1. Law of Conservation of Mass: Mass is neither created nor destroyed in a chemical reaction. Therefore in a chemical reaction, the Mass of Reactants (Left side of arrow) must equal the Mass of Products (Right side of arrow). For the general reaction: A + B → C + D where A, B, C, D represent chemicals m A + m B = m C + m D Mercury (II) Sulfide Makes Mercury + Sulfur If 216.0 g of Mercury (II) sulfide are used, and 200.0 g of Mercury are made, how many grams of Sulfur must have been made? Example: 216.0 g = 200.0 g + m S m S = 216 g – 200.0 g = 16.0 g of S made Make special note of all problems! Expect to have to do this on a quiz or test! HgS → Hg + S (reactant) (products) Click to see Solution

2. N 2 (g) + 3 H 2 (g) → 2 NH 3 (g) 1) If 28.0 g of N 2 react exactly with 6.0 g of H 2, how many grams of NH 3 are produced? 2) If 53.5 g of NH 3 are produced by the reaction of 42.42 g of N 2 with excess H 2, how many grams of are H 2 consumed? 2 NH 3 (g) → N 2 (g) + 3 H 2 (g) 3) If 128.4 g of NH 3 are reacted to produce 98.4 g of N 2, how many grams of are H 2 produced? Law of Conservation of Mass Practice N 2 (g) + 3 H 2 (g) → 2 NH 3 (g)

3. Law of Definite Proportions: A pure substance always contains the same proportions by mass of each atom in the substance regardless of where the sample of the substance was found or how much of the substance was analyzed. This law really just tells us that if we have a sample of some chemical (like water for example), then the chemical will have to have the same formula if it really is the same substance. We know that water has the formula H 2 O. No matter where we find “water” it will always have the formula H 2 O. If we find something that has the formula H 2 O 2 for example, we will know that it is not water because its formula is different. Percent by Mass (or Mass Percent) Mass % = Mass of Element Mass of Compound x 100 The Law is worded the way it is because we can not “see” molecules to determine their formulas. Instead, we calculate their formulas based on the mass percent of each element present in a sample of the substance.

4. Mass % Example: Mass% Na = m Na m NaCl x 100 Make special note of all problems! Expect to have to do this on a quiz or test! Sodium chloride (NaCl) is known to have 39.34% sodium ions (Na +1 ) and 60.66% chloride ions (Cl  1 ). If we have a 14.67 g sample of sodium chloride, how many grams of sodium ion will be present in the sample? How many grams of chloride ion would be in the sample? Given: Mass% Na = 39.34% m Na = ? Mass% Cl = 60.66% m Cl = ? m NaCl = 14.67 g Finding the mass of Cl  would be similar using: Mass% Cl = m Cl m NaCl x 100 m Na = (Mass% Na )(m NaCl ) 100 m Na = (39.34)(14.67 g) 100 = 5.771 g Na +1 Click to see Solution

5. 1) If 28.0 g of a sample of compound are found to contain 8.00 g of calcium and 20.00 g of iodine, what is the mass % of calcium in the sample? What is the mass % of iodine in the sample? 2) If a compound is known to be 22.00% carbon and 78.00% chlorine by mass, how many grams of carbon would there be in 3.721 g of the compound? How many grams of chlorine? Mass % Practice

6. Law of Multiple Proportions: If two or more compounds are composed of the same two elements, then the mass ratio(s) of one of the elements will always be a ratio of small whole numbers. Example: at least three different compounds containing just chromium and chlorine are known. Data for these three compounds is given in the table below. Using the data in the table, the law of multiple proportions says that the ratios of the masses of the chlorine will be small whole numbers. Ratio of chlorine in B to A is 1.4998 or 1.500 which is 3/2. Ratio of chlorine in C to B is 2.000 which is 2/1. Ratio of chlorine in C to A is 3.000 which is 3/1. What does this do for us in terms of our understanding of chemistry?

7. John Dalton proposed a new atomic theory based on the three Laws we have been discussing and the ideas of Democritus. See page 64 for a discussion of Daltons ideas. Dalton’s Atomic Theory (1808 A.D.)-very similar to ideas of Democritus (~400 B.C.) Each element is made up of tiny particles called atoms that can not be broken down into smaller particles. The atoms of a given element are identical; the atoms of different elements are different in some fundamental way or ways. Chemical compounds are formed when atoms combine with each other in simple whole number ratios. A given compound always has the same relative numbers and types of atoms. Chemical reactions involve reorganization of the atoms - changes in the way they are bound together. The atoms themselves are not changed in a chemical reaction. Was Dalton 100% correct? Atomic Theories (theories of what atoms are like)

8. Cathode Ray: stream of charged particles produced by an electric field. The green “glow” in the picture. The particles move from the cathode (negatively charged plate) to the anode (positively charged plate).

9. Cathode “rays” are influenced by both magnetic and electric fields (electric field shown in the diagram). Opposite charges are known to attract each other. Since the cathode “ray” was deflected towards the positively charged plate, what charge do the particles in the cathode “ray” have?

