Presentation on theme: "Stoichiometry Objectives"— Presentation transcript:
1 Stoichiometry Objectives Identify the quantitative relationships in a balanced chemical chemical equation.Determine the mole ratios from from a balanced chemical equation.Explain the sequence of steps used in solving stoichiometric problems.Use the steps to solve stoichiometric problems.Identify the limiting reactant in a chemical equation.Identify the excess reactant and calculate the amount remaining after the reaction is complete.Calculate the mass of a product when the amounts of more than one reactant are given.Calculate the theoretical yield of a chemical reaction from data.Determine the percent yield for a chemical reaction.
2 The Mole (Ch 11)Chemists need a convenient method for counting accurately the number of atoms, molecules, or formula units in a sample of a substance.-atoms and molecules are extremely small-even in the smallest sample it’s impossible to actually count each individual atom.To fix this problem chemists created their own counting unit called the mole.-mole: commonly abbreviated mol, is the SI base unit used to measure the amount of a substance
3 The MoleThrough experimentation, it has been established 1 mole = x 1023 representative particles.-representative particle: any particle such as atoms, molecules, formula units, electrons, or ions.-called Avogadro’s number in honor of the Italian physicist and lawyer Amedeo Avogadro who, in 1811, determined the volume of one mole of a gas.-we round Avogadro’s number to three significant figures— 6.02 x 1023.- If you write out Avogadro’s number, it looks like this:
5 Other Mole & Stoichiometry Vocabulary 1. What is stoichiometry?Study of quantitative relationships among amounts of reactants and products.What is a mole ratio?Ratio between the moles of any two substances in a balanced chemical equation3. What is molar mass?Mass, in grams, of one mole of pure substance.-numerically equal to its atomic mass-has the units g/mol
6 Converting Moles to Particles (11.1) Determine how many particles of sucrose are in 3.50mol of sucrose.- write a conversion factor using Avogadro’s number that relates representative particles to moles of a substance.
7 Converting Particles to Moles (11.1) Now, suppose you want to find out how many moles are represented by a certain number of representative particles, such as 4.50 x 1024 atoms of zinc.-You can use the inverse of Avogadro’s number as a conversion factor.
8 Mole Practice 1Identify and calculate the number of representative particles in each of the following quantities.moles of goldmole of nitrogen oxidemoles of potassium bromideCalculate the number of moles of the substance that contains the following number of representative particles.x 1023 atoms of bariumx 1023 molecules of carbon monoxidex 1023 formula units of potassium iodide
9 Practice-Homework p 311 # 1-3; p 312 # 4a-c Determine the number of atoms in 2.50 mol ZnDetermine the number of formula units in 3.25 mol AgNO3Determine the number of molecules in 11.5 mol H2ODetermine the number of moles ina x 1024 atoms Alb x 1024 molecules CO2c x 1023 formula units ZnCl2
10 Moles to Mass (11.2)To convert between moles and mass, you need to use the atomic mass found on the periodic table.Calculate the mass of moles of calcium.-According to the periodic table, the atomic mass of calcium is amu, so the molar mass of calcium is g/mol.
11 Mass to Moles (11.2)How many moles of copper are in a roll of copper that has a mass of 848g?
12 Practice-Homework p 316 # 11ab-12ab Determine the mass in grams of a mol Alb mol SiDetermine the number of moles ofa g Agb g S
13 Mass to Atoms (11.2)To find the number of atoms in a sample, you must first determine the number of moles.Calculate the number of atoms in 4.77 g lead.Determine moles
14 Mass to Atoms (cont.) 2. Determine atoms You can also convert from number of particles to mass by reversing the procedure above and dividing the number of particles by Avogadro’s number to determine the number of moles present.
