1. AP Chemistry Dickson County High School Mrs. Laura Peck 1

2.  Moles, mass, representative particles (atoms, molecules, formula units), molar mass and Avogadro’s number.  Molarity; preparation of solutions.  The percent composition of an element in a compound.  Balanced chemical equations; for example, for a given mass of reactant, calculate the amount of product produced.  Limiting reactants: calculate the amount of product formed when given the amounts of all of the reactants present  Reactions in solution: given the molarity and the volume of the reactants, calculate the amount of product produced or the amount of reactant required to react.  The percent yield of a reaction.  **Stoichiometric calculations of various types make up 30-40% of AP Exam!!! 2

3.  1 mol of a monatomic element = 6.02x10 23 atoms  1 mol of a molecular compound or diatomic element = 6.02x10 23 molecules  1 mol of an ionic compound = 6.02x10 23 formula units  Now lets have some fun! (no calc)  #1) If you have 5 mols of sodium chloride, how many formula units do you have?  #2) If you have that same 5 mols of sodium chloride, how many actual ATOMS do you have? 3 3.01x10 24 formula units 6.02x10 24 individual atoms

4.  Molar mass is the mass of 1 mole of an element or compound.  Example: The principle ore in the production of aluminum cans has a molecular formula of Al 2 O 3 *2H 2 O. What is the mass in grams of 2.10x10 24 formula units (f.u.) of Al 2 O 3 *2H 2 O?  2.10x10 24 f.u. x 1mol Al 2 O 3 *2H 2 O x 138g = 481g 6.02x10 23 f.u. 1mol Al 2 O 3 *2H 2 O #3) The principle chemical in TNT is NH 4 NO 3. What is the mass in grams of 3.20x10 10 f.u. of NH 4 NO 3 ? 4 Molar mass 3.20x10 10 f.u. x 1 mol NH 4 NO 3 x 2(14.007)+4(1.007)+3(15.999) = 4.25x10 -12 g 6.02x10 23 f.u. 1 mol NH 4 NO 3

5.  The number of moles of solute in 1 liter of solution is a measure of its molarity or solution concentration.  Example: Prepare 2.oL of 0.250M NaOH from solid NaOH  2.00L x 0.250mol NaOH x 40.00g NaOH = 20.0g NaOH 1 L 1 mol NaOH “Place 20.o.gNaOH in a 2L volumetric flask; add water to dissolve the NaOH, and fill to the mark with water, mixing several times along the way.  You can also use a ratio for some problems: M 1 V 1 = M 2 V 2  #4) (No calc) Prepare 2.00L of 0.250M NaOH from 1.00M NaOH 5 1.00M(V 1 ) = (0.250M)(2.00L) V 1 = 0.500L “Add 500mL of 1.00M NaOH stock solution to a 2L volumetric flask; add Distilled/deionized water in several increments, mixing until the flask is Filled to the mark on the neck of the flask.”

6.  The mass percentages of elements in a compound can be obtained by comparing the mass of each element present in 1 mol of the compound to the total mass of 1 mol of compound.  mass of element in 1 mol of compound x 100% mass of 1 mol of compound  #5)(no calc) Calculate the percent of oxygen in NaOH  #6)(no calc) Calculate the percent of oxygen in Mg(NO 3 ) 2 6 ( 16 x 1)/(16+1+23) x 100% = 40% O (16x6)/(6(16)+2(14)+24) x 100% = 65% O

7.  The empirical formula is the simplest whole number ratio of atoms in a compound.  (two step problem. 1 st % composition of the compound is determined from combustion data. 2 nd the empirical formula of the compound is determined from the % composition)  Example: A compound contains only carbon, hydrogen and oxygen. Combustion of 10.68mg of the compound yields 16.01mg CO 2 and 4.37mg of H 2 O. What is the percent composition of the compound? 7 Solution: Assume that all the carbon in the compound is converted to CO 2 and Determine the mass of carbon present in the 10.68mg sample 0.01601g CO 2 x 12.01g C = 4.369mg C 44.01gCO 2 Mass % CO2 is 4.369mgC x 100% 10.68 mg = 40.91% C 0.0437g H 2 O x 2.016gH = 0.489 H 18.02g H 2 O Mass % H is 0.489mg x 100% 10.68 mg = 4.58% H Remainder must be oxygen. 100% - (40.91%C + 4.58%H) = 54.51% O

