1. Slides Shown Friday Oct. 10 2008

2. An Acid: Any substance that increases the [H + ] in water (g) Gas (l) Liquid (s) Solid (aq) Aqueous (water) solution HCl(g): hydrogen chloride gas HCl(aq) hydrochloric acid

3. Naming Acids 1) Binary acids solutions form when certain gaseous compounds dissolve in water. For example, when gaseous hydrogen chloride (HCl) dissolves in water, it forms a solution called hydrochloric acid. Prefix hydro- + anion nonmetal root + suffix -ic + the word acid hydrochloric acid 2) Oxoacid names are similar to those of the oxoanions, except for two suffix changes: Anion “-ate” suffix becomes an “-ic” suffix in the acid. Anion “-ite” suffix becomes an “-ous” suffix in the acid. The oxoanion prefixes “hypo-” and “per-” are retained. Thus, BrO 4 - is perbromate, and HBrO 4 is perbromic acid; IO 2 - is iodite, and HIO 2 is iodous acid.

4. Naming acids

5. Names of Acids that do Contain Oxygen Acid Name HF hydrofluoric acid HCl hydrochloric acid HBr hydrobromic acid HI hydroiodic acid HCN hydrocyanic acid H 2 S hydrosulfuric acid

6. Names of some Oxygen-Containing Acids Acid Name HNO 3 nitric acid HNO 2 nitrous acid H 2 SO 4 sulfuric acid H 2 SO 3 sulfurous acid H 3 PO 4 phosphoric acid HC 2 H 3 O 2 acetic acid

7. Naming of the Oxoacids of Chlorine Acid Anion Name HClO 4 perchlor ate perchlor ic acid HClO 3 chlor ate chlor ic acid HClO 2 chlor ite chlor ous acid HClO hypochlor ite hypochlor ous acid

8. Mass spectrometer

9. Relative intensities of the signals recorded when natural neon is injected into a mass spectrometer.

10. Mass spectrum of natural copper

11. Oxygen, Isotopes 16 8 O 8 Protons 8 Neutrons 99.759% 15.99491462 amu 17 8 O 8 Protons 9 Neutrons 0.037% 16.9997341 amu 18 8 O 8 Protons 10 Neutrons 0.204 % 17.999160 amu

12. Calculating the “Average” Atomic Mass (aka the Atomic Weight) of an Element 24 Mg (78.7%) 23.98504 amu x 0.787 = 18.876226 amu 25 Mg (10.2%) 24.98584 amu x 0.102 = 2.548556 amu 26 Mg (11.1%) 25.98636 amu x 0.111 = 2.884486 amu 24.309268 amu With Significant Digits = 24.3 amu 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%).

13. Calculate the Average Atomic Mass of Zirconium, Element #40 Zirconium has five stable isotopes: 90 Zr, 91 Zr, 92 Zr, 94 Zr, 96 Zr. Isotope (% abd.) Mass (amu) (%) Fractional Mass 90 Zr (51.45%) 89.904703 amu X 0.5145 = 46.2560 amu 91 Zr (11.27%) 90.905642 amu X 0.1127 = 10.2451 amu 92 Zr (17.17%) 91.905037 amu X 0.1717 = 15.7801 amu 94 Zr (17.33%) 93.906314 amu X 0.1733 = 16.2740 amu 96 Zr (2.78%) 95.908274 amu X 0.0278 = 2.6663 amu 91.2215 amu With Significant Digits = 91.22 amu

14. Problem: Calculate the abundance of the two Bromine isotopes: 79 Br = 78.918336 g/mol and 81 Br = 80.91629 g/mol, given that the average mass of Bromine is 79.904 g/mol. Plan: Let the abundance of 79 Br = X and of 81 Br = Y and X + Y = 1.0 Solution: X(78.918336) + Y(80.91629) = 79.904 X + Y = 1.00 therefore X = 1.00 - Y (1.00 - Y)(78.918336) + Y(80.91629) = 79.904 78.918336 - 78.918336 Y + 80.91629 Y = 79.904 1.997954 Y = 0.985664 or Y = 0.4933 X = 1.00 - Y = 1.00 - 0.4933 = 0.5067 %X = % 79 Br = 0.5067 x 100% = 50.67% = 79 Br %Y = % 81 Br = 0.4933 x 100% = 49.33% = 81 Br

