Steward Fall 08. Moles of atoms/ions in a substance. Number of atoms/ions in a substance. MgCl 2(aq) + 2 AgNO 3(aq) 2 AgCl (s) + Mg(NO 3 ) 2(aq)

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Dealing with chemical stoichiometry Steward Fall 08 of Not including volumetric stoichiometry of Chapter 6.0x10 A 6.0x10 Mol/mol ratio from balanced equation B 6.0x10 6.

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.

1 Dealing with chemical stoichiometry Steward Fall 08 of Not including volumetric stoichiometry of Chapter 6.0x10 A 6.0x10 Mol/mol ratio from balanced equation B 6.0x10 6.0x10 s, Equations, and Moles: II Atoms/FU of Atoms/FU of 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 1.48 g MgCl with excess AgNO, how many grams of AgCl can be produced? Always write out and balance the chemical equation first. MgCl (aq) + AgNO (aq) AgCl (s) + Mg(NO ) (aq) 1.48g excess????? don t care Mass MgCl to moles MgCl MgCl to moles of AgCl AgCl to grams AgCl Mass to mass conversions excess reagent Mass MgCl to moles MgCl MgCl to moles of AgCl AgCl to grams AgCl Mass to mass conversions If we have grams of HCl, how much NaOH would we need to completely react with HCl (with none left over)? HCl (aq) + NaOH (aq) H O (l) + NaCl (aq) 14.69g????? don t care 1.48g MgCl 1mol MgCl 95.11g MgCl mol AgCl 1mol MgCl.7578 g AgCl 14.g AgCl 1mol AgCl Mass HCl to moles HCl HCl to moles of NaOH NaOH to grams NaOH 6

2 Grams to Grams Mass HCl to moles HCl HCl to moles of NaOH NaOH to grams NaOH 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 g HCl 1mol HCl g HCl 1mol NaOH g NaOH 1mol HCl 1mol NaOH 16.11g NaOH What are the total number of bikes I can make? What limits the number I can make? H + O H O What is the limiting reactant? 10 H + O H O What is the limiting reactant? H + O H O What is the limiting reactant? 11 1

Limiting Reagent Calculations To determine the limiting reagent, calculate the amount of products for each reactant. Two calculations are needed. H + O H O If I mix together 1.

3 Limiting Reagent Calculations To determine the limiting reagent, calculate the amount of products for each reactant. Two calculations are needed. H + O H O If I mix together 1.0g of each reactant (CaCl and Na CO ), how much CaCO will I produce? CaCl + Na CO CaCO + NaCl 1.00g 1.00g????? don t care You must do calculations one for each reactant Remember mass to mass calculations Two calculations 4 mol H mol H O mol H mol O mol HO 1mol O 4 mol H O 6 mol H O Limiting Reagent (Produces the smaller amount) 1.00g CaCl 1mol CaCl 1mol CaCO g CaCO 0.90g CaCO g CaCl 1mol CaCl 1 mol CaCO 1.00g Na CO 1 mol Na CO 1mol CaCO g CaCO 0.944g CaCO g Na CO 1mol Na CO 1 mol CaCO The answer is the smaller of the two amounts. 1 Percent Yield Now that we know the limiting reagent, we can also calculate how much excess reagent is left over limiting How much actually reacts? 1.00g CaCl 1mol CaCl excess CaCl + Na CO CaCO + NaCl 1.00g 1.00g 0.90g still don t care 1mol Na CO g Na CO Do the same mass to mass calculations you did before g Na CO g CaCl 1mol CaCl 1mol Na CO Since the reaction only used 0.955g of the 1.00g we added 1.00 g Na CO 0.955g Na CO 0.045g 0.05 Na CO 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: Percent yield Actual Yield (from experiment) Theoretical yield (from calculations) X 100 Yields of Reactions Concentration of Solutions: Molarity So, if we perform a reaction between MgCl and AgNO (like we calculated earlier) and get.1 grams of AgCl...what would be our percent yield? Percent yield.1 g.7578 g 8.11% X 100 Molarity M Moles Concentration: amount of solute present in a given amount of solvent or solution of solute Liters of solution NaCl Na + Cl - Notice that in molarity (M), the liters are for the entire solution, not just the solvent. Na + Cl - Cl - Na + 18

Dealing with chemical stoichiometry of 6.0x10 Volume (L) of A 6.0x10 Con. (mol/l) of ca Atoms of Mol/mol ratio from balanced equation Conc. (mol/l) of Volume (L) of B 6.0x10 Atoms of of 6.

