MOLE. Chapter #3 : Stoichiometry. Mole - Mass Relationships of Elements

Transcription

1 Chapter #3 : Stoichiometry MOLE The Mole is based upon the definition: The amount of substance that contains as many elementary parts (atoms, molecules, or other?) as there are atoms in exactly 12 grams of carbon Mole = x particles Mole - Mass Relationships of Elements Element Atom/Molecule Mass Mole Mass Number of Atoms 1 atom of H = amu 1 mole of H = g = x atoms 1 atom of Fe = amu 1 mole of Fe = g = x atoms 1 atom of S = amu 1 mole of S = g = x atoms 1 atom of O = amu 1 mole of O = g = x atoms 1 molecule of O 2 = amu 1 mole of O 2 = g = x molecule 1 molecule of S 8 = amu 1 mole of S 8 = g = x molecules Molecular Mass - Molar Mass ( M ) The Molecular mass of a compound expressed in amu is numerically the same as the mass of one mole of the compound expressed in grams. For water: H 2 O Molecular mass = (2 x atomic mass of H ) + atomic mass of O = 2 ( amu) amu = amu Mass of one molecules of water = amu Molar mass = ( 2 x atomic mass of H ) + atomic mass of O = 2 ( g ) g = g g H 2 O = x molecules of water = 1 mole H 2 O

2 Calculating the Number of Moles and Atoms in a Given Mass of Element Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element 3680 o C. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal? Plan: Convert mass into moles by dividing the mass by the atomic weight of the metal, then calculate the number of atoms by multiplying by Avogadro s number! Solution: Converting from mass of W to moles: Moles of W = 35.0 mg W x 1 mol W = mol g W mol NO. of W atoms = 1.90 x x 10 mol W x 23 atoms = 1 mole of W = atoms of Tungsten Calculating the Moles and Number of Formula Units in a given Mass of Cpd. Problem: Trisodium Phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample? Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients. Solution: The formula is Na 3 PO 4. Calculating the molar mass: M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen = = 3 x g/mol + 1 x g/mol + 4 x g/mol = g/mol g/mol g/mol = g/mol Converting mass to moles: Moles Na 3 PO 4 = 38.6 g Na 3 PO 4 x (1 mol Na 3 PO 4 ) g Na 3 PO 4 = mol Na 3 PO 4 Formula units = mol Na 3 PO 4 x x formula units 1 mol Na 3 PO 4 = formula units Flow Chart of Mass Percentage Calculation Moles of X in one mole of Compound M (g / mol) of X Mass (g) of X in one mole of compound Mass fraction of X Divide by mass (g) of one mole of compound Multiply by 100 Mass % of X

3 Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I Problem: Sucrose (C 12 H 22 O 11 ) is common table sugar. ( a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in g of sucrose? (a) Determining the mass percent of each element: mass of C = 12 x g C/mol = g C/mol mass of H = 22 x g H/mol = g H/mol mass of O = 11 x g O/mol = g O/mol g/mol Finding the mass fraction of C in Sucrose & % C : Total mass of C g C Mass Fraction of C = = mass of 1 mole of sucrose g Cpd = To find mass % of C = x 100% = % Calculating Mass Percents and Masses of Elements in a Sample of Compound - II (a) continued Mass % of H = mol H x M of H x 100% = 22 x g Hx 100% mass of 1 mol sucrose g = % H Mass % of O = mol O x M of O x 100% = 11 x g O x 100% mass of 1 mol sucrose g = % O (b) Determining the mass of carbon: Mass (g) of C = mass of sucrose X( mass fraction of C in sucrose) Mass (g) of C = g sucrose X g C = g C 1 g sucrose Mol wt and % composition of NH 4 NO 3 2 mol N x g/mol = g N 4 mol H x g/mol = g H 3 mol O x g/mol = g O g/mol 28.02g N %N = 2 x 100% = % N 80.05g %H = 4.032g H 2 x 100% = % H 80.05g %O = x 100% = % O 48.00g O g %