15. During a perplexing dream one evening you come across 200 atoms of Einsteinium. What would be the total mass of this substance in grams? SOLUTION: 200 Atoms of Es X 252 AMU/Atom = 5.04 x 10 4 AMU 5.04 x 10 4 AMU x (1g / 6.022 x 10 23 AMU) = 8.37 x 10 -20 g of Einsteinium

16. Atomic Definitions II: 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 1.008 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 12.00 daltons. Isotopic Mass = The relative mass of an Isotope relative to the Isotope 12 C the chosen standard. Atomic Mass = “Atomic Weight” of an element is the average of the masses of its naturally occurring isotopes weighted according to their abundances.

17. MOLE The Mole is based upon the following definition: The amount of substance that contains as many elementary parts (atoms, molecules, or other?) as there are atoms in exactly 12 grams of carbon-12. 1 Mole = 6.022045 x 10 23 particles

18. One-mole samples of copper, sulfur, mercury, and carbon

20. One mole of common Substances CaCO 3 100.09 g Oxygen 32.00 g Copper 63.55 g Water 18.02 g

21. Moles Molecules Avogadro’s Number Molecular Formula Atoms 6.022 x 10 23

22. Mole - Mass Relationships of Elements Element Atom/Molecule Mass Mole Mass Number of Atoms 1 atom of H = 1.008 amu 1 mole of H = 1.008 g = 6.022 x 10 23 atoms 1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 10 23 atoms 1 atom of S = 32.07 amu 1 mole of S = 32.07 g = 6.022 x 10 23 atoms 1 atom of O = 16.00 amu 1 mole of O = 16.00 g = 6.022 x 10 23 atoms 1 molecule of O 2 = 32.00 amu 1 mole of O 2 = 32.00 g = 6.022 x 10 23 molecule 1 molecule of S 8 = 256.56 amu 1 mole of S 8 = 256.56 g = 6.022 x 10 23 molecules

23. Calculate the Molecular Mass of Glucose: C 6 H 12 O 6 Carbon 6 x 12.01 g/mol = 72.06 g Hydrogen 12 x 1.008 g/mol = 12.096 g Oxygen 6 x 16.00 g/mol = 96.00 g 180.16 g

24. Molecular Mass - Molar Mass ( M ) 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. For water: H 2 O Molecular mass = (2 x atomic mass of H ) + atomic mass of O = 2 ( 1.008 amu) + 16.00 amu = 18.02 amu Mass of one molecules of water = 18.02 amu Molar mass = ( 2 x atomic mass of H ) + atomic mass of O = 2 ( 1.008 g ) + 16.00 g = 18.02 g 18.02 g H 2 O = 6.022 x 10 23 molecules of water = 1 mole H 2 O

25. How many carbon atoms are present in a 2.0 g tablet of Sildenafil citrate (C 28 H 38 N 6 O 11 S) ? SOLUTION: MW of Sildenafil citrate = 28 X 12 amu (C) + 38 X 1 amu (H) + 6 X 14 amu (N) + 11 X 16 amu (O) + 1 X 32 amu (S) = 666 AMU

26. 2.0 g (C 28 H 38 N 6 O 11 S) X 1 mol/666g = 3.0 X 10 -3 mol (C 28 H 38 N 6 O 11 S) 3.0 X 10 -3 mol (C 28 H 38 N 6 O 11 S) X 6.022 X 10 23 molecules / 1 mol (C 28 H 38 N 6 O 11 S) = 1.8 X 10 21 molecules of C 28 H 38 N 6 O 11 S 1.8 X 10 21 molecules of C 28 H 38 N 6 O 11 S X 28 atoms of C / 1 molecules of C 28 H 38 N 6 O 11 S = 5.1 X 10 22 Carbon Atoms