4 Dealing with chemical stoichiometry of 6.0x10 Volume (L) of A 6.0x10 Con. (mol/l) of ca Atoms of Mol/mol ratio from balanced equation Conc. (mol/l) of Volume (L) of B 6.0x10 Atoms of of 6.0x10 Steward Fall 08 Concentration of Solutions What is the concentration of a 1 L solution made by dissolving 5.0 g NaCl in water? 500 ml solution? How much NaCl (in g) do we need to make 1 L of a 0.15 M solution? Concentration of Solutions Mad Libs 1) Water-soluble ionic compound ) Number greater than 100 ) Number less than 15 4) Number between 50 and 100 5) Any water-soluble compound 6) Number greater than 100 Concentration of Solutions - Mad Libs How many grams of (1) are required to prepare a () ml solution whose concentration is ()? What is the concentration of a solution made by adding (4) grams of (5) to ml of water? Dilution Dilution: making a less concentrated solution (by adding water); number of moles of solute stays the same, amount of solvent increases The number of moles of solute stays the same but the ratio changes (lower molarity) moles M * V Dilution How would you prepare ml of a M solution of H SO 4 starting with concentrated (18 M) solution? Remember to add acid to water. How much water do you need to add to 5.0 ml of a 4.50 M solution of NaOH to make a 1.00 M NaOH solution? M conc * V conc M dil * V dil M 1 V 1 M V Example: What is the final concentration of a ml solution of.00 M HCl if the final volume is ml?

5 Solution Stoichiometry Example Consider the reaction: CaCl (aq) + Na PO 4 (aq) Ca (PO 4 ) (s) + 6NaCl(aq) Problem Solution: CaCl (aq) + Na PO 4 (aq) Ca (PO 4 ) (s) + 6NaCl(aq) 0.00 M x L mol CaCl If we mix 5.0 ml of 0.00 M CaCl solution with 50.0 ml of 0.50 M Na PO 4 solution, what mass of Ca (PO 4 ) (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! 0.50 M x L mol Na PO 4 Limiting Reactant: mol CaCl mol Ca (PO 4 ) mol Na PO mol Na PO mol CaCl x (1 mol/ mol) x (1 g / 1 mol) 0.50 g Ca (PO 4 ) 5 6 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) Titrations Solution stoichiometry: use molarity to convert between volume and moles (instead of molar mass).78 ml of M NaOH neutralized 0.00 ml of HCl. What is the concentration of HCl? Group Work How many ml of M H SO 4 are needed to neutralize 0.00 ml of M NaOH? Percent Composition of Compounds Percent by mass: percent by mass of each element in a compound H O moles H, moles O H mol x g/mol H.016 g H O mol x g/mol O.00 g O H O.016 g H +.00 g O 4.0 g %H.016 g H / 4.0 g H O x 100% 5.96% What is %O? Solve ways..

6 Percent Composition of Compounds What is the percent by mass of Cr, S, and O in Cr (SO 4 )? Empirical s Calculate the empirical formulas: 50% S, 50% O 4.64% P, 56.6% O Assume 100 g of the material: 50 g O x 1 mol / 16.0 g.15 mol O 50 g S x 1 mol /.066 g mol S mol O / mol S.15 / or SO Assume 100 g of material: 4.64 g P mol P ( mol P) 56.6 g O mol O (.55 mol O) Ratio of two moles (.5 O / 1 P); need whole numbers Molecular s We just found the empirical formula to be P O 5. This compound has a molecular mass of g/mol. We know the molecular formula has a molar mass of 8.8 g/mol. How many multiples of P O 5 do we need to reach a mass of 8.8? Practice Problem A compound contains 0.4 % nitrogen and 69.6 % oxygen. The molecular mass of the compound is 9 g/mol. What is the empirical formula of the compound? What is the molecular formula of the compound? 4 Answers to Practice Problem Assume a sample of 100 g, which then contains 0.4 g N and 69.6 g O. 0.4 g N x (1 mol / 14 g).17 mol N 69.6 g O x (1 mol / 16.0 g) 4.5 mol O ratio 4.5 mol O /.17 mol N.00 Empirical formula NO Empirical mass 46 g/mol Molecular mass / empirical mass 9 / 46 Molecular formula (NO ) N O 4 5

Dilutions 4/8/2013. Steps involved in preparing solutions from pure solids. Steps involved in preparing solutions from pure solids

Dilutions 4/8/2013. Steps involved in preparing solutions from pure solids. Steps involved in preparing solutions from pure solids Steps involved in preparing solutions from pure solids Steps involved in preparing solutions from pure solids Calculate the amount of solid required Weigh out the solid Place in an appropriate volumetric