Station 1: Significant Figures SIGNIFICANT FIGURES (SFs or Sig Figs) NOTES How to Determine the Number of Significant Figures (SF) a Measurement Has... ** Any non-zero number (1, 2, 3, 4, 5, 6, 7, 8, 9) is ALWAYS significant. ** 1. "Sandwiched" zeroes (zeroes in between two non-zero numbers) are ALWAYS significant. EXAMPLE: 103 has 3 SFs 5007 has 4 SFs 2. “Leading” zeroes (zeroes to the left of a non-zero number) are NEVER significant. EXAMPLE: 0.000375 has 3 SFs 3. “Trailing” zeroes (zeroes to the right of non-zero numbers) are significant ONLY if a decimal point is written in the number. (Note: The decimal point can be located anywhere in the measurement.) EXAMPLES: 50 has 1 SF 50. has 2 SFs 50.0 has 3 Rules For Deciding How Many SFs an Answer Should Have... ** When multiplying or dividing... Your answer must have as many SFs as the number in the problem with the fewest (lowest) number of SFs. EXAMPLE: 9.34 cm x 4.5 cm = calculator says 42.03 (3 SF) (2 SF) Must round to 2 SFs --> so, answer is 42 cm2. EXAMPLE: 2.494 m x 3.02 m x 5.125 m = calculator says 38.60085 (4 SF) (3 SF) (4 SF) Must round to 3 SF --> 38.6 m3 ** When adding or subtracting... Your answer must have as many places to the right of the decimal point as the number in the problem with the fewest number of places to the right of the decimal. EXAMPLE: 2.194 g + 25.84 g + 5.7210 g = calculator says 33.755 (3 places) (2 places) (4 places) Must round to 2 places --> 33.76 g Practice Questions 1. Determine the number of significant figures in the following measurements: a. 640 cm3 e. 10,000 L b. 200.0 mL f. 20.900 cm c. 0.5200 g g. 0.00000056g/L d. 1.005 kg h. 0.04002 kg/m3

i. 790,001 cm2 j. 665.000kg_m/s2

2. Perform the following calculations, and express the result in the correct units and number of significant figures. a. 47.0 m ÷ 2.2 s e. 300.3 L÷180. s b. 140 cm x 35 cm f. 33.00 cm2 x 2.70 cm 3 c. 5.88 kg ÷ 200 m g. 35,000 kJ ÷ 0.250 min d. 0.0050 m2 x 0.042 m 3. Perform the following calculations and express the results in the correct units and number of significant figures: a. 22.0 m + 5.28 m + 15.5 m e. 24.50 dL + 4.30 dL + 10.2 dL b. 0.042 kg + 1.229 kg + 0.502 kg f. 3,200 mg + 325 mg - 688 mg c. 170 cm2 + 3.5 cm2 - 28 cm2 g. 14,000 kg + 8,000 kg + 590 kg d. 0.003 L + 0.0048 L + 0.100 L

Answer Key 1. Determine the number of significant figures in the following measurements: a. 2 e. 1 b. 4 f. 5 c. 4 g. 2 d. 4 h. 4

i. 6 j. 6

2. Perform the following calculations, and express the result in the correct units and number of significant figures. a. 21 m/s e. 1.67 L/s b. 4,900 cm2 f. 89.1 cm3 3 c. 0.03 kg/m g. 140,000 kJ/min 3 d. 0.00021 m 3. Perform the following calculations and express the results in the correct units and number of significant figures: a. 42.8 m e. 39.0 dL b. 1.773 kg f. 2,837 mg c. 146 cm2 g. 22,590 kg d. 0.108 L

Station 2: Periodic Table & Subatomic Particles

Subatomic Particle

Charge

Location

Proton

+

Inside the nucleus

Neutron

No charge

Inside the nucleus

Electron

-

Outside the nucleus

 ATOMIC NUMBER = PROTONS = ELECTRONS (in neutral atom)  Atomic mass – protons = neutrons  Isotope: a neutrally charged atom with a different number of neutrons  different atomic mass  Ion: a positively or negatively charged atom with a unequal number of ELECTRONS and protons Practice Questions Element 1. lithium 2. carbon 3. chlorine-38 4. silver-109 5. lead+2 6. calcium-1

