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The group number in the periodic table can tell us the charge of the metal cation and the charge of the non-metal anion that will be formed in an ionic reaction. Non-metals: Charge of Anion = Group Number - 8 Metals: Charge of Cation = Group Number (1A, 2A, 3A) NON-METALS There are 2 different naming systems for transitionmetal compounds: Stock System (new) and an older one that uses “ous” and “ic” suffixes. • STOCK SYSTEM write the name of the metal indicating the charge on metal with Roman numerals. Anion follows ionicnaming rules • OLD NAMING SYSTEM: use “ous” and “ic” suffixes on metal Charge of Transition metals is not so simple--must be learned. FORMULA Ni2+ STOCK SYSTEM OLD NAME ferrous chloride FeCl2 2 Cl- -2 so Fe is +2 iron(II) chloride FeCl3 3 Cl- -3 so Fe is +3 iron(III) chloride ferric chloride chromium(III) sulfide chromic sulfide Cr2S3 Hg2CO3 3 S-2 -6 so Cr is +3 (6/2) CO3-2 -2 so Hg2 is +2 mercury(I) carbonate mercurous carbonate Polyatomic Ions Contain Multiple Covalently Bonded Cations NH+ 4 ammonium H3 O+ hydronium -1 Anions CH3 CO2 acetate CN cyanide ClO hypochlorite ClO2 chlorite ClO3 chlorate -1 Anions MnO4 NO2 permanganate nitrite NO3 nitrate carbonate CrO42 chromate Cr2 O72 dichromate O22 peroxide perchlorate H2 PO4 HCO3 HPO42 hydrogen phosphate dihydrogen phosphate SO32 sulfite HSO3 hydrogen sulfite (bisulfite) hydrogen sulfate (bisulfate) SO42 S2 O32 sulfate HSO4 OH hydroxyl Stock System Older Name thiosulfate -3 Anions PO43 phosphate Indicate where in the periodic table the following elements are found: (a) Generic name of the Group VII elements (b) Generic names of Group IA and Group IIA elements (c) Elements that commonly form cations (d) Elements that commonly form covalent bonds (e) Elements that commonly form ionic bonds (f) Elements that commonly have multiple valency (g) Elements that commonly form anions (h) Elements that are inert (unreactive) (i) Elements that are found as molecular compounds Make the periodic table your friend Ion -2 Anions CO32 ClO4 hydrogen carbonate Transition-Metal Cations and their Names Nomenclature BaCl2.2H2O CBr4 dinitrogen trioxide Silver chromate barium chloride dihydrate carbon tetrabromide N2O3 Ag2(CrO4) Ag2(Cr2O7) Fe2O3 Calcium hypochlorite Cu(NO3)2 potassium permaganate Mg(SO4).7H2O Silver dichromate Ferric oxide (Iron III) Ca(ClO)2 copper II nitrate (cupric nitrate) KMnO4 mercury(I) acetate Hg2(CH3COO)2 Fe(ClO4)3 Iron III perchlorate magnesium sulfate decahydrate Nomenclature Binary Acids “ide” to “ic” Molecular hydrogen (H2) reacts with Group VII and VI nonmetals to produce covalent compounds: H-Xi (HCl, HF, HS2 ) Oxo-acids have a non-metal element bonded to oxygen with a hydrogen atom. Like binary acids the hydrogen ionizes in water. H2 (g) + Cl2(g) ===> 2 HCl (g) hydrogen gas + chloride gas ===> hydrogen chloride gas a covalent compound These compounds disassociate in water forming ions. HCl (g) + H2O =====> H+ + Cl- (both ions) hydrochloric chloride + water ===> hydrochloric acid “ide” =====> “ic” acid Hydrofluoric Acid => HF Hydrochloric Acid => HCl Hydroiodic Acid => HI Hydrobromic Acid => HBr ACIDS OXO-ANIONS Hydrates are ionic compounds that have “bound” water molecules attached to each formula unit. Mono- 1 CaSO4 . 4H2O hydrated calcium sulfate CaSO4 anhydrous calcium sulfate When naming hydrates, the word hydrate together with a greek prefix and a “.” is used in the formula. Lithium chloride monohydrate MgSO4.5H2O BaCl2.2H2O Di-2 Tri-3 Tetra-4 Penta-5 Hexa-6 Hepta-7 Octa-8 Nonyl-9 Deca-10 LiCl.H2O magnesium sulfate pentahydrate barium chloride dihydrate A chemical formula of a compound indicates the number ratio of combining atoms. Chapter 3 Stoichiometry of Chemical Formulas And Equations Subscript: indicates the # of bonded H atoms Leading coefficients: indicates how total molecules (or moles). We can view covalent and ionic compounds at a atomic level and use special names to describe mass. The Molecular mass is the sum of the atomic masses (in amu) in a single molecule (COVALENT). 1S 2O SO2 SO2 32.07 amu NaCl 23.00 amu 35.45 amu 58.45 amu Chemistry counts atoms, ions, molecules indirectly by comparing masses of the same number of things using ratio of masses built into the periodic table. Masses 12 red marbles 12 yellow marbles 12 purple marbles 12 green marbles How do we connect the atomic world to the big world? + 2 x 16.00 amu 64.07 amu The Formula mass is the sum of atomic masses (in amu) in a single ionic formula unit (IONIC). 1Na 1Cl NaCl Chemical reactions require specific numbers of molecules or formula units to react. If we can’t see molecules to count them, then how do we do accomplish the counting task? + 2H2 4 amu + + 1 O2 32 amu ? grams + ? grams 2H2O 36 amu ? grams KEY RIFF: The periodic table tells us the mass of 1-atom of any element in amu, and the mass of 1-mole (6.02 x 1023) of an element in grams! 16 amu = 1 atom O 16 grams = 1 mol O atoms 12 red marble = 84 grams 12 yellow marble = 48 grams 84 g 48 g 40.g 20.g 1 red must be 1.75 times heavier than 1 yellow marble (84g/ 48g =1.75) The mole links 6.02 x 1023 entities (atoms, molecules or formula units) to a mass in grams . Atomic Size The scale has 1 mole (6.02 X 1023) of atoms Fe on one side and 1 mole of S atoms on the other. 