Review I (Student) Chemistry Chemistry Cheistry.pptx

Review chemistry

1. CHEM 115.3: Review I Course : General Chemistry II – Chemical Processes Term : Fall 2017, Section 01, M/W/F, 9:30-10:20 am Instructor : Dr. Adrian Clark Email: [email protected] Office: Spinks S115

2. Chapter 1 Review I: Matter, Measurement, and Problem Solving Unless stated, all figures displayed were taken from “Tro, Chemistry: A Molecular Approach” or supporting material .

3. Chapter Sections and Topics in Workbook Hours Chapter 1: Review I (Brief Review) 2 Chapter 2: Review II (Review) 3 Chapter 3: Chemical Kinetics 7 Chapter 4: Chemical Equilibrium 5 Chapter 5, 6.1 – 6.3: Acids and Bases 6 Chapter 6.4 – 6.5: Solubility Equilibria 2 Chapter 7: Thermodynamics and Free Energy 6 Chapter 8: Electrochemistry and Redox Reactions 5 Total Hours 36 Pg. 1 Page in Workbook

4. Problem Solving This is a problem solving class which means you will have to put in the work and do the problems. Always write out everything, especially the units (there are no shortcuts to getting the right answer 100% of the time). Very important to keep up with material and do the problem sets yourself. You can’t cram “understanding” how to solve problems. Get help when you need it: TAs, web service, tutorials, personal appointments, Chemistry Learning Centre.

5. Standard Units of Measure Scientists have agreed on a set of international standard units for comparing all our measurements called the SI units ( Système International = International System ). Quantity Unit Symbol length meter m mass kilogram kg time second s temperature kelvin K Pg. 4

6. Kelvin vs. Celsius The size of a “degree” on the Kelvin scale is the same as on the Celsius scale. Technically , we don’t call the divisions on the Kelvin scale degrees; we call them Kelvins ! The 0 standard on the Kelvin scale is a much lower temperature than on the Celsius scale (0 K = absolute zero). Pg. 6 K = °C + 273.15

7. Common Prefix Multipliers Pg. 7 Prefix Symbol Decimal Equivalent Power of 10 mega- M 1,000,000 Base × 10 6 kilo- k 1,000 Base × 10 3 deci - d 0.1 Base × 10 -1 centi - c 0.01 Base × 10 -2 milli - m 0.001 Base × 10 -3 micro- μ 0.000 001 Base × 10 -6 nano - n 0.000 000 001 Base × 10 -9 pico- p 0.000 000 000 001 Base × 10 -12

8. Prefixes All units in the SI system are related to the standard unit by a power of 10. The power of 10 is indicated by a prefix multiplier. The prefix multipliers are always the same, regardless of the standard unit. Report measurements with a unit that is close to the size of the quantity being measured. Pg. 7

9. Prefixes The distance between a hydrogen atom and an oxygen atom in a water molecule is 0.9584 Å ( 1Å = 10 -10 m ). How far away are the atoms in units of picometers? Dimensional Analysis:

10. Prefixes The distance between a hydrogen atom and an oxygen atom in a water molecule is 0.9584 Å ( 1Å = 10 -10 m ). How far away are the atoms in units of picometers? 0.9584 Å       95.84 pm (or 9.584 )  

11. Volume Measure of the amount of space occupied. Any unit of length when cubed becomes a unit of volume. SI unit = cubic meter (m 3 ) Commonly measure solid volume in cubic centimeters (cm 3 ) 1 m 3 = 10 6 cm 3 1 cm 3 = 10 − 6 m 3 = 0.000 001 m 3 Commonly measure liquid or gas volume in milliliters 1 L = 1 dm 3 = 1000 mL = 10 3 mL 1 mL = 0.001 L = 10 −3 L 1 mL = 1 cm 3 Pg. 8

12. Measurement and Significant Figures What is a Measurement? Quantitative observation Comparison to an agreed standard Every measurement has a number and a unit Estimation in Weighing a Pistachio certain 1. 2 g Markings every 1 g estimate

13. A Measurement The unit tells you what standard you are comparing your object to. The number tells you: 1 . What multiple of the standard the object measures 2 . The uncertainty in the measurement Scientific measurements are reported so that every digit written is certain, except the last one, which is estimated.

