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THE MOLE

THE MOLE. How Scientists Keep Track of Atoms. 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

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THE MOLE

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  1. THE MOLE

  2. How Scientists Keep Track of Atoms • 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”

  3. Counting by weighing • 1 Bean 5 grams • 5 beans 50 grams • HOW?

  4. Average Mass • Mass out 50 beans and find the average

  5. Multiple Stuff • I Bean 5 grams • 1 Mint 15 grams

  6. 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 measuringamount of a pure substance.

  7. The mole is the chemist’s six-pack or dozen. Many objects in our everyday lives come in similar counting units.

  8. 1 dozen = 12 2 doz of atoms = 24atoms 2 mols of atoms=1. 20x1024 atoms 1 mole = 6.02x1023 1 dozen eggs = 12 eggs 1 mole eggs = 6.02x1023 eggs

  9. 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: AVOGADRO’S #= 60,200,000,000,000,000,000,000,000 or6.02x1023REPRESENTATIVE PARTICLES

  10. 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 OFMOLECULESCONTAINS MORE THAN A MOLE OFATOMS

  11. REP. PARTICLE REPRESENTATIVE CHEMICAL SUBTANCE FORMULA IN 1 MOL PARTICLE

  12. 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. HOW BIG A NUMBER ARE WE TALKING?

  13. 1 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

  14. USING THE MOLE IN CALCULATIONS #1 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

  15. THE DESIRED CONVERSION IS: ATOMS MOLES 1 mole Mg 1.25x1023 atoms Mg 6.02x1023atoms Mg = .208 mol Mg

  16. 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 AREPRESENTATIVE PARTICLEOF 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 OF3 ATOMS

  17. 1 MOLE OF CARBON DIOXIDE CONTAINS AVOGADRO’S NUMBER OF CARBON DIOXIDE MOLECULES. • THUS A MOLE OF CO2 CONTAINSTHREE TIMESAVOGADRO’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 THENMULTIPLYTHAT # OF ATOMS BY AVOGADRO’S #

  18. 2 3 1 mol C H = 6.02x10 molecules C H , UNIT EQUALITIES ARE 1 molecule C3H8 = 11 atoms C3H8 3 8 3 8 USING THE MOLE IN CALCULATIONS #2 HOW MANY ATOMS ARE IN 2.12 mols OF PROPANE (C3H8)?

  19. THE DESIRED CONVERSIONS ARE: MOLES MOLECULES  ATOMS 6.02x1023 molecules C3H8 2.12 moles C3H8 = 1 mole C3H8 1.276x1024 molecules C3H8 11 atoms C3H8 = 1 molecule C3H8 1.40x1025 atoms C3H8

  20. ATOMIC MASSES • WHAT IS AN ATOM’S MASS? • IF MEASURED IN GRAMS, THE MASSES OF ATOMS WOULD BE TOO SMALL TO WORK WITH • THEREFORE, INSTEAD OF USING THE ACTUAL MASS OF A CARBON ATOM IN GRAMS, CHEMISTS USE RELATIVE ATOMIC MASSES (.00000000000000000000002g)

  21. ATOMIC MASS UNITS 6 C CARBON 12 • 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

  22. ATOMIC MASS UNITS • 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 WEIGHS1 AMU • HELIUM WEIGHS4 AMUS • AMU’S GAVE SCIENTISTS A UNIT TO WORK WITH, BUT IT STILL DESCRIBED THE MASS IN TERMS OF INDIVIDUAL ATOMS (UNUSABLE)

  23. 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 CATOMS ARE 12 TIMES HEAVIER THAN100 HATOMS

  24. 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

  25. 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

  26. Mass Mass Mass Carbon Number Number (amu) (amu) Mass Hydrogen 12amu 12 = 1amu 1 672 56 672amu 12 = (56 x 12) (56 x 1) 56amu 1 23 Avogadro's 23 23 (6.02x10 ) x (12) (6.02x10 ) x Avogadro's (6.02x10 ) number 23 (12) number x (1) (6.02x10 ) x (1)

  27. 23 18.998g F atoms=1 mole F atoms=6.02x10 F atoms • 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 WEIGH12.01 g • 1 MOLE OF HYDROGEN ATOMS WEIGH1.008 g • 1 MOLE OF TUNGSTEN ATOMS WEIGH183.8; ETC.