55.85 g Fe 6.02 x 1023 atoms S 32.07 g S 6.02 x 1023 atoms Fe 1 mol Fe atoms 1 mol S atoms MOLE molecular mass of water molar mass of water There is Avogadro’s number of atoms on both sids of the balance, but they don’t weigh the same! The mole bridges the mass of 1 atom in amu to the mass of 1 mole (6.02 X 1023) of atoms in grams. 1 12C atom = 12.00 amu 1 mole 12C atoms = 12.00 g 12C Atomic scale Human-usable scale built into periodic table WHY MOLES?? Chemical reactions require a specific numbers or whole number ratio of reactants in order to produce a specific whole number of products (Law of Definite Composition). 2 molecule H2 Special Numbers 12 H2O molecules Defined Unit 144 H2O molecules 1 gross H2O molecules 6.02 x 1023 molecules 1 mole H2O molecules + + O2 2H2O 4 amu + 32 amu 36 amu How can we connect amu to grams? + 36 g H2O 32.0 g O2 Ionic Compounds: Thinking Small and Thinking Big none 18.0 grams of H2O The Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a single molecule. 2O SO2 2H2O molecules none Covalent Compounds Thinking Small and Thinking Big 1S 1 molecule O2 Mass Connection 1 dozen H2O molecules + 2H2 4.0 g H2 Think of the mole as a counting unit having 6.02 x 1023 things. Unlike all other counting numbers it is also linked to mass in the periodic table!! SO2 32.07 amu + 2 x 16.00 amu 64.07 amu The Molar mass of a compound is the same number as the molecular mass with units of grams per mole. 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2 Do You Understand Molecular and Molar Mass? The formula mass (or forumula weight) is the sum of the atomic masses (in amu) in a formula unit. 1 Na 23.00 amu 1 Cl 35.45 amu 58.45 amu NaCl What is the molecular mass of glucose, C6H12O6 What is the molar mass of glucose, C6H12O6 The molar mass of an ionic compound is numerically equal to the formula mass, in units of grams/mole. Molar mass of NaCl = 58.45 g/mol 1 mole of NaCl(s) = 58.45 grams NaCl(s) 6.022×1023 NaCl units = 58.45 grams NaCl(s) Do You Understand Molecular and Molar Mass? What is the molecular mass of glucose, C6H12O6 Glucose’s molecular formula C6H12O6 Molecular Mass: 6 x 12.011 + 12 x 1.007 + 6 x 15.99 = 180.18 amu Molar Mass: = 180.18 grams C6H12O6 which contains 1 mol C6H12O6 or 6.022 X 1023 molecules C6H12O6 A chemical formula provides lots of information Example: C6H12O6 (MW = 180.16 g/mol) Carbon (C) Hydrogen (H) Oxygen (O) 6 atoms 12 atoms 6 atoms 6(12.01 amu) =72.06 amu 12(1.008 amu) =12.10 amu 6(16.00 amu) =96.00 amu 1023) 1023) Atoms in 1 glucose molecule Mass of 1 glucose molecule Atoms/mole of compound Moles of atoms in 1 mole of compound 1023) 6(6.022 x atoms 12(6.022 x atoms 6(6.022 x atoms 6 moles of atoms 12 moles of atoms 6 moles of atoms 72.06 g 12.10 g 96.00 g Mass/mole of compound Molecular formulas indicates the type and the ratios of combining atoms in a covalent compound. H2O(l) covalent compound There are many conversion factors in a formula. 1 molecule H2O = 1 atom O and 2 atoms of H Molecular mass H2O = (2 x 1.008)+15.99 = 18.00 amu Molar mass = 18.00 g/mol H2O 1 mole H2O = 6.022 x 1023 molecules H2O 1 mole H2O = 2 mol H atoms = 2 x 6.02 x 1023 H atoms 1 mole H2O = 1 mol O atoms = 6.02 x 1023 O atoms We must learn how to recognize conversion factors within a chemical formula, and how to convert grams to moles to molecules to atoms and vis versa. MASS(g) of compound Molar mass (g/mol) MOLES of compound Chemical Formula MOLES of elements in compound Avogadro’s Number MOLECULES or FORMULA UNITS ATOMS in a compound A chemical formula or molecular formula provides lots of information!! Al2(SO4)3 2 atoms of Al and 3 molecules of (SO4)2- = 1 formula unit Al2(SO4)3 2 moles of Al and 3 moles of (SO4)2- = 1 formula unit Al2(SO4)3 1 formula unit Al2(SO4)3 = 342.17 amu Al2(SO4)3 1 mole Al2(SO4)3 = 342.17 g Al2(SO4)3 1 mole Al2(SO4)3 = 6.022 x 1023 formula units Al2(SO4)3 1 mole Al2(SO4)3 = 2 mol Al3+ ions 1 mole Al2(SO4)3 = 3 mol (SO4)2- ions 1 mole Al2(SO4)3 = 12 mol O atoms = 12 x A.N. O atoms 1 mole Al2(SO4)3 = 3 mol S atoms = 3 x A.N. S atoms 1 mole Al2(SO4)3 = 2 mol Al atoms =2 x A.N. Al atoms Sample Problem 3.1 and 3.3 and 3.4 Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element (1.) Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342mol of Ag? (2). Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8g of Fe? (3). How many molecules of nitrogen dioxide are in 8.92 g of nitrogen dioxide? What’s In A Chemical Formula? Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element (1.) Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342mol of Ag? g Ag = ? = 0.0342 mol Ag x 107.9 g Ag = 3.69g Ag 1 mol Ag (2). Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8g of Fe? 23 Fe atoms = 95.8 g Fe x 1 mol Fe x 6.022x10 atoms Fe 55.85g Fe 1 mol Fe = 1.04x1024 atoms Fe Urea, (NH2)2CO, is a nitrogen containing compound used as a fertilizer around the globe? Calculate the following for 25.6 g of urea: a) the molar mass of urea? b) the number of moles of urea in 25.6 g urea? b) # of molecules of urea in 25.6 g of urea? c) # hydrogen atoms present in 25.6 g of urea. (3). How many molecules of nitrogen dioxide are in 8.92 g of nitrogen dioxide? 