Chem Lecture 2 and 3 Notes PDF

All rights reserved. Dr. Gulacar, Department of Chemistry, University of California, Davis. Reproducing these notes or posting them on any online source is strongly prohibited. CONTENTS Atomic Masses WHY NOT ALWAYS AN INTEGER? H: 1.007 U...

All rights reserved. Dr. Gulacar, Department of Chemistry, University of California, Davis. Reproducing these notes or posting them on any online source is strongly prohibited. CONTENTS Atomic Masses WHY NOT ALWAYS AN INTEGER? H: 1.007 U NA: 22.99U B:10.81U 1  ATOMIC MASS- NOT ALWAYS AN INTEGER How do we find atomic masses of elements?  Atomic masses are compared to a standard mass, that of an atom of C-12  Atomic mass or atomic weight is measured in amu (u)  1 atom of Carbon-12 = 12 u  For example; the atomic mass” of the element hydrogen is 0.083992 times that of a C-12 atom.  ……………………………………., the atomic mass of hydrogen WHY IS ATOMIC MASS NOT ALWAYS AN INTEGER? ANOTHER REASON? Atoms of the same element with different masses are called isotopes For example: there are 3 isotopes of hydrogen Chemically, isotopes have virtually identical properties LEARNING CHECK Naturally occurring chlorine is a mixture of two isotopes. In every sample of this element, 75.77% of the atoms are chlorine-35 and 24.23% are chlorine-37. The measured mass of chlorine-35 is 34.9689u and that of chlorine-37 is 36.9659u. Calculate the average atomic mass of Chlorine. A. 23.45 u B. 34.23 u C. 45.67 u D. 35.46 u 2  The Mole & Molar Mass Substance Atomic Molar Mass Number of Particles in One Mole Carbon (C) 12.01g 12.01 u 6.02 x 1023 C atoms Sodium (Na) 23.00ug 23.00 6.02 x 1023 Na atoms Iron (Fe) 55.90ug 55.90 6.02 x 1023 Fe atoms Mass  Particles 200 g Fe mass (Fe: 55.85g/mol) molar mass 1mol 1mol molar mass moles 1mol 6.02 x 10 23 part icles 22.4L 1mol 1mol 22.4L 6.02 x 10 23 part icles 1mol volume at STP number of representative ? Fe atoms particles 3  Learning Check Find the number of P atoms in 0.200 g of P4. (P=30.97) A. 6.02 x 1023 B. 3.89 x 1021 C. 9.70 x 1026 D. 1.12 x 1021 E. None The same number of molecules may not represent the same mass Substance Molar Mass Number of Particles in One Mole Carbon (C) 12.01g 6.02 x 1023 C atoms Sodium (Na) 23.00 g 6.02 x 1023 Na atoms Iron (Fe) 55.90 g 6.02 x 1023 Fe atoms C6H12O6 (glucose) 180.0 g 6.02 x 1023 glucose molecules C8H10N4O2 (caffeine) 194.0 g 6.02 x 1023 caffeine molecules CaCO3 (antacid) 100.1 g 6.02 x 1023 CaCO3 formula units NaF (preventative 42.00 g 6.02 x 1023 NaF formula units for dental cavities) MOLECULAR WEIGHT (MW)  A molecular weight is the sum of the atomic weights of the atoms in a molecule.  For example, the molecular weight of ethane, C2H6, would be FORMULA WEIGHT (FW)  Formula weights are generally reported for ionic compounds.  So, the formula weight of calcium chloride, CaCl2, would be 4  Mass  Particles 201 g N2O mass (N2O : 44.01g/mol) molar mass 1mol 1mol molar mass moles 1mol 6.02 x 10 23 part icles 22.4L 1mol 1mol 22.4L 6.02 x 10 23 part icles 1mol volume at STP number of representative ? molecules N2O particles LEARNING CHECK Find the mass of 2.1  1024 molecules of NaHCO3. (NaHCO3 : 84.01g/mol) A. 253g B.237g C. 356g D. 290g  Calculate the number of moles of sodium in 2.53 moles of sodium carbonate  Calculate the number of atoms of sodium in 2.53 moles of sodium carbonate 5 Learning Check  How many grams of iron are in a 15.0 g sample of iron(III) oxide? (Fe2O3 : 159.7g/mol; Fe : 55.85g/mol) A. 23.4 g B. 12.4 g C. 11.5 g D. 10.5 g E. 14.6 g Percent Composition of Compounds % element  mass of element mass of whole sample  100 6 LEARNING CHECK What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? (C: 12.01amu; H: 1.008 amu; Cl: 35.45amu) C (%) H (%) Cl (%) A 12.08.9865 87.05 B 10.06.8442 89.07 C 9.256 1.265 78.65 D 8.978.8965 90.12 E 7.956.7896 91.25 SOLUTION: Determining the Formula of a Compound 7  Empirical Formula Determination Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid? (C: 12.01g/mol; O: 16.00g/mol; H: 1.008g/mol) 1. Base calculation on grams of compounds. 2. Determine moles of each element. 3. Divide each value by the smallest of the values. 4. Multiply each number by an integer to obtain all whole numbers (C: 12.01g/mol; O: 16.00g/mol; H: 1.008g/mol) Learning Check A. N2O4 B. NO A 2.012 g sample of a compound contains 0.522 g of nitrogen and 1.490 g of oxygen. Calculate its empirical C. NO2 formula. D. N2O (N:14.01g/mol; O: 16.00g/mol) E. N2O5 8  molecular formula = (empirical formula)n [n = Mass of molecular formula/Mass of empirical formula] The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? (C: 12.01g/mol; O: 16.00g/mol; H :1.008g/mol ) 1. Find the molar mass of C3H5O2 2. Divide the molecular mass by the molar mass of the empirical formula. 