Number of Protons

Number of Neutrons

Number of Electrons

Atomic Mass

Atomic Number

Answer Key Element

Number of Protons

Number of Neutrons

Number of Electrons

Atomic Mass

Atomic Number

1. lithium

3

4

3

7

3

2. carbon

6

6

6

12

6

3. chlorine-38

17

21

17

38

17

4. silver-109

47

62

47

109

47

5. lead+2

82

125

80

207

82

6. calcium-1

20

20

21

40

20

Station 3: Mole Conversions = 6.022 x 1023 atoms or molecules = molar mass of element or compound “molar mass” = decimal # on P.T. in grams This means that when converting among “moles”, “grams”, “atoms”, and “molecules”... the number “1” ALWAYS goes in front of the unit “moles” the number “6.022 x 1023” ALWAYS goes in front of the unit “atoms” the number “6.022 x 1023” ALWAYS goes in front of the unit “molecules (mcs)” the “molar mass” (decimal # on Periodic Table) ALWAYS goes in front of the unit “grams” 1 mole

Conversion Table

__Given Units

__Units You Want

1

__Given Units

Practice Questions 1. 2. 3. 4.

How many molecules are there in 450 grams of Na2SO4? How many grams are there in 2.3 x 1024 atoms of silver? How many moles are present in 34 grams of Cu(OH)2? How many moles are present in 2.45 x 1023 molecules of CH4?

Answer Key 1. 2. 3. 4.

1.91 x 1024 molecules 412 grams 0.35 moles 0.41 moles

Station 4: Electron Arrangement Electron Configuration Examples: Se

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4

Sn 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p2 HOEL (Highest Occupied Energy Level): energy level furthest from the nucleus that contains at least one electron How to determine this using electron configuration? ~ largest non-exponent number Se 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4 HOEL = 4 Sn 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p2 HOEL = 5 Valence Electrons: electrons in the HOEL How to determine this using electron configuration? ~ add up exponents of terms in HOEL 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4 HOEL = 4 Valence electrons = 2 + 4 = 6 2 2 6 2 6 2 Sn 1s 2s 2p 3s 3p 4s 3d10 4p6 5s2 4d10 5p2 HOEL = 5 Valence electrons = 2 + 2 = 4 Noble Gas Configuration: shortcut for electron configuration How is it written? ~ [ symbol for noble gas closest to element with lower atomic # ] ~ [after brackets] next number is the period that the element is located in ~ after that number, write “s” ~ continue electron configuration in diagonal rule order until appropriate # of electrons is reached *NOTE: ending of electron configuration and noble gas configuration should be the same* Se

Se

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4 [Ar] 4s2 3d10 4p4 18

Sn

20

30

2

2

6

36

38

48

34

2

1s 2s 2p 3s 3p6 4s2 3d10 4p6 5s2 4d10 5p2 [Kr] 5s2 4d10 5p2 50

Orbital Notation: drawing of how electrons are arranged in orbitals; will only need to do this for the HOEL *NOTE: ___ = orbital or = electrons

Se

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4

Sn

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p2

QUANTUM NUMBERS NOTES(HONORS ONLY) ~ describe one specific electron ~ 1st quantum number = PRINCIPAL QUANTUM NUMBER ~ abbreviated "n" ~ tells the energy level the electron is located in ~ n = number of the energy level ~ 1st energy level: n = 1, 4th energy level: n = 4, etc. ~ 2nd quantum number = ANGULAR MOMENTUM QUANTUM NUMBER ~ abbreviated " l " ~ tells the sublevel the electron is located in ~ tells shape of orbital ~ "s" sublevel: l = 0, "p" sublevel: l = 1, "d" sublevel: l = 2, "f" sublevel: l = 3 ~ 3rd quantum number = MAGNETIC QUANTUM NUMBER ~ abbreviated "m" ~ tells which orbital the electron is in ~ tells orientation of orbital around nucleus ~ m = - l .. + l ~ 4th quantum number = SPIN QUANTUM NUMBER ~ abbreviated " s " ~ tells which electron is being described ~ tells which way electron is spinning ~ s = -1/2 or +1/2 Practice Questions Answer the following questions about the element bromine (Br). 1. What is the atomic number of bromine? 2. What is its electron configuration? 3. What is the highest occupied energy level (H.O.E.L.) in bromine? 4. What is the noble gas configuration for Br? 5. How many valence electrons does bromine have? 6. Draw the orbital notation (arrows & lines) for the highest occupied energy level of bromine. 7. In your orbital notation for #6, circle the last electron added to bromine. What are the four quantum numbers for the electron you circled? (HONORS ONLY) n= l= m= s= 8. Draw the electron dot diagram for bromine.

