The Mole Concept 12.00 g of Carbon-12 contains 6.022 x 1023 atoms (1 mole) Moles - How chemists count atoms

Chemist’s Definition of a Mole particles (atoms, molecules, ions, formula units) of a substance 6.022 x 1023

The Mole is a Number The mole represents a # of particles. A pair = 2 A dozen = 12 A mole = 6.022 x 1023 This number is often referred to as “Avogadro’s Number”

The Mole is a BIG number 6.022 x 1023 represents an incomprehensively large number! If a mole of marbles were laid on the surface of the earth, how much of the earth would be covered? Cover entire earth to a height of 50 miles!

1 mole (6.022 x 1023 atoms) of different elements 63.55 g 253.80 g 26.98 g 200.59 g 32.07 g 55.85 g Note 1 mole of elements takes up relatively little space – each individual atom is incredibly tiny. We need an a hugh number of atoms to be able to see them with our eyes.

How many particles are present? In chemistry, it is important to know the # of atoms (or molecules) that you have. Chemical formulas, like H2O, give us ratios of atoms: 2 H atoms bonded to 1 O atom.

How many particles are present? Chemical equations: 2 H2 + O2  2 H2O give us ratios of reacting particles. Both formulas and equations are needed to calculate recipes for reactions.

BIG Problem Atoms are too small to see with the naked eye! STM image of Si on 0.09 nm scale. (0.09 x 10-9 m) = 9 x 10-7 m Counting would take a lifetime!

Chemist’s Solution to Problem Counting of Atoms or Molecules The Mole Concept The mole connects the # of particles present with a property we can measure in the lab: mass. (Counting by grouping and counting by weighing) Can’t see Can measure Work HW 5-1 #1, p. 97

COUNTING BY WEIGHING How could you determine the number of marbles in the bag using a scale? Bag of marbles Scale Problem: Marbles are similar, but not exactly the same mass How can you account for small variations in the masses of individual marbles?

Determining the Average Mass of a Marble (or an atom) Marbles (like atoms) will have similar but not necessarily identical masses. Recall mass on periodic table represents average mass taking into account all of the isotopes of an atom. Determination average mass of 10 marbles: Average atomic mass 4.90 + 4.90 + 5.00 + 5.00 + 5.00 + 5.00 + 5.00 + 5.00 + 5.10 + 5.10 ÷ 10 = 5.00 g

Counting by Weighing Determine total mass of all marbles in the bag: 2000. g Let N = number of marbles N = total mass ÷ average mass 2000 / 5.00 = 400. marbles A) 10.8 amu B) 1.08 x 1010amu/ 10.8 amu/atom= 1 x 109 Work HW 5-1 #2, p. 97

Connecting Mass in Grams with Number of Atoms How many carbon atoms (N) are present in 12.01 g C? Work HW 5-1 #3, p. 97 N = Total Mass (12.01 g) ÷ Average Mass of 1 Carbon (Where N = # of Carbon Atoms)

How many C atoms are present in 12.01 g of C? N = Total Mass (12 g) ÷ Average Mass of 1 Carbon (Where N = # of Carbon Atoms) N = 12.01 g ÷ 1.99 x 10-23 g = 6.02 x 1023

Developing the Definition of a Mole (back to p.84) The definition of a mole (the # of atoms in 12 g of Carbon) was chosen to make it easy to determine the mass of a mole of atoms using the periodic table.

Connecting Atomic Mass with the Number of Atoms Masses on periodic table represent the average mass on 1 atom measured in amu. For example, the average mass of 1 carbon atom is 12.01 amu.

Counting by weighing – Connecting Mass in Grams with the Number of Atoms However, In the lab we can’t see single atoms so we have to work with very large numbers of atoms by measuring masses in grams. We Need to work in grams (large #’s we can see, not amu (tiny numbers – can’t see)

What is the pattern in the following data? Element (atomic #) # of atoms present # of moles present Mass in grams He (#2) 6.022 x 1023 1.000 4.00 g Li (#3) 6.93 g B (#5) 10.81 g C (#6) 12.01 g Cu (#29) 63.55 g Au (#79) 196.97 g

Understanding the Mole Concept Fill in the missing information Mass of 1 mole ____________________ of an element = atomic _________ in _______ (6.022 x 1023 atoms) mass grams Work HW 5-1 #4 - 6, p. 98

HW 5-1 #5 1 mole 12 g 0.25 moles 0.33 moles 3.0 g 4.0 g 0.50 moles

HW 5-1 #5 2.00 moles 2.50 moles 24 g 30 g 3.00 moles 36 g

HW 5-1, #6, p.98 Element Number of Atoms Mass in Grams Beryllium (#4) 6.022 x 1023 Magnesium (#12) Aluminum (#13) Copper (#29) 63.55 Carbon (#6) 12.01 3.011 x 1023 6.00 1.204 x 1024 9.01 g 24.31 g 26.98 g 6.022 x 1023 24.02 g

Examples of counting by grouping (Back to p. 85)

Could you go to the store and ask for 8,563 Rice Krispies please? NO! THEY SELL RICE KRISPIES BY THE BOX BECAUSE BOXES ARE EASIER TO COUNT!

