THE MOLE / COUNTING IN CHEMISTRY

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1 1 THE MOLE / COUNTING IN CHEMISTRY ***A mole is 6.0 x 10 items.*** 1 mole = 6.0 x 10 items 1 mole = 60, 00, 000, 000, 000, 000, 000, 000 items Analogy #1 1 dozen = 1 items 18 eggs = 1.5 dz. - to convert from eggs from dozen, we need to multiply by conversion factor 1dz 18eggs = 18eggs = 15. dz 1 eggs Example: How many eggs in. dozen? 1 eggs. dz =. dz = 8eggs 1dz Example: How many dozen is 18 gears? 1dz 18gears = 18gears = gears dz gears Analogy # 1 gross = 144 items Conversion factors are 1gross 144 items and 144 items gross Example: How many gross is 95 pencils? Example: How many apples is 0.47 gross? 1 mole = 6.0 x 10 items (usually ions, atoms or molecules) x 10 is called Avogadro s number and is abbreviated N A. 6.0 x 10 molecules = 1 mole molecule x 10 molecules = mole molecule

2 Example: How many moles of atoms is 7.4 x 10 1 atoms? - use conversion factor 6. 0 x10 items or 6. 0 x10 items mol 1 1 e 7. 4x10 atoms = 7. 4x10 atoms 6. 0 x10 atoms = mol atoms Example: How many moles of ions is.5 x 10 5 ions? Example: How many molecules are in 8. mol of molecules? ATOMIC MASS AND MOLAR MASS How much mass does a hydrogen atom have? 1 p + = m(h) 1.00 amu What is an amu? 1 amu = 1.66 x 10-4 g BY DEFINITION Relationship between atomic mass and molar mass - atomic mass mass of one atom - molar mass mass of a mole of atoms How much mass does a helium atom have? p + + n amu x10 g 400. amu = 664. x10 amu How much mass does a mole of helium atoms have? 4 g x10 atoms 6.64 x10 g 4.00 g = = 4.00g mol atom mol The fact that 1 He atom has 4.00 amu of mass and 1 mole of He atoms has 4.00 g of mass is not a coincidence. Definition of amu is made to ensure this coincidence. The value of 4.00 g/mol is call the molar mass of He.

3 Molar mass of the elements are found on the periodic table. Why aren t molar masses on periodic table integers? 1.) Molar masses are an average of isotopes. - most important reason.) Nuclear energy has mass..) Mass of electrons is very small, but not zero. Converting between mass and moles - a molar mass is a conversion factor! Example: How many moles of atoms are in 96. grams of carbon? Example: How much mass in grams does moles of uranium have? FORMULA MOLAR MASS a.k.a. molecular mass, molecular weight, formula weight, etc Formula Mass (Weight) sum of atomic masses in chemical formula Calculating Formula Mass Example: What is the molar mass of ethylene, C H 4? C: x g/mol = 4.00 g/mol 4 H: 4 x g/mol = g/mol g/mol M(C H 4 ) = g/mol Ethylene is used to ripen fresh fruit. It is also used to make milk jugs. Example: What is the molar mass of Ba(NO )? 1 Ba: 1 x 17. g/mol = 17. g/mol N: x g/mol = g/mol 6 O: 6 x g/mol = g/mol 61.4 g/mol M(Ba(NO ) ) = 61.4 g/mol Barium nitrate is used to color fireworks green.

4 4 Converting between mass and moles Example: How many moles are in 58 g of Ba(NO ) First calculate molar mass. 17. g/mol x g/mol 4 x g/mol 61.4 g/mol 58g = 58g =.06 mol 61.4g Example: How many grams is mol of NaCl? M(NaCl) = g/mol g/mol = g/mol 58.44g mol = mol = 8.59g NaCl Road salt is mined under the city of Detroit. SCHEME: Converting mass to moles to number Mass (g) M Molar Mass Moles (mol) NA Avogadro s Number Number (atoms or molecules) Example: How much mass in grams does 1.8 x 10 molecules of CO have? Example: How many F atoms are in 1.19 g of CaF?