Answer Key 1. 2. 3. 4. 5.

35 1s22s22p63s23p64s23d104p5 4 [Ar] 4s23d104p5 7

6. 7. n =4

8.

l=1

m=0

s = -1/2

Station 5: Periodic Trends Electronegativity/Electron Affinity (EN/EA): measure of how much an atom wants to gain an electron EN/EA Left to Right across a Period: INCREASES (not including Noble Gases) Why? * Elements on the left side of the P.T. (metals) want to lose electrons. Elements on the right side of the P.T. (nonmetals) want to gain electrons. Trend does not include Noble Gases because these elements do not want to lose or gain electrons. EN/EA Top to Bottom in a Group: Why?

DECREASES

Duncan

* This interference (and resulting decreased “hold”) is referred to as the SHIELDING EFFECT. ======================================================================= Ionization Energy (IE): amount of energy required to remove an atom’s most loosely held electron IE Left to Right across a Period: INCREASES Why? * Elements on the left side of the P.T. (metals) want to lose electrons. Therefore, it will not require much energy to remove an electron. Elements on the right side of the P.T. (nonmetals) want to gain electrons. Consequently, a lot of energy will be needed to remove (take away) an electron. IE Top to Bottom in a Group: DECREASES Why?

* ======================================================================= Atomic Radius (AR): distance from the nucleus to the H.O.E.L.

AR Left to Right across a Period: DECREASES Why?

AR Top to Bottom in a Group: INCREASES Why? * There are more occupied energy levels as you move towards the bottom of the P.T.

Metallic Character: how easily an atom will lose valence electrons (easier to lose = more metallic = more reactive METAL) Which metal loses its valence electron(s) most easily? Fr Why? * Francium has one valence electron. It is more reactive than elements at the top of Group 1 because there are many inner shell electrons that decrease the attraction the nucleus has for the valence electrons. Nonmetallic Character: how easily an atom will gain electrons (easier to gain = more nonmetallic = more reactive NONMETAL) Which nonmetal gains electron(s) most easily? F Why? * Fluorine has seven valence electrons. It is more reactive than elements at the bottom of Group 17 because there are only a few inner shell electrons. Consequently, the nucleus has a strong attraction for other electrons.

Practice Questions 1. Which outer electron configuration has the greatest electronegativity? (A) 2s2 2p3 (B) 2s1 (C) 2s2 2p5

(D) 2s2 2p6

2. The most reactive nonmetals on the Periodic Table have (A) large radii and high electronegativities (B) large radii and low electronegativities (C) small radii and high electronegativities (D) small radii and low electronegativities 3. Which electron configuration represents the element with the smallest atomic radius? (A) 1s2 2s2 2p4 (B) 1s2 2s2 2p5 (C) 1s2 2s2 2p3 (D) 1s2 2s2 2p2 4. In Period 2, as the elements are considered from left to right, there is a decrease in (A) ionization energy (B) nonmetallic character (C) electronegativity (D) atomic radius 5. As the elements in Group 1 are considered in order of increasing atomic number, the atomic radius of each successive element increases. This is primarily due to an increase in the number of (A) protons in the nucleus (B) valence electrons (C) effective nuclear charge (D) occupied energy levels 6. The radii of the atoms become smaller from left to right across Period 3. This decrease is primarily the result of (A) the shielding effect (B) increased effective nuclear charge (C) a decrease in metallic character (D) the increased number of occupied energy levels

