Chapter 3. Formulas, Equations and Moles. Part II

Dealing with chemical stoichiometry Not including volumetric stoichiometry Mass (g) of substance A Mass (g) of substance B MM (g/mol) of substance MM (g/mol) of substance Moles of atoms/ions in a substance Chemical Formula Moles of substance A 6.022x1023 Mol/mol ratio from balanced equation Moles of substance B Chemical Formula Moles of atoms/ions in a substance Chapter 3 Formulas, Equations, and Moles: II 6.022x1023 6.022x1023 6.022x1023 Chemical Chemical Number of atoms/ions in a substance Steward Fall 08 Formula Formula Molecules / Atoms/FU of substance A Molecules / Atoms/FU of substance B Number of atoms/ions in a substance Amounts of Reactants and Products How can we predict the mass of products knowing the mass of a starting reactant? We cannot go directly between grams of reactant to grams of product. We have to use moles and the balanced chemical equation. Mass to mass conversions – excess reagent If we react 1.2482 g MgCl2 with excess AgNO3, how many grams of AgCl can be produced? Always write out and balance the chemical equation first. MgCl2(aq) + 2 AgNO3(aq)
2 AgCl(s) + Mg(NO3)2(aq) 1.2482g excess ????? don’t care Steps: •Mass MgCl2 to moles MgCl2 •Moles of MgCl2 to moles of AgCl •Moles of AgCl to grams AgCl Mass to mass conversions – excess reagent Mass to mass conversions If we have 14.69 grams of HCl, how much NaOH would we need to completely react with HCl (with none left over)? Steps: •Mass MgCl2 to moles MgCl2 •Moles of MgCl2 to moles of AgCl •Moles of AgCl to grams AgCl  1.2482g MgCl 2  1 mol MgCl 2  2 mol AgCl  143.32g AgCl    95.211g MgCl  1 mol MgCl  1 mol AgCl     2  2  HCl(aq) + NaOH(aq)
H2O(l) + NaCl(aq) 14.69g ????? don’t care Steps: •Mass HCl to moles HCl •Moles of HCl to moles of NaOH = 3.7578 g AgCl •Moles of NaOH to grams NaOH 6 Grams to Grams Notice that it doesn’t matter if you are going from reactants to products or reactants to other reactants, you still use the same steps Steps: •Mass HCl to moles HCl •Moles of HCl to moles of NaOH •Moles of NaOH to grams NaOH  14.69 g HCl  1 mol HCl  1 mol NaOH  39.9971g NaOH         36.4606g HCl  1 mol HCl  1 mol NaOH  = 16.11g NaOH Limiting Reagents Limiting Reagents What are the total number of bikes I can make? What limits the number I can make? 2H2 + O2
2H2O What is the limiting reactant? 10 Limiting Reagents Limiting Reagents 2H2 + O2
2H2O What is the limiting reactant? 2H2 + O2
2H2O What is the limiting reactant? 11 12 Limiting Reagent Calculations If I mix together 1.0g of each reactant (CaCl2 and Na2CO3), how much CaCO3 will I produce? To determine the limiting reagent, calculate the amount of products for each reactant. Two calculations are needed. CaCl2 + Na2CO3
CaCO3 + 2NaCl 1.00g 2H2 + O2
2H2O 1.00g ????? don’t care You must do 2 calculations – one for each reactant Remember…mass to mass calculations Two calculations  4 mol H 2  2 mol H 2O     = 4 mol H 2 O   2 mol H 2   1.00g CaCl2  1 mol CaCl 2  1 mol CaCO3  100.087g CaCO3    110.99g CaCl  1 mol CaCl  1 mol CaCO  = 0.902g CaCO3   2  2  3  Limiting Reagent  1.00g Na 2 CO 3  1 mol Na 2 CO 3  1 mol CaCO3  100.087g CaCO3     = 0.944g CaCO3     105.99g Na 2 CO 3  1 mol Na 2 CO3  1 mol CaCO3  (Produces the smaller amount)  3 mol O 2  2 mol H 2O   = 6 mol H 2O     1 mol O 2  The answer is the smaller of the two amounts. 13 Percent Yield Now that we know the limiting reagent, we can also calculate how much excess reagent is left over limiting excess CaCl2 + Na2CO3
CaCO3 + 2NaCl 1.00g 1.00g 0.902g still don’t care Do the same mass to mass calculations you did before. How much actually reacts? The amounts calculated using stoichiometric calculations are called theoretical yields. This is the amount of products that would be formed if the reaction is 100% complete. In the real world, most reactions do not go to 100% completion. To calculate the percent yield, we use a simple formula:  1.00 g CaCl 2  1 mol CaCl 2  1 mol Na 2CO 3  105.99g Na 2CO 3     = 0.955g Na 2CO 3     110.99g CaCl 2  1 mol CaCl 2  1 mol Na 2 CO 3  Actual Yield (from experiment) Since the reaction only used 0.955g of the 1.00g we added… Percent yield = Yields of Chemical Reactions X 100 Theoretical yield 1.00 g Na 2 CO 3 − 0.955g Na 2 CO 3 = 0.045g = 0.05 Na 2CO 3 (from calculations) Concentration of Solutions: Molarity Concentration: amount of solute present in a given amount of solvent or solution So, if we perform a reaction between MgCl2 and AgNO3 (like we calculated earlier) and get 2.12 grams of AgCl...what would be our percent yield? NaCl Percent yield = 3.123 g 3.7578 g = 83.11% X 100 Molarity M = Moles of solute Liters of solution Na+ ClNa+ Notice that in molarity (M), the liters are for the entire solution, not just the solvent. Cl- ClNa+ 18 Dealing with chemical stoichiometry Volume (L) of substance A MM (g/mol) of substance Moles of atoms/ions in a substance