14. Counting Significant Figures All non- ­zero digits are significant 1.5 has 2 sig. figs. Interior zeros (trapped zeros) are significant 1.05 has 3 sig. figs. Leading zeros are NOT significant 0.00105 has 3 sig. figs. scientific notation = 1.05 10 −3   Pg. 12

15. Counting Significant Figures Trailing zeros may or may not be significant Trailing zeros after a decimal point are significant. Trailing zeros before a decimal point are significant if the decimal point is written. Zeros at the end of a number without a written decimal point are ambiguous and should be avoided by using scientific notation. Pg. 13 1.050 has 4 sig . figs . 150.0 has 4 sig. figs . If 150 has 2 sig. figs. then  1.5 10 2 However , if 150 has 3 sig. figs . then  1.50 10 2  

16. Significant Numbers and Exact Numbers A number whose value is known with complete certainty is exact – from counting individual objects – from definitions i.e. 1 cm is exactly equal to 0.01 m – from integer values in equations i.e. in the equation for the radius of a circle, the 2 is exact Exact numbers have an unlimited number of significant figures Pg. 14 radius of circle =  

17. Determining the Number of Significant Figures How many significant figures are in each of the following? a) 0.04450 m 4 sig. figs. (the digits 4 and 5, and the trailing 0) b) 5.0003 km 5 sig. figs. (the digits 5 and 3, and the interior 0’s) c) 10 dm = 1 m I nfinite N umber of sig. figs. (exact numbers) d) 1.000 × 10 5 s 4 sig. figs . (the digit 1, and the trailing 0’s) e) 0.00002 mm 1 sig. figs. (the digit 2, not the leading 0’s) f) 10,000 m (i.e. 1× 10 4 m in scientific notation) Ambiguous (generally assume 1 sig. fig .)

18. Multiplication and Division with Significant Figures When multiplying or dividing measurements with significant figures, the result has the same number of significant figures as the measurement with the lowest number of significant figures . Pg. 14 a) 5.02 × 89.665 × 0.10 = 45.011... (2 sig. figs) (5 sig. figs) (3 sig. figs) = 45 (2 sig. figs) b ) 5.892 ÷ 6.10 = 0.96590... = 0.966 (3 sig. figs) (4 sig. figs) (3 sig. figs)

19. Addition and Subtraction with Significant Figures When adding or subtracting measurements with significant figures, the result has the same number of decimal places as the measurement with the lowest number of decimal places . Only round numbers at the end of calculation ( not intermediate steps ). Pg. 15 2 . 3 4 5 . 7 + 2 . 9 9 7 5 5 . 4 1 2 5 = 5 . 4 1 5 . 9 - . 2 2 1 5 . 6 7 9 = 5 . 7 a) b )

20. Problem Solving Sort a. Identify the question b . What information do you have or may need to obtain? c. Is all the information relevant? Strategize a. Develop your plan of action b. Identify which parts might be hard c. Note important points (i.e. units) Solve a. Follow your plan. b. Take care of sig. figs. Check a . Does the answer make sense? b . Order of magnitude c . Check your math

21. Basic Principles Law of Conservation of Mass : In an isolated system, matter is neither created nor destroyed.   (i.e. in a chemical reaction, mass of reactants must equal the mass of the products) Law of Definite Proportions : All samples of a compound have the same proportions of their constituent elements. (i.e. any sample of pure water contains 11.19 % hydrogen and 88.81 % oxygen by mass) Dalton’s Atomic Theory : Compounds are composed of atoms of different elements that combine in simple numerical ratios. ( i.e. water composition is two atoms of hydrogen and one atom of oxygen, H 2 O) Law of Multiple Proportions: 1.00 g of C reacts with 1.33 g of O = CO 1.00 g of C reacts with 2.67 g of O = CO 2

22. Mass and Amounts Individual elements have known masses. i.e. C arbon has a mass of 12.011 g/mol (or 12.011 daltons ) Actually , natural carbon is made up of the stable isotopes: 98.892 % 12 C (mass = 12.000 g/mol) 1.108 % 13 C (mass = 13.0035 g/mol) Average atomic mass = p 1 m 1 + p 2 m 2 + …. So if we know the mass of one atom, we can calculate how many atoms exist in a known mass.

