Chemistry The Central Science 10 th edition Theodore

Chemistry, The Central Science, 10 th edition Theodore L. Brown; H. Eugene Le. May, Jr. ; and Bruce E. Bursten Unit 3 (Chp 1, 2, 3): Matter, Measurement, & Stoichiometry John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice Hall Inc.

Chemistry: The study of matter and the changes it undergoes. Ni + H 2 HCl nickel hydrochloric acid solid aqueous metal solution + Ni. Cl 2 hydrogen nickel(II) chloride gas solid crystals Quantitative or Qualitative

Matter Atom: simplest particle with properties of element Element: same type of atom (1 or more) HH O O H 2 O 2 C C Na C Compound: different atoms bonded molecule H 2 O CO 2 Na. Cl

chromatography distillation (boiling) Matter separate physically Mixture differences or unevenly mixed filtering Heterogeneous Mixture (suspensions/colloids) Physical changes uniform or evenly mixed Homogeneous Mixture (solutions) cannot separate physically Pure Substance separate chemically cannot separate Chemical changes Compounds Elements salt, baking soda, oxygen, iron, water, sugar hydrogen, gold Na. Cl Na. HCO 3 O 2 Fe H 2 O C 12 H 22 O 11 H 2 Au

Physical Separation: Filtration: Separates heterogeneous mixtures (solids from liquids).

Physical Separation: Distillation: Separates solution by boiling point differences.

Physical Separation: Chromatography: Separates solution by differences in solubility (attractions).

Metric Prefixes Prefix Symbol Multiplier Examples: 1, 000, 000 B 1, 000 J 1, 000 g BASE UNIT: 1 m 1 L 1 g 0. 01 m 0. 001 L (light wavelength) (atoms) (nuclei) 0. 000 001 g 0. 000 001 m

Precision in Measurements Measuring devices have different uses and different degrees of precision. (uncertainty) % Error = |Experimental – Accepted| x 100 Accepted

Significant Digits • measured digits (using marks on instrument) • last estimated digit (one digit past marks) 5. 23 cm • do not overstate the precision 5. 230 cm

Significant Zeroes 0. 0003700400 grams 0’s 1. All nonzero digits are significant. 2. Captive Zeroes between two significant figures are significant. 3. Leading Zeroes at the beginning of a number are never significant. 4. Trailing Zeroes: Sig, if there’s a decimal point. NOT, if there is no decimal point.

Scientific Notation Power of 10 is the number of places the decimal has been moved. Examples: 42000 = 4. 2 x 104 0. 0508 = 5. 08 x 10– 2 positive power: move decimal right to obtain the original # in standard notation. negative power: move decimal left to obtain the original # in standard notation.

Scientific Notation 1. Convert the numbers to scientific notation. 2. 45 x 104 (i) 24500 9. 85 x 10– 4 (ii) 0. 000985 (iii) 12002 1. 2002 x 104 2. Convert to standard notation. 420, 000 (i) 4. 2 x 105 0. 000215 (ii) 2. 15 x 10 -4 0. 003 (iii) 3 x 10 -3

Sigs Digs in Operations + or – round answers to keep the fewest decimal places 3. 48 + 2. 2 = 5. 68 x or ÷ 5. 7 round answers to keep the fewest significant digits 6. 40 x 2. 0 = 12. 8 13

Sig Digs Practice WS 1 s 1. How many sig digs are in each number? 4 (i) 250. 0 (ii) 4. 7 x 10– 5 2 (iii) 34000000 2 4 (iv) 0. 03400 2. Round the answer to the correct sig digs. (i) 34. 5 x 23. 46 809 (ii) 123/3 40 (iii) 23. 888897 + 11. 2 35. 1 (iv) 2. 50 x 2. 0 – 3 2

WARM UP (for QUIZ!!!) • Review WS 1 s #1, 3, 10 • Complete WS 1 a #1, 2, 8, 9, 10

Law of Definite Proportions • 2 H’s & 1 O is ALWAYS water. • Water is ALWAYS 2 H’s & 1 O. • 2 H’s & 2 O’s is NOT water. √ H 2 O X H 2 O 2 Ø elemental formulas (composition) of pure compounds cannot vary. H H O O H H O

Law of Conservation of Mass The total mass of substances present at the end of a chemical process is the same as the mass of substances present before the process took place. __H 2 2 + __O 2 2 2 O __H Balancing Equations!!!