  28. 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?

  29. Formula Model Formula Model SO H O 3 2 3 oxygen atoms 2 Hydrogen atoms per molecule per molecule • 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.

  30. 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 IS32.1g/mol. • THE MASS OF THREE ATOMS OF OXYGEN IS3 TIMESTHE MOLAR MASS OF A SINGLE OXYGEN ATOMS, WHICH IS (3)(16g/mol) OR48g/mol • THE TOTAL MASS OF EACH OF THE ATOMS IN 1 MOLECULE OF SO3 = 32.1g/mol + 48 g/mol = 80.1 g/mol

  31. CALCULATING MOLAR MASSES USING CHEMICAL FORMULAS MM of C6H12O6: (6C’S)(12g/mol)= 72g/mol 12g/mol (12H’S)(1 g/mol)= 96g/mol (6O’S)(16 g/mol)= 180g/mol IF WE HAD 1 MOLE OF THE COMPND C6H12O6 OR 6.02X1023 MOLECULES OF THE COMPND – IT WOULD WEIGH 180 grams

  32. 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 IS1 MOLE = __ MM OF THE SUBSTANCE

  33. 1 mol N O = (2)(14g/mol)+(3)(16g/mol) 2 3 1 mol N O = (28g/mol)+(48g/mol) 2 3 1 mol N O = 76g/mol 2 3 USING THE MOLE IN CALCULATIONS #3 HOW MANY GRAMS ARE IN 9.45 mol OF DINITROGEN TRIOXIDE (N2O3) UNIT EQUALITY TO USE

  34. THE DESIRED CONVERSION IS: MOLES  GRAMS 76 grams N2O3 9.45 mol N2O3 1 mole N2O3 = 718 g N2O3

  35. 1 mol Fe O = (2)(55.8g/mol)+(3)(16g/mol) 2 3 1 mol Fe O = (111.6g/mol)+(48g/mol) 2 3 1 mol Fe O = 159.6g/mol 2 3 USING THE MOLE IN CALCULATIONS #4 FIND THE NUMBER OF MOLES OF 92.2g OF IRON (III) OXIDE (Fe2O3) UNIT EQUALITY TO USE

  36. THE DESIRED CONVERSION IS: GRAMS  MOLES 1 mole Fe2O3 92.2 g Fe2O3 159.6 g Fe2O3 = 0.578 mol Fe2O3

  37. VOLUME 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

  38. MOLAR VOLUME 1 MOLE OF ANY GAS AT STP (0°C and 1 atm) HAS A VOLUME OF: 1 mole = 22.4 L

  39. 1 mol SO = 22.4 L 2 USING THE MOLE IN CALCULATIONS #5 DETERMINE THE VOLUME, IN LITERS, OF 0.60 molSO2 GAS AT STP. UNIT EQUALITY TO USE

  40. THE DESIRED CONVERSION IS: MOLES  LITERS 22.4 L SO2 0.60 moles SO2 1 mole SO2 = 13 L SO2

  41. 23 6.02x10 AVOGADRO’S NUMBER PARTICLES Atoms, Molecules, Formula Units Periodic Table MASS (in grams) MOLAR MASS MOLE 22.4L Volume (of gas at STP) MOLAR VOLUME

  42. GOOD EXAMPLE PROBLEM! 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 FORNUMBER OF PARTICLES • LET’S GET SOME STRATEGY

  43. WE ARE GIVEN MASS

  44. WE ARE ASKED FOR ATOMS WE ARE GIVEN MASS

  45. WE ARE ASKED FOR ATOMS WE ARE GIVEN MASS • IT’S GOING TO TAKE US 2 STEPS, WE JUST FOLLOW THE ARROWS

  46. MASS MOLES 1 mole G MM of G • 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

  47. .686 mole Cr = 1 mole Cr 35.67g Cr 52 g 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

  48. 23 6.02X10 atoms 1 mole G 6.02x1023 atoms Cr .686 mole Cr 1 mole Cr = 4.130x1023 atoms Cr

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