8.  We discovered that our unknown consisted of 40.91%C, 4.58%H and 54.51% O. So now how do we find the empirical formula from this?  1 st convert each to moles  2 nd divide by smallest number 8 40.91gC x 1mol C = 3.406 mol C 12.01gC 4.58gH x 1mol H = 4.54mol H 1.008gH 54.51gO x 1 mol O = 3.407mol O 16.00gO C 3.406 H 4.54 O 3.407 = (C 1.00 H 1.33 O 1.00 )3 = C 3 H 4 O 3 3.406 3.406 3.406 **since a whole number ratio is required, we multiply all subscripts By the lowest factor (3) to obtain the whole number. Rounding can Only occur when the subscripts are within 0.1 of the nearest whole Number.

9. Type of conversionExample G  mol (using molar mass)18g/1 mol H2O Mol  mol (using mol ratio = coefficients from balanced chemical equation) 1 mol O2/2 mol H2O Mol  L (using molarity which is a solution’s concentration equal to the number of mols of solute in 1 liter of solution) 6 mol HCl/1L 9

10.  Solution stoichiometry involves calculations for reactions that occur in aqueous solution.  Example: The compound with the empirical formula, C 3 H 4 O 3, is a monoprotic acid. A 2.20g sample of the unknown monoprotic acid is dissolved in 1.0L H 2 O. A titration required 25.0mL of 0.500M NaOH to react completely with all the acid present. What is the molecular formula of the acid? 1 st find the molar mass of the acid. 10 Solution: to find molar mass, let HA = unknown acid. Write a balanced molecular Equation: HA + NaOH  NaA + H 2 O Find the moles of HA present from the volume and molarity of NaOH: 0.250L NaOH x 0.500mol NaOH x 1mol HA = 0.0125mol HA 1 L NaOH To find the molar mass of HA, divide mass(g) of HA which is given by mol of HA 2.20gHA / 0.0125 mol HA = 176 g/mol To find the molecular formula of the compound if the empirical formula and molar Mass are known, use this equation: molar mass/empirical mass = number Then multiply number to each subscript: Determine empirical formula mass: 3(12) + 4(1) + 3(16) = 88 176/88 = 2.0 (number) (C 3 H 4 O 3 )2 = C 6 H 8 O 6

11.  In a limiting reagent problem, the amounts of all reactants are given and the amount of product is to be determined. The limiting reactant is completely consumed when the reaction goes to completion. It determines how much product is formed.  Example: what mass of precipitate can be produced when 50.0mL of 0.200M Al(NO 3 ) 3 is added to 200.0mL of 0.100M KOH? 11 Step 1: Always begin with a balanced molecular equation: Al(No 3 ) 3(aq) + 3KOH (aq)  Al(OH) 3(s) + 3KNO 3(aq) Step 2: For each reactant, determine the amount of product produced (in mol or g) 0.0500L x 0.200mol Al(NO 3 ) 3 x 1mol Al(OH) 3 = 0.0100mol Al(OH) 3 1 L 1 mol Al(NO 3 ) 3 0.200L x 0.100mol KOH x 1mol Al(OH) 3 = 0.00667 mol Al(OH) 3 1L 3mol KOH The limiting reagent produces the least amount of product. The Limiting reagent in this case is KOH. It produces only 0.00667mol Of product which is less than 0.0100mol. The mass of precipitate produced is: 0.00667mol Al(OH) 3 x 78.0g Al(OH) 3 = 0.520 g AL(OH) 3 1 mol Al(OH) 3

12.  The percent yield of a reaction is the actual yield of a product as a percentage of the theoretical yield. The actual yield is the amount of product obtained in an experiment. The theoretical yield is the amount of product calculated from the amounts of reactants used.  Example: If the reaction from the last problem has an 85.3% yield of precipitate, how much aluminum hydroxide is produced? (remember, our experiment produced 0.520 g Al(OH) 3 )  Percent yield = actual yield x 100% theoretical yield 12 Actual yield = 85.3% x 0.520g = 0.444 g Al(OH) 3 100%

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