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The Mole Concept: “How Chemist’s Measure” and Stoichiometry: “Balancing of Chemical Rxns” Ch# 3, Ch#7, Ch#9.

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Presentation on theme: "The Mole Concept: “How Chemist’s Measure” and Stoichiometry: “Balancing of Chemical Rxns” Ch# 3, Ch#7, Ch#9."— Presentation transcript:


2 The Mole Concept: “How Chemist’s Measure” and Stoichiometry: “Balancing of Chemical Rxns” Ch# 3, Ch#7, Ch#9

3 I. Chemistry = Quantitative Science A.Quantitative: Analyze/describe via definite numeric results. Ex:___________________ B.Qualitative: Analyze via nonnumeric or descriptive results (words) Ex:____________________ C.We measure the amt. of matter to do calculations relating quantities of reactants to products to chemical equations. REACTANTS PRODUCTS Yields

4 D. We often use very small and very large numbers in chemistry. E. Scientific notation is a method to express these numbers in a manageable fashion. -Thus 0.000 000 1 cm can be written 1 x 10 -7 cm. F.Scientific notation expresses a number as the product of two factors, the first falling between 1 and 10 and the second being a power of 10.

5 Format is Mantissa x Base Power We just move the decimal point around. Decimal part of original number Decimal you moved 6.02 x 10 23 602000000000000000000000

6 G. Converting into scientific notation: 1.Move decimal until there’s 1 digit to its left. Places moved = exponent. 2.If number converting is: a)Large # (>1)  positive exponent b)Small # (<1)  negative exponent 3.Only include sig. figs. 0. 0 0 0 0 2 2 0 5 2,500,080,000

7 H. Powers of 10 to know: Metric 0.000 000 001 or 0.000 001 or 0.001 or 0.01 or 0.1 1000 or

8 Symbols Used in Chemical Equations “Yields”; indicates result of reaction Used to indicate a reversible reaction A reactant or product in the solid state; also used to indicate a precipitate Alternative to (s), but used only to indicate a precipitate A reactant or product in the liquid state A reactant or product in an aqueous solution (dissolved in water) A reactant or product in the gaseous state (s)(s) (l)(l) (aq) (g)(g)

9 Additional Symbols Used in Chemical Equations Alternative to (g), but used only to indicate a gaseous product Reactants are heated Pressure at which reaction is carried out, in this case 2 atm Pressure at which reaction is carried out exceeds normal atmospheric pressure Temperature at which reaction is carried out, in this case 0 o C Formula of catalyst, in this case manganese (IV) oxide, used to alter the rate of the reaction, NOT CONSUMED Example: H 2 O 2 (aq) H 2 O (l) + O 2 (g) 2 atm pressure 0 o C MnO 2 

10 II. Measuring Matter A.Measure by: 1)Counting (How many Ss are in this class?) 2)Weighing (Au sells for $25/gram) 3)Volume (The National Ave. for gas is $3.05/gal) 4)Sometimes by each ( bananas 10¢/ea, by lb., by quart) B.Chemists needed a counting unit: mole = Amt. of Substance

11 III. The Mole (Counting) A.Matter is made of Particles: (Representative) 1)Elements = Atoms 2)Ionic Cmpds = Formula unit (3D/ Cube) 3)Molecular Cmpds = Molecules 4)1 mole is the amount of substance that contains as many particles (atoms or molecules) as there are in 12.0 g of C-12. B.Count them using Avagadro’s Number: 1)Dozen (doz) eggs = 12 eggs 2)1 mole = 6.02 X 10 23 Particles

12 Pics of Particles/Pieces

13 3. Amedeo Avagadro (1766-1856) – Italian scientist & lawyer who first tried to experimentally determine the # of particles in a mole. –never knew his own number; –it was named in his honor by a French scientist in 1909. –its value was first estimated by Josef Loschmidt, an Austrian chemistry teacher, in 1895.

14 Amadeo Avogadro (1776 – 1856) 1 mole = 602213673600000000000000 or 6.022 x 10 23 particles thousands millions billionstrillions quadrillions ? There is Avogadro's number of particles in a mole of any substance! Particles in a Mole

15 How Big is a Mole? One mole of marbles would cover the entire Earth (oceans included) for a depth of two miles. One mole of $1 bills stacked one on top of another would reach from the Sun to Pluto and back 7.5 million times. It would take light 9500 years to travel from the bottom to the top of a stack of 1 mole of $1 bills.

16 Welcome to Mole Island 1 mole = 22.4 L @ STP 1 mol = molar mass 1 mol = 6.02 x 10 23 particles

17 Mass, Volume, Mole Relationship

18 IV. Particle Conversions: A. Particles  Moles & Moles  Particles 1.How many moles of Mg is 3.01X10 22 Atoms of Mg? Unknown = Known = Conv: Soln: 3.01 x 10 22 Atoms Mg X 1 Mol Mg 6.02 x 10 23 Atoms Mg = 5.00 X 10 -2 Mol Mg

19 2. How many moles are 1.20 X 10 25 atoms of phosphorus? 3. How many atoms are in 0.750 mol of zinc?

