Chapter 2 Overview. Chapter 2 Overview

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Domenic Shon Russell
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1 Chapter 2 Overview! all matter, whether liquid, solid, or gas,consists of atoms, which form molecules! the identity of an atom and its chemical behavior is strictly defined by the number of electrons in the outer electronic sphere, which equals to the number of protons in the nucleus! the atomic weight is the combined mass of protons and neutrons in the nucleus. Each contributes one unit of mass. The mass of the electrons is negligible Chapter 2 Overview 6 electrons 6 electrons 6 n o 6 p + 7 n o 6 p + 12 C 13 C number of neutrons ISOTOPES OF CARBON N = Mass - Z atomic number number of protons

2 Periodic Table Main Group Elements:Group Number corresponds to ionic charge Chapter 2 Overview Ionic compounds: Simple salts. Charges Balanced NaCl Na + Cl LiI Li + I Cs 2 SO 4 Cs + SO 2 4 Fe 2 (CO 3 ) 3 Fe 3+ CO 2 3 (NH 4 ) 3 PO 4 NH + 4 PO 3 4

3 Chapter 2 Overview Molecular Weight = Sum of Atomic Weights MW(NaCl) = Atomic Mass of Na + At. Mass of Cl Chapter 2 Sample Problem Succinic acid is an important metabolite in biological energy production. Give the molecular formula, empirical formula, and molecular mass of succinic acid. H O H H O C C C C O H H O succinic acid H

4 Chapter 2 Sample Problem Succinic acid is an important metabolite in biological energy production. Give the molecular formula, empirical formula, and molecular mass of succinic acid. mol. formula:count all atoms: C 4 H 6 O 4 H O H H O C C C C O H H O succinic acid H Chapter 2 Sample Problem Succinic acid is an important metabolite in biological energy production. Give the molecular formula, empirical formula, and molecular mass of succinic acid. mol. formula:count all atoms: C 4 H 6 O 4 empirical formula: C 2 H 3 O 2 (whole numbers!) H O H H O C C C C O H H O succinic acid H

Chapter 2 Sample Problem 2.132. Succinic acid is an important metabolite in biological energy production. Give the mole

5 Chapter 2 Sample Problem Succinic acid is an important metabolite in biological energy production. Give the molecular formula, empirical formula, and molecular mass of succinic acid. mol. formula:count all atoms: C 4 H 6 O 4 empirical formula: C 2 H 3 O 2 (whole numbers!) H O H H O C C C C O H H O succinic acid H MW = 4! ! !16.00 = Chapter 2 Sample Problem How can iodine (Z = 53) have a higher atomic number yet lower atomic weight than tellurium (Z = 52)?

6 Stoichiometry: The Mole describes quantitative aspects of chemical reactions One mole (1 mol) is 6.022!10 23 molecules (ions, atoms, etc.) Avogadro s number M = MW molar weight (g/mol) molecular weight Stoichiometry MW(H 2 ) = 2.02 MW(NaCl) = MW(AgNO 3 ) = M = 2.02 g/mol M = g/mol M = g/mol 1 : 1 ratio of NaCl and AgNO 3 : g of NaCl and g of AgNO 3 or g of NaCl and g of AgNO 3 etc

7 Stoichiometry mass in grams molar weight (g/mol) M (g/mol) = m (g) n (mol) number of mols Stoichiometry mass in grams molar weight (g/mol) M (g/mol) = m (g) n (mol) number of mols n (mol) = m (g) M (g/mol)

8 Stoichiometry mass in grams m (g) = n (mol)! M (g/mol) number of moles (mol) molar weight (g/mol) Sample Problem 3.2 Ammonium carbonate, is a white solid that decomposes with warming. Among its many uses, it is a component of baking powder, fire extinguishers, and smelling salts. How many moles are in 41.6 g of ammonium carbonate?

