1. Quizzes and Exams Please write in non-erasable blue or black pen, no pencil! (Will lose 1 point for pencil or erasable ink) I will put periodic table on backs of quizzes, if you need them. BOX answers! If it’s not obvious where (or which) the answer is that I’m supposed to grade, it will get no credit. Remember to bring your non-programmable calculators to class on Wednesdays. Make sure your answers make sense (can there be 23 grams of zinc in a 4.75 g sample of zinc oxide?). You may only use approved calculators on quizzes; others will be confiscated for the duration of the quiz or exam.

2. Chapter 2: Atoms and Elements 1.Dalton’s Atomic Theory is based on empirical observations, formulated as Laws of Conservation of Mass Definite Proportions Multiple Proportions No detectable gain or loss of mass occurs in chemical reactions. In a compound, elements are always combined in the same proportion by mass. Whenever two elements form more than one compound, the different masses of one element that combine with the same mass of the other element are in the ratio of small whole numbers. 2. Summary of Dalton’s Theory --Matter consists of tiny particles called atoms. --Atoms are indestructible. In chemical reactions, the atoms rearrange but they do not themselves break apart. --In any sample of a pure element, all the atoms are identical in mass and other properties. --The atoms of different elements differ in mass and other properties. --In a given compound the constituent atoms are always present in the same fixed numerical ratio.

3. Sample Problems 1. A sample of caffeine contains 96.1 g of carbon for every 10.1 g of hydrogen. If another sample of caffeine contains 30.0 g of carbon, how many g of hydrogen does it contain? 2.* Phosphorus forms two compounds with chlorine. In 5.5 g of one of these compounds, there were 1.24 g of P. In 5.5 g of the other compound, there were 0.818 g of P. Explain how these compounds exhibit the law of multiple proportions.

4. Sample Problems 1. A sample of caffeine contains 96.1 g of carbon for every 10.1 g of hydrogen. If another sample of caffeine contains 30.0 g of carbon, how many g of hydrogen does it contain? Answer: 2.* Phosphorus forms two compounds with chlorine. In 5.5 g of one of these compounds, there were 1.24 g of P. In 5.5 g of the other compound, there were 0.818 g of P. Explain how these compounds exhibit the law of multiple proportions. Answer: According to the law of multiple proportions, if you compare the same amounts of phosphorus, then the chlorine masses should be in a ratio of small whole numbers. Since 1.24/0.818 = 1.516, then you should multiply the compound B quantities by 1.516. In compound A, there is (5.5 - 1.24 =) 4.26 g Cl. In compound B there is (5.5 - 0.818 =) 4.682 g Cl. Multiply that by 1.516 to get 7.10 g Cl for every 1.24 g P in compound B. Then you can check the Cl ratio: 4.26 g / 7.10 g = 0.6 = 3/5, a ratio of small whole numbers. 96.1 10.1 = 30.0 x x = 3.15 g

5. “Modern” View of an Atom nucleus has protons(+) and neutrons electrons(-) form a diffuse cloud around the nucleus Mass proton  neutron  1 amu electron  10 -4 amu (amu = 1.66 x 10 -24 g = 1/12 of mass of 12 C) ~If there is a different number of protons, it is a different ___________? ~If there is a different number of neutrons, it is a different __________? Atomic number, Z = # of protons Mass number, A = # of protons + number of neutrons Isotopes = atoms of the same element with different mass numbers Matter is neutral, so # of electrons = # of protons AEAE Z ~Where is most of the mass of an atom?

6. Sample Problems If a beryllium atom has 4 protons, then it should weigh 4 amu; but it actually weighs 9 amu! Where is the extra mass coming from? How many protons, electrons, and neutrons are in an atom of 52 Cr? 24

7. Sample Problems If a beryllium atom has 4 protons, then it should weigh 4 amu; but it actually weighs 9 amu! Where is the extra mass coming from? How many protons, electrons, and neutrons are in an atom of 52 Cr? 24 neutrons Atomic number = 24, so 24 protons Neutral, so 24 electrons 52 – 24 = 27, so 27 neutrons Answer:

8. Some Notes on Charge Two kinds of charge called + and – Opposite charges attract –+ attracted to – Like charges repel –+ repels + –– repels – To be neutral, something must have no charge or equal amounts of opposite charges

9. The Periodic Table Average among all isotopes Decides ordering of elements Inner transition elements Transition metals

10. Types of Elements Metals: Shiny, malleable, ductile solids with high mp and bp Good electrical conductors Nonmetals: Gases, liquids, or low-melting solids Non-conductors of electricity Metalloids: Intermediate properties, often semiconductors Diatomic elements: H 2, O 2, N 2, F 2, Cl 2, Br 2, I 2

11. Predicting Ionic Charges Li +1 Na +1 K +1 Rb +1 Cs +1 Mg +2 Ca +2 Sr +2 Ba +2 Al +3 O -2 S -2 Se -2 Te -2 F -1 Cl -1 Br -1 I -1 N -3 1A 2A3A7A6A5A He Ne Ar Kr Xe Rn F? Rb? N? Noble gases have very favorable electron configurations. The other elements are jealous… If an ion forms, it will commonly match the nearest noble gas.

12. Calculating Atomic Mass (Weighted Avg) Atomic Mass = ∑ (fraction of isotope n) x (mass of isotope n) = (fraction 1 ∙mass 1 ) + (fraction 2 ∙ mass 2 ) + (fraction 3 ∙ mass 3 ) + … n Example 63 Cu has mass = 62.9396, natural abundance 69.17%, and 65 Cu has mass = 64.9278 and natural abundance 30.83%. What is the atomic mass? Answer: Atomic mass = (62.9396)(0.6917) + (64.9273)(0.3083) = 63.55 amu (4 sig fig!)

13. Sample Problems 1. In a class A is worth 4 points and B is worth 3. Out of 20 grades, if you have 4 A’s (20%) and 16 B’s (80%), what is your average? If an element has two isotopes, A (Z = 4 amu) and B (Z = 3 amu), and their natural abundance is 20% A and 80% B, what is the average molecular weight? 2. Naturally occurring boron (Z = 10.8110 amu) is composed of two isotopes, 10 B and 11 B. Atoms of 10 B have a mass of 10.0129 amu and those of 11 B have a mass of 11.0093 amu. Calculate the percentages by mass of the individual isotopes, 10 B and 11 B.

14. Sample Problems 1. In a class A is worth 4 points and B is worth 3. Out of 20 grades, if you have 4 A’s (20%) and 16 B’s (80%), what is your average? If an element has two isotopes, A (Z = 4 amu) and B (Z = 3 amu), and their natural abundance is 20% A and 80% B, what is the average molecular weight? Answer: Grade average = 3.2 points Average molecular weight = 3.2 amu 2. Naturally occurring boron (Z = 10.8110 amu) is composed of two isotopes, 10 B and 11 B. Atoms of 10 B have a mass of 10.0129 amu and those of 11 B have a mass of 11.0093 amu. Calculate the percentages by mass of the individual isotopes, 10 B and 11 B. Answer: Mass percent of 10 B = 19.90% Mass percent of 11 B = 80.10%

15. The Mole; How Chemists “Count” 1. Avogadro’s Number -- The Chemists’ “Dozen” N 0 = number of atoms in exactly 12 g of carbon-12 = 6.022 x 10 23 “things” ( a very large number!) This is a conversion factor, just like 12 things per dozen, e.g. Mass of one atom of carbon-12 = (12.0 g)/6.022 x 10 23 atoms) = 1.99 x 10 -23 g/atom 2. The Mole One Mole of a substance contains an Avogadro’s number of formula units. e.g. 1 mole of H 2 O = 6.022 x 10 23 H 2 O molecules

16. Sample Problems How many atoms of gallium are there in 0.2398 moles of gallium? How many molecules of sugar are there in 0.2398 moles of sugar? How many marbles are there in 0.2398 moles of marbles?

17. Sample Problems How many atoms of gallium are there in 0.2398 moles of gallium? Answer: 1.444 x 10 23 atoms How many molecules of sugar are there in 0.2398 moles of sugar? Answer: 1.444 x 10 23 molecules How many marbles are there in 0.2398 moles of marbles? Answer: 1.444 x 10 23 marbles