15 Atoms to Mass Example problem 11-5, p 318 A party balloon has 5.50x1022 atoms of helium. What is the mass in grams of the helium?5.50x1022 atoms He x mol He = mol He6.02 x1023 atoms Hemol He x g He = g He1 mol He
16 Mole Practice 2 How many atoms are in the following samples? g cobaltg cesiumHow many grams are in the following samples?x1023 atoms of radiumx 1020 atoms of cadmium
17 Practice-Homworkp 316 # 11ab-12ab, p 318 # 13ab-14ab
18 Moles of Compounds (11.3)A mole of a compound contains as many moles of each element as are indicated by the subscripts in the formula.-For example, a mole of ammonia (NH3) consists of one mole of nitrogen atoms and three moles of hydrogen atoms.-the molar mass of the compound is found by adding the molar masses of all of the atoms in the representative particle.Molar mass of NH3 = 1(molar mass of N) + 3(molar mass of H)Molar mass of NH3 = 1( g) + 3(1.008 g) = g/mol
19 Practicep 322 # 25P 318 # 13ab-14abp32225. Determine the molar mass of each of the following:NaOH, CaCl2, Sr(NO3)2p318How many atoms are ina g Li b g PbWhat is the mass ofa x1023 atoms Bib x1024 atoms Mn
20 -Mole relationships from a formula (p 321) Determine the number of moles of aluminum ions in 1.25 moles of aluminum oxide (Al2O3).First we need the ratio of Al ions to Al2O3.2 mol Al ions1 mole Al2O31.25 mol Al2O3 x 2 mol Al ions = mol Al ions mole Al2O3
21 -Mole relationships from a formula (p 321) Determine the number of moles of chloride ions in 2.50 mol ZnCl2.Calculate the number of moles of each element in 1.25 mole glucose (C6H12O6).
22 -Mole to Mass for compounds ( p 323) What is the mass of 2.50 moles of allyl sulfide, (C3H5)2S?Calculate the mass of allyl sulfide.6(12.01 g/mol) = g/mol10(1.01 g/mol) = g/mol1(32.07 g/mol) = g/molg/mol
23 -Mole to Mass for compounds ( p 323) What is the mass of 2.50 moles of allyl sulfide, (C3H5)2S?2. Convert the moles to mass.2.50mol (C3H5)2S x g (C3H5)2S1 mol (C3H5)2S= 286g (C3H5)2S
24 Mass of Compound to Moles Calculate the number of moles of water that are in kg of water?1. Before you can calculate moles, you must determine the molar mass of water (H2O).molar mass H2O = 2(molar mass H) + molar mass O
25 2. Now you can use the molar mass of water as a conversion factor to determine moles of water. -Notice kg is converted to x 103 g
26 Practicep 323 # 27-28, p 324 # 30a&b27. What is the mass of 3.25 moles H2SO4?What is the mass of 4.35x10-2 moles of ZnCl2?Determine the number of moles present in each of the following:a g AgNO3b g ZnSO4
27 Mole Practice 3 Calculate the molar mass of the following: C2H5OH HCN What is the mass of the following:2.25 moles of KMnO41.56 moles of H2ODetermine the number of moles in the following:35.0 g HCl254 g PbCl4What is the mass in grams of one molecule of the following:7. H2SO4
28 Percent Composition, Molecular & Empirical Formulas Recall that every chemical compound has a definite composition—a composition that is always the same wherever that compound is found.The composition of a compound is usually stated as the percent by mass of each element in the compound, using the following process.
29 Percent CompositionExample: Determine the percent composition of calcium chloride (CaCl2).1. Determine mass of each ion in CaCl2.-1mol CaCl2 consists of 1mol Ca+2 ions and 2mol Cl- ions.1mol Ca+2 ions x 40.08g Ca+2 ions = 40.08g Ca+2 ions1mol Ca+2 ions2mol Cl- ions x 35.45g Cl- ions = 70.90g Cl- ions1mol Cl- ions
30 Percent CompositionExample: Determine the percent composition of calcium chloride (CaCl2).2. Calculate molar mass of CaCl2.g Ca+2 ions g Cl- ions = g CaCl21 mole CaCl mole CaCl23. Determine percent by mass of each element.
