ÐÏࡱá>þÿ ikþÿÿÿfgh€€Ñÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿì¥Á7 ø¿îÅbjbjUU "º7|7|¦ÁGÿÿÿÿÿÿl(((((((<´Û´Û´Û8ìÛ|hÝD<B2D¸ã,äë"ìììòòòÁ1Ã1Ã1Ã1Ã1Ã1Ã1$†3 ¦5ˆç1(ò1ñlòòòç1±((ìì+ü1±±±òZ (ì(ìÁ1±òÁ1±–±Gª½|((]ì¬ã ÙF,Y‘Í<xÚ´Û÷üº9]d20B2K.6±.6]±<<((((ÙTopic 3 – Stoichiometry BACKGROUND FOR STOICHIOMETRY A. Definition The study and calculation of quantitative relationships of the reactants and products in chemical reactions B. Word origin Greek Stoicheion (“element”) and Metrikos (“measure) C. Is based on The law of conservation of mass The law of constant composition The law of multiple proportions FORMULA MASS (also called the “Formula Weight”) A. Definition The sum of the atomic masses in the formula for the compound B. Procedure 1. Determine the atomic mass of each element in the formula. 2. Multiply each element’s atomic mass by its subscript. 3. Total your results. C. Examples Calculate the formula mass for C2H6 2 x C = 2 x 12.0107 amu = 24.0214 amu 6 x H = 6 x 1.00794 amu = 6.04764 amu 30.06904 amu = 30.0690 amu Calculate the formula mass for Al2(HPO4)3 2 x Al = 2 x 26.981538 amu = 53.963076 amu 3 x H = 3 x 1.00794 amu = 3.02382 amu 3 x P = 3 x 30.973761 amu = 92.921283 amu 12 x O = 12 x 15.9994 amu = 191.9928 amu 341.900979 amu = 341.9010 amu MOLES A. Terms 1. Mole a. Definition The amount of a substance that contains as many particles as the number of atoms in exactly 12 g of carbon ( 12 b. Symbol mol 2. Avogadro’s number (symbol NA) a. Definition of Avogadro’s number The number of atoms in exactly 12 g of carbon ( 12 b. Numerical value of Avogadro’s number Approximately equal to 6.0221367 x 1023 c. Symbol NA Remember Ava Gadro’s number (602) 214-1023. 3. Molar mass a. Definition The mass of one mole of a substance b. Numerical value of molar mass It is equal to the formula mass expressed in grams. c. Symbol MM B. Mole calculations 1. Calculating molar mass a. Procedure Do the calculations as you would for formula mass but substitute the unit of “g” for the unit of “amu”. b. Example Calculate the molar mass of Na2CO3. 2 x Na = 2 x 22.989770 g = 45.979540 g 1 x C = 1 x 12.0107 g = 12.0107 g 3 x O = 3 x 15.9994 g = 47.9982 g 105.988440 g = 105.9884 g 2. Converting moles to mass a. Procedure (1) Determine the molar mass of the substance. (2) Use the conversion factor: molar mass1 mol b. Example What is the mass of 2.35 moles of Na2CO3? 2.35 mol Na2CO3105.9884 g Na2CO31 mol Na2CO3 = 249 g Na2CO3 3. Converting mass to moles a. Procedure (1) Determine the molar mass (2) Use the conversion factor: 1 mol molar mass b. Example 122.56 g of Na2CO3 is equal to how many moles? 122.56 g Na2CO31 mol Na2CO3105.9884 g Na2CO3 = 1.1562 mol Na2CO3 4. Converting moles to number of particles a. Procedure Use the conversion factor: 6.02214 x 1023 particles1 mol b. Example How many molecules are in 3.013 moles of O2 molecules? 3.013 mol O26.02214 x 1023 O2 molecules1 mol O2 = 1.814 x 1024 molecules 5. Converting number of particles to moles a. Procedure Use the conversion factor: 1 mol 6.02214 x 1023 particles b. Example 4.391 x 1025 formula units of NaCl is equal to how many moles? EMBED Equation.2  4.391 x 1025 f.u. NaCl1 mol NaCl6.02214 x 1023 f.u. NaCl = 7.291 x 101 mol PERCENT COMPOSITION FROM ELEMENTAL MASSES A. Definition of percent composition The percent by mass of each element in a sample of a compound B. Procedure to calculate percent composition from elemental masses Worked as a standard percentage problem C. Example 65.000 g of a compound of Na and O was determined to contain 48.221 g of Na and 16.779 g of O. What is the percent composition of each element in this compound? GivenFindmass of sample = 65.000 g mass of Na = 48.221 g mass of O = 16.779 g % Na = ? % O = ? 1. Na % Na =  EMBED Equation.2  x 100% = 74.186% 2. O % O =  EMBED Equation.2  x 100 % = 25.814%  EMBED Equation.2  PERCENT COMPOSITION FROM A FORMULA A. Description The percent composition of an element in the formula of a compound is the parts per hundred of that element in that compound assuming that you have one molar mass of that compound. B. Procedure 1. Assume that you have exactly one mole of that compound. 2. Calculate the mass contribution of each element by multiplying its molar mass by its subscript. 3. Calculate the molar mass of the compound by adding together the mass contributions of each element.  EMBED Equation.2  4. Calculate the percent composition for each element in that compound. C. Examples Calculate the percent composition to two decimal places for each element in NaOH. 1. Mass contributions for each element Na 1 x Na = 1 x 22.989770 g = 22.989770 g  EMBED Equation.2  O 1 x O = 1 x 15.9994 g = 15.9994 g H 1 x H = 1 x 1.00794 g = 1.00794 g = 39.99711 g 2. Molar mass of NaOH = 39.9971 g 3. Percent composition for each element a. Na % Na =  EMBED Equation.2  x 100% = 57.48% b. O % O =  EMBED Equation.2  x 100% = 40.00% c. H % H =  EMBED Equation.2  x 100% = 2.52% d. Double checking total = 100.00% Calculate the percent composition to two decimal places for each element in CoCl2 ( 6 H2O. 1. Mass contributions for each element Co 1 x Co = 1 x 58.9332 g = 58.9332 g Cl 2 x Cl = 2 x 35.453 g = 70.906 g  EMBED Equation.2  O 