Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations PDF

Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations Stoichiometry  Law of Conservation of Mass “We may lay it down as an incontestable axiom ‫ ))ﺑدﯾﮭﯾﺔ ﻏﯾر ﻗﺎﺑﻠﺔ ﻟﻠﺟدل‬that, in all the operation...

Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations Stoichiometry  Law of Conservation of Mass “We may lay it down as an incontestable axiom ‫ ))ﺑدﯾﮭﯾﺔ ﻏﯾر ﻗﺎﺑﻠﺔ ﻟﻠﺟدل‬that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.” --Antoine Lavoisier, 1789 ‫أﻧطوان ﻻﻓوازﯾﯾﮫ‬ Stoichiometry 3.1: Chemical Equations Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Reactants appear on the Products appear on the right left side of the equation. side of the equation Stoichiometry  Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) The states of the reactants and products are written in parentheses to the right of each compound. Stoichiometry  Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Coefficients are inserted to balance the equation. Stoichiometry  Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in a molecule Coefficients tell the number of molecules (compounds). Stoichiometry 3.2: Some simple patterns of chemical reactivity Reaction Types Stoichiometry  1. Combination Reactions Two or more substances react to form one product Examples: N2 (g) + 3 H2 (g) ⎯⎯→ 2 NH3 (g) C3H6 (g) + Br2 (l) ⎯⎯→ C3H6Br2 (l) 2 Mg (s) + O2 (g) ⎯⎯→ 2 MgO (s) Stoichiometry  2. Decomposition Reactions One substance breaks down into two or more substances Examples: CaCO3 (s) ⎯⎯→ CaO (s) + CO2 (g) 2 KClO3 (s) ⎯⎯→ 2 KCl (s) + O2 (g) 2 NaN3 (s) ⎯⎯→ 2 Na (s) + 3 N2 (g) Stoichiometry  3. Combustion Reactions Rapid reactions that have oxygen as a reactant sometimes produce a flame Most often involve hydrocarbons reacting with oxygen in the air to produce CO2 and H2O. Examples: CH4 (g) + 2 O2 (g) ⎯⎯→ CO2 (g) + 2 H2O (g) C3H8 (g) + 5 O2 (g) ⎯⎯→ 3 CO2 (g) + 4 H2O (g) 2H2 + O2 ------- 2H2O Stoichiometry 3.3: Formula Weights Stoichiometry  The amu unit Defined (since 1961) as: 1/12 mass of the 12C isotope. 12C = 12 amu Stoichiometry  Formula Weight (FW) Sum of the atomic weights for the atoms in a chemical formula So, the formula weight of calcium chloride, CaCl2, would be Ca: 1(40.1 amu) + Cl: 2(35.5 amu) 111.1 amu These are generally reported for ionic compounds Stoichiometry  Molecular Weight (MW) Sum of the atomic weights of the atoms in a molecule For the molecule ethane, C2H6, the molecular weight would be C: 2(12.0 amu) + H: 6(1.0 amu) 30.0 amu Stoichiometry  Percent Composition One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation: (number of atoms)(atomic weight) % element = x 100 (FW of the compound) Stoichiometry  Percent Composition So the percentage of carbon and hydrogen in ethane (C2H6, molecular mass = 30.0) is: (2)(12.0 amu) 24.0 amu %C = = x 100 = 80.0% (30.0 amu) 30.0 amu (6)(1.01 amu) 6.06 amu %H = = x 100 = 20.0% (30.0 amu) 30.0 amu Stoichiometry 3.4 : Avogadro’s number and the mole Moles Stoichiometry  Atomic mass unit and the mole amu definition: 1-atom 12C = 12 amu. The atomic mass unit is defined this way. 1 amu = 1.6605 x 10-24 g How many 12C atoms weigh 12 g? 6.022x1023 12C atoms weigh 12 g. Avogadro’s number The mole #atoms in 12 g of 12C = (1 atom/12 amu)(1 amu/1.66x10-24 g)(12g) = 6.022x1023 12C weigh 12 g Stoichiometry  Therefore: Any Stoichiometry  The mole The mole is just a number of things 1 dozen = 12 things 1 pair = 2 things 1 mole = 6.022141x1023 things Stoichiometry  Molar Mass By definition, this is the mass of 1 mol of a substance (i.e., g/mol) – The molar mass of an element is the mass number for the element that we find on the periodic table – The formula weight (in amu’s) will be the same number as the molar mass (in g/mol) Stoichiometry  Using Moles Moles provide a bridge from the molecular scale to the real-world scale The number of moles correspond to the number of molecules. 1 mole of any substance has the same number of molecules. Stoichiometry  Mole Relationships One mole of atoms, ions, or molecules contains Avogadro’s number of those particles One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound Stoichiometry 3.5: Finding Empirical Formulas Stoichiometry  Combustion Analysis gives % composition C nH nO n + O 2 nCO2 + 1/2nH2O Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this – %C is determined from the mass of CO2 produced – %H is determined from the mass of H2O produced – %O is determined by difference after the C and H have been determined Stoichiometry Calculating Empirical Formulas One can calculate the empirical formula from the percent composition Stoichiometry Calculating Empirical Formulas Example The compound para-aminobenzoic acid (you may have seen it listed as PABA on bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA. Stoichiometry  Calculating Empirical