9 Sample Problem 3.2 Ammonium carbonate, is a white solid that decomposes with warming. Among its many uses, it is a component of baking powder, fire extinguishers, and smelling salts. How many moles are in 41.6 g of ammonium carbonate? divide given mass by mol weight Sample Problem 3.2 Ammonium carbonate, is a white solid that decomposes with warming. Among its many uses, it is a component of baking powder, fire extinguishers, and smelling salts. How many moles are in 41.6 g of ammonium carbonate? divide given mass by mol weight what is ammonium carbonate? n (mol) = m (g) M (g/mol)

10 Sample Problem 3.2 Ammonium carbonate, is a white solid that decomposes with warming. Among its many uses, it is a component of baking powder, fire extinguishers, and smelling salts. How many moles are in 41.6 g of ammonium carbonate? m (g) n (mol) = divide given mass by mol weight M (g/mol) what is ammonium carbonate? NH + 4,CO 2 3 " (NH 4 ) 2 CO 3 what is its mol. weight? Sample Problem 3.2 Ammonium carbonate, is a white solid that decomposes with warming. Among its many uses, it is a component of baking powder, fire extinguishers, and smelling salts. How many moles are in 41.6 g of ammonium carbonate? m (g) n (mol) = divide given mass by mol weight M (g/mol) what is ammonium carbonate? NH + 4,CO 2 3 " (NH 4 ) 2 CO 3 what is its mol. weight? 2! ! !16.00 = g/mol

11 Sample Problem 3.2 Ammonium carbonate, is a white solid that decomposes with warming. Among its many uses, it is a component of baking powder, fire extinguishers, and smelling salts. How many moles are in 41.6 g of ammonium carbonate? n (mol) = divide given mass by mol weight what is ammonium carbonate? NH + 4,CO 2 3 " (NH 4 ) 2 CO 3 what is its mol. weight? 2! ! !16.00 = m (g) M (g/mol) g/mol answer: 41.6 g : g/mol = mol Sample Problem 3.2 Ammonium carbonate, is a white solid that decomposes with warming. Among its many uses, it is a component of baking powder, fire extinguishers, and smelling salts. How many moles are in 41.6 g of ammonium carbonate? n (mol) = divide given mass by mol weight what is ammonium carbonate? NH + 4,CO 2 3 " (NH 4 ) 2 CO 3 what is its mol. weight? 2! ! !16.00 = m (g) M (g/mol) g/mol answer: 41.6 g : g/mol = mol (number of formula units = mol! 6.022!10 23 mol -1 = 2.61!10 23 )

12 Mass Percent from Chemical Formula Glucose (C 6 H 12 O 6 ) is the most important nutrient in the living cell for generating energy. What is the mass percent of each element in glucose? let s focus on carbon first Mass Percent from Chemical Formula Glucose (C 6 H 12 O 6 ) is the most important nutrient in the living cell for generating energy. What is the mass percent of each element in glucose? let s focus on carbon first mass of C in one mole(cule). mass of C 6 H 12 O 6 in one mole(cule)

13 Mass Percent from Chemical Formula Glucose (C 6 H 12 O 6 ) is the most important nutrient in the living cell for generating energy. What is the mass percent of each element in glucose? let s focus on carbon first mass of C in one mole(cule). mass of C 6 H 12 O 6 in one mole(cule) 6! = = = 40.0 mass% 6! ! ! Elemental Analysis and Molecular Formula complete burning, than analysis of oxidation products used to determine empirical formula of an unknown compound (gives an empirical formula only!) used to confirm the structure (mass spectroscopy is a superior modern technique that gives a molecular formula directly)

14 Elemental Analysis and Molecular Formula Sample Problem 3.5: During physical activity, lactic acid (M = g/mol) forms in muscle and is responsible for muscle soreness. Elemental analysis shows that it contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) determine the empirical formula of lactic acid (b) determine the molecular formula of lactic acid Elemental Analysis and Molecular Formula Sample Problem 3.5: During physical activity, lactic acid (M = g/mol) forms in muscle and is responsible for muscle soreness. Elemental analysis shows that it contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) determine the empirical formula of lactic acid (b) determine the molecular formula of lactic acid mass% # arbitrary weight (let s take 100 g) convert grams to moles

15 Elemental Analysis and Molecular Formula Sample Problem 3.5: During physical activity, lactic acid (M = g/mol) forms in muscle and is responsible for muscle soreness. Elemental analysis shows that it contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) determine the empirical formula of lactic acid (b) determine the molecular formula of lactic acid mass% # arbitrary weight (let s take 100 g) convert grams to moles 40.0 g C = g/mol = 3.33 mol H = 6.71 g = 6.66 mol g/mol 53.3 g O = = 3.33 mol g/mol Elemental Analysis and Molecular Formula Sample Problem 3.5: During physical activity, lactic acid (M = g/mol) forms in muscle and is responsible for muscle soreness. Elemental analysis shows that it contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) determine the empirical formula of lactic acid (b) determine the molecular formula of lactic acid mass% # arbitrary weight (let s take 100 g) convert grams to moles 40.0 g C = g/mol = 3.33 mol H = 6.71 g = 6.66 mol g/mol 53.3 g O = g/mol = 3.33 mol CH 2 O