10. J. J. Thompson and Millikan demonstrated that the mass of the particles in a Cathode Ray are much less than the mass of the smallest atom. How did this change our understanding of Dalton’s theory? Plum Pudding Model: Electrons are evenly distributed throughout a uniform positive charge equal to the negative charge of all of the electrons. The part with the positive charge is where most of the mass of an atom is located. J. J. Thompson proposed the Plum Pudding Model

11. Ernest Rutherford (1911) “gold foil” experiment Large, high energy particles (alpha particles, +2 charge) do not always pass straight through a thin sheet of matter. The alpha particle being deflected or reflected is like having a 30-06 bullet bounce off of a sheet of tissue paper.

12. a) Results of golf foil experiment expected if Plum Pudding Model were true. b) Results of golf foil experiment explained by new model. Rutherford proposed a nuclear model of the atom. The small, dense nucleus contains virtually all the mass of the atom and all of the positive charge while the negatively charged electrons exist apart from the nucleus. Rutherford did not know where the electrons were-they were just outside of the nucleus. How does this model fit Rutherford’s results?

13. If all atoms contain the same types of particles, what makes one atom hydrogen and another carbon? Nuclear Model of an Atom The number of protons in the nucleus determine what element an atom is. Remember that in normal atoms, the number of protons is equal to the number of electrons (they are neutral-a total charge of zero). How many times bigger is the atom compared to the nucleus? Small, very dense nucleus containing massive protons and neutrons, surrounded by small rapidly moving electrons If the nucleus is very small compared to the overall size of the atom, then the atom consists of mostly empty space!

14. Rutherford called the positive particle in the nucleus a proton. In 1932, James Chadwick discovered a neutral particle in the nucleus and called it a neutron. Summary of Subatomic Particles (mid 1900’s) Particle SymbolsRelative ChargeMass #Relative MassActual Mass Electron  1 0 1/18379.11X10  31 kg Proton +1 1 11.67X10  27 kg Neutron 0 1 11.68X10  27 kg Beta  1 0 1/18379.11X10  31 kg Alpha  +2 4 4 6.64X10  27 kg pp  1 1 H ee 11 0 e nn  0 1 n  2 4 He Other Related Particles (mid 1900’s) ee 11 0 e 11 0 

15. Atomic Number: Atomic number is the number of protons in the nucleus of an element. Since in neutral atoms the number of protons is equal to the number of electrons, atomic number also indicates the number of electrons in a single atom of an element. The Periodic Table is arranged according to atomic number. For example, hydrogen is element 1 and has 1 proton in its nucleus. Similarly, helium is element 2 and has 2 protons in its nucleus. How many protons does element number 25 contain? What is the name of this element? How many electrons does element 25 contain? What is the charge of one atom of element 25?

16. Isotopes are atoms with the same number of protons (same element) but with a different number of neutrons. Na-23 (11 + 12 = 23) Na-24 (11 + 13 = 24)

17. Element Symbol Mass Number (total number of protons and neutrons in a nucleus) Atomic Number (total number of protons in a nucleus) # Neutrons = Mass #  Atomic # or # Neutrons = A  Z Mass # = Atomic # + # Neutrons Identifying Specific Isotopes (Nuclides)

18. 18 Use a periodic table to help you fill in the missing information in the following tables.

19. 19 Unified AMU: unified atomic mass unit is a unit of mass based on 12 C. The mass of one atom of 12 C = 12 u so 1 u = 1/12 of the mass of one atom of 12 C On this scale, 1 proton = 1.007276 u 1 neutron = 1.008665 u 1 electron = 0.0005486 u The Mass of “C” on the periodic table is 12.011 u because pure carbon is made up of both 12 C (12 u exactly) and 13 C (close to 13 u). The mass of “C” is a weighted average of all isotopes of C!

20. 20 Average Atomic Mass: the weighted average of the atomic masses of all naturally occurring isotopes of an element. The weighted average is found by multiplying the natural abundance (%Abund) for a given isotope times the actual mass (m) of that isotope and then adding all of those masses together. For example: if an element (X) has three isotopes, the formula for Average atomic Mass would be: Average Mass = [(%Abund 1 )*(m 1 ) + (%Abund 2 )*(m 2 ) + (%Abund 3 )*(m 3 )] How would the formula change if an element had 5 naturally occurring isotopes? What would it look like if an element only had 2 naturally occurring isotopes?

21. 21 24 Mg 25 Mg 26 Mg IsotopeAbundanceMass 78.99% 10.00% 11.01% 23.985 u 24.986 u 25.982 u Given: Data for the known isotopes of Magnesium = [(0.7899)*(23.985 u) + (0.1000)*(24.986 u) + (0.1101)*(25.982 u)] Find the Average Mass of Magnesium (Mg) Average Mass = [(%Abund Mg-24 )*(m Mg-24 ) + (%Abund Mg-25 )*(m Mg-25 ) + (%Abund Mg-26 )*(m Mg-26 )] = 24.3049697 u = 24.30 u Click to see Solution

22. One Mole is the number of atoms in exactly 12 g of carbon-12. Avogadro’s Number is the number of particles in exactly one mole and has a value of 6.022X10 23 The definition of “mole” and Avogadro’s number allows us to “count” atoms by weighing them when we know what the mass of one mole of a substance is. Molar Mass is the mass in grams of one mole of a pure substance. When the atomic masses on the periodic table are expressed in grams (instead of “u”), the atomic mass represents the mass of one mole of that element. Important Equivalence Statements: 1 mol = 6.022X10 23 particles (atoms, molecules, ions, etc…) 1 mol = atomic mass in grams for each element on the periodic table for example: 24.3050 g Mg = 1 mol Mg and 55.845 g Fe = 1 mol Fe Remember that Equivalence Statements can be used to convert from one set of units to another!