31 Percent CompositionExample: Determine the percent composition of calcium chloride (CaCl2).3. Determine percent by mass of each element.% Ca = g Ca+2 x 100 = % Ca+2g CaCl2% Cl = g Cl x 100 = % Cl-4. Make sure your percent compositions equal 100%.36.11% Ca % Cl- = 100%
32 Practicep 331 # 43, 45Calculate the percent composition of sodium sulfate (Na2SO4).45. What is the percent composition of phosphoric acid (H3PO4).
33 Empirical FormulaYou can use percent composition data to help identify an unknown compound by determining its empirical formula.-empirical formula-simplest whole-number ratio of atoms of elements in the compound.~In many cases, the empirical formula is the actualformula for the compound.the empirical formula of sodium chloride is Na1Cl1,or NaCl, which is the true formula~sometimes, the empirical formula is not the actualformula of the compound.the empirical formula for N2O4 (the actual) is NO2.
34 Empirical FormulaExample: The percent composition of an unknown compound is found to be 38.43% Mn, 16.80% C, and 44.77% O. Determine the compound’s empirical formula.- Because percent means “parts per hundred parts,” assume that you have 100 g of the compound.1. Calculate the number of moles of each element in the 100 g of compound.
35 Empirical FormulaExample: The percent composition of an unknown compound is found to be 38.43% Mn, 16.80% C, and 44.77% O. Determine the compound’s empirical formula.1. Calculate the number of moles of each element in the 100 g of compound.
36 Empirical Formula 0.6995 mol 0.6995 mol 0.6995 mol Example: The percent composition of an unknown compound is found to be 38.43% Mn, 16.80% C, and 44.77% O. Determine the compound’s empirical formula.2. The results show the following relationship3. Obtain the simplest whole-number ratio of moles:-divide each number of moles by the smallest number of moles.mol Mn : mol C : mol Omol mol mol: :
37 Empirical FormulaExample: The percent composition of an unknown compound is found to be 38.43% Mn, 16.80% C, and 44.77% O. Determine the compound’s empirical formula.3. Obtain the simplest whole-number ratio of moles:Mn : C : O: :4. Determine the empirical formula.MnC2O4
38 Practicep 333 # 46-47A blue solid is found to contain 36.84% N and 63.16% O. What is the empirical formula?Determine the empirical formula for a compound that contains 35.98% Al and 64.02% S.
39 Molecular FormulaFor many compounds, the empirical formula is not the true formula.-Chemists have learned, though, that acetic acid is a molecule with the formula C2H4O2, which is the molecular formula for acetic acid.-molecular formula tells the exact number of atoms of each element in a molecule or formula unit of a compound.Notice the molecular formula for acetic acid (C2H4O2) has exactly twice as many atoms of each element as the empirical formula (CH2O).-The molecular formula is always a whole-number multiple of the empirical formula.
40 -composition of maleic acid is 41.39% C, 3.47% H, and Molecular FormulaExample: Determine the molecular formula for maleic acid, which has a molar mass of 116.1g/mol.1. empirical formula of the compound-composition of maleic acid is 41.39% C, 3.47% H, and55.14% O (change the % to g)
41 Molecular FormulaExample: Determine the molecular formula for maleic acid.1. empirical formula of the compound-the ratio of C:H:O is 1:1:1, making the empirical formulaCHO2. calculate the molar mass of CHO (empirical formula).-29.01g/mol3. Determine the molecular formula for maleic acid,
42 Molecular FormulaExample: Determine the molecular formula for maleic acid.3. Determine the molecular formula for maleic acid,-shows the molar mass of maleic acid is 4x that of CHO.4. Multiply CHO by 4 to get C4H4O4
43 Practicep 335 # 51, 5351. A substance has a chemical composition of 65.45% C, 5.45% H and 29.09% O. The molar mass of the molecular formula is g/mol. Determine the molecular formula.53. A compound contains % N and % O. It has a molar mass of g/mol. What is the molecular formula?
