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Topic 3: Stoichiometry

AP Chemistry Dickson County High School Mrs. Laura Peck. Topic 3: Stoichiometry. 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.

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Topic 3: Stoichiometry

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  1. AP Chemistry Dickson County High School Mrs. Laura Peck Topic 3: Stoichiometry

  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!!! What you should know for the test:

  3. 1 mol of a monatomic element = 6.02x1023 atoms 1 mol of a molecular compound or diatomic element = 6.02x1023 molecules 1 mol of an ionic compound = 6.02x1023 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? The Mole 3.01x1024 formula units 6.02x1024 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 Al2O3*2H2O. What is the mass in grams of 2.10x1024 formula units (f.u.) of Al2O3*2H2O? 2.10x1024 f.u. x 1mol Al2O3*2H2O x 138g = 481g 6.02x1023 f.u. 1mol Al2O3*2H2O #3) The principle chemical in TNT is NH4NO3. What is the mass in grams of 3.20x1010 f.u. of NH4NO3? Molar Mass Molar mass 3.20x1010 f.u. x 1 mol NH4NO3 x 2(14.007)+4(1.007)+3(15.999) = 4.25x10-12g 6.02x1023 f.u. 1 mol NH4NO3

  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: M1V1 = M2V2 #4) (No calc) Prepare 2.00L of 0.250M NaOH from 1.00M NaOH Molarity (M) 1.00M(V1) = (0.250M)(2.00L) V1 = 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(NO3)2 Percent composition (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. 1st % composition of the compound is determined from combustion data. 2nd 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 CO2 and 4.37mg of H2O. What is the percent composition of the compound? Determination of empirical formula by combustion analysis Remainder must be oxygen. 100% - (40.91%C + 4.58%H) = 54.51% O Solution: Assume that all the carbon in the compound is converted to CO2 and Determine the mass of carbon present in the 10.68mg sample Mass % CO2 is 4.369mgC x 100% 10.68 mg = 40.91% C 0.01601g CO2 x 12.01g C = 4.369mg C 44.01gCO2 0.0437g H2O x 2.016gH = 0.489 H 18.02g H2O Mass % H is 0.489mg x 100% 10.68 mg = 4.58% H

  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? 1st convert each to moles 2nd divide by smallest number Now find empirical formula 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 3.406H4.54O3.407 = (C1.00H1.33O1.00)3 = C3H4O3 3.4063.4063.406 **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. Conversions involved in stoichiometric calculations.

  10. Solution stoichiometry involves calculations for reactions that occur in aqueous solution. Example: The compound with the empirical formula, C3H4O3, is a monoprotic acid. A 2.20g sample of the unknown monoprotic acid is dissolved in 1.0L H2O. 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? 1st find the molar mass of the acid. Solution stoichiometry 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) (C3H4O3)2 = C6H8O6 Solution: to find molar mass, let HA = unknown acid. Write a balanced molecular Equation: HA + NaOH  NaA + H2O 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

  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(NO3)3 is added to 200.0mL of 0.100M KOH? Limiting reagent 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 Step 1: Always begin with a balanced molecular equation: Al(No3)3(aq) + 3KOH(aq) Al(OH)3(s) + 3KNO3(aq) Step 2: For each reactant, determine the amount of product produced (in mol or g) 0.0500L x 0.200mol Al(NO3)3x 1mol Al(OH)3 = 0.0100mol Al(OH)3 1 L 1 mol Al(NO3)3 0.200L x 0.100mol KOH x 1mol Al(OH)3= 0.00667 mol Al(OH)3 1L 3mol KOH

  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 Percent Yield Actual yield = 85.3% x 0.520g = 0.444 g Al(OH)3 100%

  13. The end…..

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