Formulas Solution Assuming 100.00 g of para-aminobenzoic acid, 1 mol C: 61.31 g x = 5.105 molg C 12.01 1 mol H: 5.14 g x = 5.091.01 molgH 1 mol N: 10.21 g x = 0.7288 14.01mol g N 1 mol O: 23.33 g x = 1.456 16.00mol gO Stoichiometry  Calculating Empirical Formulas…… Calculate the mole ratio by dividing by the smallest number of moles: 5.105 mol C: = 7.005 ≈ 7 0.7288 mol 5.09 mol H: = 6.984 ≈ 7 0.7288 mol 0.7288 mol N: = 1.000 0.7288 mol 1.458 mol O: = 2.001 ≈ 2 0.7288 mol Stoichiometry  Calculating Empirical Formulas These are the subscripts for the empirical formula: C7H7NO2 Stoichiometry Elemental Analyses Compounds containing other elements are analyzed using methods analogous to those used for C, H and O Stoichiometry  Molecular Formulas from Empirical Formulas We can obtain a molecular formula for any compound from its empirical formula if we know either the molecular weight or the molar mass of the compound. The subscripts in the molecular formula of a substance are always whole-number multiples of the subscripts in its empirical formula. This multiple can be found by dividing the molecular weight by the empirical formula weight Example Mestylene, a hydrocarbon found in crude oil, has an empirical formula C3H4 and an experimentally determined molecular weight of 121 amu. What is its molecular formula? Answer The formula weight of the empirical formula is 3(12.0 amu) + 4(1.0 amu) = 40.0 amu Whole-number multiple = molecular weight / empirical formula weight = 121 / 40.0 = 3.02 ~ 3 Molecular formula : C9H12 Stoichiometry 3.6: Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products Stoichiometry  Stoichiometric Calculations From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant) Stoichiometry  Stoichiometric Calculations Example: 10 grams of glucose (C6H12O6) react in a combustion reaction. How many grams of each product are produced? C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) 10.g ? + ? Starting with 10. g of C6H12O6… we calculate the moles of C6H12O6… use the coefficients to find the moles of H2O & CO2 and then turn the moles to grams Stoichiometry  Stoichiometric calculations C6H12O6 + 6O2 → 6CO2 + 6H2O 10.g ? + ? MW: 180g/mol 44 g/mol 18g/mol #mol: 10.g(1mol/180g) 0.055 mol 6(.055) 6(.055mol) 6(.055mol)44g/mol 6(.055mol)18g/mol #grams: 15g 5.9 g Stoichiometry 3.7: Limiting Reactants Stoichiometry How Many Cookies Can I Make? You can make cookies until you run out of one of the ingredients Once you run out of sugar, you will stop making cookies Stoichiometry  How Many Cookies Can I Make? In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can makeStoichiometry  Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount 2H2 + O2 --------> 2H2O #moles 14 7 10 5 10 Stoichiometry Left: 0 2 10  Limiting Reactants In the example below, the O2 would be the excess reagent Stoichiometry  Limiting reagent, example: Soda fizz comes from sodium bicarbonate and citric acid (H3C6H5O7) reacting to make carbon dioxide, sodium citrate (Na3C6H5O7) and water. If 1.0 g of sodium bicarbonate and 1.0g citric acid are reacted, which is limiting? How much carbon dioxide is produced? 3NaHCO3(aq) + H3C6H5O7(aq) ------> 3CO2(g) + 3H2O(l) + Na3C6H5O7(aq) 1.0g 1.0g 84g/mol 192g/mol 44g/mol 1.0g(1mol/84g) 1.0(1mol/192g) 0.012 mol 0.0052 mol (if citrate limiting) 0.0052(3)=0.016 0.0052 mol So bicarbonate limiting: 0.012 mol 0.012(1/3)=.0040mol 0.012 moles CO2 44g/mol(0.012mol)=0.53g CO 2.0052-.0040=.0012mol left Stoichiometry 0.0012 mol(192 g/mol)= 0.023 g left.  Theoretical Yield The theoretical yield is the amount of product that can be made – In other words it’s the amount of product possible from stoichiometry. The “perfect reaction.” This is different from the actual yield, the amount one actually produces and measures Stoichiometry  Percent Yield A comparison of the amount actually obtained to the amount it was possible to make Actual Yield Percent Yield = x 100 Theoretical Yield Stoichiometry  Example Benzene (C6H6) reacts with Bromine to produce bromobenzene (C6H6Br) and hydrobromic acid. If 30. g of benzene reacts with 65 g of bromine and produces 56.7 g of bromobenzene, what is the percent yield of the reaction? C6H6 + Br2 ------> C6H5Br + HBr 30.g 65 g 56.7 g 78g/mol 160.g/mol 157g/mol 30.g(1mol/78g) 65g(1mol/160g) 0.38 mol 0.41 mol (If Br2 limiting) 0.41 mol 0.41 mol (If C6H6 limiting) 0.38 mol 0.38 mol 0.38mol(157g/1mol) = 60.g Stoichiometry 56.7g/60.g(100)=94.5%=95%  Example, one more React 1.5 g of NH3 with 2.75 g of O2. How much NO and H2O is produced? What is left? 4NH3 + 5O2 --------> 4NO + 6H2O 1.5g 2.75g ? ? 17g/mol 32g/mol 30.g/mol 18g/mol 1.5g(1mol/17g)= 2.75g(1mol/32g)=.088mol.086 (If NH3 limiting):.088mol.088(5/4)=.11 O2 limiting:.086(4/5)=.086 mol.086 mol(4/5)=.086(6/5)=.069mol.069 mol.10mol.069mol(17g/mol).069mol(30.g/mol).10mol(18g/mol) 1.2g 2.75g 2.1 g 1.8g Stoichiometry