Symbols of Elements Mass Number = p’s + n’s 12 C 6 Element Symbol Atomic Number (Z) = p’s All atoms of the same element have the same number of protons (same Z), but… can have different mass numbers. HOW?

Isotopes element: same or different mass: same or different why? same # of protons (& electrons), but different # of neutrons 1 H 1 protium 2 H 1 deuterium 3 H 1 tritium

Average Atomic Mass • average atomic mass: calculated as a weighted average of isotopes by their relative abundances. • lithium-6 (6. 015 amu), which has a relative abundance of 7. 50%, and • lithium-7 (7. 016 amu), which has a relative abundance of 92. 5%. (6. 015)(0. 0750) + (7. 016)(0. 925) = 6. 94 amu Avg. Mass = (Mass 1)(%) + (Mass 2)(%) …

Mass Spectrometry atomized, ionized magnetic field element sample isotopes separated WS Atomic by Structure difference Cl (avg at. Mass) = in mass (35)(~0. 75) + (37)(~0. 25) = ? ~75% ~25%

Molecular (Covalent) Compounds Covalent compounds contain nonmetals that “share” electrons to form molecules. (molecular compounds)

Diatomic Molecules “H-air-ogens” 7 These seven elements occur naturally as molecules containing two atoms.

Binary Molecular Compounds • list less electronegative atom first. (left to right on PT) • use prefix for the number of atoms of each element. • change ending to –ide. CO 2: carbon dioxide CCl 4: carbon tetrachloride pentoxide N 2 O 5: dinitrogen ________ Cu. SO 4∙ 5 H 2 O (ionic & covalent) copper(II) sulfate pentahydrate

Cations metals lose e’s (+) charge (metal) ion Anions Ions nonmetals gain e’s (–) charge (nonmetal)ide

Ionic Bonds Attraction between +/– ions formed by metals & nonmetals transferring e–’s.

Formulas of Ionic Compounds • Compounds are electrically neutral, so the formulas can be determined by: – Crisscross the charges as subscripts (then erase) – If needed, reduce to lowest whole number ratio. Pb 4+ O 2– Pb 2 O 4 Pb. O 2

Naming Ionic Compounds 1) Cation: Write metal name (ammonium NH 4+) For transition metals with multiple charges, write charge as Roman numeral in parentheses. Iron(II) chloride, Fe. Cl 2 Iron(III) chloride, Fe. Cl 3 2) Anion: Write nonmetal name with –ide OR the polyatomic anion name. (–ate, –ite) Iron(II) sulfide, Fe. S Magnesium sulfate, Mg. SO 4

Common Polyatomic Ions * these 12 will be on Quiz 1 - all 20 Polyatomic Ions will be on Quiz 2 WS 2 d Name Symbol Charge *ammonium NH 4+ 1+ *acetate C 2 H 3 O 2– 1– – (ethanoate) (CH 3 COO ) *hydroxide OH– 1– *perchlorate Cl. O 4– 1– *chlorate Cl. O 3– 1– chlorite Cl. O 2– 1– hypochlorite Cl. O– 1– bromate Br. O 3– 1– iodate IO 3– 1– *nitrate NO 3– 1– nitrite NO 2– 1– cyanide CN– 1– *permanganate Mn. O 4– 1– *bicarbonate 1– HCO 3– (hydrogen carbonate) *carbonate CO 32– 2– *sulfate SO 42– 2– sulfite SO 32– 2– *chromate Cr. O 42– 2– dichromate Cr 2 O 72– 2– *phosphate PO 43– 3–

“Oxyanion” Names (elb. O’s) perchlorate chlorite hypochlorite Cl. O 4– Cl. O 3– Cl. O 2– Cl. O– nitrate nitrite NO 3– NO 2– sulfate sulfite SO 42– SO 32– phosphate PO 43– C Si N O P S As Se Te In Out 4 – 3 2 1 4 3 – F Cl Br I Ion Name per-___-ate ___-ite hypo-___-ite

Naming Acids Ion add H+ Acid In Out Ion Name Acid Name 4 – per-___-ate per-___-ic acid 3 4 ___-ic acid ___-ate 2 3 ___-ous acid ___-ite 1 – hypo-___-ite hypo-___-ous acid perchlorate Name Acids chlorate from these chlorite oxyanions: hypochlorite WS 2 e Cl. O 4 Cl. O 3– Cl. O 2– Cl. O– – nitrate NO 3– nitrite NO 2– sulfate SO 42– sulfite SO 32–

Anatomy of a Chemical Equation CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g)

Anatomy of a Chemical Equation CH 4(g) + 2 O 2(g) Reactants appear on the left side of the equation. CO 2(g) + 2 H 2 O(g)

Anatomy of a Chemical Equation CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g) Products appear on the right side of the equation.

