IMPORTANT TO NOTE THAT A DOZEN CUPS OF MARBLES CONTAINS MORE THAN A DOZEN MARBLES– SIMILARLY, A MOLE OF MOLECULES
CONTAINS MORE THAN A MOLE OF ATOMS
• THE TERM REPRESENTATIVE PARTICLE REFERS TO THE SPECIES PRESENT IN A SUBSTANCE– USUALLY ATOMS– MOLECULES– OR FORMULA UNITS (IONS)
• IT’S IMPORTANT TO NOTE THAT A DOZEN CUPS OF MARBLES CONTAINS MORE THAN A DOZEN MARBLES– SIMILARLY, A MOLE OF MOLECULES
CONTAINS MORE THAN A MOLE OF ATOMS
REPRESENTATIVE PARTICLES & MOLESREPRESENTATIVE PARTICLES & MOLES
ATOMIC ATOMIC NITROGENNITROGEN ATOMATOM NN 6.02x106.02x102323
NITROGEN NITROGEN GASGAS MOLECMOLEC.. NN22 6.02x106.02x102323
WATERWATER MOLEC.MOLEC. HH2200 6.02x106.02x102323
CALCIUM IONCALCIUM ION IONION CaCa2+2+ 6.02x106.02x102323
CALCIUM CALCIUM FLUORIDEFLUORIDE
FORMULA FORMULA UNITUNIT CaFCaF22 6.02x106.02x102323
HOW DO WE USE THE MOLE?
HOW DO WE USE THE MOLE?
• SINCE THE MOLE IS SUCH A HUGE NUMBER OF ITEMS, IT IS ONLY USED TO DESCRIBE THE AMOUNT OF THINGS THAT ARE VERY, VERY SMALL.
• WE’D NEVER USE THE MOLE TO DESCRIBE MACROSCOPIC OR REAL WORLD OBJECTS.
• SINCE THE MOLE IS SUCH A HUGE NUMBER OF ITEMS, IT IS ONLY USED TO DESCRIBE THE AMOUNT OF THINGS THAT ARE VERY, VERY SMALL.
• WE’D NEVER USE THE MOLE TO DESCRIBE MACROSCOPIC OR REAL WORLD OBJECTS.
HOW BIG A NUMBER ARE WE TALKING?
HOW BIG A NUMBER ARE WE TALKING?
1 mole = 6.02x10231 mole = 6.02x1023
•6.02x1023 Watermelon seeds: would be found inside a melon slightly larger than the moon.
•6.02x1023donut holes: would cover the earth and be 5 miles deep.
•6.02x1023 grains of sand: would be cover miami beach 10 ft deep
•6.02x1023bloodcells: would be more than the total # of blood cells found in every human on earth
•6.02x1023 Watermelon seeds: would be found inside a melon slightly larger than the moon.
•6.02x1023donut holes: would cover the earth and be 5 miles deep.
•6.02x1023 grains of sand: would be cover miami beach 10 ft deep
•6.02x1023bloodcells: would be more than the total # of blood cells found in every human on earth
USING THE MOLE IN CALCULATIONS #1
USING THE MOLE IN CALCULATIONS #1
HOW MANY MOLES OF MAGNESIUM IS 1.25x1023 ATOMS
OF MAGNESIUM?
HOW MANY MOLES OF MAGNESIUM IS 1.25x1023 ATOMS
OF MAGNESIUM?
OUR UNIT EQUALITY TO DO THIS CONVERSION IS
1 mol Mg = 6.02x1023 atoms Mg
OUR UNIT EQUALITY TO DO THIS CONVERSION IS
1 mol Mg = 6.02x1023 atoms Mg
6.02x1023atoms Mg
6.02x1023atoms Mg
1 mole Mg1 mole Mg
THE DESIRED CONVERSION IS: ATOMS MOLES
THE DESIRED CONVERSION IS: ATOMS MOLES
1.25x1023 atoms Mg1.25x1023 atoms Mg
= .208 mol Mg= .208 mol Mg
• NOW SUPPOSE YOU WANT TO DETERMINE HOW MANY ATOMS ARE IN A MOLE OF A COMPOUND– TO DO THIS YOU MUST KNOW HOW
MANY ATOMS ARE IN A REPRESENTATIVE PARTICLE OF THE COMPOUND.