Volume (L) of substance B Con. (mol/L) of substance cA Mass (g) of substance A Chemical Formula Moles of substance A 6.022x1023 • What is the concentration of a 1 L solution made by dissolving 5.0 g NaCl in water? Mass (g) of substance B Conc. (mol/L) of substance B Mol/mol ratio from balanced equation • 500 mL solution? • How much NaCl (in g) do we need to make 1 L of a 0.125 M solution? MM (g/mol) of substance Moles of substance B Chemical Formula Moles of atoms/ions in a substance 6.022x1023 6.022x1023 Number of atoms/ions in a substance Concentration of Solutions Steward Fall 08 Molecules / Atoms of substance A 6.022x1023 Molecules / Atoms of substance B Number of atoms/ions in a substance Concentration of Solutions – Mad Libs Concentration of Solutions - Mad Libs • 1) Water-soluble ionic compound • How many grams of (1) _______ are required to prepare a (2) _____ mL solution whose concentration is (3) _____? • 2) Number greater than 100 • 3) Number less than 15 • What is the concentration of a solution made by adding (4) ______ grams of (5) ______ to _______ mL of water? • 4) Number between 50 and 100 • 5) Any water-soluble compound • 6) Number greater than 100 Dilution Dilution • Dilution: making a less concentrated solution (by adding water); number of moles of solute stays the same, amount of solvent increases • How would you prepare 500.0 mL of a 0.500 M solution of H2SO4 starting with concentrated (18 M) solution? – Remember to add acid to water. • The number of moles of solute stays the same but the ratio changes (lower molarity) – moles = M * V – Mconc * Vconc = Mdil * Vdil – M1V1 = M2V2 • How much water do you need to add to 25.0 mL of a 4.50 M solution of NaOH to make a 1.00 M NaOH solution? • Example: What is the final concentration of a 100.0 mL solution of 3.00 M HCl if the final volume is 400.0 mL? 23 Solution Stoichiometry Example Problem Solution: • Consider the reaction: – 3CaCl2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaCl(aq) 3CaCl2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaCl(aq) • 0.200 M x 0.0250 L = 0.00500 mol CaCl2 • If we mix 25.0 mL of 0.200 M CaCl2 solution with 50.0 mL of 0.250 M Na3PO4 solution, what mass of Ca3(PO4)2(s) is formed? • Notice this is a limiting reagent problem. You need to be able to work problems of this type. Be sure to practice these! • 0.250 M x 0.0500 L = 0.0125 mol Na3PO4 • Limiting Reactant: – 0.00500 mol CaCl2
0.00167 mol Ca3(PO4)2 – 0.0125 mol Na3PO4
0.00625 mol Na3PO4 • 0.00500 mol CaCl2 x (1 mol/3 mol) x (312 g / = 0.520 g Ca3(PO4)2 1 mol) 25 Titrations • Titration: method of determining the concentration of an unknown solution by using a solution with a known concentration (standard) – moles = M * V • Equivalence point: point at which acid has completely reacted with/been neutralized by base (moles acid = moles base) • End point: point at which the indicator changes color (slight change in solution’s pH) Group Work • How many mL of 1.018 M H2SO4 are needed to neutralize 20.00 mL of 0.9989 M NaOH? 26 Titrations • Solution stoichiometry: use molarity to convert between volume and moles (instead of molar mass) • 23.78 mL of 0.2500 M NaOH neutralized 20.00 mL of HCl. What is the concentration of HCl? Percent Composition of Compounds Percent by mass: percent by mass of each element in a compound • H2O2 – 2 moles H, 2 moles O – H = 2 mol x 1.008 g/mol H = 2.016 g H – O = 2 mol x 16.00 g/mol O = 32.00 g O – H2O2 = 2.016 g H + 32.00 g O = 34.02 g – %H = 2.016 g H / 34.02 g H2O2 x 100% = 5.926% • What is %O? Solve 2 ways….. Empirical Formulas Percent Composition of Compounds • What is the percent by mass of Cr, S, and O in Cr2(SO4)3? • Calculate the empirical formulas: – 50% S, 50% O – 43.64% P, 56.36% O • Assume 100 g of the material: – 50 g O x 1 mol / 16.0 g = 3.125 mol O – 50 g S x 1 mol / 32.066 g = 1.559 mol S – mol O / mol S = 3.125 / 1.559 = 2.004 or 2 – SO2 • Assume 100 g of material: – 43.64 g P
mol P (1.4089 mol P) – 56.36 g O
mol O (3.5225 mol O) – Ratio of two moles (2.5 O / 1 P); need whole numbers 32 Molecular Formulas Practice Problem • We just found the empirical formula to be P2O5. This compound has a molecular mass of 141.9 g/mol. • A compound contains 30.4 % nitrogen and 69.6 % oxygen. The molecular mass of the compound is 92 g/mol. – What is the empirical formula of the compound? – What is the molecular formula of the compound? • We know the molecular formula has a molar mass of 283.8 g/mol. How many multiples of P2O5 do we need to reach a mass of 283.8? 34 Answers to Practice Problem • Assume a sample of 100 g, which then contains 30.4 g N and 69.6 g O. • 30.4 g N x (1 mol / 14 g) = 2.17 mol N • 69.6 g O x (1 mol / 16.0 g) = 4.35 mol O • ratio = 4.35 mol O / 2.17 mol N = 2.00 • Empirical formula = NO2 • Empirical mass = 46 g/mol • Molecular mass / empirical mass = 92 / 46 = 2 • Molecular formula = (NO2)2 = N2O4 35