1. LECTURE 3 CHM 151 ©slg TOPICS: 1. DENSITY Calculations 2. PERCENT Calculations 3. The Nuclear Atom 4. Atomic Weight, Isotopes KOTZ: 1.7, 2.1-2.6

2. Common Density Values Magnesium: 1.74 g/cm 3 Aluminum: 2.70 g/cm 3 Silver: 10.5 g/cm 3 Gold: 19.3 g/cm 3 Dry air: 1.2 g/L at 25 o C, 1 atm pressure Water:.917 g/mL at 0 o C 1.00 g/ mL at 4.0 o C.997 g/mL at 25 o C Sea Water: 1.025 g/mL at 15 o C Antifreeze: 1.1135 g/mL at 20 o C

3. SOLVING DENSITY PROBLEMS 1. Density itself can be calculated from experimental values: Density = mass of object, solution, substance volume occupied (mL, cm 3, L) Volume can be determined in several ways: a) direct measurement, liquid b) liquid displacement, solid c) measurement of dimensions, calculation

4. Volume by Displacement: A liquid in which the solid does not dissolve is suitable for this technique

5. Calculation, Volume: V = l X w X ht (rectangle) V = e 3 (cube) V =  r 2 ht (cylinder) radius = diameter / 2; all dimensions in same units diameter height l w ht

6. 2. In all other cases, where conversion of mass to volume or volume to mass is desired, density is used as a “conversion factor”: SOLVING DENSITY PROBLEMS 2.70 g Al = 1 = 1 cm 3 Al 1 cm 3 Al 2.70 g Al “conversion factors” D, Al = 2.70 g /cm 3 2.70 g Al = 1 cm 3 Al

7. Sample, Obtaining Density Value A “cup” is a volume used by cooks in US. One cup is equivalent to 237 mL. If one cup of olive oil has a mass of 205 g, what is the density of the oil in g/mL and in oz avoir / in 3 ? 1. State the question: D, olive oil = ? g/ mL = ? oz avoir / in 3 2. State formula: Density = mass, g volume, mL

8. 3. Substitute values and solve: D = mass = 205 g =.86497 g =.865 g volume 237 mL mL mL Second question: What is this density factor expressed as oz avoir/ in 3 ? 1. State Question:.865 g = ? oz avoir mL in 3

9. 2. State relationships: 453.6 g = 1 lb 1 lb = 16 oz avoir 1 mL = 1 cm 3 2.540 cm = 1 in ( 2.540 cm) 3 = (1 in) 3 16.39 cm 3 = 1 in 3.865 g = ? oz avoir mL in 3 Pathway: g  lb  oz avoir; mL  cm 3  in 3

10. 3. Setup and Solve:.865 g X 1 lb X 16 oz avoir X 1 mL X 16.39 cm 3 1 mL 453.6 g 1 lb 1 cm 3 1 in 3 =.50008 oz avoir in 3 =.500 oz avoir in 3.865 g = ? oz avoir mL in 3

11. Density as a Conversion Factor: Peanut oil has a density of.92 g/mL. If the recipe calls for 1 cup of oil (1 cup = 237 mL), what mass of oil, in lb, are you going to use? State question: 237 mL oil =? lb oil State relationship: ( D, oil=.92 g / mL) 1 mL oil =.92 g oil 453.6 g = 1 lb Pathway: mL  g  lb

12. Solution: 237 mL oil =? lb oil Pathway: mL  g  lb 237 ml oil X.92 g oil X 1 lb =.48068 lb oil 1 mL oil 453.6 g =.48 lb oil Density conversion factor

13. GROUP WORK: Setup and solve as Conversion Type Problem A gold coin is 2.75 cm in diameter and.50 cm thick. If the density of gold is 19.3 g / cm 3, what is the mass of the coin in grams? Note: Volume =  r 2 ht r = d / 2 ht = “thickness” D, Au = 19.3 g/cm 3 19.3 g Au = 1 cm 3 Au