Anatomy of a Chemical Equation CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g) States (s, l, g, aq) written in parentheses next to each compound

Anatomy of a Chemical Equation CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g) Subscripts show many atoms of each element

Anatomy of a Chemical Equation CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g) Coefficients show the amount of each particle and are inserted to balance the equation.

Reaction Types

Combination A + B → AB Demo: Mg. O 2 Mg(s) + 2→ 1 O 2(g) 2 Mg. O(s)

Decomposition AB → A + B 1→ 2 (50 milliseconds!) 2 Na. N 3(s) 2 Na(s) + 3 N 2(g)

Replacement Reactions (or “Displacement”) Single Replacement AB + C → A + CB video Ag. NO 3(aq) + Cu(s) Ag(s) + Cu. NO 3(aq) Double Replacement AB + CD → AD + CB Pb(NO 3)2(aq) + KI(aq) Pb. I 2(s) + KNO 3(aq) Demo: Pb. I 2

Combustion Cx. Hy + _O 2 _CO 2 + _H 2 O • Often involve hydrocarbons reacting with oxygen in the air WS 4 a CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g) C 3 H 8(g) + 5 O 2(g) 3 CO 2(g) + 4 H 2 O(g)

Formula Weights

Formula Weight (FW) Molecular Weight (MW) • Sum of the atomic weights for the atoms in a chemical formula • Formula Weight of calcium chloride, Ca. Cl 2, is… • Sum of the atomic weights for the atoms in a molecule or compound • Molecular Weight of ethane, C 2 H 6, is… Ca: 1(40. 08 amu) + Cl: 2(35. 45 amu) 110. 98 amu C: 2(12. 01 amu) + H: 6(1. 008 amu) 30. 07 amu

Percent Composition (# of atoms)(AW) x 100 % element = (FW) One can find the percent by mass of a compound of each element in the compound by using this equation.

Percent Composition So the percentage of carbon in ethane (C 2 H 6) is… %C = = (2)(12. 01) (30. 07) 24. 02 x 100 30. 07 = 79. 88% C

Moles

Avogadro Constant • One mole of particles contains the Avogadro constant of those particles 6. 022 x 1023

Mole Relationships • One mole of atoms, ions, or molecules contains the Avogadro constant of those particles 6. 022 x 1023 In 1 mol Na 2 CO 3 , how many… • Na atoms? • C atoms? • O atoms? • How many donuts in 1 mol of donuts? • How many boogers in 1 mol of boogers? Which has more atoms, 1 mol CH 3 or 1 mol NH 3 ? How about CH 3 CH 2 OH or H 2 SO 4 ?

Molar Mass • the mass of 1 mol of a substance (g/mol) – molar mass (in g/mol) of an element is the atomic mass (in amu) on the periodic table – formula weight (amu) of a compound same number as the molar mass (g/mol) of 1 mole of particles of that compound

Using Moles are the bridge from the particle (micro) scale to the real-world (macro) scale. macro. Mass (grams) molar mass bridge Moles micro. Avogadro constant Particles 1 mol #g (groups of (atoms) 6. 022 x 1023 (molecules) 23 particles) 6. 022 x 10 (units) 1 mol #g 1 mol 6. 022 x 1023

Using Moles 1. What is the mass of 1 mole of copper(II) bromide, Cu. Br 2? (63. 55) + 2(79. 90) = 223. 35 g = 6. 022 x 1023 particles 2. How many moles are there in 112 g of copper(II) bromide, Cu. Br 2? 1 mol Cu. Br 2 = 0. 501 mol 112 g Cu. Br 2 x 223. 35 g Cu. Br 2 3. How many particles present in each of the questions #1 & #2 above? 6. 022 x 1023 particles = 3. 02 x 1023 0. 501 mol x 1 mol particles

Stoichiometry: calculations of quantities in chemical rxns –how much reactant is consumed or –how much product is formed • Balanced chemical equations show the amount of: atoms, molecules, moles, and mass Most important are the ratios of reactants and products in moles, or… mol-to-mol ratios

