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CHM 101- Chapter 4,5,6

by: SunDevil_21

CHM 101- Chapter 4,5,6 CHM 101

Marketplace > Arizona State University > Chemistry > CHM 101 > CHM 101 Chapter 4 5 6
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Notes for chapters 4-6. CHM 101 with Richard Bauer at Arizona State University
CHM 101
Richard Bauer
Class Notes
Chemistry, Science




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This 4 page Class Notes was uploaded by SunDevil_21 on Tuesday May 10, 2016. The Class Notes belongs to CHM 101 at Arizona State University taught by Richard Bauer in Fall 2014. Since its upload, it has received 17 views. For similar materials see CHM 101 in Chemistry at Arizona State University.

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Date Created: 05/10/16
CHM 101 Exam 2  Understand concepts associated with laboratory investigations and recitation activities.  Understand material from exam 1 as it applies to exam 2 content. Chapter 4 23   Define the mole­ Contains 6.022x10 atoms, molecules, ions, or formula units this number is called  Avogadro’s number  ***Explain the physical significance of Avogadro's number­ is the same as the number of atoms in  exactly 12 g of the isotope carbon­12 or 12C.   Calculate the molar mass for a compound from the molar masses of the component elements.   Example: What is the molar mass of SO ?  2 Mass of 1 mol of S=1molx32.06g/mol = 32.06g Mass of 2 mol of O=2 molx16.00g/mol= 32.00g Mass of 1 mol of SO =232.06g+32.00g= 64.06 g/mol  Convert between moles, mass, and number of atoms, molecules, ions, or formula units.  Pages 132­135   Distinguish between empirical and molecular formulas.  Empirical is the shortened version of the molecular it expresses the simplest ratios of atoms in a  compound and is written with the smallest possible whole­number subscript. So if the molecular formula is C5H10then the empirical formula is CH  2nother example is P O 4a6 the molecular and the empirical  formula would be P O2.3  ***   Determine molecular formulas from appropriate data.  Example: If the empirical formula is the same as the molecular formula, the molar mass of HCO  sho2ld be 90.0 g/mol. We calculate the molar mass of the empirical formula by summing the weighed molar  masses: Mass of 1 mol of H= 1 mol x 1.008 g/mol= 1.008 g Mass of 1 mol of C=1 mol x 12.01 g/mol= 12.01 g Mass of 2 mol of O=2 mol x 16.00 g/mol= 32.00 g Mass of 1 mol of HCO = 12008+12.01+32.00= 45.02 g So then we take the Experimental MM/Empirical formula MM 90.0 / 45.02 = 2.00 so then the subscripts must all be multiplied by 2 so the molecular formula is H 2 O2 4  Determine the percent composition of elements in a compound  Example: Calculate the percent of copper in Chalcocite Cu S 2 For chalcocite 1 mol of compound contains 2 mol x 63.55 g/mol or 127.1 of copper. The molar  mass of the element is 159.2 g/mol %Cu= 127.1/159.2x100 SO 79.84% Copper  Distinguish between solute and solvent in a solution  Solute­ the substance that is dissolved (usually present in a lesser amount)  Solvent­ the substance doing the dissolving (usually present in a greater amount)  Calculate the concentrations of solutions in various units (percent by mass and molarity).  % Mass = (mass solute/mass solution) x 100%  Molarity = (moles of solute/liters of solution) A solution is prepared from 17.0 g of NaCl dissolved in sufficient water to give 150.0 mL of  solution. What is the molarity of the solution? (The molar mass of NaCl is 58.44 g/mol) The moles of sodium chloride can be calculated from its mass using the molar mass, as we have  done previously:  Mol NaCl= 17.0 g NaCl x (1 mol NaCl/58.44 g NaCl) = 0.291 mol NaCl Convert the volume to units of liters: Volume (L) = 150.0 mL x (1L/1000mL) = .1500 L Divide the number of moles by the volume in liters to get molarity: Molarity = (0.291 mol NaCl/0.1500 L solution) =1.94 M  Calculate quantities associated with dilutions.  Dilution is to lower the concentration of a solution. Or adding more solvent to a solution  If 85.2 mL of 2.25 M CuCl  2olution is diluted to a final volume of 250.0 mL what is the molarity of  the diluted solution of CuC2 ? The numbers of moles of copper (II) chloride are the same before and after dilution, so we can use the  equation for dilution: M dil(M conc concV dil (2.25 M x .0852 L)/.2500 L = .767 M mma REVIEW PAGE 149­151 MAY BE ON TEST Chapter 5  Write and balance chemical equations  Classify chemical reactions.  Draw pictures at an atomic/molecular level that depict the nature of various  classes of chemical reactions.  Decomposition: CD  C + D  Combination: A + B  AB  Single­Displacement: A + CD  C + AD  Double­Displacement: CD + EF  CF + ED Predicting Chemical Reactions is 178­179 pg. 189  Predict the products of reactions from information such as the periodic table, an activity series, or the  solubility rules. Chapter 6  Use quantitative relationships in chemical reactions including mass/mole and mole/mole.  Mole­Mole  Use mole ratios, which are obtained from the coefficients in the balanced equation.  If 1.14 mol of CO  2as formed by the combustion of C3H9, how many moles of H2O were also  formed?  C 3 8g) + 5O (2)  3CO (g)2+ 4H O(g)2      mol H O2= 1.14 mol CO x  42    H O = 1.22 mol H O 2           3 mol C2  Identify the "limiting reactant" in a reaction and calculate the amount of product present when the  reaction is complete.  Limiting Reactant: the reactant that reacts completely. It limits the amount of the other reactant that  can ract, and it limits the amount of product that can fom   If the reactants mixed together are not present in the exact mole ration, then there is too much of one  reactant, and not enough of another. On reactant reacts completely, while some of the other is left  unreacted.  N 2g)  +  3H 2g)    2NH (3)  Find the limiting reactant when   2.0 mol N  2nd 5.0 mole H 2  How many moles of NH  can be produced in the case? 3  ((2.0 mol N 2(2 mol NH ))/31 mol N = 42  mol NH 3  Hydrogen is the limiting reactant  Use quantitative relationships in chemical reactions including mass/mole and mole/mole for limiting  reactant and percent yield problems.  Percent Yield = actual yield     x100                        theoretical yield    2Na(s)  +  Cl2(g)   2NaCl(s)  If 0.450 mol chlorine reacts with excess sodium, and 0.385 mol NaCl are produced, what is the percent  yield for the reaction?  ((.450 mol Cl)(2 mol NaCl)) / 1 mol Cl = .90 mol NaCl  %Yield = (.385 mol NaCl / .90 mol NaCl ) x 100 = 43%  Understand energy changes as they apply to chemical reactions.  Distinguish between exothermic and  endothermic reactions.  Exothermic: energy released to surroundings  Endothermic: energy obtained  Use quantitative relationships related to energy including specific heat and calorimetry.  1 cal = 4.184 J  1 Cal – 100 cal = 1 kcal  1 serving of cereal has 190 Calories  What is the energy content in units of Joules?    190 Cal | 1000 Cal | 4.184 J      =     794,960 Joules              |  1 Cal       | 1 cal


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