14. d = 2.75 cm, r = 2.75 cm / 2 = 1.375 cm ht =.50 cm V=? =  r 2 ht = 3.1416 X (1.375 cm) 2 X.50 cm V= 2.9698 cm 3 = 3.0 cm 3 First step: solve for volume Second step: solve for mass 3.0 cm 3 Au = ? g Au 19.3 g Au = 1 cm 3 Au 3.0 cm 3 Au X 19.3 g Au = 57.9 g Au = 58 g Au 1 cm 3

15. Percent Composition by Mass “Percent composition” is a convenient way to describe a mixture or solution or compound in terms of mass of the part contained in 100 mass units of the whole: “This solution is 15% salt” means that for every 15 g of salt there is 100 g of solution or: 15 g salt = 100 g solution “The brass alloy is 15 % tin and 45% copper” 100 g alloy = 15 g tin = 45 g copper

16. Like Density, Percent Composition is: calculated from experimental values used as a conversion factor. Calculation of % by mass (g): % Part = g part X 100% g whole A brass alloy weighing 79.456 g was found to contain 34.29 g of copper. What is the % of Cu in the alloy? %Cu = 34.29 g Cu X 100% = 43.16% Cu 79.456 g alloy = 43.16 g Cu in 100 g alloy

17. Percent as a Conversion Factor What mass in grams of a brass alloy would contain 25.00 g Cu if the alloy is 43.16% Cu? Question: 25.00 g Cu = ? g brass Relationship: 43.16 g Cu = 100 g brass Setup and solve: 25.00 g Cu X 100 g brass = 57.92 g brass 43.16 g Cu % factor

18. Density/Percent Solution Problems Automobile batteries are filled with sulfuric acid, which is a solution of liquid H 2 SO 4 in water. What is the mass (in grams) of the H 2 SO 4 in 500. mL of the battery acid solution, if the density of the solution is 1.285 g /mL and the solution is 38.08% H 2 SO 4 by mass?

19. Analysis of Problem 1.“Describing the Scene”: Automobile batteries are filled with sulfuric acid, which is a solution of liquid H 2 SO 4 in water. 3.“Giving the Relationships”: if the density of the solution is 1.285 g /mL and the solution is 38.08% H 2 SO 4 by mass? 2. “Stating the Question”: What is the mass (in grams) of the H 2 SO 4 in 500. mL of the battery acid solution

20. Automobile batteries are filled with sulfuric acid, which is a solution of liquid H 2 SO 4 in water. What is the mass (in grams) of the H 2 SO 4 in 500. mL of the battery acid solution, if the density of the solution is 1.285 g /mL and the solution is 38.08% H 2 SO 4 by mass? Question: 500. mL soltn = ? g H 2 SO 4 Relationships: 1.285 g soltn = 1 mL soltn 38.08 g H 2 SO 4 = 100 g soltn Pathway: mL soltn  g soltn  g H 2 SO 4

21. Question: 500. mL soltn = ? g H 2 SO 4 Relationships: 1.285 g soltn = 1 mL soltn 38.08 g H 2 SO 4 = 100 g soltn Pathway: mL soltn  g soltn  g H 2 SO 4 Setup and solve: 500. mL soltn X density factor X % factor = g H 2 SO 4 500. mL soltn X 1.285 g soltn X 38.08 g H 2 SO 4 1 mL soltn 100 g soltn = 244.664 g H 2 SO 4 = 245 g H 2 SO 4

22. Group Work Automobile batteries are filled with sulfuric acid, which is a solution of liquid H 2 SO 4 in water. How many mL of the acid solution would contain 15.00 g H 2 SO 4, if the density of the solution is 1.285 g /mL and the solution is 38.08% H 2 SO 4 by mass? Question: 15.00 g H 2 SO 4 = ? mL soltn Relationships: 1.285 g soltn = 1 mL soltn 38.08 g H 2 SO 4 = 100 g soltn Pathway: g H 2 SO 4  g soltn  mL soltn % D