6 x O = 6 x 15.9994 g = 95.9964 g H 12 x H = 12 x 1.00794 g = 12.09528 g = 237.93088 g 2. Molar mass of CoCl2 ( 6 H2O = 237.931 3. Percent composition for each element a. Co % Co =  EMBED Equation.2  x 100% = 24.77% b. Cl % Cl =  EMBED Equation.2  x 100% = 29.80% c. O % O =  EMBED Equation.2  x 100% = 40.35% d. H % H =  EMBED Equation.2  x 100% = 5.08% e. Double checking total = 100.00% PERCENT COMPOSITION BY ELEMENTAL ANALYSIS A. The process involves decomposition reactions yielding products that can be collected, identified, and quantitatively analyzed. B. Examples 1. At very high temperatures 0.8000 g of an oxide of tin are allowed to react with pure hydrogen gas. The oxygen in the tin oxide is converted quantitatively to water vapor which gets flushed out with the excess hydrogen. The solid residue that remains is pure tin. The mass of the pure tin is 0.6301 g. What is the percent composition for each element? GivenFindmass of Sn and O = 0.8000 g mass of Sn = 0.6301 g mass of O = ? % comp of Sn = ? a. Finding the mass of O Since the sample is made up only of tin and oxygen then the difference between the mass of tin remaining and the mass of the original sample must equal the mass of oxygen. mass of (Sn + O) ( mass of Sn = mass of O 0.8000 g ( 0.6301 g = 0.1699 g b. Finding the % comp for Sn % Sn =  EMBED Equation  x 100% = 78.76% c. Finding the % comp for O % O =  EMBED Equation  x 100% = 21.24% DETERMINING FORMULAS A. Definition The formula with the lowest whole number ratio of elements in a compound and is written with the smallest whole number subscripts 1. Determining the formula of a hydrated salt by dehydration and mass difference a. Procedure (1) Determine the mass of the waters of hydration. (2) Convert the mass of the water and the mass of the anhydrous salt to moles. (3) Determine the ratio of the moles of water to the moles of anhydrous salt. (4) Write the formula. b. Example 4.132 g of the hydrated salt of CaSO4 were heated in a crucible until all the water of hydration was driven off. The mass of the anhydrous salt was 3.267 g. What is the formula of the hydrate? GivenFindmass of hydrate = 4.132 g mass of anhydrous = 3.267g mass of H2O = ? mol H2O = ? mol CaSO4 = ? Determine the mass of the waters of hydration: mass of water = mass of hydrated salt – mass of anhydrous salt = 4.132 g – 3.267g = 0.865 g Convert the mass of the water and the mass of the anhydrous salt to moles: H2O 0.865 g H2O1 mol H2O18.0153 g H2O = 0.0480 mol H2O CaSO4 3.267g CaSO41 mol CaSO4136.141 g CaSO4 = 0.0240 mol CaSO4 Determine the ratio of the water to the anhydrous salt:  EMBED Equation  =  EMBED Equation  Write the formula: CaSO4• 2 H2O 2. Determining an empirical formula from elemental analysis a. Procedure (1) Determine the mass of each element in a given mass of a sample. (2) Convert the mass of each element to the number of moles of that element. (3) Determine the ratios of the elements by dividing each of the number of moles by the smallest number of moles. (4) If all the ratios are within 5 % of being integers, then round to the nearest integer. Examples:  EMBED Equation.2  =  EMBED Equation.2   EMBED Equation.2  =  EMBED Equation.2  (5) If the ratios vary from being integers by more than 5%, then consider ratios of integers where the denominator is a value other than one. Examples:  EMBED Equation.2  =  EMBED Equation.2   EMBED Equation.2  =  EMBED Equation.2  (6) Write the empirical formula using the smallest whole number ratios. b. Examples (1) Determine the empirical formula of a compound if a 42.44 g sample contains 8.59 g of aluminum and 33.85 g of chlorine GivenFindmass of sample = 42.44 g mass of Al = 8.59 g mass of Cl = 33.85 g mol Al = ? mol Cl = ?  EMBED Equation  or  EMBED Equation  = ? formula is? (a) Convert the mass of each element to the number of moles of that element, and carry over an unwarranted significant digit. Al 8.59 g Al 1 mol Al26.981538 g Al = 0.3184 mol Al Cl 33.85 g Cl1 mol Cl35.453 g Cl = 0.95479 mol Cl (b) Determine the ratio.  EMBED Equation  =  EMBED Equation  =  EMBED Equation  (c) Write the empirical formula. AlCl3 (2) Determine the empirical formula of a compound if a 26.29 g sample contains 11.47 g of phosphorus and 14.81 g of oxygen. GivenFindmass of sample = 26.29 g mass of P = 11.47 g mass of O = 14.81 g mol P = ? mol O = ?  EMBED Equation  or  EMBED Equation  = ? formula is? (a) Convert the mass of each element to moles. P 11.47 g P1 mol P30.973762 g P = 0.37031 mol P O 14.81 g O1 mol O15.9994 g O = 0.92566 mol O (b) Determine the ratio.  EMBED Equation  =  EMBED Equation  =  EMBED Equation  =  EMBED Equation  (c) Write the empirical formula. P2O5 3. Determining an empirical formula from percent composition a. Procedure (1) Assume that you have a 100.00 g sample of the compound. (2) Convert the percent of each element to the mass of that element in a 100.00 g sample of that compound. (3) Convert the mass of each element to the number of moles of that element. (4) Determine the ratios of the elements by dividing each of the number of moles by the smallest number of moles. (5) Write the empirical formula. b. Examples (1) Determine the empirical formula of potassium chromate which is 43.88% potassium, 29.18% chromium, and 26.94% oxygen. GivenFindmass of sample = 100.00 g % K = 43.88% % Cr = 29.18% % O = 26.94% mass K = ? mass Cr = ? mass O = ? mol K = ? mol Cr = ? mol O = ? ratios = ? formula is? (a) Convert the percent of each element to its mass in a 100.00 g sample. 