Lecture 5 Outline. Derived from the Greek stoicheion ( element ) and metron ( measure )

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1 Lecture 5 Outline 5.1 Stoichiometry,, the mole etc. 5.2 Chemical Equations 5.3 Molarity 5.4 Limiting reagents and yields 5.5 Reaction enthalpies and Gibbs free energy 5.6 Catalyst Lecture 5 Stoichiometry Derived from the Greek stoicheion ( element ) and metron ( measure ) Stoichiometry is based on atomic masses and on the Law of Conservation of mass: The total mass of all substances present after a chemical reaction is the same as the total mass before the reaction. 1

2 Stoichiometry Reminder!!! mole - the amount of a substance that contains the same number of entities as there are atoms in exactly 12g of carbon x10 23 = mole This amount is 6.022x The number is called Avogadro s number and is abbreviated as N. One mole (1 mol) contains 6.022x10 23 entities (to four significant figures) Counting Objects of Fixed Relative Mass Oxygen g 55.85g Fe = x atoms Fe 32.07g S = x atoms S CaCO g Water g Copper g 2

3 Summary of Mass Terminology Term Definition Unit Isotopic mass Mass of an isotope of an element amu Atomic mass (also called atomic weight) Average of the masses of the naturally occurring isotopes of an element weighted according to their abundance amu Molecular (or formula) mass (also called molecular weight) Sum of the atomic masses of the atoms (or ions) in a molecule (or formula unit) amu Molar mass (Mm) (also called grammolecular weight) Mass of 1 mole of chemical entities (atoms, ions, molecules, formula units) g/mol Mass (g) Number of moles (n) = Molar Mass (g mol-1) M = m/n n = m M m = n M No. of entities = no. of moles x 6.022x1023 entities 1 mol 3

4 Sample Problem PROBLEM: Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element (a) Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in mol of Ag? (b) Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8g of Fe? (a) SOLUTION: m = n M mol x g mol = 3.69g 3.69 g Ag (b) SOLUTION: n = m/m 95.8 g x mol g = 1.72 mol 1.72mol x 6.022x10 23 atoms mol = 1.04x10 24 atoms 1.04x10 24 Fe atoms Mass % Mass % of element X = moles of X in formula x molar mass of X (amu) molecular (or formula) mass of compound (amu) x 100 4

5 Sample Problem Determining the Empirical Formula from Masses of Elements PROBLEM: PLAN: SOLUTION: Elemental analysis of a sample of an ionic compound gave the following results: 2.82g of Na, 4.35g of Cl, and 7.83g of O. What are the empirical formula and name of the compound? Once we find the relative number of moles of each element, we can divide by the lowest mol amount to find the relative mol ratios (empirical formula). n = m/m 2.82g Na mol Na 22.99g Na = mol Na Na 1 Cl 1 O 3.98 NaClO g Cl mol Cl 35.45g Cl = mol Cl NaClO 4 is sodium perchlorate. 7.83g O mol O O = mol O Sample Problem 3.5 Determining a Molecular Formula from Elemental Analysis and Molar Mass PROBLEM: Trick: During physical activity lactic acid (M=90.08g/mol) forms in muscle tissue 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. Assume an amount: e.g. 100g amount (mol) of each element preliminary formula convert to integer subscripts molecular formula divide mol mass by mass of empirical formula to get a multiplier empirical formula 5

6 Sample Problem continued Determining a Molecular Formula from Elemental Analysis and Molar Mass SOLUTION: Assuming there are 100.0g of lactic acid, the constituents are n = m/m 40.0g mol 6.71g mol 53.3g 12.01g 1.008g mol 16.00g 3.33 mol C 6.66 mol H 3.33 mol O C 3.33 H 6.66 O CH 2 O empirical formula M(CH 2 O) = g/mol M(lactic acid ) = g/mol 3 C 3 H 6 O 3 is the molecular formula Chemical Equations The formation of HF gas on the macroscopic and molecular levels reactants products 6

7 The Chemical Reaction Important aspects 1) The stoechiometric factor specifies how many entities of a chemical species are involved in a chemical reaction 2) Distinguishing between reactants (left from the arrow) and products (on the right) 3) Conservation of mass 4) Neutral Charge Sample Problem Balancing Chemical Equations PROBLEM: Within the cylinders of a car s engine, the hydrocarbon octane (C 8 H 18 ), one of many components of gasoline, mixes with oxygen from the air and burns to form carbon dioxide and water vapor. Write a balanced equation for this reaction. PLAN: translate the statement SOLUTION: C 8 H 18 + O 2 CO 2 + H 2 O balance the atoms C 8 H /2O 2 8 CO H 2 O adjust the coefficients 2C 8 H O 2 16CO H 2 O check the atom balance 2C 8 H O 2 16CO H 2 O specify states of matter 2C 8 H 18 (l) + 25O 2 (g) 16CO 2 (g) + 18H 2 O (g) 7