• TO DETERMINE THE NUMBER OF ATOMS IN A PARTICLE REQUIRES KNOWING THE CHEMICAL FORMULA– FOR EXAMPLE, EACH MOLECULE OF
CARBON DIOXIDE (CO2) IS COMPOSED OF 3 ATOMS
• NOW SUPPOSE YOU WANT TO DETERMINE HOW MANY ATOMS ARE IN A MOLE OF A COMPOUND– TO DO THIS YOU MUST KNOW HOW
MANY ATOMS ARE IN A REPRESENTATIVE PARTICLE OF THE COMPOUND.
• TO DETERMINE THE NUMBER OF ATOMS IN A PARTICLE REQUIRES KNOWING THE CHEMICAL FORMULA– FOR EXAMPLE, EACH MOLECULE OF
CARBON DIOXIDE (CO2) IS COMPOSED OF 3 ATOMS
• 1 MOLE OF CARBON DIOXIDE CONTAINS AVOGADRO’S NUMBER OF CARBON DIOXIDE MOLECULES.– THUS A MOLE OF CO2 CONTAINS
THREE TIMES AVOGADRO’S NUMBER OF ATOMS
• TO FIND THE # OF ATOMS IN A MOL OF A COMPND, – YOU 1ST DETERMINE THE # OF ATOMS
IN A REPRESENTATIVE PARTICLE OF THAT COMPND
– AND THEN MULTIPLY THAT # OF ATOMS BY AVOGADRO’S #
• 1 MOLE OF CARBON DIOXIDE CONTAINS AVOGADRO’S NUMBER OF CARBON DIOXIDE MOLECULES.– THUS A MOLE OF CO2 CONTAINS
THREE TIMES AVOGADRO’S NUMBER OF ATOMS
• TO FIND THE # OF ATOMS IN A MOL OF A COMPND, – YOU 1ST DETERMINE THE # OF ATOMS
IN A REPRESENTATIVE PARTICLE OF THAT COMPND
– AND THEN MULTIPLY THAT # OF ATOMS BY AVOGADRO’S #
USING THE MOLE IN CALCULATIONS #2
USING THE MOLE IN CALCULATIONS #2
HOW MANY ATOMS ARE IN 2.12 mols OF PROPANE (C3H8)?
HOW MANY ATOMS ARE IN 2.12 mols OF PROPANE (C3H8)?
UNIT EQUALITIES ARE
1 molecule C3H8 = 11 atoms C3H8
UNIT EQUALITIES ARE
1 molecule C3H8 = 11 atoms C3H8
6.02x1023
molecules C3H8
6.02x1023
molecules C3H8
1 mole C3H81 mole C3H8
THE DESIRED CONVERSIONS ARE: MOLES MOLECULES
ATOMS
THE DESIRED CONVERSIONS ARE: MOLES MOLECULES
ATOMS
2.12 moles C3H8
2.12 moles C3H8
1.276x1024 molecule
s C3H8
1.276x1024 molecule
s C3H8
==
11 atoms C3H811 atoms C3H8
1 molecule C3H81 molecule C3H8==
1.40x1025 atoms C3H81.40x1025 atoms C3H8
–THEREFORE, INSTEAD OF USING THE ACTUAL MASS OF A CARBON ATOM IN GRAMS, CHEMISTS USE RELATIVE ATOMIC MASSES
–THEREFORE, INSTEAD OF USING THE ACTUAL MASS OF A CARBON ATOM IN GRAMS, CHEMISTS USE RELATIVE ATOMIC MASSES
•WHAT IS AN ATOM’S MASS?–IF MEASURED IN GRAMS, THE MASSES OF ATOMS WOULD BE TOO SMALL TO WORK WITH
•WHAT IS AN ATOM’S MASS?–IF MEASURED IN GRAMS, THE MASSES OF ATOMS WOULD BE TOO SMALL TO WORK WITH
ATOMIC MASSESATOMIC MASSES
ATOMIC MASS UNITSATOMIC MASS UNITS• IN DETERMINING RELATIVE
MASSES, ONE ATOM IS ARBITRARILY CHOSEN AS THE STANDARD
•THE MASS OF ALL THE OTHER ATOMS ARE THEN EXPRESSED IN RELATION TO THIS STANDARD VALUE
• IN DETERMINING RELATIVE MASSES, ONE ATOM IS ARBITRARILY CHOSEN AS THE STANDARD
•THE MASS OF ALL THE OTHER ATOMS ARE THEN EXPRESSED IN RELATION TO THIS STANDARD VALUE
•FOR THE RELATIVE MASS OF AN ATOM CHEMISTS AGREED UPON THE CARBON-12 ATOM
•FOR THE RELATIVE MASS OF AN ATOM CHEMISTS AGREED UPON THE CARBON-12 ATOM
•A SINGLE CARBON-12 ATOM WAS ASSIGNED THE VALUE OF 12 ATOMIC MASS UNITS (AMU).