1 mol NO2 6.02x1023 molec NO2 molcs NO2=8.92g NO2 46.01g NO2 1 mol NO2 = 1.17x1023 molecules NO2 Solution To Urea Problem 1. Calculate the molar mass (MM) of urea, (NH2)2CO MM(NH2)2CO = 2 x MN + 4 x MH + MC + MO MM(NH2)2CO = (2 x 14.07g) + 4 (1.007g) + 12.01g + 15.99g MM(NH2)2CO = 60.06 g 2. # moles of (NH2)2CO in 25.6 g M ol (NH2 )2 CO = 25.6 g (NH2 )2 CO 1 mol (NH2 )2 CO = 0.426 mol (NH2 )2 CO 60.06 g (NH2 )2 CO 3. # molecules of (NH2)2CO in 25.6 g M olec urea = 0.426 mol urea 6.02 1023 molecu urea = 2.57 1 mol urea 1023 molecu urea Borax is the common name of a mineral sodium tetraborate, an industrial cleaning adjunct, Na2B4O7. You are given 20.0 g of borax...... (a) what is the formula mass of Na2B4O7 (b) how many moles of borax is 20.0 g? (c) how many moles of boron are present in 20.0 g Na2B4O7? (d) how many grams of boron are present in 20.0 g Na2B4O7? (e) how many atoms of B are present in 20.0g? (f) how many atoms of O are present in 20.0g? (g) how many grams of atomic oxygen are present? 4. # H atoms in 25.6 g (NH2)2CO #H atoms = 0.426 mol urea 4 mol Hatoms 1 mol urea 6.02 1023 H atoms = 1.03 1 mol urea 1024 atoms Borax, Na2B4O7, is the common name of a sodium tetraborate, an industrial cleaning adjunct. Suppose you are given 20.0 g of borax: Solution: (a) The formula mass of Na2B4O7 is (2 × 23.0) + (4 × 10.8) + (7 × 16.0) = 201.2 g. b) # mol borax = (20.0 g borax) / (201.2 g mol–1) = 0.9940 = 0.01 mol of borax, A chemical formula determines the % mass of each element in a compound. n x molar mass of element x 100% molar mass of compound n is the number of moles of an element in 1 mole of the compound. Example: C2H4O2 Molar mass = 60.05 g/mole c) # mol B = 20.0 g borax)/(201.2 g mol–1) X 4 mol B/ 1 mol borax = 0.40 mol of B. d) # g B = (0.40 mol) × (10.8 g B/ mol B) = 4.3 g B e) # atoms B = 0.40 mol B x 6.02 X 1023 atoms B/1 mol B = 2.41 X 1023 atoms B f) # g O = 20.0 g borax)/(201.2 g mol–1) X 7 mol O/ 1 mol borax x 15.99 g O/1 mol O = 11.1 g O 2 x (12.01 g) x 100% = 40.0% 60.05 g 4 x (1.008 g) %H = x 100% = 6.714% 60.05 g 2 x (16.00 g) %O = x 100% = 53.28% 60.05 g %C = C2H4O2 40.0% + 6.71% + 53.2% = 100.0% An empirical formula shows the simplest most reduced whole-number ratio of the atoms in a compound. A molecular formula shows the whole number ratio of an actual known molecule. molecular empirical H 2O H 2O Announcements Chapter 3: 2, 6, 7, 11, 10, 13, 15, 17, 23, 25, 27, 33, 36, 37, 39, 42, 44, 50, 54, 58, 66, 68, 70, 74, 93, 107 Quiz 2 on Chapter 2 today C6H12O6 CH2O Exam 1: Thursday Evening July 17 C12H24O12 CH2O Time: C18H36O18 CH2O O3 N 2H 4 O NH2 N8H16 NH2 6:00PM - 7:30PM Coverage: Chapter 1-4 Don’t fall behind! If we know the chemical formula of a compound, we can determine the % mass of its elements and vis versa. All of the compounds below have the same empirical formula and the same % mass ! Empirical Formula % Mass 40.0% C 6.71% H 53.3% O CH2O % Mass Element %C = a% %H = b% %O = c% Empirical Formula Molecular Mass Molecular Formula If we know the % mass of elements in a compound we can determine the empirical formula. With molar mass we get molecular formula Molecular formula GIVEN Molar Mass Assume 100 g sample Calculate mole ratio Molar Mass Name formaldehyde acetic acid lactic acid erythrose ribose glucose Molecular Formula CH2O C2H4O2 C3H6O3 C4H8O4 C5H10O5 C6H12O6 Whole-Number M Multiple (g/mol) 1 2 3 4 5 6 30.03 60.05 90.09 Use or Function disinfectant; biological preservative acetate polymers; vinegar(5% soln) sour milk; forms in exercising muscle 120.10 part of sugar metabolism 150.13 component of nucleic acids and B2 180.16 major energy source of the cell Determining a Molecular Formula from Elemental Analysis and Molar Mass Dibutyl succinate is an insect repellent used against household ants and roaches. Elemental analysis or analysis indicates that % mass of the composition is 62.58% C, 9.63% H and 27.79% O. Its experimentally determined molecular mass is 230 u. What are the empirical and molecular formulas of dibutyl succinate? Step 1: Determine the mass of each element. Assume a 100 gram sample and use the given data we have: C 62.58 g H 9.63 g Step 5: Convert to a small whole number ratio. Multiply by 2 to get C5.98H10.98O2 O 27.79 g The empirical formula is C6H11O2 Empirical formula mass = 6(12.01) + 1.008(11) + 2(16.00) = 115 u. Step 2: Convert masses to amounts in moles. Step 6: Now using the empirical formula mass and molecular mass together determine the molecular formula. Empirical formula mass is 115 u. Molecular formula mass is 230 u. Step 3: Write a tentative empirical formula. C5.21H9.55O1.74 n = Molecular mass/empirical mass = 230 amu/115 amu = 2 Step 4: Convert to small whole numbers. Divide by smallest number of moles C2.99H5.49O Determining a Molecular Formula from Elemental Analysis and Molar Mass During physical activity, lactic acid (M = 90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness at fatigue. Elemental analysis shows that this compound contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) Determine the empirical formula of lactic acid. (b) Determine the molecular formula. Understand what is asked: What is the formula CxHyOz Mass Percent Empirical Formula Molecular Mass Molecular Formula Calculating the Mass Percents and Masses of Elements in a Sample of Compound The molecular formula is C12H22O4 Determining a Molecular Formula from Elemental Analysis and Molar Mass 1. Assume there are 100. g of lactic acid then use % mass: 40.0 g C 1 mol C 12.01g C 6.71 g H 1 mol H 1.008 g H 3.33 mol C 6.66 mol H C3.33 H6.66 O3.33 3.33 3.33 3.33 3.33 mol O CH2O empirical formula 3. The molecular formula will be a whole number multiple of the empirical formula determined BY THE MOLAR MASS ratio molar mass of lactate mass of CH2O 6 mol C x 12 mol H x (a) What is the mass percent of each element in glucose? 