Lecture 3: (Lec3A) Atomic Theory

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1 Lecture 3: (Lec3A) Atomic Theory Mass of Atoms Sections (Zumdahl 6 th Edition) The Concept of the Mole Outline: The mass of a mole of atoms and the mass of a mole of molecules The composition of a molecule by number and weight. Ch 3 Problems: , 3.27, , , 3.36, , , 3.48, 3.52 Discussion: 3:1-2, , 3.14**, 3.19, 3.21 **Tough but cool problem

2 How do we know how much mass an atom contains? 3.1)

3 Determine the Average Atomic mass of Cu from the Mass Spec analyzer

4 MOLE just a number, like a dozen The Mole is a number like a dozen (12 items) Or a gross (144 items); it is Avogadro s number of items. One mole of anything then is N A of that thing. 1 Mole of items = items 1Gross of eggs = eggs 1 Mole of eggs = eggs 1 = eggs = Mole of eggs eggs Mole of eggs N A = = mole The Mole is exactly the number of carbon-12 (i.e., 12 C o ) atoms that weigh 12 grams. Therefore the Mole is that number of items, of any type of thing. We can have a mole of stars for example or a mole of Carbon-12 atoms, or a mole of H 2 molecules.

5 Using N A Have 2 moles of water, how many molecules? This is correct: x = 2 moles This is incorrect: x moles = 2 This is correct: N = ( ) A 23 mole of molecules = molecules therefore 23 1 = N A = molecules mole ( of molecules) 23 x = 2 moles () 1 = 2 moles ( N ) = ( molecules) A 23 molecules moles of molecules The value of x changes only as the units of x change. How big (volume) is 2 moles of water?

6 Dalton: Atomic Mass Unit, amu One 12 C atom weighs 12 amus One mole of 12 C atoms weighs 12 grams 12 mass mole C = 12 gram 12 mass C = 12 amu Divide (Ratio the two) 12 mass mole C 12 gram gram = = 1 12 mass C amu amu Cancel Units 12 gram 1mole = 1 1gram = 1mole amu amu gram = 1 mole N A = 1 mole amu mole gram = gram= amu + 23 amu

7 Summary: Avogadro s Number N A = Is a conversion factor between items (molecules) and moles AND between grams and amu (or Daltons): mol

8 y The point of Avogadro s number N A is so useful because we can talk about molecules reacting and having weights in UAMUs (or Daltons), and then that means we can talk about moles of molecules reacting having weights in grams. 1 gram = m ole of M olecules M olecules 1 ( ) = amu grams amu mole + 23 = x m ole gram M olecule = amu x M olecule grams 5 = 5 m ole amu M olecule If you have one mole of a substance that weights 5 grams, then one molecule of the substance weights 5 Daltons (UAMU).

9 Compare Avogadro s Number with the number of stars in the universe How many stars are there in the universe? -- There are about 10 billion (10^10) stars in an average galaxy, and there are about 10 billion galaxies that we can observe in the universe, so the answer is 100 billion-billion or 10^20. John Hawley (Argonne) How many universes do we need to get Avogadro's number of stars? (6,000). Avogadro s number of anything is truly inconceivably large!!! How many nucleons are there in our Sun (roughly)? 33 Our sun is about 210 g

10 The Mass/Mole relationship 12 red 7g each = 84g 55.85g Fe => x atoms Fe 12 yellow each=48g 32.07g S => x atoms S Each element has its own atomic mass, on the periodic table. Why is Carbon not 12.00? Reminder: Generally the atomic mass is very close to the mass number. Why is Chlorine so far from an atomic mass number?

11 A Reminder: Average Atomic Mass of an Element Problem: Calculate the average atomic mass of Magnesium! Magnesium Has three stable isotopes, 24 Mg ( 78.7%); 25 Mg (10.2%); 26 Mg (11.1%). 24 Mg (78.7%) amu x = amu 25 Mg (10.2%) amu x = amu 26 Mg (11.1%) amu x = amu amu With Significant Digits = 24.3 amu

12

13 Mass/Mole Relationship Atoms/Elements Atomic masses (or weights) are conversion factors to go from number to mass (for atoms). ( H) = ( ) 1atom H = 1.008amu AW = 1 = amu = amu atom amu atom atoms mole amu gram N atom N A A A W ( H) = 1 = amu = atom g mol 1 atom Fe = amu 1 mol Fe = g A W ( ) ( ) Fe ( ) = 55.85amu = g atom mol