•THEREFORE, 1 ATOMIC MASS UNIT IS EXACTLY 1/12 OF THE MASS OF A CARBON-12 ATOM–HYDROGEN THEN WEIGHS 1 AMU–HELIUM WEIGHS 4 AMUS
•AMU’S GAVE SCIENTISTS A UNIT TO WORK WITH, BUT IT STILL DESCRIBED THE MASS IN TERMS OF INDIVIDUAL ATOMS (UNUSABLE)
•A SINGLE CARBON-12 ATOM WAS ASSIGNED THE VALUE OF 12 ATOMIC MASS UNITS (AMU).
•THEREFORE, 1 ATOMIC MASS UNIT IS EXACTLY 1/12 OF THE MASS OF A CARBON-12 ATOM–HYDROGEN THEN WEIGHS 1 AMU–HELIUM WEIGHS 4 AMUS
•AMU’S GAVE SCIENTISTS A UNIT TO WORK WITH, BUT IT STILL DESCRIBED THE MASS IN TERMS OF INDIVIDUAL ATOMS (UNUSABLE)
ATOMIC MASS UNITSATOMIC MASS UNITS
• SCIENTISTS MUST FIGURE OUT A WAY TO WORK WITH A COLLECTION OF PARTICLES THAT AREN’T HANDLED INDIVIDUALLY
• AND THAT CAN STILL BE THOUGHT OF IN TERMS OF A RELATIVE (OR COMPARED) MASS– AN AVERAGE C ATOM WITH AN
ATOMIC MASS OF 12.0 amu IS 12 TIMES HEAVIER THAN AN AVERAGE H ATOM WITH AN ATOMIC MASS OF 1.0 amu
– THEREFORE, 100 C ATOMS ARE 12 TIMES HEAVIER THAN 100 H ATOMS
• SCIENTISTS MUST FIGURE OUT A WAY TO WORK WITH A COLLECTION OF PARTICLES THAT AREN’T HANDLED INDIVIDUALLY
• AND THAT CAN STILL BE THOUGHT OF IN TERMS OF A RELATIVE (OR COMPARED) MASS– AN AVERAGE C ATOM WITH AN
ATOMIC MASS OF 12.0 amu IS 12 TIMES HEAVIER THAN AN AVERAGE H ATOM WITH AN ATOMIC MASS OF 1.0 amu
– THEREFORE, 100 C ATOMS ARE 12 TIMES HEAVIER THAN 100 H ATOMS
•ANY NUMBER OF C ATOMS IS 12 TIMES HEAVIER THAN THE SAME # OF H ATOMS–SO, IF WE HAD IF WE HAD A PILE OF CARBON ATOMS THAT WEIGHED 12g AND A PILE OF HYDROGEN ATOMS THAT WEIGHED 1g,
–BOTH PILES SHOULD CONTAIN THE SAME NUMBER OF ATOMS
•ANY NUMBER OF C ATOMS IS 12 TIMES HEAVIER THAN THE SAME # OF H ATOMS–SO, IF WE HAD IF WE HAD A PILE OF CARBON ATOMS THAT WEIGHED 12g AND A PILE OF HYDROGEN ATOMS THAT WEIGHED 1g,
–BOTH PILES SHOULD CONTAIN THE SAME NUMBER OF ATOMS
•THE GRAM ATOMIC MASSES OF ANY 2 ELEMENTS (SINCE THEY ARE RELATIVE TO CARBON) MUST CONTAIN THE SAME NUMBER OF ATOMS
•A PILE OF ANY ATOM THAT CORRESPONDS TO ITS AVERAGE ATOMIC MASS FROM THE PT CONTAINS EXACTLY 6.02x1023 ATOMS OF THAT ELEMENT.–ALSO CALLED A MOLE
•THE GRAM ATOMIC MASSES OF ANY 2 ELEMENTS (SINCE THEY ARE RELATIVE TO CARBON) MUST CONTAIN THE SAME NUMBER OF ATOMS
•A PILE OF ANY ATOM THAT CORRESPONDS TO ITS AVERAGE ATOMIC MASS FROM THE PT CONTAINS EXACTLY 6.02x1023 ATOMS OF THAT ELEMENT.–ALSO CALLED A MOLE
CARBON ATOMSCARBON ATOMS HYDROGEN HYDROGEN ATOMSATOMS MASS RATIOMASS RATIO
1212 11
•WHAT THIS ALLOWS US TO DO IS TO USE THE MASS OFF OF THE PERIODIC TABLE TO REPRESENT HOW MUCH 1 MOLE OF THAT ELEMENT WEIGHS–1 MOLE OF CARBON ATOMS WEIGH 12.01 g
–1 MOLE OF HYDROGEN ATOMS WEIGH 1.008 g
–1 MOLE OF TUNGSTEN ATOMS WEIGH 183.8; ETC.