20 4. How many molecules are there in 4.00 mol of glucose, C 6 H 12 O 6 ? Atoms, F. U.’s, or Molecules

21 B. Analogy: Have a doz. packages of pencils. How many pencils do you have? 8 in a package = 8 x 12 = 96 C. To find the # of atoms in a Mol of a cmpd, must determine the # of atoms in a representative formula of that cmpd. 1)How many Fluoride ions are in 1.46 mol of aluminum fluoride?

22 2)How many ammonium ions are in 0.036 mol ammonium phosphate? 3)How many C atoms are in a mixture of 3.00 mol C 2 H 2 and 0.700 mol CO?

23 V. Mole (Mass/Weight) A.Learned - mass of a single atom is given in amu (atomic mass units). B.Gram Atomic Mass (gam) – # of grams of an element numerically equal to amu (we will use the entire #). Ex. C (gam) = 12.01115 g/mol (grams/mole) H (gam) = 1.0079 g/mol C.The gam = 1 mole = 6.02 x 10 23 particles for a mono-atomic element!!!

24 1 Mole of Particles Do They look alike ? Do you think they will have the same mass if 1 mole?

25 Ex. What is the mass of 1 mol of each of these monatomic elements? a.Sodiumb. Arsenicc. Uranium D.What is the mass of 1 mole of a cmpd? 1)Formula of a cmpd tells us the # of atoms of each element in a rep. particle of that cmpd. 2)Calculate the gram molecular mass (gmm) or gram formula mass (gfm) by adding together the atomic masses of the atoms making up a molecule or F.U.

26 3)Examples: Sulfur Trioxide: Ammonium Phosphate: Aluminum Sulfate:

27 E.gam, gfm, & gmm are commonly referred to as molar mass in chemistry. (1 mol= mass from pd table= 6.02x10 23 parts) F.Mole  Mass (g) Conversions use Molar Mass 1)Molar mass of an element or cmpd is used to convert grams of a substance into moles and vice versa. 2)Examples: a)How many grams are in 7.20 mol of dinitrogen trioxide?

28 b)Find the number of moles in 922g of iron (III) oxide. c)What is the mass of 2.65 x 10 24 molecules of Carbon Tetrachloride?

29 d)A sample of chlorous acid has a mass of 3500.1 grams, how many representative particles are present in this sample?

30 Each substance is a mole, so why do they look so different?

31 . One mole of NaCl(the solid) mass = 58.45 g One mole Water, mass= 18.0 g occupies a volume of 18.0 mL. One mole of the O 2 gas has a mass of 32.0 g & 22.4L volume One mole each of a solid, a liquid, and a gas :

32 VI. Mole (Volume of A Gas) A.Volumes of moles of different substances can be different. B.Vol of a mole of gas is much more predictable. C.Vol of a gas is measured at STP (Standard Temperature & Pressure) 1.Standard Temp = 0 o C 2.Standard Pressure = 1 atm (atmosphere) How do we measure pressure?

33 D.At STP, 1 mol of any gas occupies a vol of 22.4 L aka molar volume of a gas. 1 mol = 6.02 x 10 23 P = Molar Mass ? = 22.4L E.Problems: 1. Determine the volume, in liters, of 0.600 mol of SO 2 gas at STP.

34 2.Determine the number of moles in 333.6 mL of He gas.

35 3.What is the volume at STP for 3.25 x 10 24 molecules of CO 2 gas?

36 Homework Problems: 1. What is the vol at STP of 0.960 mol CH 4 ? 2. Determine the vol (in L) of 127.80 g of Titanium (II) Oxide gas.

37 VII. Calculating Percent Composition A.When a new cmpd is made in a laboratory, we need to determine its formula. B.So we use percent composition which is simply finding the percent (%) by mass of each element in a compound. C.The % comp includes as many %’s as there are elements in the cmpd.

38 For Example: K 2 CrO 4 % comp: 40.3% K, 26.8% Cr, & 32.9% O (∑=100%) D. Percent Composition Formula: is the # of grams of the element divided by the grams of the cmpd, multiplied by 100% Or % Comp = grams of element x 100% grams of cmpd We Can Use This Formula 2 Ways!!!!

39 Most Common Use:

40 E.Example Problems: (given g amts 1 mol  use #’s directly) 1.An 8.20 g piece of Mg combines completely w/ 5.40 g of O to form a cmpd. What is the % comp of this cmpd?