4 Calculate the Percent Composition of Sulfuric Acid H 2 SO 4 Molar Mass of Sulfuric acid = 2(1.008g) + 1(32.07g) + 4(16.00g) = g/mol 2(1.008g H %H = 2 ) x 100% = % H 98.09g 1(32.07g S) %S = x 100% = % S 98.09g 4(16.00g O) %O = x 100% = % O g Check = % Empirical and Molecular Formulas Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis! The smallest set of whole numbers of atoms. Molecular Formula - The formula of the compound as it exists, it may be a multiple of the Empirical formula. Steps to Determine Empirical Formulas Mass (g) of Element M (g/mol ) Moles of Element use no. of moles as subscripts Preliminary Formula change to integer subscripts Empirical Formula

5 Some Examples of Compounds with the same Elemental Ratio s Empirical Formula Molecular Formula CH 2 (unsaturated Hydrocarbons) C 2 H 4, C 3 H 6, C 4 H 8 OH or HO H 2 O 2 S S 8 P P 4 Cl Cl 2 CH 2 O (carbohydrates) C 6 H 12 O 6 Determining Empirical Formulas from Masses of Elements - I Problem: The elemental analysis of a sample compound gave the following results: 5.677g Na, g Cr, and g O. What is the empirical formula and name of the compound? Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound. Solution: Finding the moles of the elements: 1 mol Na Moles of Na = g Na x g Na = mol Na Moles of Cr = g Cr x 1 mol Cr g Cr = mol Cr 1 mol O Moles of O = g O x g O = mol O Determining Empirical Formulas from Masses of Elements - II Constructing the preliminary formula: Na Cr O Converting to integer subscripts (dividing all by smallest subscript): Na 1.99 Cr 1.00 O 4.02 Rounding off to whole numbers: Na 2 CrO 4 Sodium Chromate

6 Determining the Molecular Formula from Elemental Composition and Molar Mass - I Problem: The sugar burned for energy in cells of the body is Glucose (M = g/mol), elemental analysis shows that it contains mass % C, mass % H, and mass % O. (a) Determine the empirical formula of glucose. (b) Determine the Molecular formula. Plan: We are only given mass %, and no weight of the compound so we will assume 100g of the compound, and % becomes grams, and we can do as done previously with masses of the elements. Solution: Mass Carbon = 40.00% x 100g/100% = g C Mass Hydrogen = 6.719% x 100g/100% = 6.719g H Mass Oxygen = 53.27% x 100g/100% = g O g Cpd Determining the Molecular Formula from Elemental Composition and Molar Mass - II Converting from Grams of Elements to moles: Moles of C = Mass of C x Moles of H = Mass of H x Moles of O = Mass of O x 1 mole C g C 1 mol H g H 1 mol O g O = moles C = moles H = moles O Constructing the preliminary formula C 3.33 H 6.67 O 3.33 Converting to integer subscripts, divide all subscripts by the smallest: C 3.33/3.33 H / 3.33 O 3.33 / 3.33 = CH 2 O Determining the Molecular Formula from Elemental Composition and Molar Mass - III (b) Determining the Molecular Formula: The formula weight of the empirical formula is: 1 x C + 2 x H + 1 x O = 1 x x x = M of Glucose Whole-number multiple = = empirical formula mass = = 6.00 = Therefore the Molecular Formula is: C 1 x 6 H 2 x 6 O 1 x 6 = C 6 H 12 O 6

7 Adrenaline is a very Important Compound in the Body - I Analysis gives : C = 56.8 % H = 6.50 % O = 28.4 % N = 8.28 % Calculate the Empirical Formula! Adrenaline - II Assume 100g! C = 56.8 g C/(12.01 g C/ mol C) = 4.73 mol C H = 6.50 g H/( g H / mol H) = 6.45 mol H O = 28.4 g O/(16.00 g O/ mol O) = 1.78 mol O N = 8.28 g N/(14.01 g N/ mol N) = mol N Divide by = C = 8.00 mol C = 8.0 mol C or H = 10.9 mol H = 11.0 mol H O = 3.01 mol O = 3.0 mol O C 8 H 11 O 3 N N = 1.00 mol N = 1.0 mol N Ascorbic acid ( Vitamin C ) - I contains C, H, and O Upon combustion in excess oxygen, a 6.49 mg sample yielded 9.74 mg CO 2 and 2.64 mg H 2 O Calculate it s Empirical formula! C: 9.74 x10-3 g CO 2 x(12.01 g C/44.01 g CO 2 ) = g C H: 2.64 x10-3 g H 2 O x (2.016 g H 2 /18.02 gh 2 O) = g H Mass Oxygen = 6.49 mg mg mg = mg O