•WHAT THIS ALLOWS US TO DO IS TO USE THE MASS OFF OF THE PERIODIC TABLE TO REPRESENT HOW MUCH 1 MOLE OF THAT ELEMENT WEIGHS–1 MOLE OF CARBON ATOMS WEIGH 12.01 g
–1 MOLE OF HYDROGEN ATOMS WEIGH 1.008 g
–1 MOLE OF TUNGSTEN ATOMS WEIGH 183.8; ETC.
• THIS NEW VERSION OF MASS FROM THE PERIODIC TABLE IS CALLED THE GRAM MOLAR MASS, OR MOLAR MASS.– MOLAR MASS = MASS OF 1 MOLE OF
ATOMS/MOLECULES/OR FORMULA UNITS IN GRAMS
– SYMBOL = MM – UNITS = GRAMS/MOLE
• SO HOW DO WE FIGURE OUT THE MASS OF A MOLE OF A COMPOUND RATHER THAN JUST 1 ELEMENT?
• THIS NEW VERSION OF MASS FROM THE PERIODIC TABLE IS CALLED THE GRAM MOLAR MASS, OR MOLAR MASS.– MOLAR MASS = MASS OF 1 MOLE OF
ATOMS/MOLECULES/OR FORMULA UNITS IN GRAMS
– SYMBOL = MM – UNITS = GRAMS/MOLE
• SO HOW DO WE FIGURE OUT THE MASS OF A MOLE OF A COMPOUND RATHER THAN JUST 1 ELEMENT?
•TO ANSWER THAT QUESTION YOU MUST HAVE THE FORMULA OF THE COMPOUND.–THE FORMULA OF A COMPND TELLS YOU HOW MANY ATOMS OF EACH ELEMENT COMBINE TO MAKE THE REPRESENTATIVE PARTICLE OF THAT COMPND.
•TO ANSWER THAT QUESTION YOU MUST HAVE THE FORMULA OF THE COMPOUND.–THE FORMULA OF A COMPND TELLS YOU HOW MANY ATOMS OF EACH ELEMENT COMBINE TO MAKE THE REPRESENTATIVE PARTICLE OF THAT COMPND.
