Matter, Measurement, & Stoichiometry

Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, 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. (has mass and takes up space) Ni + HCl H 2 + NiCl 2 nickel hydrochloric acid hydrogen nickel(ii) chloride solid aqueous gas solid crystals metal solution Quantitative or Qualitative Matter Atom: simplest particle retaining properties. Element: same type of atom (1 or more) H H C O O C H 2 O 2 C Compound: molecule C C different atoms bonded. H 2 O CO 2 NaCl Na 1

Heterogeneous Mixture distillation (boiling) separate physically differences orunevenly mixed filtering Mixture uniform orevenly mixed Homogeneous Mixture Matter Physical changes separate chemically Compounds salt, baking soda, water, sugar (suspensions/colloids) (solutions) NaCl NaHCO 3 cannot separate physically Pure Substance Chemical changes cannot separate Elements oxygen, iron, hydrogen, gold H 2 O C 12 H 22 O 11 H 2 Au O 2 Fe Density: ratio of mass to volume or matter to space occupied d = m V Units: g/ml g/cm 3 kg/l Chemical Property OR Physical Property Why? Changes of Matter Physical Changes: do not change the composition of a substance. temperature, changes of state, amount, etc. Chemical Changes: result in new substances. combustion, oxidation, decomposition, etc. 2

Chemical Separation: Compounds can be decomposed into elements. Physical Separation: Filtration: Separates heterogeneous mixtures (solids from liquids). Physical Separation: Distillation: Separates solution by boiling point differences. 3

Physical Separation: Chromatography: Separates homogeneous mixture by differences in solubility (attractions). Scientific Notation Power of 10 is the number of places the decimal has been moved. Examples: 42000 = 4.2 x 10 4 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. (i) 24500 2.45 x 10 4 (ii) 0.000985 9.85 x 10 4 (iii) 12002 1.2002 x 10 4 2. Convert to standard notation. (i) 4.2 x 10 5 420,000 (ii) 2.15 x 10-4 0.000215 (iii) 3 x 10-3 0.003 4

Metric Prefixes Prefix Symbol Multiplier Examples: BASE UNIT: 1 m 1 L 1 g (light wavelength) (atoms) (nuclei) 1,000,000,000 GB 1,000,000 MJ 1,000 kg 0.01 cm 0.001 ml 0.000 001 µg 0.000 000 001 nm Uncertainty in Measurements Measuring devices have different uses and different degrees of precision. (uncertainty) % Error = Accepted Experimental x100 Accepted Uncertainty 5.23 cm (uncertain) 5

Significant Figures measured digits. last digit is estimated, but IS significant. 5.23 cm do not overstate the precision 5.230 cm Significant Zeroes 0 s 0.0003700400 grams 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 at end AND a decimal point. NOT, if there is no decimal point. + or Significant Figures 3.48 + 2.2 = 5.68 5.7 round answers to keep the fewest decimal places x or 6.40 x 2.0 = 12.8 13 round answers to keep the fewest significant figures 6

Significant Figures WS 1s 1. How many sig figs are in each number? (i) 250.0 4 (ii) 4.7 x 10 5 2 (iii) 34000000 (iv) 0.03400 2 4 2. Round the answer to the correct sig figs. (i) 34.5 x 23.46 809 (ii) 123/3 (iii) 23.888897 + 11.2 40 35.1 (iv) 2.50 x 2.0 3 2 WARM UP (for QUIZ!!!) Review WS 1s #1, 3, 10 Complete WS 1a #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 H O H elemental formulas (composition) of pure compounds cannot vary. O H H O 7

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 H 2 2 O Balancing Equations!!! Symbols of Elements Mass Number = p s + n s 12 6 C 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? element: mass: why? Isotopes same or different same or different same # of protons (& electrons), but different # of neutrons 1 1 H 2 1 H 3 1 H protium deuterium tritium 8

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) 9

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 N 2 O 5 : dinitrogen pentoxide CuSO 4 5H 2 O (ionic & covalent) copper(ii) sulfate pentahydrate Cations metals lose e s positive (+) (metal) ion Ions Anions nonmetals gain e s negative ( ) (nonmetal)ide 10

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 PbO 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, FeCl 2 Iron(III) chloride, FeCl 3 2) Anion: Write nonmetal name with ide OR the polyatomic anion name. ( ate, ite) Iron(II) sulfide, FeS Magnesium sulfate, MgSO 4 11

