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Chapter 3. Formulas, Equations, and Moles. Balancing Chemical Equations. Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations (chemical reactions) reactants products limestone quicklime + gas Calcium carbonate calcium oxide + carbon dioxide
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Chapter 3 Formulas, Equations, and Moles
Balancing Chemical Equations • Alphabet – elemental symbols • Words – chemical formulas • Sentences – chemical equations (chemical reactions) reactants products limestone quicklime + gas Calcium carbonate calcium oxide + carbon dioxide CaCO3(s) CaO(s) + CO2(g)
Balancing Chemical Equations • Chemical reactions include • Reactants • Products • Balanced – Law of Conservation of Mass • # of atoms of an element on the reactant side must equal the # of atoms of that element on the product side. • Indicate the state of matter of each chemical in the reaction (Chapter 4)
Balancing Chemical Equations • Write the equation without coefficients • List the elements in each equation • Secret: if the same polyatomic ion exists on both sides, keep it together • Determine the # of each kind of atom on both sides • Balance atoms one element at a time by adjusting coefficients • DO NOT ALTER THE FORMULA OF THE COMPOUND!!!!! • Only coefficients can be altered • Secret: • Balance atoms appearing only once on each side first. • Save compounds comprised of only one type of element till last. • Reduce to lowest terms if necessary
Examples • Balance the following equations: • Al(s) + Fe2O3(s) → Al2O3 (s) + Fe (l) • Solid copper reacts with aqueous silver nitrate to form aqueous copper (II) nitrate and silver solid • H3PO4 (l) → H2O (l) + P4O10 (s) • C4H10(g) + O2 (g) → CO2(g) + H2O (g)
Avogadro’s Number and the Mole • Meaning of a chemical reaction 2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O (g) • 2 molecules of C4H10(g) reacts with 13 molecules of O2(g) to form 8 molecule of CO2(g) and 10 molecules of H2O(g)
Avogadro’s Number and the Mole • Molecule’s mass = the sum of the atomic masses of the atoms making up the molecule. • m(C2H4O2) = 2·mC + 4·mH + 2·mO • = 2·(12.01) + 4·(1.01) + 2·(16.00) • = 60.06 amu
Avogadro’s Number of the Mole • One mole (mol) of any substance contains 6.02 x 1023 (Avogadro’s Number) units of that substance. • One mole (mol) of a substance is the gram mass value equal to the amu mass of the substance. • Calculated the same as amu’s for a molecule
Avogadro’s Number and the Mole • Calculate the molar mass of the following: • Fe2O3 (Rust) • C6H8O7 (Citric acid) • C16H18N2O4 (Penicillin G)
Avogadro’s Number and the Mole • Methionine, an amino acid used by organisms to make proteins, is represented below. Write the formula for methionine and calculate its molar mass. (red = O; gray = C; blue = N; yellow = S; ivory = H)
Stoichiometry • 4 Conversion units • Chemical formula • Balanced chemical equation • Coefficients can read as; • # of molecules • # of moles of that molecules • Allows conversion between compounds in an equation • Avogadro’s # - 6.02 x 1023 of X = 1 mole of X • Molar mass – how many grams of a substance = 1 mole of that substance
Avogadro’s Number and the Mole • How many grams of oxygen are present in 5.961 x 1020 molecules of KClO3? How many atoms of oxygen are present?
Avogadro’s Number and the Mole • Calculate the number of oxygen atoms in 29.34 g of sodium sulfate, Na2SO4. • A. 1.244 × 1023 O atoms • B. 4.976 × 1023 O atoms • C. 2.409 × 1024 O atoms • D. 2.915 × 1024 O atoms • E. 1.166 × 1025 O atoms
Problem • Potassium dichromate, K2Cr2O7, is used in tanning leather, decorating porcelain and water proofing fabrics. Calculate the number of chromium atoms in 78.82 g of K2Cr2O7. • A. 9.490 × 1025 Cr atoms • B. 2.248 × 1024 Cr atoms • C. 1.124 × 1024 Cr atoms • D. 3.227 × 1023 Cr atoms • E. 1.613 × 1023 Cr atoms
Stoichiometry: Equation Arithmetic • Balance the following, and determine how many moles of CO will react with 0.500 moles of Fe2O3. Fe2O3(s) + CO(g) → Fe(s) + CO2(g)
Stoichiometry: Chemical Arithmetic • Aqueous sodium hydroxide and chlorine gas are combined to form aqueous sodium hypochlorite (household bleach), aqueous sodium chloride and liquid water. • How many grams of NaOH are needed to react with 25.0 g of Cl2?