27. Calculating the Number of Moles and Atoms in a Given Mass of Element Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element 3680 o C. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal? Plan: Convert mass into moles by dividing the mass by the atomic weight of the metal, then calculate the number of atoms by multiplying by Avogadro’s number! Solution: Converting from mass of W to moles: Moles of W = 35.0 mg W x = 0.00019032 mol 1.90 x 10 - 4 mol NO. of W atoms = 1.90 x 10 - 4 mol W x = = 1.15 x 10 20 atoms of Tungsten 1 mol W 183.9 g W 6.022 x 10 23 atoms 1 mole of W

28. Calculating the Moles and Number of Formula Units in a given Mass of Cpd. Problem: Trisodium Phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample? Plan: We need to determine the formula, 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 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol = 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/mol Converting mass to moles: Moles Na 3 PO 4 = 38.6 g Na 3 PO 4 x (1 mol Na 3 PO 4 ) 163.94 g Na 3 PO 4 = 0.23545 mol Na 3 PO 4 Formula units = 0.23545 mol Na 3 PO 4 x 6.022 x 10 23 formula units 1 mol Na 3 PO 4 = 1.46 x 10 23 formula units

29. Flow Chart of Mass Percentage Calculation Moles of X in one mole of Compound Mass % of X Mass fraction of X Mass (g) of X in one mole of compound M (g / mol) of X Divide by mass (g) of one mole of compound Multiply by 100

30. Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I Problem: Sucrose (C 12 H 22 O 11 ) is common table sugar. ( a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in 24.35 g of sucrose? (a) Determining the mass percent of each element: mass of C = 12 x 12.01 g C/mol = 144.12 g C/mol mass of H = 22 x 1.008 g H/mol = 22.176 g H/mol mass of O = 11 x 16.00 g O/mol = 176.00 g O/mol 342.296 g/mol Finding the mass fraction of C in Sucrose & % C : Total mass of C 144.12 g C mass of 1 mole of sucrose 342.30 g Cpd = 0.421046 To find mass % of C = 0.421046 x 100% = 42.105% Mass Fraction of C = =

31. Calculating Mass Percents and Masses of Elements in a Sample of Compound - II (a) continued Mass % of H = x 100% = x 100% = 6.479% H Mass % of O = x 100% = x 100% = 51.417% O (b) Determining the mass of carbon: Mass (g) of C = mass of sucrose X( mass fraction of C in sucrose) Mass (g) of C = 24.35 g sucrose X = 10.25 g C mol H x M of H 22 x 1.008 g H mass of 1 mol sucrose 342.30 g mol O x M of O 11 x 16.00 g O mass of 1 mol sucrose 342.30 g 0.421046 g C 1 g sucrose

32. Mol wt and % composition of NH 4 NO 3 2 mol N x 14.01 g/mol = 28.02 g N 4 mol H x 1.008 g/mol = 4.032 g H 3 mol O x 15.999 g/mol = 48.00 g O 80.05 g/mol %N = x 100% = 35.00% N 28.02g N 2 80.05g %H = x 100% = 5.037% H 4.032g H 2 80.05g %O = x 100% = 59.96% O 48.00g O 2 80.05g 99.997%

33. Calculate the Percent Composition of Sulfuric Acid H 2 SO 4 Molar Mass of Sulfuric acid = 2(1.008g) + 1(32.07g) + 4(16.00g) = 98.09 g/mol %H = x 100% = 2.06% H 2(1.008g H 2 ) 98.09g %S = x 100% = 32.69% S 1(32.07g S) 98.09g %O = x 100% = 65.25% O 4(16.00g O) 98.09 g Check = 100.00%