Common Polyatomic Ions * these 12 will be on Quiz 1 - all 20 Polyatomic Ions will be on Quiz 2 WS 2d Name Symbol Charge *ammonium NH + 4 1+ *acetate (ethanoate) C 2 H 3 O 2 (CH 3 COO ) 1 *hydroxide OH 1 *perchlorate ClO 4 1 *chlorate ClO 3 1 chlorite ClO 2 1 hypochlorite ClO 1 bromate BrO 3 1 iodate IO 3 1 *nitrate NO 3 1 nitrite NO 2 1 cyanide CN 1 *permanganate MnO 4 1 *bicarbonate HCO 3 1 (hydrogen carbonate) *carbonate CO 2 3 2 *sulfate SO 2 4 2 sulfite SO 2 3 2 *chromate CrO 2 4 2 dichromate Cr 2 O 2 7 2 *phosphate PO 3 4 3 Oxyanion Names (elbo s) perchlorate ClO 4 chlorate ClO 3 chlorite ClO 2 hypochlorite ClO nitrate NO 3 nitrite NO 2 sulfate SO 2 4 sulfite SO 2 3 phosphate PO 4 3 C N O F Si P S Cl As Se Br Te I In Out Ion Name 4 3 4 per- -ate -ate 2 3 -ite 1 hypo- -ite Naming Acids Ion add H + Acid In Out Ion Name Acid Name 4 per- -ate per- -ic acid 3 4 -ate -ic acid 2 3 -ite -ous acid 1 hypo- -ite hypo- -ous acid Name Acids from these oxyanions: WS 2e perchlorate ClO 4 chlorate ClO 3 chlorite ClO 2 hypochlorite ClO nitrate NO 3 nitrite NO 2 sulfate SO 4 2 sulfite SO 3 2 12

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) CO 2 (g) + 2 H 2 O(g) Reactants appear on the left side of the equation. 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. 13

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 how 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. 14

Reaction Types Combination Demo: MgO A + B AB 2 1 2 Mg(s) + O 2 (g) 2 MgO(s) Decomposition AB A + B 1 2 (50 milliseconds!) 2 NaN 3 (s) 2 Na(s) + 3 N 2 (g) 15

Replacement Reactions (or Displacement ) Single Replacement AB + C A + CB video clip (aq) + (s) (s) + (aq) Double Replacement AB + CD AD + CB Pb(NO 3 ) 2 (aq) + KI(aq) PbI 2 (s) + KNO 3 (aq) Demo: Combustion C x H y + _O 2 _CO 2 + _H 2 O Often involve hydrocarbons reacting with oxygen in the air WS 4a 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 16

Formula Weight (FW) Sum of the atomic weights for the atoms in a chemical formula Formula Weight of calcium chloride, CaCl 2, is Ca: 1(40.08 amu) + Cl: 2(35.45 amu) 110.98 amu Molecular Weight (MW) Sum of the atomic weights for the atoms in a molecule or compound Molecular Weight of ethane, C 2 H 6, is C: 2(12.01 amu) + H: 6(1.008 amu) 30.07 amu Percent Composition (# of atoms)(aw) % element = (FW) x 100 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 17

Moles Avogadro Constant One mole of particles contains the Avogadro constant of those particles 6.022 x 10 23 Mole Relationships One mole of atoms, ions, or molecules contains the Avogadro constant of those particles 6.022 x 10 23 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? 18

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 Moles are the bridge from the particle (micro) scale to the real-world (macro) scale. Mass (grams) molar mass 1 mol # g # g 1 mol Moles (groups of 6.022x10 23 particles) Avogadro constant 6.022x10 23 1 mol 1 mol 6.022x10 23 macrobridge micro- Particles (atoms) (molecules) (units) Using Moles 1.What is the mass of 1 mole of copper(ii) bromide, CuBr 2?(63.55) + 2(79.90) = 223.35 g = 6.022 x 10 23 particles 2.How many moles are there in 112 g of copper(ii) bromide, CuBr 2? 1 mol CuBr 112 g CuBr 2 x 2 = 0.501 mol 223.35 g CuBr 2 CuBr 2 3.How many particles present in each of the questions #1 & #2 above? 0.501 mol x 6.022 x 1023 particles 1 mol = 3.02 x 10 23 particles 19

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: A(aq) + 2 B(aq) C(aq) + 2 D(aq) g A??? g B OR molar mass A g A 1 mol A 1 mol A g A g B 1 mol B molar mass B mol A mol-to-mol ratio mol B Coefficients of balanced equation OR 2 mol B 1 mol A 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. grams A x _ mol B x grams B mol A 1 mol B = 1) molar mass 2) mole ratio 3) molar mass 20

Example : g of A g of B 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) MgCl 2 (aq) + H 2 (g) 18.0 g HCl x g of A Stoichiometric Calculations 1 mol HCl 1 mol H x 2 g H x 2.016 2 36.46 g HCl 2 mol HCl 1 mol H 2 molar mass A mole ratio B/A HW p. 114 #58 molar mass B = 0.498 g Hg 2 H 2 Finding Empirical Formulas Types of Formulas Empirical formulas: the lowest ratio of atoms of CH each element in a compound. 3 Molecular formulas: C 2 H 6 the total number of atoms of each element in a compound. C 2 H 4 O C 6 H 12 O 3 molecular mass = emp. form. empirical mass multiple 21