Problem • Sulfur dioxide reacts with chlorine to produce thionyl chloride (used as a drying agent for inorganic halides) and dichlorine monoxide (used as a bleach for wood, pulp and textiles).SO2(g) + 2Cl2(g) → SOCl2(g) + Cl2O(g)If 0.400 mol of Cl2 reacts with excess SO2, how many moles of Cl2O are formed? • A. 0.800 mol • B. 0.400 mol • C. 0.200 mol • D. 0.100 mol • E. 0.0500 mol
Problem • Nitrogen gas and hydrogen gas are combined to form ammonia (NH3), an important source of fixed nitrogen that can be metabolized by plants, using the Haber process.How many grams of nitrogen are needed to produce 325 grams of ammonia? • A. 1070 g • B. 535 g • C. 267 g • D. 178 g • E. 108 g
Lab Homework • MISC 486 Problem Set 2 – Due
Yields of Chemical Reactions • Yields of Chemical Reactions:If the actual amount of product formed in a reaction is less than the theoretical amount, we can calculate apercentage yield.
Yield of Chemical Reactions • Dichloromethane (CH2Cl2) is prepared by reaction of methane (CH4) with chlorine (Cl2) giving hydrogen chloride as a by-product. How many grams of dichloromethane result from the reaction of 1.85 kg of methane if the yield is 43.1%?
Problem • What is the percent yield for the reactionPCl3(g) + Cl2(g) → PCl5(g)if 119.3 g of PCl5 ( MM = 208.2 g/mol) are formed when 61.3 g of Cl2 ( MM = 70.91 g/mol) react with excess PCl3? • A. 195% • B. 85.0% • C. 66.3% • D. 51.4% • E. 43.7%
Reactions with Limiting Amounts of Reactants • Limiting Reagents: The extent to which a reaction takes place depends on the reactant that is present in limiting amounts—the limiting reagent. • Process • Convert each reactant into a single product • The one that forms the least is the limiting reactant • Complete all other calculations using the limiting reactant
Reactions with Limiting Amounts of Reactants • Limiting Reagent Calculation: Lithium oxide is a drying agent used on the space shuttle. If 80.0 kg of water is to be removed and 65 kg of lithium oxide is available, which reactant is limiting? Li2O(s) + H2O(l) 2 LiOH(s) • MM(Li2O) = 29.88 g/mol • MM(H2O) = 18.02 g/mol
Reactions with Limiting Amounts of Reactants • Limiting Reagent Calculation: Cisplatin is an anti-cancer agent prepared as follows: K2PtCl4 + 2 NH3 Pt(NH3)2Cl2 + 2 KCl If 10.0 g of K2PtCl4 and 10.0 g of NH3 are allowed to react: (a) which is the limiting reagent? (b) How many grams of the excess reagent are consumed? (c) How many grams of cisplatin are formed? MM(K2PtCl4) = 415.08 g/mol MM(NH3) = 18.04 g/mol MM [Pt(NH3)2Cl2] = 299.9 g/mol
Problem • What is the percent yield of H2O if 97.2 g CH4S reacts with 183 g of O2 to produce 58.5 g H2O according to the following chemical reaction? (Reaction may or may not be balanced.) CH4S + O2 → CO2 + H2O + SO3 • Write the balanced chemical reaction? • Calculate the molar mass of CH4S? • Calculate the molar mass of H2O? • Calculate the molar mass of O2? • Calculate the theoretical yield of H2O? • What is the limiting reagent? • Calculate the percent yield of H2O?
Lab Homework • MISC 486 Problem Set 4 - Due
Moles of solute = Molarity (M) Liters of solution Concentrations of Reactants in Solution: Molarity • Molarity:The most common way of expressing the amount of a substance dissolved in a solution • Conversion factor between moles and volume • It is important to note that the final volume of solution must be used, not volume of solvent.
Solution Stoichiometry • Known: molarity of solution, volume of solution, and balanced chemical equation
Concentrations of Reactants in Solution: Molarity • How many moles of solute are present in 125 mL of 0.20 M NaHCO3? • How many grams of solute would you use to prepare 500.0 mL of 1.25 M NaOH?