16 Elemental Analysis and Molecular Formula Sample Problem 3.5: During physical activity, lactic acid (M = g/mol) forms in muscle and is responsible for muscle soreness. Elemental analysis shows that it contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) determine the empirical formula of lactic acid (b) determine the molecular formula of lactic acid Empirical formula: CH 2 O " M = g/mol Actual M is 3 (three) times greater, therefore: Molecular formula: C 3 H 6 O 3 Chemical Equations Typical Structure: Starting Material(s) # Products(s) H 2 + Cl 2 # 2HCl H H + = Cl Cl Cl H Cl H

17 Chemical Equations n (mol) = H 2 + Cl 2 # 2HCl m (g) M (g/mol) mol H mol Cl 2 2!0.124 mol HCl mol H mol Cl 2 2!23.12 mol HCl mol H mol Cl 2 Chemical Equations n (mol) = H 2 + Cl 2 # 2HCl m (g) M (g/mol) mol H mol Cl 2 2!0.124 mol HCl mol H mol Cl 2 2!23.12 mol HCl mol H mol Cl 2 2!0.745 mol HCl

18 Chemical Equations n (mol) = H 2 + Cl 2 # 2HCl m (g) M (g/mol) Global Strategy: Input (g, ml, kg, etc) " moles (starting) " moles (final) " Output (g, ml, kg, etc) Balancing Chemical Equations Starting Material(s) # Products(s) space for coefficients! the same number of atoms of each element on both sides

19 Balancing Chemical Equations Unbalanced equation: PF 3 + HCl # PCl 3 + HF Balancing Chemical Equations Unbalanced equation: PF 3 + HCl # PCl 3 + HF PF 3 + HCl # PCl 3 + 3HF

20 Balancing Chemical Equations Unbalanced equation: PF 3 + HCl # PCl 3 + HF PF 3 + HCl # PCl 3 + 3HF PF 3 + 3HCl # PCl 3 + 3HF Balancing Chemical Equations Unbalanced equation: PF 3 + HCl # PCl 3 + HF PF 3 + HCl # PCl 3 + 3HF Balanced equation!: PF 3 + 3HCl # PCl 3 + 3HF Check

21 Balancing Chemical Equations Unbalanced equation: C 8 H 18 + O 2 # CO 2 + H 2 O Balancing Chemical Equations Unbalanced equation: C 8 H 18 + O 2 # CO 2 + H 2 O Balanced equation: C 8 H 18 + O 2 # 8CO 2 + H 2 O C 8 H 18 + O 2 # 8CO 2 + 9H 2 O C 8 H O 2 # 8CO 2 + 9H 2 O 16+9=25 2C 8 H O 2 # 16CO H 2 O Check

22 Stoichiometric Equivalents C 3 H 8 + 5O 2 # 3CO 2 + 4H 2 O (propane) 1 mol of C 3 H 8 is equivalent to 4 mol of H 2 O 3 mol of CO 2 is equivalent to 5 mol of O 2 3 mol of CO 2 is equivalent to 4 mol of H 2 O Stoichiometric Equivalents C 3 H 8 + 5O 2 # 3CO 2 + 4H 2 O In the combustion of propane, how many moles of carbon dioxide are produced along with 10.0 moles of water? 10 4 = !3 = 7.5 " 7.5 moles of CO 2

16) How many grams of glucose are produced during photosynthesis that

23 Stoichiometric Equivalents 6CO 2 + 6H 2 O # C 6 H 12 O 6 + 6O 2 glucose Stoichiometric Equivalents 6CO 2 + 6H 2 O # C 6 H 12 O 6 + 6O 2 glucose (MW = ) How many grams of glucose are produced during photosynthesis that generates 25 m 3 of oxygen (at normal conditions, d = 1.34 g/l).