Notes p. 85 To avoid working with extremely large numbers, we count by grouping in units of 6.022 x 1023 – called 1 mole. Write 1 mole instead of 6.022 x 1023 Write 0.500 mole instead of 0.500 (6.022 x 1023) or 3.011 x 1023 2 mole = 2 (6.022 x 1023) = 1.204 x 1024 3 mole = 3 (6.022 x 1023) = 1.81 x 1024

HW 5-1, #7, p. 99 Count the atoms by drawing a box around each set of 12 atoms. How total sets are present? How many total atoms? Ans: 3 boxes; 36 atoms

HW 5-1, #8 Group Number of objects 1 Pair of Aces 2 2 Pairs of Aces 1 Dozen Doughnuts 12 3 Dozen Doughnuts Dozen Doughnuts 60 1.5 Doughnuts 1 Mole of B atoms 6.022 x 1023 atoms 2(6.022 x 1023) atoms 3(6.022 x 1023) atoms 4(6.022 x 1023) atoms ½ ( 6.022 x 1023) atoms 4 36 5 18 2 Moles of B atoms 3 Moles of B atoms 4 Moles of B atoms 1/2 Moles of B atoms

HW 5-1, # 9, p. 99 Formula to calculate # of moles present in a given number of atoms (N)? Moles = N atoms ÷ 6.022 x 1023 atoms/mole

HW 5-1, #10 1.204 x 1024 atoms B ÷ 6.022 x1023 atom/mole = 2.000 moles B

HW 5-1, #11 Element Number of Atoms Number of Moles He 6.022 x 1023 1.00 Li Ne Ar Cu 3.01 x 1023 0.50 1.20 x 1024 1.81 x 1024 Zn 3.00 1.00 6.022 x1023 2.00 3.00 0.50 1.81 x1024 2.00

HW 5-1, #11 Element Number of Atoms Number of Moles Ag 0.500 2.00 1.81 x 1024 3.01 x1023 1.20 x1024 3.00

Relationship between # of particles,“packing boxes” and mass If each Has a mass of 10 mass units, (Ignore mass of “packing box”) HW 5-1 #12A p. 100 ? MASS UNITS WEIGHES ? MASS UNITS 1 “PACKING BOX” 12 particles Ignoring the container mass, how much would 1 box weigh? ANS: 120 mass units (12 particles per box x 10 mass units per particles)

HW 5-1, #12, p. 100 11B) 600 mass units ÷ 120 mass units/box = 5 boxes 5 boxes x 12 atom/box = 60 atoms 11C) 1260 mass units ÷ 120 mass units/box = 10.5 boxes 10.5 boxes x 12 atom/box = 126 atoms

HW 5-1, #13 moles Number of atoms Element Mass in grams   3.011 x 1023 He 6.022 x 1023 1.204 x 1024 1.806 x 1024 0.5000 Ne 1.000 2.000 3.000 0.5000 2.00 g 1.000 4.00 g 2.000 8.00 g 3.000 12.00 g 3.011 x 1023 10.09 g 6.022 x 1023 20.18 g 1.204 x 1024 40.36 g 1.806 x 1024 60.54 g

HW 5-1, #13 moles Number of atoms Element Mass in grams   6.022 x 1023 26.98 1 39.95 1 Al 6.022 x 1023 Ar

HW 5-1 #14 Which of the following contains the same # of atoms as 12.01 g of C? Equal masses does NOT mean equal # of atoms because each type of atom has a different mass. Need to compare # of particles – chemists unit for # of particles = mole How many moles are present in 12.01 g? Ans: 1 mole C = 12 amu He = 4 amu

HW 5-1 #14 Question now becomes which of the following also represents 1 mole? A) (1 mole = 6.94 g); 13.88 g Li ≈ 2 moles B) (1 mole = 4.00 g); 12.0 g He ≈ 3 moles C) (1 mole = 10.8 g); 5.4 g of B ≈ 1/2 moles D) 20.18 g Ne = mass on periodic table = 1 mole

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