44 Empirical Formula from Mass You can also calculate the empirical formula of a compound from mass of individual elements.Example: Determine the empirical formula for ilmenite, which contains 5.41g Fe, 4.64g Ti and 4.65g O.1. Multiply the mass by molar mass to get moles5.41g Fe x 1 mol Fe = mol Fe55.85 g Fe4.64g Ti x 1 mol Ti = mol Ti47.88g Ti4.65g O x 1 mol O = mol O16.00g O
45 Empirical Formula from Mass Example: Determine the empirical formula for ilmenite, which contains 5.41g Fe, 4.64g Ti and 4.65g O.2. Multiply by the smallest number to get the mole ratio.mol Fe : mol Ti : mol Omol mol mol: :3. Calculate the empirical formula.FeTiO3
48 Stoichiometry Practice – Conservation of Mass For the following balanced chemical equations, determine all possible mole ratios.HCl(aq) + KOH(aq) KCl(aq) + H2O(l)2. 2Mg(s) + O2(g) 2MgO(s)2HgO(s) 2Hg(l) + O2(g)
49 Stoichiometric Calculations Many times we need to determine a certain amount of product from a reaction or want to know how much product will form from a given amount of reactant.To do this you need:1. Balanced chemical equations2. Mole ratiosMoles of known x moles of unknown =moles of known
50 Stoichiometric Calculations: mole-mole Example: If you put mol of K into water, how much hydrogen gas will be produced?2K(s) + 2H2O(l) → 2KOH(aq) +H2(g)Mole ratio between K and KOH is 2 mol K or 1 mol H21 mol H mol Kmoles of known x moles of unknownmoles of knownmol K x 1 mol H2 =2 mol Kmol H2
51 Stoichiometric Calculations Practice How many moles of carbon dioxide are produced when 10.0 moles of propane (C3H8) are burned in excess oxygen in a gas grill. Water is also a product.2. Sulfuric acid is formed when sulfur dioxide reacts with oxygen and water. Write the balanced chemical equation for the reaction. If 12.5 mol SO2 reacts, how many moles H2SO4 can be produced? How many mole O2 is needed?
52 Practicep 357 #3a , p 359 #103a. Balance the following and determine the possible moleratios: ZnO(s) + HCl(aq) ZnCl2(aq) + H2O(l)10. CH4(g) + S8(s) CS2(l) + H2S(g)a. Balance the equation.b. Calculate moles of CS2 produced when 1.50 mol S8 isused.c. How many mol H2S is produced?
53 Stoichiometric Calculations: mole-mass Example: If you put mol of K into water, how many grams of hydrogen gas will be produced?2K(s) + 2H2O(l) → 2KOH(aq) +H2(g)Mole ratio between K and KOH is 2 mol K or 1 mol H21 mol H mol Kmol K x 1 mol H2 =2 mol Kmol H2mol H2 x 2.02 g H2 =1 mol H2g H2
54 Practicep 360 #11-12If you begin with 1.25 mol TiO2, what mass of Cl2 isneeded? TiO2 + C + Cl2 TiCl4 + CO212. Sodium chloride is decomposed into the elements sodium and chlorine by means of electrical energy. How many grams of chlorine gas are produced from 2.50 mol sodium chloride?
55 Stoichiometric Calculations: mass-mass Example: If you put 15g of K into water, how many grams of hydrogen gas will be produced?2K(s) + 2H2O(l) → 2KOH(aq) +H2(g)15 g K x 1 mol K = mol Kg KMole ratio between K and KOH is 2 mol K or 1 mol H21 mol H mol K0.384 mol K x 1 mol H2 =2 mol K0.192 mol H20.192 mol H2 x 2.02 g H2 =1 mol H20.388 g H2
56 Practicep 361 #13-14Determine the mass of N2 produced if 100.0g NaN3 isdecomposed. NaN3(s) Na(s) + N2(g)14. If 2.50 g sulfur dioxide reacts with excess oxygen and water, how many grams of sulfuric acid are produced?