24 Stoichiometric Equivalents 6CO 2 + 6H 2 O # C 6 H 12 O 6 + 6O 2 glucose (MW = ) How many grams of glucose are produced during photosynthesis that generates 25 m 3 of oxygen (at normal conditions, d = 1.34 g/l). Fundamental Strategy: moles (initial) " moles (final) " Input (g, ml, kg, etc) " moles (starting) " moles (final) " Output (g, ml, kg, etc) Stoichiometric Equivalents 6CO 2 + 6H 2 O # C 6 H 12 O 6 + 6O 2 glucose (MW = ) How many grams of glucose are produced during photosynthesis that generates 25 m 3 of oxygen (at normal conditions, d = 1.34 g/l). 1 m 3 = 1000 L, 1kg = 1000 g, d = 1.34 kg/m 3 d = m V

25 Stoichiometric Equivalents 6CO 2 + 6H 2 O # C 6 H 12 O 6 + 6O 2 glucose (MW = ) How many grams of glucose are produced during photosynthesis that generates 25 m 3 of oxygen (at normal conditions, d = 1.34 g/l). 1 m 3 = 1000 L, 1kg = 1000 g, d = 1.34 kg/m 3 d = m V m(o 2 ) = 25 m 3!1.34 kg/m 3 = 33.5 kg = g n = m M n(o 2 ) = g g/mol = 1047 mol Stoichiometric Equivalents 6CO 2 + 6H 2 O # C 6 H 12 O 6 + 6O 2 glucose (MW = ) How many grams of glucose are produced during photosynthesis that generates 25 m 3 of oxygen (at normal conditions, d = 1.34 g/l). n(o 2 ) = 1047 mol n(c 6 H 12 O 6 ) = (1/6)!n(O 2 ) = (1/6)!1047 mol = mol n = m M

26 Stoichiometric Equivalents 6CO 2 + 6H 2 O # C 6 H 12 O 6 + 6O 2 glucose (MW = ) How many grams of glucose are produced during photosynthesis that generates 25 m 3 of oxygen (at normal conditions, d = 1.34 g/l). n(o 2 ) = 1047 mol n(c 6 H 12 O 6 ) = (1/6)!n(O 2 ) = (1/6)!1047 mol = mol n = m M m(c 6 H 12 O 6 ) = mol! g/mol = g Stoichiometric Equivalents 6CO 2 + 6H 2 O # C 6 H 12 O 6 + 6O 2 glucose (MW = ) How many grams of glucose are produced during photosynthesis that generates 25 m 3 of oxygen (at normal conditions, d = 1.34 g/l). n(o 2 ) = 1047 mol n(c 6 H 12 O 6 ) = (1/6)!n(O 2 ) = (1/6)!1047 mol = mol n = m M m(c 6 H 12 O 6 ) = mol! g/mol = g

27 Example 2H 2 + O 2 # 2H 2 O 16.0 g 16.0 g? g n (mol) = m (g) M (g/mol) Example 2H 2 + O 2 # 2H 2 O 16.0 g 16.0 g? g 7.94 mol 0.5 mol n (mol) = m (g) M (g/mol)

28 Example 2H 2 + O 2 # 2H 2 O 16.0 g 16.0 g? g 7.94 mol 0.5 mol 1.0 mol n (mol) = m (g) M (g/mol) Example 2H 2 + O 2 # 2H 2 O 16.0 g 16.0 g? g 7.94 mol 0.5 mol 1.0 mol " g n (mol) = m (g) M (g/mol)

29 Yields: Theoretical, Percent 2A + B # A 2 B + c + d g should give g actually, g was obtained Yield = Y = g g!100% = 84% n (mol) = m (g) M (g/mol) Yields: Theoretical, Percent 2A + B # A 2 B + c + d mol should give mol actually, mol was obtained Yield = Y = mol mol!100% = 84% n (mol) = m (g) M (g/mol)

30 Yields: Overall 90% 75% 85% A # B # C # D Overall Y = 0.90! 0.75! 0.85! 100% = 57% 57% A # D Solutions Solute Solvent concentration: a quantitative description of a solution

31 Solutions Molarity = moles of solute liters of solution M = n solute (mol) Vsolvent (L) 3 M means 3 mol/l Sample Problem 3.14 Isotonic saline is 0.15 M aqueous solution of NaCl that simulates the total concentration of ions found in many cellular fluids. Its uses range from a cleansing rinse for contact lenses to a washing medium for red blood cells. How do you prepare 0.80 L of isotonic saline from 6.0 M stock solution of NaCl? M = n V