23. Mass and Amounts However, we are left with very large numbers – simpler to define a constant: 1 mole = 6.022×10 23 = Avogadro’s Number For example: 1 mole of c arbon consists of 6.022×10 23 carbon atoms. mass (g) = MW (g/mol) × n (mol ) i.e. What is weight of 1.000 mol of carbon? mass = ( 12.01 g/mol) × (1.000 mol) = 12.01 g Pg. 19

24. Sort You are given the amount of copper in moles and asked to find the number of copper atoms. Given: 2.45 mol Cu  Find : Cu atoms Strategize Convert between number of moles and number of atoms by using Avogadro ’ s number as a conversion factor . 6.022 × 10 23 = 1 mol (Avogadro ’ s number) Pg. 22

25. Solve Follow the conceptual plan to solve the problem. Begin with 2.45 mol Cu and multiply by Avogadro ’ s number to get to the number of Cu atoms . Check Since atoms are small, it makes sense that the answer is large. The given number of moles of copper is almost 2.5, so the number of atoms is almost 2.5 times Avogadro ’ s number. Pg. 22 ×      

26. Moles Example: 40 K has a natural abundance of 0.012%. How many moles of 40 K atoms do you ingest in a cup of milk containing 125 mg of potassium? (Note : Usual isotope 39 K ). Mass of (all) K atoms = 125 mg = 0.125 g Moles of (all) K atoms = mass K atoms / atomic mass of K = 0.125 g / (39.10 g/mol) = 3.1969309 10 -3 mol = 3.19 7 10 -3 moles Moles of 40 K atoms is 0.012% of this: = (0.00012 )(3.19 7 10 -3 moles) = 3.8 10 - 7 moles     (%  divide by 100%)

27. Stoichiometry Molecular f ormula : actual number of atoms of each of the elements present in one molecule. i.e. Glucose: C 6 H 12 O 2 Molecular mass (or molecular weight, MW) is simply the sum of the atomic masses. i.e. Water, H 2 O: MW (H 2 O) = 2 MW (H ) + MW (O) = (2 1.008 g/mol) + (15.999 g/mol) = 18.015 g/mol   Pg. 23

28. Stoichiometry Example : What % of the total mass of water is hydrogen? and oxygen ? MW (H 2 O ) = 18.015 g/mol MW (H ) = 1.008 g/mol Mass % H ydrogen = 100 % = 100 % = 11.19 % Mass % Oxygen = 100% – 11.19% = 88.81 %   Pg. 23 2

29. Balancing Chemical Equations To show that a reaction obeys the Law of Conservation of Mass the chemical equation must be balanced. Example: Combustion of Methane: Pg. 33 H H C H H O O O O O O C H O H H O H CH 4 ( g ) + O 2 ( g ) → CO 2 ( g ) + H 2 O ( g ) + → + 2 2 1 - C atoms 4 - H atoms 4 - O atoms 1 - C atoms 4 - H atoms 4 - O atoms

30. Balancing Chemical Equations Example: Fertilizer Chemistry: N 2 ( g ) + H 2 ( g ) → NH 3 ( g ) 2 3 1 molecule 3 molecules 2 molecules 6.022×10 23 3(6.022×10 23 ) 2(6.022×10 23 ) 1 mol 3 mol 2 mol 28 g/mol 2 g/mol 17 g/mol 28 g 6 g 34 g Conservation of Mass! = +

31. Balancing Chemical Equations You are expected to: Be able to balance chemical reactions. Work out mass and mole relationships between reactants and products (converting mass and molarity to moles, etc.). Decide which reactant is limiting and which is in excess. Calculate theoretical yield.