23. When working on problems, we will use “n” to represent the variable of moles in a problem (just like we may use “l” to represent length in a math problem). When working on problems, I will try to use “MW” to represent the variable of molar mass. The book often uses “M” for molar mass, but later in the year the “M” will stand for molarity which means something entirely different. When working on problems, we use “m” to represent the variable of mass in a problem. Please note that variable and units mean different things. When working on problems, sometimes we will use the equivalence statement approach and other times we will use the equation approach. They both do the same math, but they use a slightly different kind of logic to solve.

24. Mass to Moles and Moles to Mass Example Problems: What is the mass in grams of 0.0695 moles of silver? First we will solve this using the equivalence statement method: Start with what was given and then use the equivalence statement from the periodic table: 1 mole Ag = 107.9 g Ag 1 mol Ag 107.9 g Ag ( ) 0.0695 moles Ag ( ) = 7.50 g Ag This is a conversion factor based on the equivalence statement from the periodic table. Now we can solve the same problem using the equation method: Given: n = 0.0695 mol MW = 107.9 g/mol (form periodic table) m = nMW m = (0.0695 mol Ag)(107.9 g Ag/mol Ag) = 7.50 g Ag Click to see Solution

25. Which method should we use? Both methods (equation versus equivalence statement) work well, but the equivalence statement method will work in many instances where we do not have an obvious equation to use. Ultimately, you may use any method you want, but you need to learn both because there are times when really only one of the two methods will work. Can you decide which method to use with a given problem? If a reaction is producing water at a rate of 3.5 grams per second, how many moles of water per hour will be produced. Use the Equivalence Statement method Use the Equation method Lead has a density of 13.2 g/mL. If you have a sample of lead that has a mass of 276.4 g, what is the volume of the sample?

26. = 2.003 mol Cu First we will solve this using the equivalence statement method: Start with what was given and then use the equivalence statement from the periodic table: 1 mole Cu = 63.55 g Cu 63.55 g Cu 1 mol Cu ( ) 127.3 g Cu ( ) = 2.003 mol Cu Now we can solve the same problem using the equation method: Click to see Solution How many moles of copper (Cu) are present in a sample of copper that has a mass of 127.3g?

27. Particles to Moles and Moles to Particles How many atoms of Zn are contained in 0.0087 moles of Zn? How many moles of copper are present if you have a sample containing 3.8X10 21 atoms of copper? Given: 0.0087 moles Zn 1 mole Zn = 6.022X10 23 atoms Zn 1 mol Zn 6.022X10 23 atoms Zn ( ) 0.0087 moles Zn ( ) = 5.2X10 21 atoms Zn Given: 3.8X10 21 atoms Cu 1 mole Cu = 6.022X10 23 atoms Cu 1 mol Cu 6.022X10 23 atoms Cu ( ) 3.8X10 21 atoms Cu ( ) = 0.0063 mol Cu Click to see Solution

28. 1 mol Cu 6.022X10 23 atoms Cu ( ) 7.430X10 22 atoms Cu ( ) = 7.841 g Cu ( ) 1 mol Cu 63.55 g Cu Given: 7.430X10 22 atoms Cu 1 mole Cu = 6.022X10 23 atoms Cu 1 mol Cu = 63.55 g Cu Given: 14.3 g Zn 1 mole Zn = 6.022X10 23 atoms Zn 1 mol Zn = 65.38 g Zn 1 mol Zn 6.022X10 23 atoms Zn ( ) 14.3 g Zn ( ) = 1.32X10 23 atoms Zn ( ) 1 mol Zn 65.38 g Zn Particles to Mass and Mass to Particles How many atoms of Zn are contained in 14.3 g of Zn? How many grams of copper are present if you have a sample containing 7.430X10 22 atoms of copper? Click to see Solution

29. The extension is that we need to convert from mg to g before we start the rest of the problem. 1 mol Si 6.022X10 23 atoms Si ( ) 84.39 mg Si ( ) = 1.809X10 21 atoms Si ( ) 1 mol Si 28.09 g Si ( ) 1000 mg Si 1 g SiExample with an extension A chemist has a sample of silicon (Si) that has a mass of 84.39 mg. How many atoms of silicon are in the sample? Given: 84.39 mg Si 1 g Si = 1000 mg Si 1 mole Si = 6.022X10 23 atoms Si 1 mol Si = 28.09 g Si Click to see Solution