14 Molecular Mass vs Molar Mass ( M W ) The Molecular mass of a compound expressed in amu is numerically the same as the mass of one mole of the compound expressed in grams, called its molar mass. For water: H 2 O Molecular mass = (2 x atomic mass of H ) + atomic mass of O = 2 ( amu) amu = amu Mass of one molecules of water = amu Molar mass = ( 2 x molar mass of H ) + (1 x molar mass of O) = 2 ( g ) g = g g H 2 O = x molecules of water = 1 mole H 2 O

15 Use Atomic Masses to obtain Molecular Mass Use atomic masses to compute the mass of a molecule. The periodic table gives us the average mass of each atom type ( atomic weight ) We need to use the molecular formulas and learn how to compute the molecular masses from them. ( ) = ( ) + ( ) M HO A H A O 2 W 2 W W

16 Calculate the Molecular Mass of Glucose: C 6 H 12 O 6 ( ) = 6 ( ) + 12 ( ) + 6 ( ) M C H O A C A H A O M W W W W W ( C H O ) Carbon = g = amu mol molecule 6 x amu = amu Hydrogen 12 x amu = amu Oxygen 6 x amu = amu amu A shortcut: carbohydrate formula is (CH 2 O) 6. What does one mole of glucose weigh?

17 Masses of 1 mole of Common substances CaCO g Oxygen g Copper g Note: This balloon of oxygen (O 2 ) gas is a bit too small. It should be >20 L unless highly pressurized. Water g

18 Know Atomic and Molecular Masses We have reproduced Cannizzaro s solution to the puzzle of molecular formulae. If we know the masses then we can infer the molecular formula/structure (the stoichiometry) If we know the molecular formula we can infer the atomic masses.

19 The Percent Composition of Molecules Know the masses of each type of atom So you know the total mass Can compute the fractional or percentage of each atom type in the molecule. Eg: Glucose, C 6 H 12 O 6, done above is 180 Da total, 72 Da C, 12 Da H, 96 Da O. So the percent of this molecule that is carbon, by weight (called the mass percent): m C = 6 AW C = 6 12 g = 72 g mol mol mc ( ) 72 = = = 0.40 C M Glucose ( ) ( ) f C W = 40% C by mass

20 The RYP molecule Atomic Masses given on each atom What is the molecular Formula: R 3 Y 3 P 2 Red, Yellow, Purple

21 Check: Mass Fraction and Mass % Total Mass = Mass Red +Mass Purple + Mass Yellow Mass of Red Balls = Mass Fraction Red = Mass % Red = Mass Fraction Purple = Similarly, mass fraction yellow =

22 Mass Fraction and Mass % Mass of Red Balls = 3 balls x 3.0 g/ball = 9.0 g Mass Fraction Red = 9.0 g / 16.0 g total = 0.56 Mass % Red = 0.56 x 100% = 56% red Mass Fraction Purple = 2 balls x 2.0 g/ball / (16.0 g total) = 0.25 = 25% Similarly, mass fraction yellow = 3x1.0/16.0 = 0.19 Check: 56% + 25% + 19% = 100%

23 Why do we need to go between moles and grams? Molecules go together as atoms (or moles) Molecules can only be weighed to determine amount. We need to convert back and forth because reactions preserve atoms (or moles of atoms) Molecules represent the rearrangement and joining together of numbers of atoms of different types (or moles of atoms).

24 Extra Information More Information and extra problems. When working extra problems (or problems in the text): Be sure to cover up the solution before you start the problem.

25 Calculate M and % composition of NH 4 NO 3. 2 mol N x 4 mol H x 3 mol O x Molar mass = M = %N = g N x 100% = g cpd %H = g H x 100% = g cpd %O = x 100% = g O g cpd Check: 100% total?