5 5 MASS COMPOSITION OF A COMPOUND - Mass composition tells us percentage of mass for each element in compound. - Mass of molecule equals the sum of the masses of the atoms. - Given a chemical formula, one is able to find the percent mass of each element. Strategy to find percent mass of compound 1. Assume 1 mole of substance. Calculate the total mass of one mole of molecules, i. e., find the molar mass of the compound.. Calculate the mass of a single element by multiplying number of atoms by atomic weight 4. Divide the mass of single element by total mass and multiply by 100% to get percent mass. 5. Repeat for all elements. 6. Adding all percentages should equal 100%. Example: What the mass percentages of the elements in C H 6? 1. Assume 1 mol. M(C H 6 ) = x g/mol + 6 x g/mol = g/mol g/mol = g/mol g For % C. x M(C) = x g/mol = g/mol g. 4. % C 6 045g.. g 100% = % For % H. 6 x M(H) = 6 x g/mol = g/mol g. 4. % H g.. g 100% = % As a check % % = % C H 6 is propylene which is used to make polypropylene. Polypropylene is used to make wash bottles, plastic sheet protectors, long underwear, rope and Tupperware.

6 6 EMPIRICAL FORMULAS - chemical formula with lowest possible ratio of atoms Molecular Formula Empirical Formula C H 6 CH P 4 O 10 P O 5 SnCl SnCl C 6 H 1 O 6 CH O Calculating Empirical Formulas from Mass Percentages Given: Percent Mass Composition Find: Empirical Formula Strategy: 1) Assume 100 g of matter. ) Multiply 100 g by mass percent to find amount of each element. ) Convert mass of each element to moles using molar mass. 4) Find whole number ratios by dividing each number of moles by lowest number of moles. 5) Use these ratios to find empirical formula. Example: Find the empirical formula for a compound with the following mass percentages: 69.6 % O 0.4 % N 1 Assume 100 g m(o) = g x = 69.6 g m(n) = g x 0.04 = 0.4 g O: 69. 6g g = 45. mol O N: 0. 4g g = 17. mol N 4 mol 4.5mol mol = =.00 = mol.17 mol O O O N N N 5 Empirical Formula is NO Nitrogen dioxide is a component of automotive exhaust.

7 7 Example: Find the empirical formula for a compound with the following mass percentages: 4.6 % H 40.9 % C 54.5 % O Empirical Formula is H 4 C O H 4 C O is the empirical formula for ascorbic acid, H 8 C 6 O 6, the chemical name for vitamin C.

8 8 THEORETICAL STOICHIOMETRY - coefficients of balanced equations relate moles of reactants to moles of products - Comparisons in chemistry must be done by comparing numbers of molecules to each other, i. e., comparing number of moles of each substance. Example: N (g) + H (g) NH (g) mole of N is stoichiometrically equivalent to moles of NH. - in other words, for every 1 mole of N reacted, moles of NH are produced. - N mol NH - equivalence is only true for specific chemical reaction - equivalence can be considered a conversion factor N N mol NH mol NH or mol NH N - other equivalences are N mol H mol H mol NH Example: a) What are all of the stoichiometric equivalences for the reaction C H (g) + 5 O (g) 4 CO (g) + H O (g)? molch 5molO C H molco CH H O 5molO 4molCO 5molO mol H O mol CO H O b) How many moles of carbon dioxide are formed when 5 moles of acetylene (C H ) is combusted? 5mol CH molco = 10mol CH c) How many moles of oxygen are needed to fully burn 9.8 moles of acetylene (C H )? CO Acetylene is a welder s fuel.

9 9 Example: For the reaction Pb(NO ) 4 (aq) + 4 KCl (aq) PbCl 4 (s) + 4 KNO (aq), How many moles of KCl are needed to form 15.6 moles of PbCl 4? PRACTICAL STOICHIOMETRY - can t measure moles directly in the real world. - must measure amount of substance with grams. ***Cannot compare substances stoichiometrically by mass, must convert to moles.*** SCHEME: Mass of reactant (g) Mass of product (g) M Molar Mass Moles of reactant (mol) Balanced Equation Moles of product (mol) M Molar Mass Example: For the reaction, NH (g) + (g) NH 4 Cl (s), a) how much NH is needed to react with 9. g of? 1) First convert grams of reactant to moles of reactant 9. g = 5. mol 6. 5g ) Compare moles of one reactant to other reactant. NH 5. mol = 5. mol ) Convert moles of other reactant to grams g NH 5. molnh = 4. 0g mol NH NH NH

10 b) How much ammonium chloride is produced when 9. g of is fully reacted? - Note with factor-label method, we can do problems all on one line. 9. g 55. g NH 4Cl NH 4Cl = 15g 65. g mol NH Cl The reaction of ammonia with hydrogen chloride gas is used to a smokescreen. 4 NH Cl Example: For the reaction 4BaCO (s) + Y (CO ) (s) + 6 CuCO (s) YBa Cu O 7 (s) + 1 CO (g) + O (g) a) Calculate how many grams of CuCO is needed to fully react with g of BaCO, 4 10 b) Calculate how many grams of YBa Cu O 7 is formed from g of BaCO fully reacting. Yttrium barium copper oxide (YBCO) is a superconducting ceramic. It is superconducting below a temperature of 95 K. Example: For the reaction SiO (s) + 6 HF (aq) H SiF 6 (aq) + H O (aq), a) Calculate how many grams of HF is needed to fully react with g of silicon dioxide, b) How much H SiF 6 is made when 1.4 g of HF fully reacts? Hydrofluoric acid, HF(aq), is used to frost glass, SiO.