6 mol O x (b) How many grams of carbon are in 16.55g of glucose? 12.01 g C mol C 1.008 g H mol H 16.00 g O mol O 90.08 g 30.03 g 3 C3H6O3 is the molecular formula = 72.06 g C = 12.096 g H M = 180.16 g/mol = 96.00 g O (b) mass percent of C = We have to find the total mass of glucose and the masses of the constituent elements in order to relate them. 1 mol O 16.00 g O 2. The red numbers are the number of moles of atoms in lactic acid. This is what we use in the formula C3.33 H6.66 O3.33 Glucose (C6H12O6) is the most important nutrient in the living cell for generating chemical potential energy. PLAN: 53.3 g O 72.06 g C = 0.3999 x 100 = 39.99 mass %C 180.16 g glucose 12.096 g H mass percent of H = 180.16 g glucose = 0.06714 x 100 = 6.714 mass %H mass percent of O = 96.00 g O = 0.5329 x 100 = 53.29 mass %O 180.16 g glucose Chemical equations are symbolic representations of “what happens” in a chemical reaction. CH4(g) + O2(g) CO2(g) + H2O(g) Unbalanced => No mass conservation => WRONG Balanced Equation = Mass Conservation = CORRECT CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) Only balanced chemical equations contain useful stoichiometric conversion factors. NH3 + O2 ===> NO + H2 Not Balanced & NOT TRUE! 1 moles NH3 = 1 moles O2 6NH3 + 3O2 ===> 6NO + 9H2 Balance first and IT’s VALID! Many Correct Stoichiometric Conversion Factors + Useful stoichiometric coefficients A balanced chemical equation contains a lot of very useful chemical information. Learn it and everything gets easy! C3H8(g) + 5O2(g) 6 mol NH3 = 3 mol O2 6 mol NH3 = 9 mol H2 3 mol O2 = 9 mol H2 6 mol NH3 = 6 mol NO 3 mol O2 = 6 mol NO 6 mol NO = 9 mol H2 We balance equations using trial and error! 3CO2(g) + 4H2O(g) Little molecules 1 molecule C3H8 + 5 molecules O2 3 molecules CO2 + 4 molecules H2O __Na3PO4(aq) + __HCl(aq) ==>__H3PO4(aq) + __NaCl(aq) mass of atoms 44.09 amu C3H8 + 160.00 amu O2 132.03 amu CO2 + 72.06 amu H2O __Ba(OH)2(aq) + __HCl(aq) ==> __H2O(l) + __BaCl2(aq) Big (mol) 1 mol C3H8 + 5 mol O2 3 mol CO2 + 4 mol H2O mass (g) 44.09 g C3H8 + 160.00 g O2 132.03 g CO2 + 72.06 g H2O total mass (g) 204.09 g __C7H14 + __O2 ===> __H2O(l) + __CO2(g) 204.09 g Solutions Na3PO4(aq) + 3HCl(aq) ====> H3PO4(aq) + 3NaCl(aq) To learn how to balance equations quickly for Exams you have to practice! Try it. Na2SO4 (s) + C(s) Na + H2O Na2S (s) + CO(g) NaOH + H2 (g) Ba(OH)2(aq) + 2HCl(aq) ====> 2H2O(l) + BaCl2(aq) 2C7H14 + 21O2 ====> 14H2O(l) + 14CO2(g) Mg3N2(s) + H2O(l) Mg(OH)2 (s) + NH3(g) H2S (g) + SO2(g) S (s) + H2O(l) CO2 (g) + KOH(s) K2CO3 (g) + H2O(s) Some equations are harder than others..practice! Example: Ethane, C2H6, reacts (is combusted are key words) with O2 to form CO2 and H2O. Write a balanced equation for this reaction. 1. Write the correct formula(s) for reactants and products. C2H6 + O2 CO2 + H2O 2. Start by balancing those elements that appear in only one reactant and one product. C2H6 + O2 2 carbon on left CO2 + H2O 1 carbon on right start with C or H but not O C2H6 + O2 2CO2 + H2O 6 hydrogen on left C2H6 + O2 2 hydrogen on right 2CO2 + 3H2O 4. Balance those elements that appear in two or more reactants or products. C2H6 + O2 2CO2 + 3H2O 2 O on left C 2 H 6 + 7 O2 2 multiply CO2 by 2 How To Master Stoichiometry! 1. Always write a balanced chemical equation. 2. Work in moles---not masses....we need to count atoms and molecules using the mole connection to mass. 3. Use dimensional analysis correctly. multiply H2O by 3 4O +3O 2CO2 + 3H2O = 7 oxygen on right multiply O2 by 7 2 remove fraction 2C2H6 + 7O2 4CO2 + 6H2O multiply by 2 Iron III oxide reacts with carbon monoxide as shown below. How many CO molecules are required to react with 25 formula units of Fe2O3 as shown below in the balanced equation? Fe2O3 (s) + CO(g) Fe(s) + CO2(g) Translate words Fe2O3 (s) + 3CO(g) to formula 2Fe(s) + 3CO2(g) Balance First!! Grams of Reactant Molar Mass Moles of Reactant Grams of Product balanced equation Molar Mass Moles of Product How many CO molecules are required to react with 25 formula units of Fe2O3 as shown below in the balanced equation? Fe2O3 (s) + 3CO(g) THE MAGIC 1. Balance chemical equation first! 