32 Sample Problem 3.14 Isotonic saline is 0.15 M aqueous solution of NaCl that simulates the total concentration of ions found in many cellular fluids. Its uses range from a cleansing rinse for contact lenses to a washing medium for red blood cells. How do you prepare 0.80 L of isotonic saline from 6.0 M stock solution of NaCl? We ll need: n(nacl) = 0.15 mol/l! 0.80 L = 0.12 mol M = n V Sample Problem 3.14 Isotonic saline is 0.15 M aqueous solution of NaCl that simulates the total concentration of ions found in many cellular fluids. Its uses range from a cleansing rinse for contact lenses to a washing medium for red blood cells. How do you prepare 0.80 L of isotonic saline from 6.0 M stock solution of NaCl? We ll need: n(nacl) = 0.15 mol/l! 0.80 L = 0.12 mol V(6M_soln) = 0.12 mol 6.0 mol/l = 0.02 L M = n V 0.78 L

33 Example * ml of a solution of NaOH with an unknown concentration reacts with an excess of an acid solution of oxalic acid prepared by dissolving g of oxalic acid dihydrate (H 2 C 2 O 4 2H 2 O) in water. The excess of oxalic acid then reacts completely with 9.65 ml of M of stock NaOH solution. What is the concentration of the unknown NaOH solution. 2NaOH + H 2 C 2 O 4 # Na 2 C 2 O 4 + 2H 2 O M = n V Example * ml of a solution of NaOH with an unknown concentration reacts with an excess of an acid solution of oxalic acid prepared by dissolving g of oxalic acid dihydrate (H 2 C 2 O 4 2H 2 O) in water. The excess of oxalic acid then reacts completely with 9.65 ml of M of stock NaOH solution. What is the concentration of the unknown NaOH solution. MW oad = NaOH + H 2 C 2 O 4 # Na 2 C 2 O 4 + 2H 2 O MW NaOH = M = n V

34 Example * ml of a solution of NaOH with an unknown concentration reacts with an excess of an acid solution of oxalic acid prepared by dissolving g of oxalic acid dihydrate (H 2 C 2 O 4 2H 2 O) in water. The excess of oxalic acid then reacts completely with 9.65 ml of M of stock NaOH solution. What is the concentration of the unknown NaOH solution. MW oad = NaOH + H 2 C 2 O 4 # Na 2 C 2 O 4 + 2H 2 O MW NaOH = n(h 2 C 2 O 4 ) = 0.588g g/mol = 4.66 mmol M = n V Example * ml of a solution of NaOH with an unknown concentration reacts with an excess of an acid solution of oxalic acid prepared by dissolving g of oxalic acid dihydrate (H 2 C 2 O 4 2H 2 O) in water. The excess of oxalic acid then reacts completely with 9.65 ml of M of stock NaOH solution. What is the concentration of the unknown NaOH solution. MW oad = NaOH + H 2 C 2 O 4 # Na 2 C 2 O 4 + 2H 2 O MW NaOH = M = n V n(h 2 C 2 O 4 ) = 0.588g g/mol = 4.66 mmol n(naoh-known) = 9.65 ml! 0.115mol/L = 1.11 mmol

35 Example * ml of a solution of NaOH with an unknown concentration reacts with an excess of an acid solution of oxalic acid prepared by dissolving g of oxalic acid dihydrate (H 2 C 2 O 4 2H 2 O) in water. The excess of oxalic acid then reacts completely with 9.65 ml of M of stock NaOH solution. What is the concentration of the unknown NaOH solution. MW oad = NaOH + H 2 C 2 O 4 # Na 2 C 2 O 4 + 2H 2 O MW NaOH = n(h 2 C 2 O 4 ) = 0.588g g/mol = 4.66 mmol n(naoh-known) = 9.65 ml! 0.115mol/L = 1.11 mmol M = n V n(oxalic-unknown) = /2 = 4.11 mmol Example * ml of a solution of NaOH with an unknown concentration reacts with an excess of an acid solution of oxalic acid prepared by dissolving g of oxalic acid dihydrate (H 2 C 2 O 4 2H 2 O) in water. The excess of oxalic acid then reacts completely with 9.65 ml of M of stock NaOH solution. What is the concentration of the unknown NaOH solution. MW oad = NaOH + H 2 C 2 O 4 # Na 2 C 2 O 4 + 2H 2 O MW NaOH = n(naoh-unknown) = 4.11mmol!2 = 8.22 mmol M = n V M = 8.22 mmol 50 ml = M Look at equation

Chapter 2 Overview. Chapter 2 Overview

Chapter 2 Overview. Chapter 2 Overview Chapter 2 Overview! all matter, whether liquid, solid, or gas,consists of atoms, which form molecules! the identity of an atom and its chemical behavior is strictly defined by the number of electrons in