43.88% K x 100.00 g = 43.88 g K 29.18% Cr x 100.00 g = 29.18 g Cr 26.94% O x 100.00 g = 26.94 g O (b) Convert the mass of each element to moles. K 43.88 g K1 mol K39.0983 g K = 1.1223 mol K Cr 29.18 g Cr1 mol Cr51.9961 g Cr = 0.56120 mol Cr O 26.94 g O1 mol O15.9994 g O = 1.6838 mol O (c) Determine the ratios.  EMBED Equation  =  EMBED Equation  =  EMBED Equation   EMBED Equation  =  EMBED Equation  =  EMBED Equation  (d) Write the empirical formula. K2CrO3 (TAKE NOTE !) (2) Determine the empirical formula of vitamin C which is 40.92% carbon, 4.5785% hydrogen, and 54.50% oxygen. GivenFindmass of sample = 100.00 g % C = 40.92% % H = 4.5785% % O = 54.50% mass C = ? mass H = ? mass O = ? mol C = ? mol H = ? mol O = ? ratios = ? formula is? (a) Convert the percent of each element to its mass in a 100.00 g sample. 40.92% C x 100.00 g = 40.92 g C 4.578% H x 100.00 g = 4.578 g H 54.50% O x 100.00 g = 54.50 g O (b) Convert the mass of each element to moles. C 40.92 g C1 mol C12.0107 g C = 3.4068 mol C H 4.578 g H1 mol H1.00794 g H = 4.5418 mol H O 54.50 g O1 mol O15.9994 g O = 3.4063 mol O (c) Determine the ratios.  EMBED Equation  =  EMBED Equation  =  EMBED Equation   EMBED Equation  =  EMBED Equation  =  EMBED Equation  (d) Write the empirical formula. C3H4O3 4. Determining an empirical formula of a organic compound from combustion analysis a. Procedure (1) Determine the mass of the sample. (2) Assume that this combustion will be in pure oxygen present in large excess. (3) Assume that all of the carbon present in the sample winds up as CO2, and all of the hydrogen present winds up as H2O. (4) Convert mass of CO2 to mol CO2 and then to mol C. (5) Convert mass of H2O to mol H2O and then to mol H. Don’t forget that there are 2 mol H atoms to 1 mol H2O. (6) Convert mol C to mass C and mol H to mass H, then compare the total of the mass of C and the mass of H to the mass of the sample. Any difference, unless otherwise specified, is oxygen. If it is present, convert the mass O to mol O. (7) Determine the ratios of the elements by dividing each of the number of moles by the smallest number of moles. (8) Write the empirical formula. b. Example containing only C and H A 11.50 mg sample of cyclopropane undergoes complete combustion to produce 36.12 mg of CO2 and 14.70 mg of H2O. What is the empirical formula of this compound? (1) Convert mass of CO2 to mol CO2 and then to mol C. 36.12 mg CO21 g CO21 mol CO21 mol C1000 mg CO244.0095 g CO2 1 mol CO2 = 8.2073 x 10(4 mol C (2) Convert mass of H2O to mol H2O and then to mol H. 14.70 mg H2O1 g H2O1 mol H2O2 mol H1000 mg H2O18.0153 g H2O 1 mol H2O = 1.6319 x 10(3 mol H (3) Convert mol C to mass C and mol H to mass H, then compare the total of the mass of C and the mass of H to the mass of the sample. Mass C 8.2073 x 10(4 mol C12.0107 g C1000 mg C1 mol C1 g C = 9.8575 mg C Mass H 1.6319 x 10(3 mol H1.00794 g H1000 mg H1 mol H1 g H = 1.6448 mg H Mass C + Mass H = Mass sample ? 9.8575 mg C + 1.6448 mg H = 11.5023 mg C+H = 11.50 mg ( (4) Determine the ratios  EMBED Equation  =  EMBED Equation  =  EMBED Equation  (5) Write the empirical formula. CH2 c. Example containing C, H, and O A 25.50 mg sample of 2-propanol undergoes complete combustion to produce 56.11 mg of CO2 and 30.58 mg of H2O. What is the empirical formula of this compound? 1) Convert mass of CO2 to mol CO2 and then to mol C. 56.11 mg CO21 g CO21 mol CO21 mol C1000 mg CO244.0095 g CO2 1 mol CO2 = 1.2750 x 10(3 mol C (2) Convert mass of H2O to mol H2O and then to mol H. 30.58 mg H2O1 g H2O1 mol H2O2 mol H1000 mg H2O18.0153 g H2O 1 mol H2O = 3.3949 x 10(3 mol H (3) Convert mol C to mass C and mol H to mass H, then compare the total of the mass of C and the mass of H to the mass of the sample. If present, convert the mass O to mol O. Mass C 1.2750 x 10(3 mol C12.0107 g C1000 mg C1 mol C1 g C = 15.314 mg C Mass H 3.3949 x 10(3 mol H1.00794 g H1000 mg H1 mol H1 g H = 3.4218 mg H mass C + mass H = mass sample ? 15.314 mg C + 3.4218 mg H = 18.736 mg C+H = 25.50 mg NO!! Mass of O !! 25.50 mg ( 18.736 mg C+H = 6.764 mg O Mass O to mol O 6.764 mg O1 g O1 mol O1000 mg O15.9994 g O = 4.228 x 10(4 mol O (4) Determine the ratios  EMBED Equation  =  EMBED Equation  =  EMBED Equation   EMBED Equation  =  EMBED Equation  =  EMBED Equation  (5) Write the empirical formula. C3H8O B. Molecular formulas ( the formula that shows the actual number of atoms of each element present in a compound 1. The molar mass will be some whole number multiple “n” of the empirical formula mass for that compound. Molar mass = n x empirical formula mass 2. The molecular formula will be some whole number multiple “n” of the empirical formula for that compound. Molecular formula = n x empirical formula 3. In both cases “n” will be the same. 4. Determining a molecular formula from an empirical formula. a. Procedure for determining a molecular