Problem • A 0.150 M sodium chloride solution is referred to as a physiological saline solution because it has the same concentration of salts as normal human blood. Calculate the mass of solute needed to prepare 275.0 mL of a physiological saline solution. • A. 41.3 g • B. 31.9 g • C. 16.1 g • D. 8.77 g • E. 2.41 g
Concentrations of Reactants in Solution: Molarity • Solution Preparation • Determine the mass of solid needed to obtain the desired # of moles • Mass the solid • Add it to a volumetric flask • Add water to the volumetric flask until it is about half-full • Cap and shake to dissolve the solid • Add water to the line • Cap and shake
Diluting Concentrated Solutions • Dilution: process of reducing a solution’s concentration by adding more solvent. • Key concept: # of moles remains constant
Diluting Concentrated Solutions Concentrated solution + Solvent Dilute solution Moles of solute (mol) = molarity (M) x volume (V) Mconcentrated x Vconcentrated = Mdilute x Vdilute only use if the initial solution and the final solution are the same substance
Diluting Concentrated Solutions • What volume of 18.0 M H2SO4 is required to prepare 250.0 mL of 0.500 M aqueous H2SO4? • What is the final concentration if 750 mL of 3.50 M glucose is diluted to a volume of 400.0 mL?
Problem • Calcium chloride is used to melt ice and snow on roads and sidewalks and to remove water from organic liquids. Calculate the molarity of a solution prepared by diluting 165 mL of 0.688 M calcium chloride to 925.0 mL. • A. 3.86 M • B. 0.743 M • C. 0.222 M • D. 0.123 M • E. 0.114 M
Titration • Titration: A technique for determining the concentration of a solution. • Process – • a carefully measured volume of an unknown solution is allowed to react with of a standard solution (concentration is known). • The volume of the known is measured. • Stoichiometry calculations are performed
Titration • What is the molarity of a sulfuric acid solution if a 25.0 mL sample is titrated to equivalence with 50.0 mL of 0.150 M potassium hydroxide solution? H2SO4(aq) + KOH(aq) K2SO4(aq) + H2O(l)
Problem • How many milliliters of 1.58 M HCl are needed to react completely with 23.2 g of NaHCO3 ( MM = 84.02 g/mol)? HCl(aq) + NaHCO3(s) → NaCl(s) + H2O(l) + CO2(g) • A. 638 mL • B. 572 mL • C. 536 mL • D. 276 mL • E. 175 mL
Lab Homework • MISC 486 Problem Set 5 - Due
Percent Composition and Empirical Formulas • Percent Composition: Identifies the elements present in a compound as a mass percent of the total compound mass. • The mass percent is obtained by dividing the mass of each element by the total mass of a compound and converting to percentage.
Percent Composition and Empirical Formulas • Sugar is 42.1% C, 6.4% H, and 51.5% O • Meaning – out of 100 g of sugar 42.1 g of it is due to C
Percent Composition and Empirical Formulas • From the percent composition empirical formulas are developed • empirical formula gives the smallest whole number ratio of the atoms of each element in a compound. • Same as ionic formula for ionic compounds
Percent Composition and Empirical Formulas • CompoundFormulaEmpirical Formula • Hydrogen H2O2 OH peroxide • Benzene C6H6 CH • Ethylene C2H4 CH2 • Propane C3H8 C3H8
Percent Composition and Empirical Formulas • A compound’s empirical formula can be determined from its percent composition. • A compound’s molecular formula is determined from the molar mass and empirical formula.
Percent Composition and Empirical Formulas • A compound was analyzed to be 82.67% carbon and 17.33% hydrogen by mass. An osmotic pressure experiment determined that its molar mass is 58.11 g/mol. What is the empirical formula and molecular formula for the compound?
Problem • Gadolinium oxide, a colorless powder which absorbs carbon dioxide from the air, contains 86.76 mass % Gd. Determine its empirical formula. • A. Gd2O3 • B. Gd3O2 • C. Gd3O4 • D. Gd4O3 • E. GdO
Problem • Hydroxylamine nitrate contains 29.17 mass % N, 4.20 mass % H, and 66.63 mass O. If its molar mass is between 94 and 98 g/mol, what is its molecular formula? • A. NH2O5 • B. N2H4O4 • C. N3H3O3 • D. N4H8O2 • E. N2H2O4