8 Limiting Reagents The limiting reagent determines the yield of the product An Ice Cream Sundae Analogy for Limiting Reactions What does this mean in chemistry? Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe Which is the limiting reagent? n = m/m!!! g Al mol Al mol Fe 2 O 3 needed g Fe 2 O 3 needed 124 g Al = 4.59 mol Al 27.0 g/mol Al 2 mol Al react with 1 mol Fe 2 O /2 = 2.30 mol Fe 2 O 3 Mass (g) Fe 2 O 3 = 2.30 mol x g mol -1 = 367 g Fe 2 O 3 Start with 124 g Al need 367 g Fe 2 O 3 Have more Fe 2 O 3 (601 g) so Al is limiting reagent 8

9 Yield of a Reaction Theoretical is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield x 100 Fundamentals of Solution Stoichiometry Solute: The substance that dissolves in the solvent Solvent: The substance in which the solute(s) dissolve Molarity is the number of moles of solute per liter of solution. c = Number of moles Volume mol L = M also called: concentration: Symbol: c or [ ] (sometimes M) c(h 2 O) = 5 mol/l [H 2 O] = 5 mol/l c = n V 9

10 Sample Problem Calculating Mass of Solute in a Given Volume of Solution PROBLEM: How many grams of solute are in 1.75L of 0.460M sodium monohydrogen phosphate? c = n/v n n = m/m m SOLUTION: 1.75L 0.460moles 1 L = 0.805mol Na 2 HPO mol Na 2 HPO g Na 2 HPO 4 mol Na 2 HPO 4 = 114g Na 2 HPO 4 Converting a Concentrated Solution to a Dilute Solution molarity or concentration 10

11 Sample Problem Calculating Amounts of Reactants and Products for a Reaction in Solution PROBLEM: PLAN: Specialized cells in the stomach release HCl to aid digestion. If they release too much, the excess can be neutralized with antacids. A common antacid contains magnesium hydroxide, which reacts with the acid to form water and magnesium chloride solution. As a government chemist testing commercial antacids, you use 0.10M HCl to simulate the acid concentration in the stomach. How many liters of stomach acid react with a tablet containing 0.10g of magnesium hydroxide? Write a balanced equation for the reaction; find the grams of Mg(OH) 2 ; determine the mol ratio of reactants and products; use mols to convert to molarity. 1 mass Mg(OH) 2 mol Mg(OH) 2 n=m/m 3 mol HCl V( HCl) 2 look at reaction equation; molar ratio c=n/v 11

12 Sample Problem 3.15 continued Calculating Amounts of Reactants and Products for a Reaction in Solution SOLUTION: Mg(OH) 2 (s) + 2HCl(aq) MgCl 2 (aq) + 2H 2 O(l) n=m/m n(mg(oh) 2 ) = 0.10g 58.33g/mol = 1.7x10-3 mol Mg(OH) 2 The reaction ratio is: 1 mole of Mg(OH) 2 reacts with 2 moles HCl 1.7x10-3 mol x 2 = 3.4x10-3 mol c = n/v 3.4x10-3 mol 0.10mol/L = 3.4x10-2 L HCl Chemical Reactions.. Energetic Effects.. more from the physical chemistry section!!!! Reaction heat of a chemical reaction that takes place at constant pressure is called reaction enthalpie ( H). H positive: endotherm heat is released H positive: exotherm heat is used/needed 12

13 Reaction enthalpies H Stoechimetric factors have to be considered H =ΣH reactants ΣH products exothermic reaction endothermic reaction reactants products Energy products Energy reactants If just 1 mol: ½ H H: does not tell us if a reaction occurs freely on its own.. description of the state of a chemical system Gibbs-Helmholtz Equation free reaction enthalpie entropy all systems strive for a maximum disorder; Increase of particles during a reaction G < 0: spontaneous reaction G = 0: equilibrium G < 0: not a spontaneous reaction exergonic reaction endergonic reaction 13

14 Entropy changes are especially high if during the reaction if the number of moles of gases increases. A positive G value can get compensated if S is positive Thermal Energy Light Energy Electrical Energy Velocity of Reactions. Kinetic a) Reactions can occur with different velocity b) Higher temperature higher reaction velocity c) Some reactions are irreversible, some are reversible Increase of the reaction velocity catalysis Activation Energy without catalyst Energy Activation Energy with catalyst The catalyst is not used during the reaction!! reactants products 14

15 Example: V 2 O 5 catalysis of the H 2 SO 4 Synthesis S + O 2 SO 2 sulphuric acid SO 2 + ½O 2 SO 3 H 2 O Decomposes at high temperatures but reaction has a high activation energy H 2 SO 4 Catalytic Process: 15

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