2. What does the question want? 3. Find stoichiometric factors 4. Use the factor-label method and solve 5. Be mindful of significant figures 6. Check answer Iron and carbon dioxide form by reaction between iron(III) oxide and carbon monoxide. How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide? 2Fe(s) + 3CO2(g) Fe2O3 (s) + 3CO(g) 2Fe(s) + 3CO2(g) How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide? Fe2O3 (s) + 3CO(g) 2Fe(s) + 3CO2(g) What mass of CO is required to react completely with 146 g of iron (III) oxide? Fe2O3 (s) + 3CO(g) 2Fe(s) + 3CO2(g) # Fe atoms = = 2.50 ⇥ 105 f u Fe2 O3 ⇥ 2 F e atoms = 5.00 ⇥ 105 F e atoms 1 f u Fe2 O3 What mass of CO is required to react completely with 146 g of iron (III) oxide? Fe2O3 (s) + 3CO(g) 2Fe(s) + 3CO2(g) What mass of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams? key riff! Excess tells you one reactant is in excess and the other is not! Fe2O3 (s) + 3CO(g) How many grams of iron (III) oxide react with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams? Fe2O3 (s) + 3CO(g) 2Fe(s) + 3CO2(g) What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide? key riff! It tells you one reactant is in excess and the other is not! 2Fe(s) + 3CO2(g) will run out first excess Fe2O3 (s) + 3CO(g) 2Fe(s) + 3CO2(g) What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide? Fe2O3 (s) + 3CO(g) 2Fe(s) + 3CO2(g) Another stoichiometry example from Silberberg Copper metal is obtained from copper(I) sulfide containing ores in multistep-extractive process. After grinding the ore into fine rocks, it is heated with oxygen gas forming powdered cuprous oxide and gaseous sulfur dioxide. (0) Write a balanced chemical equation for this process (a) How many moles of molecular oxygen are required to fully roast 10.0 mol of copper(I) sulfide? (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted? (c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide? (0) Write a balanced chemical equation for this process Cu2S(s) + O2(g) Cu2O(s) + SO2(g) 2Cu2S(s) + 3O2(g) unbalanced 2Cu2O(s) + 2SO2(g) (a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide? 3 mol O2 = 15.0 mol O mol O2 = ? = 10.0 mol Cu2S x 2 2 mol Cu2S 2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g) (c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide? kg O2 = 2.86 kg Cu2O x 103g Cu2O 1 mol Cu2O x 3mol O2 x x kg Cu2O 143.10g Cu2O 2mol Cu2O kg O2 = 0.959 = 1 kg O2 32.00g O2 x 1000 g O2 1 mol O2 (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted? g SO2 = 10.0 mol Cu2S x 2mol SO2 x 64.07g SO2 = 641g SO2 2mol Cu2S 1 mol SO2 The limiting reagent is the reactant that is runs out (or is consumed) and dictates the amount of product that can be formed. We deal with limiting reagents all the time, we just don’t give it a buzzword like chemists do. Reactants 1 CB + 4T ===> CBT4 Which reactant is the limiting reagent? It’s the same in chemistry. We must “count” the number of each reactant (via moles) and see which is limiting--it’s not how much they weigh! Number of Copies Possible 87 copies 83 copies 168/2 = 84 328/4 = 82 Product Do You Understand Limiting Reagent? If 25.0 g CH4 is combusted with 40.0 g O2, which reactant is the limiting reactant? Note the absence of “excess” which tells us it is a limiting reagent problem! There are two ways you can calculate the answer. Method 1: For both reactants, use balanced equation & stoichiometric factors and compute the amount of any product formed for both (I teach this method!) Method 2: Pick one of the reactant and compute how much of the other reactant you need and compare with the given amount. Methanol,CH3OH, burns when ignited in air. Assuming excess O2 is added and 209. g of methanol is combusted, what mass of water and CO2 is produced? NOTE THE KEY WORDS EXCESS O2 MAKES THIS EASIER. IT SAYS THE OTHER REACTANT IS THE LIMITING REAGENT AND LIMITS THE EXTENT OF THE REACTION! • If 25.0 g CH 4 is combusted with 40.0 g O2, which reactant is the limiting reactant? CH4(g) + 2 O2(g) ""→ CO2(g) + 2 H2O(g) Method 1 (Calculating the # moles of product formed by each reactant to determine which reactant makes the least amount) 1. Let’s use the amount of CO2 formed as our “yardstick” of how much product can be made (we could chose H2O). mol CO2 = 25.0 g CH4 mol CO2 = 40.0 g O2 1 mol CH4 1 mol CO2 = 1.559 mol CO2 16.04 g CH4 1 mol CH4 1 mol O2 1 mol CO2 = 0.625 mol CO2 32.00 g O2 2 mol O2 O2 must be the limiting reagent as the amount of CO2 produced is the least amount of product! Methanol, CH3OH, burn when ignited in air. Assuming excess O2 is added and 209 g of methanol is combusted what mass of water and CO2 is produced? 2CH3OH + 3O2 2CO2 + 4H2O grams CH3OH moles CH3OH moles H2O grams H2O grams CH3OH moles CH3OH moles CO2 grams CO2 209 g CH3OH x 4 mol H2O 18.0 g H2O 1 mol CH3OH = x x 32.0 g CH3OH 2 mol CH3OH 1 mol H2O = 235 g H2O Do You Understand Limiting Reagents? Phosphorus trichloride is a commercially important compound used in the manufacture of pesticides. It is made by the direct combination of phosphorus P4 and gaseous molecular chlorine. Suppose 323 g of chlorine is combined with 125 g P4. Determine the amount of phosphorous trichloride that can be produced when these reactants are combined. P4 (s) + Cl2 (g) → PCl3 (l) Step 1. Convert words to formulas P4 (s) + 6Cl2 (g) → 4PCl3 (l) Step 2. Balance Step 3. Determine reactant that produces the least product. mol PCl3 = 323 g Cl2 1 mol Cl2 70.91 g Cl2 4 mol PCl3 = 3.04mol PCl3 6 mol Cl2 mol PCl3 = 125 g P4 1 mol P4 123.88 g P4 4 mol PCl3 = 4.04mol PCl3 1 mol P4 NOTE: 2 Reactants With Masses => Limiting Reagent Step 4. Cl2 is the limiting reagent and determines the amount of PCl3 g PCl3 = 125 g P4 1 mol P4 123.88 g P4 4 mol PCl3 6 molCl2 137.32 g PCl3 = 417.5 g PCl3 1 mol PCl3 Do You Understand Limiting Reagents? 