8 Vitamin C combustion - II C = 2.65 x 10-3 g C / ( g C / mol C ) = = mol C H = x 10-3 g H / ( g H / mol H ) = = mol H O = 3.54 x 10-3 g O / ( g O / mol O ) = = mol O Divide each by C = 1.00 Multiply each by 3 = 3.00 = 3.0 H = 1.32 = 3.96 = 4.0 O = 1.00 = 3.00 = 3.0 C 3 H 4 O 3 Determining a Chemical Formula from Combustion Analysis - I Problem: Erythrose (M = 120 g/mol) is an important chemical compound as a starting material in chemical synthesis, and contains Carbon Hydrogen, and Oxygen. Combustion analysis of a mg sample yielded g CO 2 and g H 2 O. Plan: We find the masses of Hydrogen and Carbon using the mass fractions of H in H 2 O, and C in CO 2. The mass of Carbon and Hydrogen are subtracted from the sample mass to get the mass of Oxygen. We then calculate moles, and construct the empirical formula, and from the given molar mass we can calculate the molecular formula. Determining a Chemical Formula from Combustion Analysis - II Calculating the mass fractions of the elements: mol C x M of C Mass fraction of C in CO 2 = = mass of 1 mol CO 2 = 1 mol C x g C/ 1 mol C = g C / 1 g CO g CO 2 Mass fraction of H in H 2 O = mol H x M of H = mass of 1 mol H 2 O 2 mol H x g H / 1 mol H = = g H / 1 g H 2 O g H 2 O Calculating masses of C and H Mass of Element = mass of compound x mass fraction of element

9 Determining a Chemical Formula from Combustion Analysis - III Mass (g) of C = g CO 2 x g C 1 g CO g H Mass (g) of H = g H 2 O x 1 g H 2 O = g C = g H Calculating the mass of O: Mass (g) of O = Sample mass -( mass of C + mass of H ) = g g C g H = g O Calculating moles of each element: C = g C / g C/ mol C = mol C H = g H / g H / mol H = mol H O = g O / g O / mol O = mol O C H O = CH 2 O formula weight = 30 g / formula 120 g /mol / 30 g / formula = 4 formula units / cpd = C 4 H 8 O 4 Atoms Molecular Formula Molecules Avogadro s Number Moles Moles x Chemical Equations Qualitative Information: Reactants Products States of Matter: (s) solid (l) liquid (g) gaseous (aq) aqueous 2 H 2 (g) + O 2 (g) 2 H 2 O (g)

10 Balanced Equations mass balance (atom balance)- same number of each element (1) start with simplest element (2) progress to other elements (3) make all whole numbers (4) re-check atom balance 1 CH 4 (g) + O 2 (g) 1 CO 2 (g) + H 2 O (g) 1 CH 4 (g) + O 2 (g) 1 CO 2 (g) + 2 H 2 O (g) 1 CH 4 (g) + 2 O 2 (g) 1 CO 2 (g) + 2 H 2 O (g) charge balance (no spectator ions) Ca 2+ (aq) + 2 OH - (aq) + Na + Ca(OH) 2 (s) + Na + Information Contained in a Balanced Equation Viewed in Reactants Products terms of: 2 C 2 H 6 (g) + 7 O 2 (g) = 4 CO 2 (g) + 6 H 2 O (g) + Energy Molecules 2 molecules of C 2 H molecules of O 2 = 4 molecules of CO molecules of H 2 O Amount (mol) 2 mol C 2 H mol O 2 = 4 mol CO mol H 2 O Mass (amu) amu C 2 H amu O 2 = amu CO amu H 2 O Mass (g) g C 2 H g O 2 = g CO g H 2 O Total Mass (g) g = g Balancing Chemical Equations - I Problem: The hydrocarbon hexane is a component of Gasoline that burns in an automobile engine to produce carbon dioxide and water as well as energy. Write the balanced chemical equation for the combustion of hexane (C 6 H 14 ). Plan: Write the skelton equation from the words into chemical compounds with blanks before each compound. begin the balance with the most complex compound first, and save oxygen untill last! Solution: C 6 H 14 (l) + O 2 (g) CO 2 (g) + H 2 O (g) + Energy Begin with one Hexane molecule which says that we will get 6 CO 2 s! 1 C 6 H 14 (l) + O 2 (g) 6 CO 2 (g) + H 2 O (g) + Energy

