Noteworthy Book sections not on the exam: 2.8,2.9,3.9,3.10 but these will be covered later. Problems in the text worth reading( all of the problems in the text are worth your time to read, the following are just more worthy): –2.92,2.93,2.104,2.110,2.111 –In Chapter 3, it would be a good idea to read all problems 3.58 to 3.77 –3.120,3.121,3.130 Tutoring available: –Chemistry:Room 221 SB1 most days 9-4 –Campus Tutor Program:

From Tuesday Chemical formulas –mass % composition –empirical (simplest) vs molecular formula you can always determine the empirical formula from the analysis the molecular formula requires knowledge of the gmw –Remember, since the composition of the compound is intrinsic, any complete set of mass data will yield the formula. The chemical equation –descriptive –balanced Stoichiometry –all mass/mole relationships produced by the balanced equation are direct –complete treatment from the mass/moles of one species determine the quantities in terms of mass and moles of all the others –focused from one get one

Use of general algorithms All problems involving the comparison of “conditions(states)” for a pair of directly related variables fit the following equation: The solution of a problem “simply” involves the correct “insertion” of the data into the equation

For the following chemical reaction, provide a complete treatment for the reaction of 9.34g of NH 3 NH 3 + CO 2 => C 3 H 6 + NO 3 1. Balance the equation, if needed 2NH 3 + 3CO 2 => C 3 H 6 + 2NO 3 2. It is useful to have a method of organizing the needed info. One method is to write the gfws above all of the species in the equation and the reaction masses (stoichiometric masses) which are the products of the coefficients and the gfws underneath the species NH 3 + 3CO 2 => C 3 H 6 + 2NO The stoichiometric masses must obey the mass conservation law. This will verify the validity of the data. For a given problem, you generally will not need all of this. `34+132=166= Be certain to convert the “given” into moles 9.34/17.0=.549moles 3. The solution can then be presented in a tabular fashion MolesGrams CO 2 : (3/2)*0.549=0.823 *44 = 36.2 C 3 H 6 :(1/2)*0.549=0.274*42 = 11.5 NO 3 :(2/2)*0.549=0.549*62 = 34.0 The final masses must follow mass conservation = =45.5

Focused Questions Most questions involving balanced chemical equations are binary and have the form: How many _____ of _____ would be _______ (by the reaction of ____ ____ of ____) or (if ____ _____ of ____ are _____)? The above should look very familiar. P H 2 O => 4H 3 PO H 2 How many moles of H 3 PO 4 would be produced by the reaction of 94.0g of H 2 O?

Reality 101:% yield and the limiting reagent We are seldom able to perform a reaction in the precise fashion shown by the balanced chemical equation. Many reactions stop short of completion and the quantitative isolation of products is often difficult. It is impossible and often not desirable to mix reactants in perfect stoichiometric ratios.

% yield % yield is defined as: –(actual yield/theoretical yield)*100=%yield –another 3 variable equation, though the theoretical yield is actually determined from basic stoichiometry –If the reaction of 12.3 of C 2 H 4 produces 35.6g of CO 2, what is the % yield? C 2 H 4 + 3O 2 => 2CO 2 + 2H 2 O

Limiting Reagent If the reagents are not present in exact stoichiometric ratios, then one will limit how far the reaction proceeds. There are a number of algorithms for identifying the limiting reagent. What you are doing is comparing the actual ratio with the stoichiometric one. C 2 H 4 + 3O 2 => 2CO 2 + 2H 2 O If 23.4g of C 2 H 4 are reacted with 74g of O 2, what is the limiting reagent?

Method 1-mole based The question is really, how does a mixture of 23.4g to 74g compare with a mixture of 1 mole to 3 moles. –convert to like dimensions: –23.4/28=0.84moles –74/32=2.31moles –2.31/0.84 vs 3/1 ? –If you’re unsure, divide 2.31 by 0.84(2.75) or multiply 0.84 by 3(2.52) –What’s the limiting reagent? Method 2-mass based How many grams of O 2 will react with 23.4g of C 2 H –Using standard approaches –(96/28)*23.4=80.2g of O 2 –the reaction mixture has 74g. Which is limiting?

Putting it all together The reaction of 21.4g of PH 3 with 17.2g of water yields 34.2g of P 2 O 5. What is the % yield. PH 3 + H 2 O => P 2 O 5 + H 2 What are the steps ( and in what sequence) needed to solve this question. 1.Balance the equation 2. Identify the limiting reagent 3. Use the limiting reagent to calculate the theoretical yield of P 2 O 5 4. Calculate the % yield It is strongly recommended that you do this problem. The answer will be posted next week.

Solution Basics What’s a solution? –any homogeneous mixture; solutions are not limited to the liquid phase. A true solution does not scatter light. Solvent and solute are commonly used terms with no real rigid definitions. It’s common to call the component present in greatest amount the solvent and all other species solutes. There’s this implicit “I dissolved x in y” at work. When ionic compounds dissolve, it is common that the ions dissociate. This is the “stoichiometry of dissolution” and can result in the ions having different concentrations. –NaCl(s) =>Na + (aq) + Cl - (aq) –Ca(NO 3 ) 2 => Ca 2+ (aq) + 2NO 3 - (aq)

Concentrations There are numerous systems for describing the composition of a solution and a number of these will be discussed as appropriate. Since the current focus is on chemical reactions with the mole as the key measure, a system which readily describes the number of moles of solute (presumed to be the chemically active component) present in a given quantity of solution is needed. Molarity: M=(moles solute)/(L of solution) Quite astonishingly, it’s a three variable equation. “Problems” in this arena break down into three obvious categories –What is the molarity of a solution prepared…. –How many _____ of solute are present in a…. –What volume of solution would contain… There is a critical issue here. Remember that your solute is most likely going to be weighed out, as mass is the most precise measure available. Thus, use of moles=mass/gfw is commonly needed in conjunction with molarity computations. An improperly prepared solution is likely to doom an experiment. You should assume that the name of the solute, rather than its formula is going to be provided.

Dilution It is common practice to have available a set of stock solutions from which solutions of lesser concentrations are prepared. Further, one often adds a solution of known concentration into a larger volume as part of a procedure. This process is known as dilution. Consider two solutions, with the second having been prepared by a quantitative dilution. That means that the initial M and V are known and the final V is known. In the process of adding a solution into pure solvent (or adding solvent to the solution, what must be conserved?) How are M and V related? What general equation fits the above? M 1 *V 1 =M 2 *V 2

Recalling general algorithms for inversely related pairs: There is the usual standard set of problems in this area. If ____ (L/mL) of ______M solution are diluted to a final volume of ______(L/mL) what is the resultant M? What volume in (L/mL) of _____M solution should be dissolved to a final volume of ______(L/mL) if you want a _____M solution? _____(L/mL) of a _____M solution are diluted to a new concentration of ______M. What is the new volume? _____(L/mL) of a solution are diluted to a final volume of _____(L/mL) and the concentration is found to be _____M. What was the initial concentration?