Unit 3 (Chp 1,2,3): 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.

Matter H2O CO2 NaCl Atom: Element: same type of atom (1 or more) H2 O2 simplest particle with properties of element Element: same type of atom (1 or more) C C H H O O Na C H2 O2 C C Compound: different atoms bonded H2O CO2 NaCl molecule

Heterogeneous Mixture chromatography distillation Matter (boiling) separate physically cannot separate physically Physical changes Pure Substance Mixture separate chemically cannot separate differences or unevenly mixed uniform or evenly mixed filtering Chemical changes Classification of Matter (flowchart handout) Sugar and Water make Lemonade Sugar and Water and Fat make Milk Sugar and Salt and Water and Fat and Iron and Oxygen make Blood Heterogeneous Mixture Homogeneous Mixture Compounds Elements salt, baking soda, water, sugar oxygen, iron, hydrogen, gold (suspensions/colloids) (solutions) NaCl NaHCO3 H2O C12H22O11 O2 Fe H2 Au

Precision in Measurements Measuring devices have different uses and different degrees of precision. (uncertainty) % Error = |Experimental – Accepted| x100 Accepted

0.0003700400 grams Significant Digits Box & Dot nonzero trailing 0’s if decimal 0.0003700400 grams 500 500. 0.0500 Nonzero digits Captive Zeroes (between two sig digs) Leading Zeroes (at the beginning of a number) are NEVER significant Trailing Zeroes: Significant ONLY if there’s a decimal 0’s Exact numbers are infinitely significant. How many boy students in here? Plus girls? How many sig figs? What if 500 is significant to the 10’s digit? 5.0 x 102

Sigs Digs in Operations + or – round answers to keep the fewest decimal places round answers to keep the fewest significant digits 3.48 + 2.2 = 5.68 5.7 x or ÷ 6.40 x 2.0 = 12.8 13

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

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

Average Atomic Mass Avg. Mass = (Mass1)(%) + (Mass2)(%) … 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%. amu = 1/12 mass of C-12 1amu = 1.66053X10^-24 grams (6.015)(0.0750) + (7.016)(0.925) = 6.94 amu Avg. Mass = (Mass1)(%) + (Mass2)(%) …

7 Diatomic Molecules “H-air-ogens” Seems like all gases, why not noble gases? (have octet already) 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. CO2: carbon dioxide CCl4: carbon tetrachloride N2O5: ________________ Where are the less electro negative elements(direction)? The more? dinitrogen pentoxide CuSO4∙5H2O copper(II) sulfate pentahydrate (ionic & covalent)

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

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. Do Al and SO4 on board Pb4+ O2– Pb2O4 PbO2

Nick and his Polyatomic Ions Nick the Camel ate a Clam for Supper in Phoenix with Mandy. NO31– (nitrate) CO32– (carbonate) ClO31– (chlorate) SO42– (sulfate) PO43– (phosphate) MnO41– (permanganate) #consonants = O’s #vowels = charge

Naming Polyatomic Ions & Acids Ion Name Acid Name 4 – per-___-ate 3 ___-ate 2 ___-ite 1 hypo-___-ite per-___-ic acid ___-ic acid ___-ous acid hypo-___-ous acid perchlorate ClO4– chlorate ClO3– chlorite ClO2– hypochlorite ClO– perchloric acid HClO4 chloric acid HClO3 chlorous acid HClO2 hypochlorous acid HClO sulfate SO42– sulfite SO32– sulfuric acid H2SO4 sulfurous acid H2SO3

Reaction Types Synthesis Decomposition Single Replacement A + B → AB 2 → 1 Decomposition AB → A + B 1 → 2 Single Replacement AB + C → A + CB Double Replacement AB + CD → AD + CB Zn in acid releases H2 gas Combustion CxHy + O2 → CO2 + H2O 16

% by mass of each element in a compound Percent Composition % element = (# of atoms)(AW) (FW) x 100 % by mass of each element in a compound So the percentage of carbon in ethane (C2H6) is… %C = (2)(12.01) (30.07) x 100 24.02 30.07 = x 100 = 79.88% C

(groups of 6.022x1023 particles) Using Moles Moles are the bridge from the particle (micro) scale to the real-world (macro) scale. (from PT) (equation sheet) bridge micro- macro- molar mass Avogadro constant Moles (groups of 6.022x1023 particles) Particles (atoms) (molecules) (units) Mass (grams) 1 mol g 6.022x1023 1 mol g 1 mol 1 mol 6.022x1023