formula from an empirical formula (1) Determine the empirical formula. (2) Determine the molar mass by experiment. (It will be provided for these problems) (3) Calculate the empirical formula mass the same way as a molar mass. (4) Divide the molar mass by the empirical formula mass to determine n. (5) Multiply the empirical formula by the factor n. (6) Write the molecular formula. b. Examples (1) When vitamin C was analyzed, its empirical formula was found to be C3H4O3. In another experiment its molar mass was determined to be about 180 g/mol. Determine its molecular formula. (a) Calculate the empirical formula mass. 3 x C = 3 x 12.0107 g = 36.0321 g 4 x H = 4 x 1.00794g = 4.03176 g 3 x O = 3 x 15.9994 g = 47.9982 g 88.06206 g /mol = 88.0621g/mol (b) Divide the molar mass by the empirical formula mass to get n. n =  EMBED Equation  n = 2.04401 n = 2 (c) Multiply the empirical formula by the factor n. 2(C3H4O3) (d) Write the molecular formula. C6H8O6 (2) When glucose was analyzed its empirical formula was found to be CH2O. Its molar mass was found to be about 180 g/mol. Determine its molecular formula. (a) Calculate its empirical formula mass. 1 x C = 1 x 12.0107 g =12.0107 g 2 x H = 2 x 1.00794 g = 2.01598 g 1 x O = 1 x 15.9994 g = 15.9994 g 30.02608 g/mol = 30.0261 g/mol (b) Divide the molar mass by the empirical formula mass to get n. n =  EMBED Equation  n = 5.99478 n = 6 (c) Multiply the empirical formula by the factor n. 6(CH2O) (d) Write the molecular formula. C6H12O6 STOICHIOMETRY A. Definition and description of stoichiometry 1. Definition of stoichiometry The calculation of the quantities of reactants and products involved in a chemical reaction 2. Description of stoichiometry a. Deals with numerical relationships in chemical reactions b. Involves the calculation of the quantities of substances involved in chemical reactions c. Uses the coefficients of a balanced molecular equation. B. Relationships that can be determined from a balanced molecular equation such as: N2 (g) + 3 H2 (g) ( 2 NH3 (g) 1. Particles ( atoms, molecules, and formula units 1 molecule of N2 reacts with 3 molecules of H2 to produce 2 molecules of NH3. This ratio 1 N2 : 3 H2 : 2 NH3 will always hold true for this reaction. Likewise any multiple of this ratio will react: 10 molecules of N2 will react with 30 molecules of H2 to form 20 molecules of NH3. 2. Moles 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Likewise any multiple of this ratio willreact: 3 moles of N2 will react with 9 moles of H2 to form 6 moles of NH3. 3. Mass 1 molar mass of N2 reacts with 3 molar masses of H2 to produce 2 molar masses of NH3. 1 x (28.01 g/mol) of N2 reacts with 3 x (2.016 g/mol) of H2 to produce 2 x (17.03 g/mol) of NH3. Likewise any multiple of this ratio will react: 0.25 x (28.01 g/mol) of N2 will react with 0.75 x (2.016 g/mol) of H2 to produce 0.50 x (17.03 g/mol) of NH3. 4. Volume 1 molar volume of N2 reacts with 3 molar volumes of H2 to produce 2 molar volumes of NH3 At a temperature of 0(C and a pressure of 1 atmosphere 1 mole of a gas has a volume of 22.4 L. 1 x (22.4 L) of N2 reacts with 3 x (22.4 L) of H2 to produce 2 x (22.4 L) of NH3. Likewise any multiple of the ratio will react: 0.2 x (22.4 L) of N2 will react with 0.6 x (22.4 L) of H2 to produce 0.4 x (22.4 L) of NH3. MOLE-MOLE CALCULATIONS A. There are four possible mole-mole conversions for the general equation: aA + bB ( cC + dD 1. Moles of reactant ( moles reactant 2. Moles of reactant ( moles product 3. Moles of product ( moles reactant 4. Moles of product ( moles product B. All mole-mole conversions are based on mole ratios determined from the coefficients of the balanced molecular equation. 1. These conversion factors will take the form of a ratio of the moles of the two substances, called a “mole ratio.”  EMBED Equation  2. Example ( from the equation above  EMBED Equation  C. Mole-mole conversions 1. Procedure a. Set up the given and the find. b. Draw a map. c. Determine the mole ratios needed for conversion factor/s. d. Use the “big, long line” method. 2. Examples a. For the reaction N2 (g) + 3 H2 (g) ( 2 NH3 (g) how many moles of NH3 are formed when 0.45 moles of N2 react with excess H2? GivenFindbalanced equation mol N2 = 0.45 mol excess H2 mol NH3 = ? Map: mol N2 ( mol NH3 Mole ratio:  EMBED Equation  Big, long line: 0.45 mol N22 mol NH31 mol N2 = 0.90 mol NH3 b. For the reaction N2 (g) + 3 H2 (g) ( 2 NH3 (g) how many moles of H2 are needed to completely react with 1.25 moles of N2? GivenFindbalanced equation mol N2 = 1.25 mol mol H2 = ? Map: mol N2 ( mol H2 Mole ratio:  EMBED Equation  Big, long line: 1.25 mol N23 mol H21 mol N2 = 3.75 mol H2 MASS-MASS CALCULATIONS A. There are four possible mass-mass conversions for the general equation aA + bB ( cC + dD 1. Mass of reactant ( mass reactant 2. Mass of reactant ( mass product 3. Mass of product ( mass reactant 4. Mass of product ( mass product B. All mass-mass conversions 1. Are based on mole ratios determined from the coefficients of the balanced molecular equation 2. Use the molar mass for both substances C. Mass-mass conversions 1. Procedure a. Set up the given and the find b. Draw a map Mass of A Mass