32. Balancing Chemical Equations Tips to Balancing Equations : Never introduce extra atoms to balance ( i.e. “O ”). Never change a formula for the purpose of balancing an equation (i.e . no such thing as CO 3 ). Balance elements that occur in only one compound on each side first. Balance free elements last. Balance unchanged polyatomics as groups (i.e. SO 4 2- ­). Fractional coefficients are acceptable and can be cleared at the end by multiplication (i.e. 1/2 O 2 ).

33. Limiting Reagents Step 1 : Set up equation Step 2 : Balance equation Step 3 : Calculate moles of each reactant Step 4 : Decide what is limiting reagent and what is in excess Step 5 : Calculate maximum number of moles of product that can be made if all of limiting reagent is used up Step 6 : Calculate mass of product using molecular weight of product Step 7 : D etermine the mass of leftover excess reactant Pg. 39

34. Limiting Reagents Example : If you have 500. kg of N 2 and 100. kg of H 2 , what is the maximum mass (in kg) of NH 3 that can be made ? Step 1 : Set up equation Step 2 : Balance equation Step 3 : Calculate moles of each reactant N 2 ( g ) + H 2 ( g ) → NH 3 ( g ) 2 3 Mass (H 2 ) = 100. kg = 1.00 × 10 5 g Moles (H 2 ) = ( 1.00 × 10 5 g)/( 2.016 g/mol) = 4.96 × 10 4 moles Mass ( N 2 ) = 500. kg = 5.00 × 10 5 g Moles ( N 2 ) = ( 5.00 × 10 5 g)/( 28.01 g/mol) = 1.78 5 × 10 4 moles

35. Limiting Reagents Example : If you have 500. kg of N 2 and 100. kg of H 2 , what is the maximum mass (in kg) of NH 3 that can be made? Step 4 : Decide what is limiting reagent and what is in excess Moles ( N 2 ) = 1.78 5 × 10 4 moles Moles (H 2 ) = 4.96 × 10 4 moles From the Chemical equation: The reaction is in a 1 N 2 : 3 H 2 ratio Therefore 1.78 5 × 10 4 moles N 2 will require 5.35 5 × 10 4 moles H 2 to react completely B ut we only have 4.96 × 10 4 moles H 2 ∴ Limiting reagent is H 2 N 2 ( g ) + 3 H 2 ( g ) → 2 NH 3 ( g )

36. Limiting Reagents Example : If you have 500. kg of N 2 and 100. kg of H 2 , what is the maximum mass (in kg) of NH 3 that can be made? Step 5 : Calculate maximum number of moles of product that can be made if all of limiting reagent is used up Limiting reagent is H 2 , (4.96 × 10 4 moles)   N 2 ( g ) + 3 H 2 ( g ) → 2 NH 3 ( g )   4.96 × 10 4 mol H 2 = 3.30 7 × 10 4 mol NH 3 Step 6 : Calculate mass of product using molecular weight of product 3.30 7 × 10 4 mol NH 3   = 563 kg NH 3

37. Limiting Reagents Example : If you have 500. kg of N 2 and 100. kg of H 2 , what is the maximum mass (in kg) of NH 3 that can be made? Step 7 : Determine the mass of leftover excess reactant (optional) 4.96 × 10 4 mol H 2   = 1.65 3 × 10 4 mol N 2 Leftover N 2 = N 2 used up: - 1.65 3 × 10 4 mol N 2 1.78 5 × 10 4 mol N 2 = 1.32 × 10 3 mol N 2 1.32 × 10 3 mol N 2   = 37.0 kg N 2   N 2 ( g ) + 3 H 2 ( g ) → 2 NH 3 ( g )

38. Yields Theoretical yield is the expected yield from a reactant determined by finding which reactant is limiting reagent and thus how much product is expected to form. Actual yield is the amount of product actually produced experimentally. Percent Yield = × 100%   Several Reasons for <100 % yield: 1. Experimental errors and losses 2. Incomplete reaction of all reactants 3. Side reactions or by-products >100 % yield : Often seen with incomplete solvent drying or impurities Pg. 40