26 Calculate M and % composition of NH 4 NO 3. 2 mol N x g/mol = g N 4 mol H x g/mol = g H 3 mol O x g/mol = g O This is the molecular weight: g/mol %N = x 100% = %H = x 100% = %O = x 100% =

27 Calculate M and % composition of NH 4 NO 3. 2 mol N x g/mol = g N 4 mol H x g/mol = g H 3 mol O x g/mol = g O g/mol %N = 28.02g N x 100% = 35.00% 80.05g %H = 4.032g H x 100% = 5.037% 80.05g %O = x 100% = 59.96% 48.00g O 80.05g %

28 2 H 2 (g) + O 2 (g) 2 H 2 O(g) 2 dozen H 2 molecules react with exactly 1 dozen O 2 molecules to give exactly 2 dozen H 2 O molecules. 2 mole of H 2 molecules react with exactly 1 mole of O 2 molecules to give exactly 2 moles of H 2 O molecules. Why do we do this? Because these last sizes are in the gram range and easy to weigh. Conventions: 1 mole of 12 C atoms weighs 12 g exactly. 1 atom of 12 C weighs 12 Da (amu) exactly. (amu = atomic mass unit = ~mass of a proton or neutron)

29 2 H 2 (g) + O 2 (g) 2 H 2 O(g) 2 dozen H 2 molecules react with exactly 1 dozen O 2 molecules to give exactly 2 dozen H 2 O molecules. 2 mole of H 2 molecules 4 x g = g 1 mole of O 2 molecules 2 x g = g 2 moles of H 2 O molecules 2 x (2x1.008 g g) = g Why do we do this? Because these last sizes are in the gram range and easy to weigh. Conventions: 1 mole of 12 C atoms weighs 12 g exactly. 1 atom of 12 C weighs 12 amu exactly. (amu = atomic mass unit = ~mass of a proton or neutron)

30 Summary: AMU, Dalton, 12 C Std. Atomic mass Unit (AMU) = 1/12 the mass of a carbon - 12 atom on this scale Hydrogen has a mass of AMU. Dalton (D) = The new name for the Atomic Mass Unit, one dalton = one Atomic Mass Unit on this scale, 12 C has a mass of daltons. Isotopic Mass = The relative mass of an Isotope relative to the 12 C isotope standard = amu. Atomic Mass or Atomic Weight of an element = the average of the masses of its naturally occurring isotopes weighted according to their abundances.

31 The Average Atomic Mass of an Element Problem: Calculate the average atomic mass of Magnesium! Magnesium Has three stable isotopes, 24 Mg ( 78.7%); 25 Mg (10.2%); 26 Mg (11.1%). Data from Mass Spec. 24 Mg (78.7%) amu 25 Mg (10.2%) amu 26 Mg (11.1%) amu Total = With Significant Digits = amu

32 .

33 Calculating the Moles and Number of Molecules or Formula Units in a given Mass of a known Cpd. Problem: Sodium Phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample? Plan: We determine the formula from the name, and the molecular mass from the atomic masses of each element multiplied by the coefficients. Solution: The formula is Na 3 PO 4. Calculating the molar mass: M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen = = 3 x g/mol + 1 x g/mol + 4 x g/mol = g/mol g/mol g/mol = g/mol Converting mass to moles: Moles Na 3 PO 4 = 38.6 g Na 3 PO 4 x (1 mol of Na 3 PO 4 ) g Na 3 PO 4 = mol Na 3 PO 4 Number of molecules (Formula units) = mol Na 3 PO 4 x x formula units/mole 1 mol Na 3 PO 4 = 1.46 x formula units

34 g g = M m m n n= N m A

35 .

36 Problem: Estimate the abundance of the two Bromine isotopes, given that the average mass of Br is amu. Since exact masses of isotopes not given, estimate from: mass in amu = #p + + #n: 79 Br = 79 g/mol and 81 Br = 81 g/mol (approximately). Plan: Let the abundance of 79 Br = X and of 81 Br = Y and X + Y = 1.0 Solution: X(79) + Y(81) = X + Y = 1.00 therefore X = Y ( Y)(79) + Y(81) = Y + 81Y = Y = or 1 with/ sig. figs. so Y = 0.5 X = Y = = 0.5 %X = % 79 Br = 0.5 x 100% = 50% (Actual: 50.67% = 79 Br) %Y = % 81 Br = 0.5 x 100% = 50% (Actual: 49.33% = 81 Br)

37 Mole - Mass Relationships of Atoms/Elements Element Atomic Mass Molar Mass Number of Atoms 1 atom of H = Da 1 mole of H = g = x atoms 1 atom of Fe = u 1 mole of Fe = g = x atoms 1 atom of S = 32 Da 1 mole of S = g = atoms 1 atom of O = 16 u 1 mole of O = g = atoms Molecular mass: 1 molecule of O 2 = amu 1 mole of O 2 = g = molecules 1 molecule of S 8 = amu 1 mole of S 8 = g = molecules

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