of B  Molar Molar Mass A Mass B  Moles of A Moles of B Mole Ratio c. Determine the necessary conversion factors. (1) Molar masses to convert (a) From mass ( moles (b) From moles ( mass (2) Mole ratios d. Use the “big, long line” method 2. Examples a. For the reaction N2 (g) + 3 H2 (g) ( 2 NH3 (g) how many grams of NH3 will be produced when 5.40 g of H2 react with excess N2? GivenFindbalanced equation mass H2 = 5.40 g MM H2 = 2.01588 g/mol MM NH3 = 17.0305 g/mol mole ratio =  EMBED Equation  mass of NH3 = ? Mass of H2 Mass of NH3  molar molar mass mass H2 NH3  Moles of H2 Moles of NH3 mole ratio  EMBED Equation  5.40 g H21 mol H22 mol NH317.0305 g NH32.01588 g H23 mol H21 mol NH3 = 30.4134 g NH3 = 30.4 g NH3 b. For the reaction N2 (g) + 3 H2 (g) ( 2 NH3 (g) how many grams of N2 are needed to produce 30.4 g of NH3? GivenFindbalanced equation mass NH3 = 30.4 g MM NH3 = 17.0305g/mol MM N2 = 28.0134 g/mol mole ratio =  EMBED Equation  mass of N2 = ? Mass of NH3 Mass of N2 molar molar mass mass NH3 N2  Moles of NH3 Moles of N2 mole ratio  EMBED Equation  30.4 g NH31 mol NH31 mol N228.0134 g N217.0305g NH32 mol NH31 mol N2 = 25.0024 g N2 = 25.0 g N2 LIMITING REACTANT A. Definitions 1. Limiting reactant also called “limiting reagent” 2. Limiting reactant The reactant that is entirely used up in a reaction and that determines the amount of product formed. 3. Excess reactant A reactant present in quantity that is more than sufficient to react with the limiting reactant, in other words, it is any reactant that remains after the limiting reactant has been used up. B. Analogy Making a Double-cheese Cheeseburger Recipe: one bun one beef patty two cheese slices 1. How many double-cheese cheeseburgers can be made from 2 buns, 2 patties, and 2 slices of cheese? 1 bun + 1 patty + 2 cheese slices = 1 double-cheese cheeseburger 1 bun is left over. 1 beef patty is left over. All of the cheese has been used up. Only 1 double-cheese cheeseburger can be made from that amount of ingredients. In this case: Cheese slices is the limiting reactant. Buns and patties are the excess reactants. 2. How many double-cheese cheeseburgers can be made from 21 buns, 21 beef patties, and 40 cheese slices? 21 buns x  EMBED Equation  = 21 burgers 21 patties x  EMBED Equation  = 21 burgers 40 cheese slices x  EMBED Equation = 20 burgers A little thought will show you that the greatest number of complete double-cheese cheeseburgers is only 20. There will be buns and patties left over. In this case: Cheese slices is the limiting reactant. Patties and buns are excess reactants. C. Limiting reactant problems using moles 1. Procedure a. Convert the moles of each reactant into moles of product. b. The reactant that produces the least product is the limiting reactant. c. If requested, determine the moles of excess reactants used up and the moles remaining. 2. Example Sodium metal reacts with chlorine gas according to the equation: 2 Na (s) + Cl2 (g) ( 2 NaCl (s) 6.70 moles of sodium are mixed with 3.20 moles of chlorine and are allowed to react. a. What is the limiting reactant? b. How many moles of NaCl are produced? c. How many moles of the excess reactant will be used up? d. How many moles of the excess reactant will be left over? GivenFindmol Na = 6.70 mol Na mol Cl2 = 3.20 mol Cl2mol NaCl from Na = ? mol NaCl from Cl2 = ? mol excess used = ? mol excess left = ? 6.70 mol Na2 mol NaCl= 6.70 mol NaCl2 mol Na 3.20 mol Cl22 mol NaCl= 6.40 mol NaCl1 mol Cl2 answers a. Cl2 is the limiting reactant because it produces the least product. b. 6.40 mol NaCl will be produced. c. 6.40 mol NaCl2 mol Na= 6.40 mol Na2 mol NaCl 6.40 mol Na will be used up. d. 6.70 mol Na ( 6.40 mol Na = 0.30 mol Na 0.30 mol Na will be left over. D. Limiting reactant problems using mass 1. Procedure a. Convert the mass of each reactant into mass of product. b. The reactant that produces the least product is the limiting reactant. c. If requested, determine the mass of excess reactants used up and the mass remaining. 2. Example When heated, copper metal reacts with powdered sulfur to form copper (I) sulfide according to the equation: 2 Cu (s) + S (s) ( Cu2S (s) 80.0 g of copper are heated with 25.0 g of sulfur. a. What is the limiting reactant? b. How many grams of Cu2S are produced? c. How many grams of the excess reactant will be used up? d. How many grams of the excess reactant will be left over? GivenFindmass Cu = 80.0 g Cu mass S = 25.0 g S  mass Cu2S from Cu = ? mass Cu2S from S = ? mass excess used = ? mass excess left = ? 80.0 g Cu1 mol Cu1 mol Cu2S159.158 g Cu2S63.546 g Cu2 mol Cu1 mol Cu2S = 100.2 g Cu2S 25.0 g S1 mol S1 mol Cu2S159.158 g Cu2S32.066 g S1 mol S 1 mol Cu2S = 124.1 g Cu2S answers a. Cu is the limiting reactant because it produces the least product. b. 1.00 x 102 g of Cu2S will be produced. c. 20.2 g of S will be used up. 1.00 x 102 g Cu2S1 mol Cu2S 1 mol S 32.066 g S159.158 g Cu2S1 mol Cu2S1 mol S = 20.15 g S d. 25.0 g S ( 20.15 g S = 4.85 g S = 4.8 g S 4.8 g of S will be left over. THEORETICAL YIELD AND PERCENT YIELD A. Definitions 1. Theoretical yield The quantity of product that is calculated to form when all of the limiting reactant reacts 2. Actual yield The quantity of product that is actually produced in a given experiment 3. Percent yield The ratio of the actual (experimental) yield of a product to its theoretical (calculated) yield, multiplied by 100%. B. Reasons why the theoretical yield and the actual yield may differ 1. Reasons why the actual may be larger. a. Contaminants in product b. Product is still wet 2. Reasons why the actual may be smaller. a. Impure reactants b. Not all of the reactant actually reacted c. Competing side reactions d. Product lost during purification C. Procedure 1. Obtain the actual yield by experiment. 