11 LIMITING REAGENTS - Often starting materials are not available in proper stoichiometric proportions. - Given unbalanced amounts of reactants, we would like to know how much product can be produced. Analogy: Bicycle Factory The equation to make a bicycle is wheels + 1 frame + 1 handlebar 1 bicycle 11 If the parts inventory is as follows: 40 wheels 150 frames we ask ourselves - What reactant limits production? - How much product can be produced? 15 handlebars, 40 wheel frame + 15 handlebar?? bicycle 1bicycle 40wheels = 10bicycles wheels 1bicycle 150frames = 150bicycles 1frame 1bicycle 15handlebars = 15bicycles 1handlebars Limiting reactant: wheels Production: 10 bicycles In a limiting reactant problem, amounts of two (or more) reactant are given. Calculate how much product is produced by each. The reactant that yields the lowest amount of product is the limiting reactant. ***In limiting reagent problems, we need to compare moles to moles*** Example: For the reaction SO (g) + O (g) + H O (l) H SO 4 (aq), if 5.6 mol of SO, 4.8 mol of O and 6.0 mol of H O are reacted together, how many moles of H SO 4 are produced? HSO4 For the SO : 56. molso = 56. mol HSO4 SO molhso4 For the O : 48. molo = 96. mol For the H O: O 6.0 mol = HSO4 HSO4 HO 6.0 molh SO4 HO SO is limiting reactant and therefore 5.6 moles of H SO 4 is produced. Sulfur dioxide is a pollutant from burning coal that is a contributor to acid rain. Sulfur dioxide is removed from air with calcium oxide. SO (g) + CaO (s) CaSO (s)

12 1 Example: For the reaction Zn (s) + CuCl (aq) ZnCl (aq) + Cu (s), what mass of copper metal is produced from the reaction of.00 g of Zn and.00 g of CuCl? Find limiting reactant by comparing moles to moles Must convert mass to moles using molar mass as conversion factor. Thus CuCl is the limiting reactant and the amount of copper produced is REACTION YIELDS - We have been calculating theoretical yields by assuming that the reactions proceed perfectly. - An actual chemical process is rarely perfect; therefore, the actual yield is always less than the theoretical yield. - We compare the actual yield to the theoretical yield by calculating percent yield. Definition of percent yield %yield actual yield = 100% theoretical yield Example: For the reaction Cr O (s) + Al (s) Cr (s) + Al O (s) 18.7 g of Chromium (III) oxide reacts to form 10.8 g of chromium metal. What the percent yield of this process? Theoretical yield of chromium metal is Thus the percent yield is 108. gcr % yield = 100% = 84. 4% 1. 8g Cr

13 1 BOND ENTHALPIES When two atoms bond together, the chemical energy of the system decreases. Consider an energy level diagram of before bonding and after bonding. H H H H before The energy of the bonded system is lower than the unbonded system. The energy released when two unbonded atoms become bonded is called the bond enthalpy. Aside: Enthalpy is another word for heat. The bond enthalpy increases as atoms are more strongly bonded together. As the strength of the bond increases, the distance between the atoms decreases. Bond enthalpies are an experimentally found quantity; i. e., we can t predict bond enthalpies from periodic table. Bond Enthalpies and Chemical Changes after **All chemical changes involve the breaking and creation of bonds.** If we can understand what bonds are breaking and what bonds are forming, then we can use bond enthalpies to estimate the energy (technically, enthalpy) change of the reaction. To break a bond, we input (add) the bond enthalpy. When a bond is formed, the bond enthalpy is released (subtracted).

14 Example: Given the table of bond enthalpies below, calculate the energy change, when two molecules of hydrogen and one molecule of oxygen change into two molecules of water. 14 H H H H + O O O H H O H H TABLE OF BOND ENTHALPIES Bond E (kj/mol) Bond E (kj/mol) C H 41 H H 46 C C 48 N N 16 C O 58 N = N 418 C = C 614 N N 941 C C 89 O H 46 C = O 107 O = O 495 Breaking two H H bonds means inputting x 46 kj/mol. Breaking one O = O bond means inputting 495 kj/mol. Forming four O H bonds means releasing 4 x 46 kj/mol. Overall the energy change is

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