41 2.Calculate the % comp of ethane, C 2 H 6. (Given no #’s so you assume 1 mol & use molar mass)

42 VIII. Determining Formulas from % Comp: A.Empirical Formula: formula showing the smallest whole number ratio of elements present. C 6 H 12 O 6 = Molecular Formula CH 2 O = Empirical Formula

43 B. Finding Empirical Formula: 1. Find % Composition: -Unknown has 63% Mn & 37% O 2. Convert % to moles:(Assume 100g sample) Mn:63 g Mn x 1 mole Mn = 1.1 mole Mn 54.9 g Mn O:

44 3. Convert Moles to Whole Numbers: Method: divide all molar amounts by the smallest molar amount Mn: 1.1 mol Mn = 1 mol Mn 1.1 MnO 2 O: 2.3 mol O = 2.1 mol O 1.1 Rule: round #’s to nearest whole # unless…. If answer is 0.4, 0.5, 0.6 then double ALL!

45 Practice Problem 1 Given the following data, find the correct empirical formula: 49.0% C, 2.70% H, 48.2% Cl

46 Practice Problem 2 Given the following data find the empirical formula: N = 26.2%, H = 7.50%, Cl = 66.4%

47 C. Molecular Formula: 1.First find the empirical formula. 2.Then compare the formula mass for your empirical formula with the formula mass for the molecular formula. 3.You may have to multiply the subscripts in your empirical formula by some factor.

48 Example #1: Given: 38.7% C, 9.70% H, 51.6% O and a Molecular mass of (gmm=62.0g) find the true molecular formula: 1.Empirical Formula: 2. gmm of E.F.: 3. Divide Masses: 4. Multiply E.F. by Factor:

49 Practice Problem #2 Given the following data, find the correct molecular formula: 24.3% C, 4.1% H, 71.6% Cl: gmm = 99.0g

50 Practice #3 Given the following data, find the correct molecular formula: 54.6% C, 9.00% H, 36.4% O and gmm: 176g

51 IX. Balancing Chemical Equations A.All equations we have written have been correct in Qualitative sense but not Quantitative. B.Equation must be balanced so that they are quantitatively correct. --balanced equation: each side of the equation has the same number of atoms of each element. Law of Conservation of Mass!!! Atoms can’t be created or destroyed just REARRANGED!

52 C. Sometimes we write equations and it will already be balanced: C (s) + O 2 (g)  CO 2 (g) D. Other equations we must correct or balance the quantities of reactant atoms to product atoms: C (s) + O 2 (g)  CO (g) H 2 (g) + O 2 (g) -----> H 2 O (l) Pt

53 E.The numbers in front are called STOICHIOMETRIC COEFFICIENTS –These tell us the # of moles/subscripts =________ F.Guidelines For Balancing: Mostly Trial and Error Do Not Change Subscripts!!!! 1.Count the number of atoms on each side 2.Polyatomic ion unchanged (consider it one unit and balance first) 3.Even / odd (make all even) 4.Single Elements last 5.Make sure all coefficients are in simplest (lowest) ratio AgNO 3 (aq) + Cu(s)  Cu(NO 3 ) 2 (aq) + Ag (s)

54 More Examples: (notice no physical states) (NH 4 ) 3 PO 4 + Mg(OH) 2  Mg 3 (PO 4 ) 2 + NH 4 OH AlCl 3 + Li 2 CO 3  Al 2 (CO 3 ) 3 + LiCl P + O 2  P 2 O 5

55 X. Mole Ratio is The KEY A.Balanced equations show proportions: Synthesis of Water – 2H 2 + O 2  2H 2 O For Every 2 H molecules we need 1 O molecule! B. Calculations involved in chemical rxns use the proportions to find the quantity of reactant and product. - Assume all Rxns go to completion

56 C.Since Avogadro’s # relates molecules to moles the coefficients also tells us the # of moles needed for a rxn to progress successfully. D.Stoichiometry-the proportional relationship between 2 or more substances during a chemical rxn.

57 E. Converting Between Amounts in Moles 1.Identify the amount in moles you know. 2.Using coefficient from eqn., set up mole ratio with Known on Bottom 3.Multiply the original amount by the mole ratio and finish problem. Known Unknown

58 Example: Consider the rxn for the commercial prep of Ammonia. N 2 + 3H 2  2NH 3 How many moles of Hydrogen are needed to prepare 312 moles of ammonia?

59 Example: What mass of NH 3 can be made from 1221 g H 2 and excess N 2 ? Plan:

60 Use: Fe 2 O 3 + 2Al  2Fe + Al 2 O 3 1.How many g of Al are needed for 135g Fe 2 O 3 ? 2.How many g of Al 2 O 3 can form when 23.6g Al with excess Fe 2 O 3 ? 3.How many g of Fe 2 O 3 react with excess Al to make 475g Fe? 4.How many g of Fe will form when 97.6g Al 2 O 3 form?

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