2. Calculate the theoretical yield using stoichiometry. 3. Calculate the percent yield using the equation: % yield =  EMBED Equation  x 100% D. Example Calcium carbonate decomposed when heated to form calcium oxide and carbon dioxide according to the equation: CaCO3 (s) ( CaO (s) + CO2 (g) What is the percent yield if 24.8 g of CaCO3 are heated and 13.1 g of CaO are produced? GivenFindmass CaCO3 = 24.8 g CaCO3 actual yield = 13.1 g CaO  MM CaCO3 = ? MM of CaO = ? theor. yield = ? g CaO % yield = ? MM CaCO3 = 100.087 g/mol MM CaO = 56.077 g/mol 24.8 g CaCO31 mol CaCO31 mol CaO 56.077 g CaO100.087 g CaCO31 mol CaCO31 mol CaO theoretical yield = 13.90 g CaO percent yield =  EMBED Equation x 100% = 94.2446 % = 94.2 % WORKING WITH SOLUTIONS A. Definitions 1. Solution A homogeneous mixture with uniform composition of solvent and solute 2. Solvent The medium that does the dissolving, it is normally present in the greater amount 3. Solute The substance that is dissolved in a solvent to form a solution, normally present in the smaller amount 4. Concentration The quantity of solute present in a given quantity of solvent or solution 5. Concentrated solution A solution containing a large amount of solute per given quantity of solvent or solution 6. Dilute solution A solution containing a small amount of solute per given quantity of solvent or solution 7. Strong and weak Refer to the degree of ionization of the solute not to the amount of solute present 8. Dilution The process of adding more solvent to a solution to reduce its concentration B. Molarity 1. Definition The concentration of a solution expressed as moles of solute per liter of solution not per liter of solvent 2. Symbol ( M 3. Equation M =  EMBED Equation  where V is in liters 4. Determining the molarity of a solution a. Procedure (1) Determine the identity and mass of the solute (2) Determine the final volume of the resulting solution in liters. (3) Calculate the molar mass of the solute. (4) Since molarity is the ratio of solute to solution begin with  EMBED Equation  and use conversion factors to reach the desired units of molarity. b. Example What is the molarity of a solution made by dissolving 23.4 g of Na2SO4 in enough water to reach a final volume of 125 mL? GivenFindm = 23.4 g Na2SO4 V = 125 mL MM Na2SO4 = ? V (in L) = ? M = ? molar mass = 142.0421 g/mol Treat this like a “ratio of units” conversion: Convert (23.4 g/125 mL) to (mol/L) map:  EMBED Equation  ((  EMBED Equation  ((  EMBED Equation  solution: 23.4 g Na2SO41 mol Na2SO41000 mL125 mL142.0421 g Na2SO41 L = 1.3179 mol/L = 1.32 M 5. Making a solution a. Procedure (1) Determine the identity of the solute. (2) Determine the desired volume and molarity of the final solution. (3) Convert the desired volume to liters, if necessary. (4) Calculate the molar mass of the solute. (5) Calculate the mass of solute. (6) Describe how to make the solution. b. Example How would you make 500.0 mL of a 0.250 M Na2SO4 solution? GivenFindV = 500.0 mL M = 0.250 M mass Na2SO4 = ? V(M ( mol ( mass 500.0 mL0.250 mol1 L142.043 g Na2SO4L1000 mL1 mol Na2SO4 = 17.8 g Na2SO4 actually making the solution Weigh 17.8 g of Na2SO4 and put it into a 500.0 ml volumetric flask. Fill the flask about half-full of distilled water and dissolve the solute. Add enough distilled water to bring the final volume up to 500.0 mL Mix thoroughly. 6. Determining the needed volume of solution a. Procedure (1) Determine the identity and molarity of the solution. (2) Determine the amount of solute desired. (3) Calculate the molar mass of the solute. (4) Calculate the volume that contains the desired amount of solute. b. Example What volume of a 0.250 M Na2SO4 solution would be needed to provide 33.6 g of Na2SO4? GivenFindM = 0.250 M mass = 33.6 g Na2SO4 V = ?  mass ( mol ( volume 33.6 g Na2SO41 mol Na2SO41 L142.043 g Na2SO40.250 mol Na2SO4 = 0.946 L or 946 mL C. Normality 1. Definitions a. Normality The concentration of a solution expressed as equivalents of solute per liter of solution b. Equivalent (1) For acid-base reactions One equivalent is the amount of acid that supplies 1 mole of H+ or the amount of base that reacts with 1 mole of H+. (2) For redox reactions One equivalent is the amount of substance that will gain or lose one mole of electrons. 2. Usefulness of normality One equivalent of a reactant will exactly react with one equivalent of another reactant, but this is not true for moles. 