11 Balancing Chemical Equations - II The H atoms in the hexane will end up as H 2 O, and we have 14 H atoms, and since rach water molecule has two H atoms, we will get a total of 7 water molecules. 1 C 6 H 14 (l) + O 2 (g) 6 CO 2 (g) + 7 H 2 O (g) + Energy Since oxygen atoms only come as diatomic molecules (two O atoms, O 2 ),we must have even numbers of oxygen atoms on the product side. We do not since we have 7 water molecules! Therefore multiply the hexane by 2, giving a total of 12 CO 2 molecules, and 14 H 2 O molecules. 2 C 6 H 14 (l) + O 2 (g) 12 CO 2 (g) + 14 H 2 O (g) + Energy This now gives 12 O 2 from the carbon dioxide, and 14 O atoms from the water, which will be another 7 O 2 molecules for a total of 19 O 2! 2 C 6 H 14 (l) + 19 O 2 (g) 12 CO 2 (g) + 14 H 2 O (g) + Energy Chemical Equation Calc - I Atoms (Molecules) Avogadro s Number Reactants 6.02 x Molecules Products Chemical Equation Calc - II Atoms (Molecules) Avogadro s Number Reactants 6.02 x Molecules Moles Molecular Weight Mass g/mol Products

12 Sample Problem: Calculating Reactants and Products in a Chemical Reaction - I Problem: Given the following chemical reaction between Aluminum Sulfide and water, if we are given g of Al 2 S 3 : a) How many moles of water are required for the reaction? b) What mass of H 2 S & Al(OH) 3 would be formed? Al 2 S 3 (s) + 6 H 2 O (l) 2 Al(OH) 3 (s) + 3 H 2 S (g) Plan: Calculate moles of Aluminum Sulfide using it s molar mass, then from the equation, calculate the moles of Water, and then the moles of Hydrogen Sulfide, and finally the mass of Hydrogen Sulfide using it s molecular weight. Solution: a) molar mass of Aluminum Sulfide = g / mol moles Al 2 S 3 = g Al 2 S 3 = moles Al 2 S g Al 2 S 3 / mol Al 2 S 3 Calculating Reactants and Products in a Chemical Reaction - II a) cont moles Al 2 S 3 x 6 moles H 2 O = moles H 2 O 1 mole Al 2 S 3 b) moles Al 2 S 3 x 3 moles H 2 S = moles H 2 S 1 mole Al 2 S 3 molar mass of H 2 S = g / mol mass H 2 S = moles H 2 S x g H 2 S = g H 2 S 1 mole H 2 S moles Al 2 S 3 x 2 moles Al(OH) 3 = moles Al(OH) 3 1 mole Al 2 S 3 molar mass of Al(OH) 3 = g / mol mass Al(OH) 3 = moles Al(OH) 3 x g Al(OH) 3 = 1 mole Al(OH) 3 = g Al(OH) 3 Calculating the Amounts of Reactants and Products in a Reaction Sequence - I Problem: Calcium Phosphate could be prepared in the following reaction sequence: 4 P 4 (s) + 10 KClO 3 (s) 4 P 4 O 10 (s) + 10 KCl (s) P 4 O 10 (s) + 6 H 2 O (l) 4 H 3 PO 4 (aq) 2 H 3 PO 4 (aq) + 3 Ca(OH) 2 (aq) 6 H 2 O (aq) + Ca 3 (PO 4 ) 2 (s) Given: 15.5 g P 4 and sufficient KClO 3, H 2 O and Ca(OH) 2. What mass of Calcium Phosphate could be formed? Plan: (1) Calculate moles of P 4. (2) Use molar ratios to get moles of Ca 3 (PO 4 ) 2. (3) Convert the moles of product back into mass by using the molar mass of Calcium Phosphate.