3. Multiply the empirical formula by this number to get the molecular formula. COMBUSTION ANALYSIS 9  Example: Isopropyl alcohol, sold as rubbing alcohol, is composed of C, H, and O. Combustion of 0.255 g of isopropyl alcohol produces 0.561 g of CO2 and 0.306 g of H2O. Determine the empirical formula of isopropyl alcohol. (CO2=44.0g/mol; H2O=18.0g/mol; C=12.0g/mol; O=16.0g/mol; H=1.01g/mol) Example Cont’d Chemical Reactions and Chemical Equations 10 Chemical Reactions As reactants are converted to products we observe:  …  …  …  …. Some Simple Patterns of Chemical Reactivity TYPES OF CHEMICAL REACTIONS Combination Reactions In combination reactions two or more substances react to form one product. Examples: – 2Mg(s) + O2(g)  2MgO(s) – N2(g) + 3H2(g)  2NH3(g) – C3H6(g) + Br2(l)  C3H6Br2(l) 11  Decomposition Reactions In a decomposition reaction one substance breaks down into two or more substances. Examples: – CaCO3(s)  CaO(s) + CO2(g) – 2KClO3(s)  2KCl(s) + O2(g) – 2NaN3(s)  2Na(s) + 3N2(g) Combustion Reactions Combustion reactions are generally rapid reactions that produce a flame. Combustion reactions most often involve hydrocarbons reacting with oxygen in the air. Examples: – CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) – C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) C3H8(g) + 7/2 O2(g)  3 CO (g) + 4H2O(g) + energy  C3H8(g) + 2 O2(g)  3 C (s) + 4 H2O(g) + energy  Chemical Equations  Chemical equations describe what happens in chemical reactions  Hydrogen and oxygen combine to form water 2 H2 + O2  2 H2O  Hydrogen and oxygen are called reactants  Water is called the product  Reactants are separated from products with “”  “” is like an equal sign  It is sometimes useful to include the physical state of reactants and products  For solids use s, liquids use l, gases use g, and for aqueous solutions use aq. 12  Balancing Chemical Equations Balancing Example Aluminum and copper(II) chloride react to form copper and aluminum chloride. Learning Check Ammonia and oxygen gas react to form nitrogen monoxide and water. What is the coefficient of NO in the balanced equation? A. 6 B. 3 C. 4 D. 5 13 Balancing Example (without a structure) __C6H12O6 + __KClO3 → __CO2 + __H2O + __KCl 40 Stoichiometric Calculations: Amounts of Reactants and Products Stoichiometry N2 + 3H2 → 2NH3 14 Stoichiometry Problems  How many grams of silver will be formed from 12.0 g copper? (Cu: 63.55g/mol; Ag: 107.8g/mol) Cu + 2AgNO3 2Ag + Cu(NO3)2 Water in Space In the space shuttle, the CO2 that the crew exhales is removed from the air by a reaction within canisters of lithium hydroxide. On average, each astronaut exhales about 20.0 mol of CO2 daily. What volume of water will be produced when this amount of CO2 reacts with an excess of LiOH? (Hint: The density of water is about 1.00 g/mL.) (H2O: 18.00g/mol) CO2(g) + 2 LiOH(s)  Li2CO3(aq) + H2O(l) 20.0 mol excess x mL Calculations Involving a Limiting Reactant 15  the reactant in shortest supply limits the amount of Grilled Cheese product that can form the computed amount Sandwich of product is always based on the limiting reagent Bread + Cheese  ‘Sandwich’ Any reagent that is not The reactant that is consumed first completely consumed during is called the limiting reactant the reactions is called excess reactant EXAMPLE 79.1 g of zinc react with 1.36 x 1024 HCl molecules. Identify the limiting and excess reactants. How many grams of hydrogen are formed? (Zn: 65.39g/mol; H2: 2.016g/mol) SOLUTION 16  Percent Yield The predicted amount of product is not always obtained experimentally Side reactions reduce the percent yield. By-products are formed by side reactions. measured in lab actual yield % yield   100 theoretica l yield calculated on paper EXAMPLE (K2CO3 : 138g/mol; KCl:74.5g/mol) When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K2CO3 + 2HCl 2KCl + H2O + CO2 17  SOME COMMON TERMS ABOUT REACTIONS  Multistep synthesis involves more than one step reaction.  When substances react independently and at the same time the reaction is a simultaneous reaction.  Reactions carried out in sequence are called consecutive reactions.  An intermediate is a substance produced in one step and consumed in another during a multistep synthesis.  The overall reaction is a chemical equation that expresses all the reactions occurring in a single overall equation. Slid e 52 of 24 18