ModelModelFormulaFormula ModelModelFormulaFormula
• YOU CAN CALCULATE THE MASS OF A MOLECULE OF SO3 BY ADDING THE MOLAR MASSES OF THE ATOMS THAT MAKE UP THE MOLECULE– FROM THE PERIODIC TABLE,
THE MASS OF SULFUR IS 32.1g/mol.– THE MASS OF THREE ATOMS OF
OXYGEN IS 3 TIMES THE MOLAR MASS OF A SINGLE OXYGEN ATOMS, WHICH IS (3)(16g/mol) OR 48g/mol
• THE TOTAL MASS OF EACH OF THE ATOMS IN 1 MOLECULE OF SO3 = 32.1g/mol + 48 g/mol = 80.1 g/mol
• YOU CAN CALCULATE THE MASS OF A MOLECULE OF SO3 BY ADDING THE MOLAR MASSES OF THE ATOMS THAT MAKE UP THE MOLECULE– FROM THE PERIODIC TABLE,
THE MASS OF SULFUR IS 32.1g/mol.– THE MASS OF THREE ATOMS OF
OXYGEN IS 3 TIMES THE MOLAR MASS OF A SINGLE OXYGEN ATOMS, WHICH IS (3)(16g/mol) OR 48g/mol
• THE TOTAL MASS OF EACH OF THE ATOMS IN 1 MOLECULE OF SO3 = 32.1g/mol + 48 g/mol = 80.1 g/mol
MM of C6H12O6:MM of
C6H12O6:(6C’S)(12g/mol)=(6C’S)(12g/mol)=
180g/mol180g/mol
CALCULATING MOLAR MASSES USING CHEMICAL
FORMULAS
CALCULATING MOLAR MASSES USING CHEMICAL
FORMULAS
(12H’S)(1 g/mol)=(12H’S)(1 g/mol)=(6O’S)(16 g/mol)=(6O’S)(16 g/mol)=
IF WE HAD 1 MOLE OF THE COMPND C6H12O6 OR 6.02X1023 MOLECULES OF THE COMPND – IT WOULD WEIGH 180
grams
IF WE HAD 1 MOLE OF THE COMPND C6H12O6 OR 6.02X1023 MOLECULES OF THE COMPND – IT WOULD WEIGH 180
grams
72g/mol72g/mol12g/mol12g/mol96g/mol96g/mol
•WE CAN USE THE MOLAR MASS OF AN ELEMENT OR COMPOUND AS A CONVERSION FACTOR TO CONVERT BETWEEN GRAMS AND MOLES OF A SUBSTANCE.–THE UNIT EQUALITY IS 1 MOLE = __ MM OF THE SUBSTANCE
•WE CAN USE THE MOLAR MASS OF AN ELEMENT OR COMPOUND AS A CONVERSION FACTOR TO CONVERT BETWEEN GRAMS AND MOLES OF A SUBSTANCE.–THE UNIT EQUALITY IS 1 MOLE = __ MM OF THE SUBSTANCE
USING THE MOLE IN CALCULATIONS #3
USING THE MOLE IN CALCULATIONS #3
HOW MANY GRAMS ARE IN 9.45 mol OF DINITROGEN TRIOXIDE
(N2O3)
HOW MANY GRAMS ARE IN 9.45 mol OF DINITROGEN TRIOXIDE
(N2O3)
UNIT EQUALITY TO USEUNIT EQUALITY TO USE
1 mole N2O31 mole N2O3
76 grams N2O3
76 grams N2O3
THE DESIRED CONVERSION IS:
MOLES GRAMS
THE DESIRED CONVERSION IS:
MOLES GRAMS
9.45 molN2O3
9.45 molN2O3
= 718 g N2O3= 718 g N2O3
USING THE MOLE IN CALCULATIONS #4
USING THE MOLE IN CALCULATIONS #4
FIND THE NUMBER OF MOLES OF 92.2g OF IRON (III) OXIDE
(Fe2O3)
FIND THE NUMBER OF MOLES OF 92.2g OF IRON (III) OXIDE
(Fe2O3)
UNIT EQUALITY TO USEUNIT EQUALITY TO USE
1 mole Fe2O31 mole Fe2O3
159.6 g Fe2O3
159.6 g Fe2O3
THE DESIRED CONVERSION IS:
GRAMS MOLES
THE DESIRED CONVERSION IS:
GRAMS MOLES
92.2 g Fe2O3
92.2 g Fe2O3
= 0.578 mol Fe2O3= 0.578 mol Fe2O3
VOLUME AND THE MOLEVOLUME AND THE MOLE•UNDER THE SAME CONDITIONS,
EQUAL VOLUMES OF GASES CONTAIN THE SAME NUMBERS OF PARTICLES.–OR 1 MOLE OF A GAS WILL OCCUPY THE SAME VOLUME AS 1 MOLE OF ANY OTHER GAS UNDER THE SAME CONDITIONS.
•IT’S KNOWN AS THE MOLAR VOLUME OF A GAS
•UNDER THE SAME CONDITIONS, EQUAL VOLUMES OF GASES CONTAIN THE SAME NUMBERS OF PARTICLES.–OR 1 MOLE OF A GAS WILL OCCUPY THE SAME VOLUME AS 1 MOLE OF ANY OTHER GAS UNDER THE SAME CONDITIONS.