Calculating Empirical Formulas from Mass % Composition Steps (rhyme) Percent to Mass assume 100 g Mass to Mole Divide by Small Times till Whole MM from PT moles by smallest to get mole ratio of atoms x (if necessary) to get whole numbers of atoms 75 % C 75 g C 6.2 mol C 25 % H 25 g H 24.8 mol H CH 4 1 C 4 H 1) Percent to Mass 2) Mass to Mole 4) Times till 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 12.01 g C = 1 1 C x 2 6.883 mol = 2 C 17.34 g H x 1 mol H = 17.20 mol H 1.008 g H 6.883 mol 3) Divide by Small = 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 #43a, 48 molecular mass 58 C 4 H 29.06 = 2 2 (C 2H 5 ) = empirical mass 10 Calculating Empirical Formulas Percent to Mass Mass to Mole Divide by Small Times till Whole 22

Combustion Analysis Hydrocarbons with C and H are analyzed through combustion with O 2 in a chamber. g C is from the g CO 2 produced g H is from the g H 2 O produced g X is found by subtracting (g C + g H) from g sample 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. Combustion Analysis The acid contains only C, H, and O. What is the empirical formula of the acid? 0.579 g CO 2 x 1 mol CO 2 1 mol C 12.01 g C x x 44.01 g CO 2 1 mol CO 2 1 mol C? g C = 0.158 g C 1 mol H 2 O x 2 mol H 1.008 g H 0.142 g H 2 O x x 18.02 g H 2 O 1 mol 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 23

0.158 g C x 1 mol C = 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 0.126 g O x 1 mol O = 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? 6.274 g CO 2 x 1 mol CO 2 1 mol C 12.01 g C x x 44.01 g CO 2 1 mol CO 2 1 mol C? g C = 1.712 g C 1 mol H 2 O x 2 mol H 1.008 g H 3.212 g H 2 O x x 18.02 g H 2 O 1 mol 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 24

1.712 g C x 1 mol C = 0.1425 mol C = 12.01 g C 0.07131 mol 1 mol H 0.3593 g H x = 0.3564 mol H = 1.008 g H 0.07131 mol 2 C 5 H 1 mol Cl 2.528 g Cl x = 0.07131 mol Cl = 1 Cl 35.45 g Cl 0.07131 mol If the compound has a C 2 H 5 Cl MW of 193 g mol 1, what MW 193 is the molecular formula? EW 64.51 = 3 HW p. 114 #52b 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 O2 After Which is limiting? 2 H 2 + O 2 2 H 2 O Initial: 10? mol? 7 mol? 0 mol Change: 10 5 +10 End: 0 mol 2 mol 10 mol limiting excess 25

Before H 2 O2 After 2 H 2 + O 2 2 H 2 O Initial: 10? mol? 7 mol? 0 mol Change: 10 5 +10 End: 0 mol 2 mol 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 Solid aluminum metal is reacted with aqueous copper(ii) chloride in solution. 2 Al(s) + 3 CuCl 2 (aq) 3 Cu(s) + 2 AlCl 3 (aq) demo If 0.030 g Al are reacted with 0.0060 mol CuCI 2, which is the limiting reactant? How much product will be produced? Limiting Reactant 1 mol Al 3 mol Cu 0.030 g Al x x = 0.0017 mol Cu 26.98 g Al 2 mol Al 3 mol Cu 0.0060 mol CuCl 2 x = 0.0060 mol Cu 3 mol CuCl 2 2 Al(s) + 3 CuCl 2 (aq) 3 Cu(s) + 2 AlCl 3 (aq) Al is limiting HW p. 115 #72 26

Theoretical Yield theoretical yield: the maximum amount of product that can be formed calculated by stoichiometry (using LR only) 1 mol Al 3 mol Cu 0.030 g Al x x = 0.0017 mol Cu 26.98 g Al 2 mol Al This is different from the actual yield, the amount one actually produces and measures (or experimental) Percent Yield A comparison of the amount actually obtained to the amount it was possible to make Actual %Yield = x 100 Theoretical NOT % Error: (calculate using the LR only) % Error = Accepted Experimental x100 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? 27

Percent Yield HW p. 116 #79 1mol Al 23.4 g Alx x 26.98 g Al 4 Al + 3 O 2 2 Al 2 O 3 2 mol Al 2 O 3 4 mol Al 39.3 g of aluminum oxide are formed. What is the percentage yield? 39.3 g %Yield = x 100 44.2 g 101.96 g Al 2 O x 3 1 mol Al 2 O 3 = 44.2 g Al 2 O 3 88.9 % 28