13 Calculating the Amounts of Reactants and Products in a Reaction Sequence - II Solution: 1 mole P 4 moles of Phosphorous = g P 4 x g P 4 = mol P 4 For Reaction #1 [ 4 P 4 (s) + 10 KClO 4 (s) 4 P 4 O 10 (s) + 10 KCl (s) ] For Reaction #2 [ 1 P 4 O 10 (s) + 6 H 2 O (l) 4 H 3 PO 4 (aq) ] For Reaction #3 [ 2 H 3 PO Ca(OH) 2 1 Ca 3 (PO 4 ) H 2 O] moles P 4 x 4 moles P 4 O 10 x 4 moles H 3 PO 4 x 1 mole Ca 3 (PO 4 ) 2 4 moles P 4 1 mole P 4 O 10 2 moles H 3 PO 4 = moles Ca 3 (PO 4 ) 2 Calculating the Amounts of Reactants and Products in a Reaction Sequence - III Molar mass of Ca 3 (PO 4 ) 2 = g mole mass of product = moles Ca 3 (PO 4 ) 2 x g Ca 3 (PO 4 ) 2 1 mole Ca 3 (PO 4 ) 2 = = g Ca 3 (PO 4 ) 2 a A + b B + c C Steps to solve Limiting Reactant Problems d D + e E + f F 1) Identify it as a limiting Reactant problem - Information on the: mass, number of moles, number of molecules, volume and molarity of a solution is given for more than one reactant! 2) Calculate moles of each reactant! 3) Divide the moles of each reactant by the coefficient (a,b,c etc...)! 4) Which ever is smallest, that reactant is the limiting reactant! 5) Use the limiting reactant to calculate the moles of product desired then convert to the units needed (moles, mass, volume, number of atoms etc...)!

14 Limiting Reactant Problem: Sample Problem 3.11 ( p ) Problem: A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine (N 2 H 4 ) and dinitrogen tetraoxide (N 2 O 4 ). They ignite on contact ( hypergolic!) to form nitrogen gas and water vapor. How many grams of nitrogen gas form when exactly 1.00 x 10 2 g N 2 H 4 and 2.00 x 10 2 g N 2 O 4 are mixed? Plan: First write the balanced equation. Since amounts of both reactants are given, it is a limiting reactant problem. Calculate the moles of each reactant, and then divide by the equation coefficient to find which is limiting and use that one to calculate the moles of nitrogen gas, then calculate mass using the molecular weight of nitrogen gas. Solution: 2 N 2 H 4 (l) + N 2 O 4 (l) 3 N 2 (g) + 4 H 2 O (g) + Energy Sample Problem 3.11 cont. molar mass N 2 H 4 = ( 2 x x ) = g/mol molar mass N 2 O 4 = ( 2 x x ) = g/mol 1.00 x 10 Moles N 2 H 4 = 2 g = 3.12 moles N 2 H g/mol x 10 Moles N 2 O 4 = 2 g = 2.17 moles N 2 O g/mol 4 dividing by coefficients 3.12 mol / 2 = 1.56 mol N 2 H mol / 1 = 2.17 mol N 2 O 4 Limiting! 3 mol N Nitrogen yielded = 3.12 mol N 2 H 4 = 2 = 4.68 moles N 2 mol N 2 2 H 4 Mass of Nitrogen = 4.68 moles N 2 x g N 2 / mol = g N 2 Acid - Metal Limiting Reactant - I 2Al (s) + 6HCl (g) 2AlCl 3(s) + 3H 2(g) Given 30.0g Al and 20.0g HCl, how many moles of Aluminum Chloride will be formed? 30.0g Al / 26.98g Al/mol Al = 1.11 mol Al 1.11 mol Al / 2 = g HCl / 36.5gHCl/mol HCl = mol HCl O.548 mol HCl / 6 = HCl is smaller therefore the Limiting reactant!