•IT’S KNOWN AS THE MOLAR VOLUME OF A GAS
MOLAR VOLUMEMOLAR
VOLUME1 MOLE OF ANY GAS AT
STP (0°C and 1 atm) HAS A VOLUME OF:
1 MOLE OF ANY GAS AT STP (0°C and 1 atm) HAS A
VOLUME OF:
1 mole = 22.4 L1 mole = 22.4 L
USING THE MOLE IN CALCULATIONS #5
USING THE MOLE IN CALCULATIONS #5
DETERMINE THE VOLUME, IN LITERS, OF 0.60 molSO2 GAS AT STP.
DETERMINE THE VOLUME, IN LITERS, OF 0.60 molSO2 GAS AT STP.
UNIT EQUALITY TO USEUNIT EQUALITY TO USE
1 mole SO21 mole SO2
22.4 L SO222.4 L SO2
THE DESIRED CONVERSION IS:
MOLES LITERS
THE DESIRED CONVERSION IS:
MOLES LITERS
0.60 moles SO2
0.60 moles SO2
= 13 L SO2= 13 L SO2
MOLEMOLEMOLEMOLEMASS MASS (in grams)(in grams)MASS MASS (in grams)(in grams)
MOLAR VOLUMEMOLAR VOLUMEVolume Volume (of gas at STP)(of gas at STP)Volume Volume (of gas at STP)(of gas at STP)
MOLAR MASSMOLAR MASS
AVOGADRO’S NUMBERAVOGADRO’S NUMBER
GOOD EXAMPLE PROBLEM!GOOD EXAMPLE PROBLEM!
IF YOU HAVE A 35.67g PIECE OF CHROMIUM METAL ON YOUR CAR, HOW MANY ATOMS OF CHROMIUM
DO YOU HAVE?
IF YOU HAVE A 35.67g PIECE OF CHROMIUM METAL ON YOUR CAR, HOW MANY ATOMS OF CHROMIUM
DO YOU HAVE?
• YOU ARE GIVEN MASS AND ASKED FOR NUMBER OF PARTICLES
• LET’S GET SOME STRATEGY
• YOU ARE GIVEN MASS AND ASKED FOR NUMBER OF PARTICLES
• LET’S GET SOME STRATEGY
WE ARE WE ARE GIVEN MASSGIVEN MASS
WE ARE WE ARE GIVEN MASSGIVEN MASS
WE ARE WE ARE ASKED FOR ASKED FOR
ATOMSATOMS
•IT’S GOING TO TAKE US 2 STEPS, WE JUST FOLLOW THE ARROWS
•IT’S GOING TO TAKE US 2 STEPS, WE JUST FOLLOW THE ARROWS
WE ARE WE ARE GIVEN MASSGIVEN MASS
WE ARE WE ARE ASKED FOR ASKED FOR
ATOMSATOMS
•THE FIRST STEP IS TO CONVERT OUR GIVEN GRAMS INTO MOLES
•TO DO THIS WE USE THE MOLAR MASS (MM) OF CHROMIUM WHICH ON THE PT IS 52g/mol
•THE FIRST STEP IS TO CONVERT OUR GIVEN GRAMS INTO MOLES
•TO DO THIS WE USE THE MOLAR MASS (MM) OF CHROMIUM WHICH ON THE PT IS 52g/mol
35.67g Cr
35.67g Cr
52 g Cr52 g Cr
1 mole Cr1 mole Cr== .686
mole Cr.686 mole Cr
• THE SECOND STEP WE ARE GOING TO TAKE OUR NEWLY CALCULATED MOLES OF Cr AND CONVERT IT TO THE NUMBER OF ATOMS OF Cr
• WE HAVE TO REMEMBER THAT IF WE HAD 1 MOLE OF Cr ATOMS WE WOULD HAVE 6.02X1023 ATOMS
• THE SECOND STEP WE ARE GOING TO TAKE OUR NEWLY CALCULATED MOLES OF Cr AND CONVERT IT TO THE NUMBER OF ATOMS OF Cr
• WE HAVE TO REMEMBER THAT IF WE HAD 1 MOLE OF Cr ATOMS WE WOULD HAVE 6.02X1023 ATOMS
.686 mole Cr.686 mole Cr
1 mole Cr1 mole Cr
6.02x1023 atoms Cr6.02x1023 atoms Cr
= 4.130x1023 atoms Cr= 4.130x1023 atoms Cr