3. Symbol ( N  EMBED Equation.2  4. Equations a. For all solutions N =  EMBED Equation.2   EMBED Equation.2 where V is in liters b. For an acid HaA (1) eq = a(mol) the number of equivalents is equal to a EMBED Equation  times the number of moles EMBED Equation.2  Examples: H1Cl a = 1  EMBED Equation  H2SO4 a = 2  EMBED Equation  H3PO4 a = 3  EMBED Equation  (2) N = a(M) the normality is equal to a EMBED Equation  times the molarity Examples: An HCl solution with a molarity of 1 M would have a normality of 1 N An H2SO4 solution with a molarity of 1 M would have a normality of 2 N c. For a base M(OH)a (1) eq = a(mol) the number of equivalents is equal to a EMBED Equation  times the number of moles Examples: NaOH a = 1  EMBED Equation  Ba(OH)2 a = 2  EMBED Equation  (2) N = a(M) the normality is equal to a EMBED Equation  times the molarity Examples: A NaOH solution with a molarity of 1 M would have a normality of 1 N A Ba(OH)2 solution with a molarity of 1 M would have a normality of 2 N d. For a oxidizing agent or reducing agent M + ae( ( M(a or M ( M+a + ae( eq = a(mol) N = a(M) The stoichiometry of redox reactions will be covered later 5. Determining the normality from the molarity a. Procedure (1) Determine the value of the subscript “a”. (2) Multiply the molarity by the value of a. b. Examples for acids (1) What is the normality of a 1.50 M HCl solution? For HCl (H1Cl) a = 1 EMBED Equation  N = 1 EMBED Equation  (1.50 M) = 1.50 N (2) What is the normality of a 1.50 M H2SO4 solution? For H2SO4 a = 2 EMBED Equation  N = 2 EMBED Equation  (1.50 M) = 3.00 N c. Examples for bases (1) What is the normality of a 0.0200 M NaOH solution? For NaOH Na(OH)1 a = 1 EMBED Equation  N = 1 EMBED Equation  (0.0200 M) = 0.0200 N (2) What is the normality of a 0.0200 M Ba(OH)2 solution? For Ba(OH)2 a = 2 EMBED Equation  N = 2 EMBED Equation  (0.0200M) = 0.0400 N 6. Determining the normality of a solution a. Procedure (1) Determine the identity and mass of the solute. (2) Determine the final volume of the resulting solution in liters. (3) Calculate the molar mass of the solute. (4) Determine the value of “a” from the molecular formula (the number of equivalents per mole) (5) Calculate the normality. b. Example What is the normality of 0.987 g of Ba(OH)2 dissolved in 345 mL of water? GivenFindm = 0.987 g Ba(OH)2 V = 345 mL MM Ba(OH)2 = ? a = ? N = ? molar mass = 171.342 g/mol a = 2 Treat this like a “ratio of units” conversion: Convert (0.987 g/345 mL) to (eq/L) map:  EMBED Equation  ((  EMBED Equation  ((  EMBED Equation  ((  EMBED Equation  solution: 0.987 g Ba(OH)21 mol Ba(OH)2 1000 mL2 eq Ba(OH)2 345 mL171.342 g Ba(OH)21 L1 mol Ba(OH)2 = 0.0334 N Ba(OH)2 7. Making a solution of a given normality a. Procedure (1) Determine the identity of the solute. (2) Determine the desired volume and normality of the final solution. (3) Convert the desired volume to liters, if necessary. (4) Calculate the molar mass of the solute. (5) Calculate the moles of solute. (6) Determine the number of equivalents per mole from the molecular formula. (7) Calculate the equivalents of solute and the mass of solute. (8) Describe how to make the solution. b. Example How would you make 500.0 mL of a 0.250 N Ba(OH)2 solution? GivenFindV = 500.0 mL or 0.5000 L N = 0.250 M eq = ? a = ? MM Ba(OH)2 = ? mass Ba(OH)2 = ? a = 2 molar mass = 171.342 g/mol V(N ( eq ( mol ( mass 0.5000 L0.250 eq1 mol171.342 g Ba(OH)2L2 eq1 mol Ba(OH)2 = 10.708875 g Ba(OH)2 = 10.7 g Ba(OH)2 actually making the solution Weigh 10.7 g of Ba(OH)2and put it into a 500.0 ml volumetric flask. Fill the flask about half-full of distilled water and dissolve the solute. Add enough distilled water to bring the final volume up to 500.0 mL Mix thoroughly. DILUTING SOLUTIONS A. Uses the fact that the number of moles of solute in a solution does not change when additional solvent is added Rearranging M =  EMBED Equation  to give mol = VM It is the initial volume and the initial molarity that determine the numbers of moles in that sample. B. Procedure Use V1M1 = V2M2 The product of the first volume and molarity must equal the product of the second volume and molarity. C. Diluting a stock solution 1. Definition of stock solution A large volume of a common reagent at a standardized concentration 2. Procedure a. Determine the molarity of the stock solution. b. Determine the desired volume and the desired molarity of the new solution. c. Calculate the volume of the stock solution that must be measured out. d. Describe how to make the new solution. Note: This method also works when concentration is in units of normality and in percent, both v/v and m/v. 3. Examples a. How would you make 100.00 mL of 0.500 M HCl from a stock solution of 6.00 M HCl? GivenFindV1 = 100.00 mL M1 = 0.500 M M2 = 6.00 M V2 = ? V1M1 = V2M2 V2 =  EMBED Equation  V2 =  EMBED Equation  = 8.33 mL actually making the solution Measure out 8.33 mL of the stock 6.00 M HCl solution. Pour it into a 100.00 mL volumetric flask. Fill the flask about half-full with distilled water and mix thoroughly. Add enough distilled water to bring the final volume to 100.00 mL. Mix thoroughly. b. What is the final molarity of 250.0 mL of a 1.00 M NaCl solution to which 100.0 mL of water has been added? GivenFindV1 = 250.0 mL volume added = 100.0 mL M1 = 1.00 M V2 = ? M2 = ? V2 = 250.0 mL + 100.0 mL = 350.0 mL V1M1 = V2M2 M2 =  EMBED Equation  M2 =  EMBED Equation  = 0.714 M GRAVIMETRIC ANALYSIS A. Definition A procedure that determines the amount of a species in a natural material by converting it to a product which can be quantitatively isolated and weighed B. Approach 1. If the material is already in solution, then measure out a specified volume of sample. 