57 Stoichiometric Review Why is a balanced chemical equation needed in solving stoichiometric calculations?When solving stoichiometric problems, how is the correct mole ratio expressed?List the steps in solving stoichiometric equations.How many grams of zinc (II) chloride can be obtained in 50.0 g of zinc reacts completely in the following reaction:Zn + 2HCl → ZnCl2 + H2How many grams of KCl is produced when mol of KClO3 decomposes in the following reaction:2KClO3 + heat → 2KCl + 3O2
60 - The left-over reactants are called excess reactants Limiting ReactantsRarely are the reactants in a chemical reaction present in the exact mole ratios specified in the balanced equation.-usually, one or more of the reactants are present in excess, and the reaction proceeds until all of one reactant is used up.-the reactant that is used up is called the limiting reactantThe limiting reactant limits the reaction and, thus, determines how much of the product forms.- The left-over reactants are called excess reactants
61 Limiting ReactantsIn the reaction below, 40.0 g of sodium hydroxide (NaOH) reacts with 60.0 g of sulfuric acid (H2SO4).a. Calculate the limiting reactantb. Determine the reactant in excessc. Calculate the mass of water producedd. Calculate the mass of reactant in excess.
62 Limiting ReactantsIn the reaction below, 40.0 g of sodium hydroxide (NaOH) reacts with 60.0 g of sulfuric acid (H2SO4).a. Calculate the limiting reactant1. calculate the actual number of moles of reactants
63 Limiting ReactantsIn the reaction below, 40.0 g of sodium hydroxide (NaOH) reacts with 60.0 g of sulfuric acid (H2SO4).a. determine the limiting reactant2. look at the actual ratio to the available ratio.~actual ratio = 1.00 mol NaOH0.612 mol H2SO4~available ratio = 2.00 mol NaOH = mol NaOH1.00 mol H2SO mol H2SO4You can see that when mol H2SO4 has reacted, all of the 1.00 mol of NaOH would be used up.-NaOH is the limiting reactant
64 Limiting ReactantsIn the reaction below, 40.0 g of sodium hydroxide (NaOH) reacts with 60.0 g of sulfuric acid (H2SO4).b. determine the reactant in excess-since NaOH is the limiting reactant, H2SO4 is the reactant in excessc. calculate the mass of sodium sulfate produced-use the limiting reactant1.00mol NaOH x 1.00mol Na2SO4 x 142g Na2SO4 = 71.0g Na2SO42.00mol NaOH mol Na2SO4
65 Limiting ReactantsIn the reaction below, 40.0 g of sodium hydroxide (NaOH) reacts with 60.0 g of sulfuric acid (H2SO4).d. calculate the mass of reactant in excess-use NaOH, the limiting reactant to determine moles and mass of H2SO4 used~1.00mol NaOH x 1.00mol H2SO4 x 98.09g H2SO = 49.05g H2SO42.00mol NaOH mol H2SO4-subtract the mass needed from the mass available.60.0g H2SO4 – 49.05g H2SO4 = 11.0g H2SO4 in excess
67 Percent YieldMost reactions never succeed in producing the predicted amount of product.-not every reaction goes cleanly or completely●liquids may stick to surfaces of containers●liquids may vaporize/evaporate●solids may be left behind on filter paper●solids may be lost in the purification process●sometimes unintended products formThe amount you have been calculating so far has been the theoretical yield, the maximum amount of product that can be produced from a given amount of reactant.
68 Percent YieldA chemical reaction rarely produces the theoretical yield.-actual yield is the amount of product produced in the chemical reactionWe can measure the efficiency of the reaction by calculating the percent yield.-percent yield (% yield) of the product is the ratio of actual yield to the theoretical yield, expressed as a percent.% yield = actual yield (from the experiment) x 100theoretical yield (from calculations)
69 Percent YieldWhen potassium chromate is added to a solution containing g of silver nitrate, solid silver chromate is formed.a. Determine the theoretical yield of silver chromateb. If g of silver chromate is actually obtained,calculate the percent yield.