Stoichiometric Calculations Rxn: A(aq) + 2 B(aq)  C(aq) + 2 D(aq) molar mass A g A mol A Coefficients of balanced equation ??? g A 1 mol A mol-to-mol ratio 2 mol B 1 mol A g B 1 mol B molar mass B g B mol B 19

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) 1 mol A . grams A _ mol B mol A grams B 1 mol B grams A x x x = 1) molar mass 2) mole ratio 3) molar mass

Types of Formulas Empirical Formulas: Molecular Formulas: CH3 C2H4O the lowest ratio of atoms of each element in a compound. Molecular Formulas: the total number of atoms of each element in a compound. CH3 C2H4O C2H6 C6H12O3 molecular mass = emp. form. empirical mass multiple

Calculating Empirical Formulas from Mass % Composition Steps (rhyme) Percent to Mass Mass to Mole Divide by Small Times ‘til Whole assume 100 g MM from PT ÷ moles by smallest to get mole ratio of atoms CH4 x (if necessary) to get 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

C2H5 C4H10 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. If molecular mass is 58 g∙mol–1, what is the Molecular Formula? 82.66 g C 17.34 g H 82.66 g C x = 6.883 mol C 17.34 g H x = 17.20 mol H 1 mol C 12.01 g C 1 mol H 1.008 g H = 1  1 C = 2.499  2.5 H x 2 = 2 C 6.883 mol 6.883 mol x 2 = 5 H C2H5 58 29.06 molecular mass empirical mass = 2 2 (C2H5) = C4H10

Step 1 is “combustion to mass” Combustion Analysis Hydrocarbons with C and H are analyzed through combustion with O2 in a chamber. g C is from the g CO2 produced g H is from the g H2O produced g X is found by subtracting (g C + g H) from g sample Step 1 is “combustion to mass” Is this chemical or physical separation?

1 mol CO2 44.01 g CO2 1 mol C 1 mol CO2 12.01 g C 1 mol C 0.579 g CO2 x x x ? g C = 0.158 g C Step 1: “combustion to mass” 1 mol H2O 18.02 g H2O 2 mol H 1 mol H2O 1.008 g H 1 mol H x x 0.142 g H2O x ? g H = 0.0159 g H 0.3000 g sample – (0.158 g C) – (0.0159 g H) = ? g O Then: Steps 2-4 to get empirical formula = 0.126 g O

Before After 2 H2 + O2 2 H2O H2 O2 No! Initial: ? mol ? mol ? mol Change: End: 10 7 –10 –5 +10 0 mol 2 mol In other words, it’s the reactant you’ll run out of first Which is the limiting here (H or O)? How do you know? 10 mol Does limiting mean smallest amount of reactant? No! O2 is in smallest amount, but… H2 is in smallest “stoichiometric” amount

Find the Limiting Reactant Convert reactant A to reactant B Compare B available to B needed If available < needed (limiting) If available > needed (excess) Solid aluminum metal is reacted with aqueous copper(II) chloride in solution 2 Al + 3 CuCl2  2 AlCl3 + 3 Cu 54.0 g Al 4.50 mol CuCl2 (Which is limiting?) DEMO – 25.361 g CuCl2-2H2O in 200mL H2O makes 0.774 M CuCl2, then… Add 0.100g Al to 10mL CuCl2 for Al as limiting, then… Add 0.200g Al more to CuCl2 as limiting and Al as excess 1 mol Al 26.98 g Al 3 mol CuCl2 2 mol Al 54.0 g Al x x = 3.00 mol CuCl2 (4.50 mol CuCl2) available > needed (3.00 mol CuCl2) CuCl2 is excess Al is limiting 27

Theoretical Yield theoretical yield: the maximum possible amount of product that can be formed calculated by stoichiometry limited by LR (use LR only to calculate) different from actual yield (or experimental), amount recovered in the experiment limiting 1 mol Al 26.98 g Al 3 mol Cu 2 mol Al 63.55 g Cu 1 mol Cu 191 g 54.0 g Al x x x = Cu produced

(calculate using the LR only) Percent Yield (actual yield) A comparison of the amount actually obtained to the amount it was possible to make (theoretical yield) %Yield = x 100 Actual Theoretical (calculate using the LR only) NOT % Error: % Error = |Accepted – Experimental| x100 Accepted