15 Acid - Metal Limiting Reactant - II since 6 moles of HCl yield 2 moles of AlCl moles of HCl will yield: mol HCl / 6 mol HCl x 2 moles of AlCl 3 = mol of AlCl 3 Ostwald Process Limiting Reactant Problem What mass of NO could be formed by the reaction 30.0g of Ammonia gas and 40.0g of Oxygen gas? 4NH 3 (g) + 5 O 2 (g) 4NO (g) + 6 H 2 O (g) 30.0g NH 3 / 17.0g NH 3 /mol NH 3 = 1.76 mol NH mol NH 3 / 4 = 0.44 mol NH g O 2 / 32.0g O 2 /mol O 2 = 1.25 mol O mol O 2 / 5 = 0.25 mol O 2 Therefore Oxygen is the Limiting Reagent! 1.25 mol O 2 x 4 mol NO = 1.00 mol NO 5 mol O 2 mass NO = 1.00 mol NO x 30.0 g NO = g NO 1 mol NO Chemical Reactions in Practice: Theoretical, Actual, and Percent Yields Theoretical yield: The amount of product indicated by the stoichiometrically equivalent molar ratio in the balanced equation. Side Reactions: These form smaller amounts of different products that take away from the theoretical yield of the main product. Actual yield: The actual amount of product that is obtained. Percent yield: (%Yield) % Yield = x 100 Actual Yield Theoretical Yield

16 Percent Yield Problem: Problem: Given the chemical reaction between Iron and water to form the iron oxide, Fe 3 O 4 and Hydrogen gas given below. If 4.55g of Iron is reacted with sufficent water to react all of the Iron to form rust, what is the percent yield if only 6.02g of the oxide are formed? Plan: Calculate the theoretical yield and use it to calculate the percent yield, using the actual yield. Solution: 3 Fe (s) + 4 H 2 O (l) Fe 3 O 4 (s) + 4 H 2 (g) 4.55 g Fe g Fe mol Fe = mol = mol 1 mol Fe mol Fe x 3 O 4 = mol Fe 3 O 3 mol Fe g Fe mol Fe 3 O 4 x 3 O 4 = 6.30 g Fe 3 O 1 mol Fe 4 3 O 4 Percent Yield = Actual Yield x 100% = 6.02 g Fe 3 O 4 x 100% = Theoretical Yield 6.30 g Fe 3 O 4 % Percent Yield / Limiting Reactant Problem - I Problem: Ammonia is produced by the Haber Process using Nitrogen and Hydrogen Gas. If 85.90g of Nitrogen are reacted with g Hydrogen and the reaction yielded g of ammonia what was the percent yield of the reaction. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reagent then calculate the theoretical yield, and then the percent yield. Solution: Moles of Nitrogen and Hydrogen: moles N 2 = g N g N 2 1 mole N 2 = mol N g H moles H 2 = g H 2 = mol H 2 1 mole H 2 Divide by coefficient to get limiting: g N 2 = g H 2 3 = Percent Yield/Limiting Reactant Problem - II Solution Cont. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) We have moles of Nitrogen, and it is limiting, therefore the theoretical yield of ammonia is: 2 mol NH mol N 2 x 3 = mol NH 1 mol N 3(Theoretical 2 Yield) g NH mol NH 3 x 3 = g NH 1 mol NH 3 3 (Theoretical Yield) Actual Yield Percent Yield = x 100% Theoretical Yield g NH Percent Yield = 3 x 100% = % g NH 3

17 Molarity (Concentration of Solutions)= M M = Moles of Solute = Liters of Solution Moles L solute = material dissolved into the solvent In air, Nitrogen is the solvent and oxygen, carbon dioxide, etc. are the solutes. In sea water, Water is the solvent, and salt, magnesium chloride, etc. are the solutes. In brass, Copper is the solvent (90%), and Zinc is the solute(10%) Preparing a Solution - I Prepare a solution of Sodium Phosphate by dissolving 3.95g of Sodium Phosphate into water and diluting it to ml or l! What is the Molarity of the salt and each of the ions? Na 3 PO 4 (s) + H 2 O (solvent) = 3 Na + (aq) + PO 4-3 (aq) Preparing a Solution - II Mol wt of Na 3 PO 4 = g / mol 3.95 g / g/mol = mol Na 3 PO 4 dissolve and dilute to ml M = mol Na 3 PO 4 / l = M Na 3 PO 4 for PO 4-3 ions = M for Na + ions = 3 x M = M