Stoichiometric Calculations Rxn: g. A A(aq) + 2 B(aq) C(aq) + 2 D(aq) molar mass A g A 1 mol A ? 1 mol A OR ? g. A ? g B 1 mol B molar g. B mass B mol A Coefficients of balanced equation OR mol-to-mol 2 mol B 1 mol A ratio mol B 1 mol A 2 mol B

Stoichiometric problems have 1 -3 Steps: (usually) 1) Convert grams to moles (if necessary) using the molar mass (from PT) 2) Convert moles (given) to moles (wanted) using the mol ratio (from coefficients) 3) Convert moles to grams (if necessary) using the molar mass (from PT) grams A x 1 mol A x _ mol B x grams A mol A. grams B = 1 mol B 1) molar mass 2) mole ratio 3) molar mass

Stoichiometric Calculations Example : g of A g of B HW p. 114 #58 Solid magnesium is added to an aqueous solution of hydrochloric acid. What mass of H 2 gas will be produced from completely reacting 18. 0 g of HCl with magnesium metal? Mg(s) + 2 HCl(aq) Mg. Cl 2(aq) + H 2(g) mole ratio B/A molar mass B 2. 016 g H 2 1 mol HCl x 1 mol H 2 x 18. 0 g HCl x 36. 46 g HCl 2 mol HCl 1 mol H 2 g of A molar mass A = ____ g Hg 2 H 2 0. 498

Finding Empirical Formulas

Types of Formulas • Empirical formulas: the lowest ratio of atoms of CH 3 each element in a compound. C 2 H 4 O • Molecular formulas: C 2 H 6 C 6 H 12 O 3 the total number of atoms of each element in a compound. molecular mass = emp. form. empirical mass multiple

Calculating Empirical Formulas Steps (rhyme) from Mass % Composition Percent to Mass assume 100 g Mass to Mole MM from PT ÷ moles by smallest to CH 4 Divide by Small get mole ratio of atoms x (if necessary) to get Times ‘til Whole whole numbers of atoms 75 % C 75 g C 6. 2 mol C 1 C 25 % H 25 g H 24. 8 mol H 4 H

1) Percent to Mass 3) Divide by Small 2) Mass to Mole 4) Times ’til Whole Butane is 17. 34% H and 82. 66% C by mass. Determine its empirical formula. 82. 66 g C x 1 mol C = 6. 883 mol C = 1 1 C x 2 12. 01 g C 6. 883 mol =2 C 17. 34 g H x 1 mol H = 17. 20 mol H 1. 008 g H 6. 883 mol = 2. 499 2. 5 H C 2 H 5 x 2 =5 H If molecular mass is 58 g∙mol– 1, what is the Molecular Formula? HW p. 113 #43 a, 48 molecular mass empirical mass 58 =2 29. 06 2 (C 2 H 5) = C 4 H 10

Calculating Empirical Formulas Percent to Mass to Mole Divide by Small Times ‘til Whole

Combustion Analysis • Hydrocarbons with C and H are analyzed through combustion with O 2 in a chamber. Step 1 is “combustion Ø g H is from the g H 2 O produced to mass” Ø g C is from the g CO 2 produced Ø g X is found by subtracting (g C + g H) from g sample

Combustion Analysis Example 1 When 4 -ketopentenoic acid is analyzed by combustion, a 0. 3000 g sample produces 0. 579 g of CO 2 and 0. 142 g of H 2 O. The acid contains only C, H, and O. What is the empirical formula of the acid?

1 mol CO 2 1 mol C 12. 01 g C 0. 579 g CO 2 x x x 44. 01 g CO 2 1 mol C ? g. C = 0. 158 g C Step 1: “combustion to mass” 1 mol H 2 O x 2 mol H x 1. 008 g H 0. 142 g H 2 O x 18. 02 g H 2 O 1 mol H ? g. H = 0. 0159 g H 0. 3000 g sample – (0. 158 g C) – (0. 0159 g H) = ? g. O = 0. 126 g O

1 mol C 0. 158 g C x = 0. 0132 mol C = 1. 67 C 12. 01 g C 0. 00788 mol x 3=5 C 1 mol H 0. 0159 g H x = 0. 0158 mol H = 2 H 1. 008 g H 0. 00788 mol x 3=6 H 1 mol O 0. 126 g O x = 0. 00788 mol O = 1 O 16. 00 g O 0. 00788 mol x 3=3 O C 5 H 6 O 3

Combustion Analysis Example 2 A sample of a chlorohydrocarbon with a mass of 4. 599 g, containing C, H and Cl, was combusted in excess oxygen to yield 6. 274 g of CO 2 and 3. 212 g of H 2 O. Calculate the empirical formula of the compound. If the compound has a MW of 193 g∙mol– 1, what is the molecular formula?