2. If the material is not already in solution, then weigh the solid sample and create a solution of the material to be analyzed. 3. Select and run the appropriate precipitation reaction to separate the species being measured as a precipitate. 4. Filter, dry, and weigh the precipitate formed in the reaction from the species being measured. C. Existing solutions 1. Procedure a. Write and balance the appropriate precipitation reaction. b. Convert the mass of the precipitate to the mass of the species being measured. c. Use the volume of the sample to calculate the concentration. 2. Example A 1.000 L sample of water from a brackish estuary was tested for chloride by precipitating it as AgCl. If 10.96 g of AgCl precipitated, what was the mass of Cl ( in a liter of that water? GivenFindmass of AgCl = 10.96 g MM AgCl = 143.321 g/mol molar mass of Cl ( = 35.453 g/mol (The mass is the same as that of atomic chlorine because the mass of the extra electron is negligible) mass Cl ( = ? Mass of AgCl Mass of Cl(  molar molar mass mass AgCl Cl(  Moles of AgCl Moles of Cl( mole ratio  EMBED Equation  10.96 g AgCl1 mol AgCl1 mol Cl(35.453 g Cl(143.321g AgCl1 mol AgCl1 mol Cl( = 2.711 g Cl( D. Solids 1. Procedure a. Write and balance the appropriate precipitation reaction. b. Convert the mass of the precipitate to the mass of the species being measured. c. Use the mass of the solid sample to calculate the percent by mass. 2. Example Nickel, an important strategic metal, is found in the ore pentlandite. 1000.00 g of the ore pentlandite is digested and put into solution. When “dmg” is added 24.5998 g of Ni (dmg)2 is precipitated. What is the mass of Ni in the sample? What is its mass percentage? “dmg” = C4H7N2O2( = 115.1127 g/mol and Ni (dmg)2 = 288.9188 g/mol 24.5998 g Ni (dmg)21 mol Ni (dmg)2 1 mol Ni 58.6934 g Ni 288.9188 g Ni (dmg)21 mol Ni (dmg)21 mol Ni = 4.99741 g Ni  EMBED Equation  x 100 % = 0.499741 % VOLUMETRIC ANALYSIS A. Definitions 1. Volumetric analysis Quantitative analysis using accurately measured titrated volumes of standard chemical solutions 2. Titration The process of reacting a solution of unknown concentration with a standard 3. Standard Either a carefully measured amount of solid, or more commonly, a solution of precisely known concentration (called a standard solution) 4. Equivalence point The point in a titration when stoichiometrically equivalent quantities have been combined, that is, where the added solute has reacted completely with the solute present in solution. 5. Indicator A substance, usually a dye, added to a solution to indicate by color change when there has begun to be an excess of the solute added. 6. End point Is determined either visually or spectrophotometrically, and is the point in the titration where the indicator changes color. If the indicator has been chosen well, then the end point is the same as the equivalence point. B. Approach 1. Titration is commonly used with acid-base reactions (but it is also used with certain redox reactions) 2. Choose an indicator whose end point is as close as possible to the equivalence point. 3. Using a buret, add measured amounts of the unknown solution to the standard until the end point has been reached. 4. Assuming that the end point is the same as the equivalence point, the number of equivalents in the unknown solution added must be equal to the number of equivalents in the standard. 5. For a standard solution use VuNu = VsNs, and for a solid standard use VuNu = eqs to calculate the concentration of the unknown solution. C. For a solid standard 1. Procedure a. Calculate the number of equivalents in the mass of the solid standard. b. Using VuNu = eqs calculate the normality of the unknown (and the molarity, if required) 2. Example 24.71 mL of a NaOH solution of unknown normality are required to titrate 0.5026 g potassium hydrogen phthalate, abbreviated “KHP”, with a molecular formula of HKC8H4O4. What is the normality and the molarity of the NaOH solution? GivenFindVu = 24.71 mL mass KHP = 0.5026 g MM KHP = 204.225 g/mol for KHP a = 1 eq/mol for NaOH a = 1eq/mol  eqs = ? Nu = ? M = ? finding the equivalents of KHP eqs =0.5026 g KHP1 mol KHP 1 eq KHP204.225 g KHP1 mol KHP eqs = 2.4610 x 10(3 eq KHP finding the normality of the NaOH solution Nu=eqs=2.4610 x 10(3 eq1000 mLVu24.71 mL1 L = 0.099595 N = 0.09960 N NaOH finding the molarity of the NaOH solution N = a(M) a =  EMBED Equation  M = N x  EMBED Equation  =0.09960 eq1 mol1 L1 eq = 0.09960 M D. For a standard solution 1. Procedure use VuNu = VsNs 2. Example 25.12 mL of a 0.09960 N NaOH solution (standard) are required to titrate 25.00 mL of an H2SO4 solution. What is its normality and molarity? GivenFindVs = 25.12 mL Ns = 0.09960 N Vu = 25.00 mL for H2SO4 a = 2 eq/mol Nu = ? 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