18 Make a Solution of Potassium Permanganate Potassium Permanganate is KMnO 4 and has a molecular mass of g / mole Problem: Prepare a solution by dissolving 1.58 grams of KMnO 4 into sufficient water to make ml of solution. 1 mole KMnO 1.58 g KMnO 4 x 4 = moles KMnO g KMnO 4 4 Molarity = moles KMnO liters = M Molarity of K + ion = [K + ] ion = [MnO 4 - ] ion = M Dilution of Solutions Take ml of the M KMnO 4 Dilute the ml to l - What is the resulting Molarity of the diluted solution? # moles = Vol x M l x M = Moles Mol / 1.00 l = M Chemical Equation Calc - III Atoms (Molecules) Avogadro s Number Reactants 6.02 x Molecules Moles Molecular Weight Mass g/mol Products Molarity moles / liter Solutions

19 Calculating Mass of Solute from a Given Volume of Solution Volume (L) of Solution Molarity M = (mol solute / Liters of solution) = M/L Moles of Solute Molar Mass (M) = ( mass / mole) = g/mol Mass (g) of Solute Calculating Amounts of Reactants and Products for a Reaction in Solution Al(OH) 3 (s) + 3 HCl (aq) 3 H 2 O (l) + AlCl 3 (aq) Mass (g) of Al(OH) 3 Moles of Al(OH) 3 Moles of HCl M (g/mol) molar ratio Given 10.0 g Al(OH) 3, what volume of 1.50 M HCl is required to neutralize the base? 10.0 g Al(OH) 3 = mol Al(OH) g/mol mol Al(OH) 3 x 3 moles HCl = moles Al(OH) Moles HCl Vol HCl = 1.00 L HCl x Moles HCl M ( mol/l) 1.50 Moles HCl Volume (L) of HCl Vol HCl = L = 256 ml Solving Limiting Reactant Problems in Solution - Precipitation Problem - I Problem: Lead has been used as a glaze for pottery for years, and can be a problem if not fired properly in an oven, and is leachable from the pottery. Vinegar is used in leaching tests, followed by Lead precipitated as a sulfide. If ml of a M solution of Lead nitrate is added to ml of a M solution of Sodium sulfide, what mass of solid Lead Sulfide will be formed? Plan: It is a limiting-reactant problem because the amounts of two reactants are given. After writing the balanced equation, determine the limiting reactant, then calculate the moles of product. Convert moles of product to mass of the product using the molar mass. Solution: Writing the balanced equation: Pb(NO 3 ) 2 (aq) + Na 2 S (aq) 2 NaNO 3 (aq) + PbS (s)

20 Multiply by M (mol/l) Molar Ratio Volume (L) of Pb(NO 3 ) 2 solution Amount (mol) of Pb(NO 3 ) 2 Amount (mol) of PbS Volume (L) of Na 2 S solution Amount (mol) of Na 2 S Amount (mol) of PbS Multiply by M (mol/l) Molar Ratio Choose the lower number of PbS and multiply by M (g/mol) Mass (g) of PbS Multiply by M (mol/l) Volume (L) of Pb(NO 3 ) 2 solution Volume (L) of Na 2 S solution Multiply by M (mol/l) Devide by equation coeficient Amount (mol) of Pb(NO 3 ) 2 Smallest Molar Ratio Amount (mol) of PbS Amount (mol) of Na 2 S Devide by equation coeficient Mass (g) of PbS Solving Limiting Reactant Problems in Solution - Precipitation Problem - II Moles Pb(NO 3 ) 2 = V x M = L x ( Mol/L) = = Mol Pb +2 Moles Na 2 S = V x M = L x (0.095 Mol/L) = mol S -2 Therefore Lead Nitrate is the Limiting Reactant! Calculation of product yield: Moles PbS = Mol Pb +2 1 mol PbS x = Mol Pb 1 mol Pb(NO +2 3 ) Mol Pb +2 = Mol PbS Mol PbS x = g PbS g PbS 1 Mol PbS