1 mol CO 2 1 mol C 12. 01 g C 6. 274 g CO 2 x x x 44. 01 g CO 2 1 mol C ? g. C = 1. 712 g C Step 1: “combustion to mass” 1 mol H 2 O x 2 mol H x 1. 008 g H 3. 212 g H 2 O x 18. 02 g H 2 O 1 mol H ? g. H = 0. 3593 g H 4. 599 g sample – (1. 712 g C) – (0. 3593 g H) = ? g Cl = 2. 528 g Cl

1 mol C 1. 712 g C x = 0. 1425 mol C = 2 C 12. 01 g C 0. 07131 mol H 0. 3593 g H x = 0. 3564 mol H = 5 H 1. 008 g H 0. 07131 mol Cl 2. 528 g Cl x = 1 Cl = 35. 45 g Cl 0. 07131 mol C 2 H 5 Cl If the compound has a MW of 193 g∙mol– 1, what MW 193 =3 is the molecular formula? EW 64. 51 HW p. 114 #52 b C 6 H 15 Cl 3

How Many Cookies Can I Make? • Which ingredient will run out first? • If out of sugar, you should stop making cookies. • Sugar is the limiting ingredient, because it will limit the amount of cookies you can make.

Before H 2 After O 2 Which is limiting? 2 H 2 + O 2 Initial: 10 ? mol Change: – 10 End: 0 mol limiting ? 7 mol – 5 2 mol excess 2 H 2 O ? 0 mol +10 10 mol

Before H 2 After O 2 2 H 2 + O 2 Initial: 10 ? mol Change: – 10 End: 0 mol ? 7 mol – 5 2 mol 2 H 2 O ? 0 mol +10 10 mol Does limiting mean smallest amount of reactant? No! O 2 is in smallest amount, but… H 2 is in smallest “stoichiometric” amount

Limiting Reactant • convert reactant A to reactant B to compare • If available < needed (limiting) • If available > needed (excess) Solid aluminum metal is reacted with aqueous copper(II) chloride in solution 2 Al + 3 Cu. Cl 2 2 Al. Cl 3 + 3 Cu 54. 0 g Al 4. 50 mol Cu. Cl 2 (Which is limiting? ) 54. 0 g Al x 1 mol Al x 3 mol Cu. Cl 2 = 3. 00 mol Cu. Cl 2 26. 98 g Al 2 mol Al (4. 50 mol Cu. Cl 2) available > needed (3. 00 mol Cu. Cl 2) Cu. Cl 2 is excess Al is limiting

Theoretical Yield theoretical yield: the maximum amount of product that can be formed – calculated by stoichiometry – limited by LR (use LR only to calculate) limiting 54. 0 g Al x 1 mol Al x 3 mol Cu x 63. 55 g Cu = 191 g Cu 26. 98 g Al 2 mol Al 1 mol Cu produced • different from actual yield (or experimental), amount recovered in the experiment HW p. 115 #72

Percent Yield A comparison of the amount actually obtained to the amount it was possible to make Actual %Yield = x 100 Theoretical (calculate using the LR only) NOT % Error: % Error = |Accepted – Experimental| x 100 Accepted

Percent Yield Aluminum will react with oxygen gas according to the equation below 4 Al + 3 O 2 2 Al 2 O 3 • In one such reaction, 23. 4 g of Al are allowed to burn in excess oxygen. 39. 3 g of aluminum oxide are formed. What is the percentage yield?

Percent Yield 4 Al + 3 O 2 HW p. 116 #79, 77 2 Al 2 O 3 1 mol Al 2 O 3 101. 96 g Al 2 O 3 23. 4 g Al x x x 26. 98 g Al 4 mol Al 1 mol Al 2 O 3 = 44. 2 g Al 2 O 3 39. 3 g of aluminum oxide are formed. What is the percentage yield? 39. 3 g %Yield = 44. 2 g x 100 88. 9 %