124.0 g of solid Al metal is reacted with 601.0 g of iron(III) oxide to produce iron metal and aluminum oxide. Calculate the mass of aluminum oxide formed. 124 g of Al are reacted with 601 g of Fe2O3 2Al + Fe2O3 g Al Al2O3 + 2Fe mol Al2SO3 produced mol Al 124 g Al x 1 mol Al 27.0 g Al x 1 mol Al2O3 2 mol Al 1. Write a balanced equation for all problems 2. Two reactant masses and “no excess” = limiting reagent. 3. Work in moles (grams => moles => equation stoichiometry) 4. Determine maximum theoretical amount of product for both reactants. The limiting reagent is the one that produces the least. Try it At Home:Answer is in your book! Hydrazine(N2H4) and dinitrogen tetraoxide(N2O4), ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00 x 102 g of N2H4 and 2.00 x 102 g of N2O4 are mixed? g Fe2O3 mol Fe2O3 601 g Fe2O3 x 1 mol Fe2O3 mol Al2SO3 produced 1 mol Al2O3 101.96 g Al2O3 x = 383 g Al2O3 1 mol Fe2O3 1 mol Al2O3 Try it At Home: Hydrazine(N2H4) and dinitrogen tetraoxide(N2O4), ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00 x 102 g of N2H4 and 2.00 x 102 g of N2O4 are mixed? 2 N2H4(l) + N2O4(l) BALANCE! mol N2H4 2.00 x 102 g N2O4 X 3 mol N2 2 mol N2H4 1 mol N2O4 92.02 g N2O4 mol N2 = 2.17 mol N2O4 X 3 N2(g) + 4 H2O(l) = 3.12mol N2H4 32.05g N2H4 mol N2 = ? = 3.12 mol N2H4 X Real reactions have side-reactions which reduce the amount of product obtained vs the “theoretical amount. We call that experimentally-determined value. 101.96 g Al2O3 = 234 g Al2O3 1 mol Al2O3 Al make the least and is the limiting reagent! 234 g Al2O3 can be produced 1.00 x 102g N2H4 When we do chemical reactions calculations in the class, they are “ideal and give 100% product. This is the called the “theoretical value”. x 160. g Fe2O3 x = 4.68 mol N2 = 2.17 mol N2O4 3 mol N2 1 mol N2O4 N2H4 is the limiting reactant because it produces less product, N2, than does N2O4. = 6.51 mol N2 4.68mol N2 28.02g N2 mol N2 = 131g N2 The % Yield of a chemical reaction is the ratio of product mass obtained in the lab over the theoretical (i.e. calculated) X 100. experimentally determined in lab % Yield = Actual Amount x 100 from limiting Theoretical Amount reagent calculation A + B C (main product) (reactants) D (side products) Actual Amount actual amount of product obtained from a reaction in the lab. It’s given in a word problem. Theoretical Amount is the amount of product that is calculated assuming all the limiting reagent reacts. Putting it all together: Limiting/Percent Yield A student reacts 30.0 g benzene, C6H6, with 65.0 g bromine, Br2, to prepare bromobenzene and hydrogen bromide, C6H5Br in the lab (MW C6H6 = 78.102, C6H5Br 156.99, Br2 = 159.808). After the reaction was complete, the student recovered 56.7 g C6H5Br. Determine the limiting reagent, the theoretical yield, and the overall % yield? Learning Check: Calculating Percent Yield A student reacts 30.0 g benzene, C6H6, with 65.0 g bromine, Br2, to prepare bromobenzene, C6H5Br. (MW C6H6 = 78.102, C6H5Br 156.99, Br2 = 159.808). After the reaction was complete, the student recovered 56.7 g C6H5Br. Determine the limiting reagent, the theoretical yield, and the overall % yield? Step 1. Converts words to formulas and balance the equation C6H6 + Br2 C6H5Br + HBr Step 2. Determine the reactant that produces the least product. 1. Write a balanced equation 2. Use stoichiometry in balanced equation 3. Find g product predicted limiting reagent actual yield/theoretical yield x 100 4. Calculate percent yield g C6 H5 Br = 30.0 g C6 H6 g C6 H5 Br = 65.0 g Br2 1 mol C6 H6 78.1 g C6 H6 1 mol Br2 159.8 g Br2 1 mol C6 H5 Br 1 molC6 H6 1 mol C6 H5 Br 1 mol Br2 156.9 g C6 H5 Br = 60.3 g C6 H5 Br 1 mol C6 H5 Br 156.9 g C6 H5 Br = 63.8 g C6 H5 Br 1 mol C6 H5 Br Step 3. C6H6 is the limiting reagent. 60.3 g C6H5Br is theory yield % Yield = g Actual g Theoretical 100 = 56.7 g 60.3 g 100 = 94.0% Learning Check: Calculating Percent Yield Learning Check: Calculating Percent Yield Silicon carbide (SiC) is made by reacting silicon dioxide with powdered carbon (C) under high temperatures. Carbon monoxide is also formed as a by-product. Suppose 100.0 kg of silicon dioxide is processed in the lab and 51.4 kg of SiC is recovered What is the percent yield of SiC from this process? SiO2(s) + C(s) Notice the word excess is missing and the amount of the other reactant. Must assume other reactant is excess! SiO2(s) + 3C(s) •Distribution of solute in solvent is uniform (mixed) •Components do not separate on standing •Not separable by filtration •Solute/Solvent mix in ratios up to the solubility limit of solute SiC(s) + 2CO(g) 2. Balance 3. We are given one reactant, we must assume excess of the other (C) mol SiO2 = 102 kg SiO2 103 g SiO2 1 kg SiO2 kg SiC = 1664 mol SiCl 1 mol SiO2 60.09 g SiO2 40.10 g SiC 1 mol SiC 1 mol SiC = 1664 mol SiC 1 mol SiO2 1 kg SiC = 66.73 kg SiC 1000 g SiC = 66.73 kg SiC 4. We get the actual from experiment and theoretical from calculation and plug them into the yield equation (units of mass should be the same) % Yield = Section 3.5 Fundamentals of Solution Stoichiometry SiC(s) + CO(g) 1. Converts words to formulas kg Actual kg Theoretical 100 = 51.4 kg 66.73 kg 100 = = 77.0% 77.0% A solution is a homogenous mixture of a solvent plus a solute (focus is aqueous). The solute is the substance(s) dissolved in the solvent. A solvent is the substance present in the larger amount----typically water in aqueous solutions. Solution is the solute + solvent. g solution = g solute + g solvent Section 3.5 Fundamentals of Solution Stoichiometry Chemists express the concentration of a solution by declaring the amount of solute present in a given quantity of solution. Molar Mass M = molarity = moles of solute total liters of solution Solute + Solvent Molar Mass grams solute We can visualize solute particles in a solvent (solute) and solvent molecules not shown. moles solute Molarity 4-Basic Problem Types Using Molarity 1. Preparing a solution with a certain molarity. Add Solvent Increasing Volume Decreases Concentration Concentrated Solution More solute particles per unit volume Volume of solution 2. Mass-Mole-Number of Molecules Involving Solutions Dilute Solution More solute particles per unit volume 3. Stoichiometry and Chemical Reactions With Solutions moles of solute M = molarity = total liters of solution 4. Dilution Problems (M1V1 = M2V2) Preparing Solutions in the Laboratory Preparing Solutions in the Laboratory Weigh the solid needed How you would make up 1.00 L of 0.100M CuSO4? (How many grams of CuSO4 is needed) Add the solvent to final volume in glassware. Dissolve the solid Dissolve completely 1L mark A •Weigh the solid needed. •Transfer the solid to a volumetric flask that contains about half the final volume of solvent. C Add solvent until the solution reaches its final volume. B Dissolve the solid thoroughly by swirling. ?g CuSO4 Dilute to mark 0.100M CuSO4 Preparing Solutions in the Laboratory How you would make up 1.00 L of 0.100M CuSO4? Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M calcium nitrate. # moles CuSO4 = M x V = 1.0 L soln x 0.100 mol CuSO4 = 0.100 mol x CuSO4.5H2O g CuSO4 = 0.100 mol CuSO4 ⇥ 1L mark 159.62 g CuSO4 ·5 H2 O = 15.96 g CuSO4 1 mol CuSO4 Dissolve completely Dilute to mark 0.100M CuSO4 15.96 g CuSO4 Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M. How many grams of KI is required to make 500. mL of a 2.80 M KI solution? volume KI M KI grams KI = 500. mL x moles KI 1L 1000 mL x M KI 2.80 mol KI 1 L soln grams KI x 166 g KI 1 mol KI = 232 g KI How many grams of KI is required to make 500. mL of a 2.80 M KI solution? Calculate the molarity of a solution that contains 12.5 g of pure sulfuric acid (H2SO4) in 1.75 L of total solution. Calculate the molarity of a solution that contains 12.5 g of pure sulfuric acid (H2SO4) in 1.75 L of solution (Molar Mass H2SO4 = 98.1g) M = molarity = moles of solute total liters of solution Identify the solute here? Calculating the Molarity of a Solution Sucrose has a molar mass of 342.29 g/mol. It is a fine, white, odorless crystalline powder with a pleasing, sweet taste. What is the molarity of an aqueous solution made by placing 75.5 g of 85% pure sugar in 500.00 mL of water? Calculating the Molarity of a Solution Sucrose (pure sugar) has a molar mass of 342.29 g/mol. What is the molarity of pure sugar in an aqueous solution made by placing 75.5 g of unrefined sugar which is 85% w/w pure sugar in 500.00 mL of water? How many grams of solute are in 1.75 L of 0.460 M sodium monohydrogen phosphate? Molar Mass: 141.96 g/mol Na2HPO4 fix this slide for ang 85 g pure sug 100. g impure Mol sugar = 75.5 g impure Msugar = 1 mol sug = 0.18748 mol sug 342.29 g sug mol pure sugar 0.18748 =. = 0.375 total volume solution 0.50000 L How many grams of solute are in 1.75 L of 0.460 M sodium monohydrogen phosphate? g of Na2HPO4 = ? = 1.75 L soln X 0.460 moles Na2HPO4 X 141.96 g Na2HPO4 1 mol Na2HPO4 1 L soln = 114 g Na2HPO4 Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Remember this formula: MiVi Moles of solute in Volume = = MfVf Moles of solute after dilution (f) How would you prepare 500.0 mL of 0.500M HCl from a stock solution of 12.1 M HCl? Vi How many mL of stock are required? Look at the picture carefully and you should see that the number of moles taken from the blue cube on the left (the concentrated stock solution) is the number of moles that ends up in the light white. Mi × V i Mf × V f Vf Stock 12.1M HCl Mi mark mol HCl before dilution n V M= dilute to 500.00 mL of 0.500M HCl Mf How would you prepare 0.80L of 0.150M saline from a 6.0M stock solution? Moles of solute in Volume MiVi = Moles of solute after dilution (f) = MfVf How would you prepare 0.80L of 0.150M saline from a 6.0M stock solution? Given: Mi = 6.00M Mf = 0.15M Vf = 0.80 L Vi = ? L MiVi = MfVf Vi = MfVf Mi 0.15 M x 0.80M L = 6.00 M HNO3 = 0.020 L 20 mL of stock + 780 mL of water = 800.0 mL of solution How would you prepare 60.0 mL of 0.2 M HNO3 from a stock solution of 4.00 M HNO3? How would you prepare 60.0 mL of 0.2 M HNO3 from a stock solution of 4.00 M HNO3? Given: Mi = 4.00 Mf = 0.200 Vf = 0.06 L Vi = ? L MiVi = MfVf Vi = MfVf Mi 0.200 M x 0.060 L = 4.00 M HNO3 = 0.003 