23. Solution Question: 15.00 g H 2 SO 4 = ? mL soltn Relationships: 1.285 g soltn = 1 mL soltn 38.08 g H 2 SO 4 = 100 g soltn Pathway: g H 2 SO 4  g soltn  mL soltn Setup and Solve: 15.00 g H 2 SO 4 X 100 g soltn X 1 mL soltn = 38.08 g H 2 SO 4 1.285 g soltn = 30.65 mL soltn

24. Chapter 2: Atoms and Elements 1. History of discovery of the atomic nature of all matter and the various particles found within the atom: Kotz, 2.1-2.2, excellent reading on all topics. Saunders CD-ROM: animated film clips Reading, viewing assignment

25. Atomic Structure, Atomic Number, Mass All matter (anything that has mass and occupies volume) can be classified as: an element (basic building blocks of nature: H, O, Au ) a compound (made up of two or more elements) a mixture (any physical combination of the above) Elements, the simplest forms of matter, are composed of unique tiny particles called atoms.

26. Atomic Structure The atom itself is composed of three types of “subatomic particles”, the proton (p), the neutron (n), and the electron (e). Each element has its own unique pairings of these three particles. It is the number and the placement of these particles which gives rise to the different properties exhibited by each element.

27. Comparative Mass, Charge: Nuclear Particles

28. Nuclear Particle Location Within the Atom 1. Protons and Neutrons: “Nucleus of Atom” compact positive mass in center of atom “all” of the mass, negligible volume (10 -2 pm) 2. Electrons: “Outside the Nucleus” cloud of negative charge “all” of the volume (10 3 pm), negligible mass

29. THE “NUCLEAR” ATOM Nucleus, C atom Electron cloud.

30. ATOMIC SYMBOLS Each atom of an element can be represented by a symbol that describes how many protons, neutrons and electrons are contained in this basic unit: X A Z Mass number, A: total #, p + n Atomic Number, Z: #p (equals #e) Elemental symbol

31. Accordingly, A, the mass number of the atom, represents both the total number of nuclear particles (p + n) and the approximate mass of the atom (in amu’s) Because atoms (and the subatomic particles) are so tiny, a relative mass scale was setup to describe atomic weights in a convenient numerical range. The atom of Carbon which contains 6 protons and 6 neutrons in its nucleus is assigned the weight of 12.00 amu (atomic weight units), which essentially makes the mass of each proton and neutron 1.00 amu. The Mass Number, A

32. Z, The Atomic Number All known elements are listed in the familiar “Periodic Table of the Elements” in order of increasing atomic number, found usually in the upper right hand corner above each element’s symbol. The atomic number represents the number of protons in the nucleus of every atom of that element. Since every atom is electrically neutral, the number of protons in the nucleus represents the total positive charge of the nucleus, which is exactly balanced by the total number of negatively charged electrons outside the nucleus.

33. Group Work: Atomic Symbols

35. Isotopes of the Elements For a given element, the number of protons and electrons is a fixed value and determines which element is being described. The number of neutrons in the atom of a given element is not fixed, and several different atoms of a given element are generally found, varying in atomic mass due to differing numbers of neutrons. The different atoms of a given element which vary by neutron count and by mass are described as “isotopes” of that element.

36. Isotopic Symbols Isotopic symbols represent specific isotopic forms of an element. Consider hydrogen:

37. Atomic Mass, revisited: The atomic mass of any isotope of an element can be approximated by a simple addition of the number of p’s and n’s in the nucleus. However, naturally occurring samples of any element generally include several different isotopes of different atomic masses. The atomic mass value given for each element (as found in your Periodic Table) is a weighted average of all the isotopes.

38. The given atomic mass for any element will be closest to the most abundant element in most cases. Consider below the calculation of the atomic mass of magnesium, 24.305 amu: Average Mass Element, amu = (mass, isotope A X % abundance A) + (mass, isotope B X % abundance B) + (mass, isotope C X % abundance C)...... General Method of Calculation:

39. Calculation of average atomic mass of Magnesium: (23.9850 x.7899) + (24.9858 x.1000) + (25.9826 x.1101) = (18.95) + (2.499) + (2.861) = 24.31 amu