39. Yields Example : How many grams of N 2 ( g ) can be made from 9.05 g of NH 3 ( g ) reacting with 45.2 g of CuO ( s )? 2 NH 3 ( g ) + 3 CuO ( s ) → N 2 ( g ) + 3 Cu( s ) + 3 H 2 O( l ) If 4.61 g of N 2 are made experimentally, what is the percent yield? Nearly equivalent amount of moles of each reactant, however in the reaction 2 NH 3 : 3 CuO ratio. Therefore the CuO is the limiting reagent . Chemical equation is already setup and balanced. Moles of each reactant : 9.05 g NH 3   = 0.531 4 mol NH 3 45.2 g CuO   = 0.568 1 mol CuO = 5.31 g N 2 (Theoretical Yield)       45.2 g CuO Overall: 4.61 g N 2 (Actual Yield) Percent Yield = × 100%   = × 100%   = 86.8 %

40. Solutions Molarity (M) is a measure of the concentration of a substance (solute) dissolved in a solvent. Units are in moles/L (or M) C ommon method used to express concentrations Pg. 44 Molarity (M) =  

41. Solutions Other common ways of expressing solution concentrations : % by weight = % by volume = parts per million (ppm) ppm = × 10 6 Molality =   Not used as often nowadays

42. Pg. 46

43. Solutions Example: What i s the molarity of a solution containing 3.4 g of NH 3 ( MW = 17.03g/mol) in 200.0 mL of solution? 3.4 g NH 3   = 0.2000 L 200.0 mL   = 0.19 96... mol NH 3     M = 1.0 M

44. Dilutions Often, solutions are stored as concentrated stock solutions. To make solutions of lower concentrations from these stock solutions, more solvent is added. (the amount of solute doesn’t change, just the volume of solution) moles solute in Solution 1 = moles solute in Solution 2 The concentrations and volumes of the stock and new solutions are inversely proportional. M 1 ∙V 1 = M 2 ∙V 2

45. Dilutions Example : What is the concentration of a solution prepared by diluting 45.0 mL of 8.25 M HNO 3 to 135.0 mL? M 1 ∙V 1 = M 2 ∙V 2 (8.25 M) (0.045 L) = M 2 ∙V 2 (8.25 M) (0.045 L) = M 2 (0.135 L) ( 8.25 M) (0.045 L ) = M 2 (0.135 L) M 2 = 2.75 M

46. Pg. 49 Sort You are given the volume and concentration of a Pb(NO 3 ) 2 solution. You are asked to find the volume of KCl solution (of a given concentration) required to react with it. Given : 0.150 M KCl solution 0.150 L of a 0.175 M Pb(NO 3 ) 2 solution Find : volume KCl solution (in L )?

47. Pg. 49 Strategize Use the molar concentrations of the KCl and Pb (NO 3 ) 2 solutions as conversion factors between the number of moles of reactants in these solutions and their volumes. Use the stoichiometric coefficients from the balanced equation to convert between number of moles of Pb (NO 3 ) 2 and number of moles of KCl .

48. Pg. 49 Strategize

49. Pg. 49 Solve Begin with liters of Pb(NO 3 ) 2 solution and follow the conceptual plan to arrive at liters of KCl solution. 0.150 L Pb(NO 3 ) 2     ×   = 0.350 L KCl solution

50. Pg. 49 Check The final units (L KCl solution) are correct. The magnitude ( 0.350 L ) is reasonable because the reaction stoichiometry requires 2 mol of KCl per mole of Pb(NO 3 ) 2 . Since the concentrations of the two solutions are not very different ( 0.150 M compared to 0.175 M), the volume of KCl required is roughly two times the 0.150 L of Pb(NO 3 ) 2 given in the problem.

51. End of Review I For additional practice, try the For Practice questions or Problems by Topic questions at the end of the Chapter (see the solutions provided in the Workbook).