L = 3.0 mL 3 mL of acid + 57 mL of water = 60.0 mL of solution What does a 3.5 M FeCl3 mean? What does 3.5 M FeCl3 mean? 3.5 M FeCl3 = 3.5 moles FeCl3 1 Liter solution (1) a homogeneous solution of 3.5 moles of dry 100% pure FeCl3 dissolved in 1.00 Liter total solution volume (not 1 L of liquid!). (3) Note: It does not mean 3.5 moles of FeCl3 is dissolved in 1.00 liter of water! (4) [Fe3+] = 3.5M and [Cl-] = 3 x 3.5 M (5) It can be used as a conversion factor Given 2 liters of 0.20M Al2(SO4)3, Given 2 liters of 0.20M Al2(SO4)3, (a) what is the molarity of aluminum sulfate? (a) what is the molarity of aluminum sulfate? [Al2(SO4)3] = 0.20M (b) what is the molarity of Al3+ in this solution (b) what is the molarity of Al3+ in this solution (c) what is the molarity of SO4 solution? 2- in [Al3+] = 2 X .20M = 0.40M this (c) what is the molarity of SO42- in this solution? (d) how many moles of aluminum ions are there in 2L of this solution? [SO42-] = 3 X .20M = 0.60M (d) how many Al3+ ions are there in 2L of this solution? How many milliliters of 0.125 M NaHCO3 solution are needed to neutralize 18.0 mL of 0.100 M HCl? Solution Stoichiometry How many milliliters of 0.125 M NaHCO3 are needed to neutralize (react with) 18.0 mL of 0.100 M HCl? HCl(aq) + NaHCO3(aq) => NaCl(aq) + H2O(l) + CO2(g) HCl(aq) + NaHCO3(aq) => NaCl(aq) + H2O(l) + CO2(g) # mol HCl = MHCl x L = 0.100M HCl X 0.018L = 1.80 X 10-3 mol HCl Volume of HCl Volume of NaHCO3 Molarity HCl Mole to Mole ratio Molarity NaHCO3 # mol NaHCO3 = = 1.80 10 3 mol HCl 1 mol NaHCO3 1 miol HCl 1 L solution = .0144 L solution 0.125 M NaHCO3 Limiting-Reactant Problems in Solution Mercury and its compounds have many uses, from fillings for teeth (as an alloy with silver, copper, and tin) to the industrial production of chlorine. Suppose 0.050L of 0.010M mercury(II) nitrate reacts with 0.020L of 0.10M sodium sulfide forming mercury(II) sulfide. How many grams of mercury(II) sulfide can form? Step 1: Words to balanced chemical equation Suppose 0.050L of 0.010M mercury(II) nitrate reacts with 0.020L of 0.10M sodium sulfide forming mercury(II) sulfide. How many grams of mercury(II) sulfide can form? Hg(NO3)2(aq) + Na2S(aq) g HgS = 0.050 L Hg(NO3 )2 Step 2: Is it limiting reagent? Yes or No Step 3: Use given information, balanced equation and good dimensional analysis technique and compute g HgS formed. g HgS = 0.020 L Na2 S HgS(s) + 2NaNO3(aq) 1 mol HgS 1 mol Hg(NO3 )2 0.010 mol Hg(NO3 )2 1 L Hg(NO3 )2 0.10 mol Na2 S 1 L Na2 S 1 mol HgS 1 mol Na2 S 232.7 g HgS = 0.12 g HgS 1 mol HgS 232.7 g HgS = 0.47 g HgS 1 mol HgS Step 4: Does it make sense? Hg(NO3)2(aq) is the limiting reagent and 0.12 g HgS can be produced from this reaction. Stoichiometry in Solution Specialized cells in the stomach release HCl(aq) to aid in digestion. If too much acid is released, the excess can cause “heartburn”. The excess acid is sometimes neutralized with “antacids”. A common antacid is magnesium hydroxide, which reacts with the hydrochloric acid to form water and magnesium chloride solution. How many liters of “stomach acid” react with a tablet containing 0.10g of magnesium hydroxide? 1. Always translate nomenclature to chemical equation Mg(OH)2(s) + HCl(aq) Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + H2O(l) unbalanced MgCl2(aq) + 2H2O(l) balanced 2. Work in moles and use the stoichiometric factors. As a government biologist testing commercial antacids, suppose you use 0.10M HCl to simulate the acid concentration in the stomach. How many liters of “stomach acid” react with a tablet containing 0.10g of magnesium hydroxide? One of the solids present in photographic film is silver bromide. It is prepared by mixing silver nitrate with calcium bromide giving insoluble silver bromide and soluble calcium nitrate . L HCl = ? = L HCl = 0.10 g Mg(OH)2 1 mol Mg(OH)2 58.33 g Mg(OH)2 2 mol HCl 1 mol Mg(OH)2 1 L HCl = 3.4 0.1 mol HCl 10 = 3.4 x 10-2 L HCl One of the solids present in photographic film is silver bromide. It is prepared by mixing silver nitrate with calcium bromide giving insoluble silver bromide and soluble calcium nitrate . CaBr2 + 2AgNO3 ==> 2AgBr + Ca(NO3)2 How many milliliter of 0.125M CaBr2 must be used to react with the solute in 50.0 mL of 0.115M AgNO3? How many milliliter of 0.125M CaBr2 must be used to react with the solute in 50.0 mL of 0.115M AgNO3? mL CaBr2 = 0.050 L soln 0.115 mol AgNO3 1 L soln 1 mol CaBr2 2 mol AgNO3 = 23.0 mL CaBr2 1 L CaBr2 0.125 mol CaBr2 1000 mL} = 23.0 mLCaB 1L Sample Problem 3.18 Visualizing Changes in Concentration The beaker and circle represents a unit volume of solution. Solvent molecule as blue background Sample Problem 3.18 Visualizing Changes in Concentration The beaker and circle represents a unit volume of solution. Draw the solution after each of these changes: (a) For every 1 mL of solution, 1 mL of solvent is added. Mdil x Vdil = Mconc x Vconc Draw a new picture after each of these changes: (a) For every 1 mL of solution, 1 mL of solvent is added. (b) One third of the solutions volume is boiled off. 8M x 1dil = ?conc x 2conc ?conc = 4conc (b) One third of the solutions volume is boiled off. Mdil x Vdil = Mconc x Vconc 8M x 1dil = ?conc x (1 x 2/3) ?conc = 12M