Answer Key 1. C 2. C 3. B 4. D 5. D 6. B

Station 6: Bonding Types of chemical bonding 1. Ionic Bonds: results from attraction between large numbers of cations and anions; involves the transfer of electrons 2. Covalent Bonds: results from sharing of e- pairs between two atoms a. Polar Covalent: type of bond resulting from the unequal sharing of electron pairs b. Nonpolar Covalent: type of bond resulting from the equal sharing of electron pairs Determining Ionic or Covalent Bond (Official Way) 1. Remember electronegativity? (how much an atom wants to gain an e-) 2. Difference in EN tells whether bond is ionic or covalent a. EN difference less than 1.7 = COVALENT 1. NONPOLAR COVALENT: EN difference 0 to 0.3 2. POLAR COVALENT: EN difference 0.3 to 1.7 b. EN difference greater than 1.7 = IONIC Determining Ionic or Covalent Bond (Easy Way) 1. Ionic Bond: metal & nonmetal in formula 2. Covalent Bond: 2 or more nonmetals a. Polar Covalent: 2 different nonmetals b. Nonpolar Covalent: 2 of the same nonmetal MOLECULAR POLARITY vs. BOND POLARITY BOND POLARITY ~ refers to the equal (nonpolar) or unequal (polar) sharing of electrons ~ What makes a bond polar (covalent)? * If the bond occurs between two different nonmetals MOLECULAR POLARITY ~ refers to the symmetry (nonpolar) or asymmetry (polar) of a molecule ~ What makes a molecule asymmetrical? * If there are unshared pairs of electrons around the central atom. and/or * If the surrounding atoms are different elements Practice Questions Draw the Lewis structures for the three below. 1. BF3 2. H2CS

Molecular Polarity: Class: Shape:

Molecular Polarity: Class: Shape:

3. CH2F2

Molecular Polarity: Class: Shape:

Answer Key 1. BF3

2. H2CS

3. CH2F2

Molecular Polarity: Nonpolar (no unshared pair of electrons and the same surrounding atoms) Class: AB3 Shape: Trigonal planar

Molecular Polarity: Polar (because the surrounding atoms are not the same) Class: AB3 Shape: Trigonal planar

Molecular Polarity: Polar (because the surround atoms are not the same) Class: AB4 Shape: Tetrahedral

Station 7: Writing Formulas & Naming Compounds

Practice Questions 1) 2) 3) 4) 5) 6) 7)

NaF K2CO3 P4S5 MgCl2 Be(OH)2 SeF6 SrS

8) 9) 10) 11) 12) 13) 14)

potassium fluoride dinitrogen trioxide ammonium sulfate magnesium iodide copper (II) sulfite phosphorus triiodide aluminum phosphate

Answer Key 1) 2) 3) 4) 5) 6) 7)

sodium fluoride potassium carbonate tetraphosphorus pentasulfide magnesium chloride berrylium hydroxide selenium hexafluoride strontium sulifde

9) 10) 13)

8) KF N2O3 (NH4)2SO4 11) MgI2 12) CuSO3 PI3 14) AlPO4

Station 8: Empirical and Molecular Formulas TO SOLVE EMPIRICAL FORMULA PROBLEMS: A sample of a compound is found to contain 36.0 % calcium and 64.0 % chlorine. Calculate the empirical formula. Step 1: Rewrite % as grams. 36.0 g Ca 64.0 g Cl Step 2: Find moles of each element. Ca: 36.0 g Ca | 1 mole Ca = 0.898 moles Ca Cl: 64.0 g Cl | 1 mole Cl = 1.80 moles Cl | 40.1 g Ca | 35.5 g Cl Step 3: Find mole ratio. (Divide by smallest number of moles.) Ca: 0.898 moles = 1 Cl: 1.80 moles = 2 0.898 0.898 * These whole numbers are subscripts in formula.* Step 4: Write the formula. Ca1Cl2 ====> CaCl2 TO SOLVE MOLECULAR FORMULA PROBLEMS: To find the molecular formula, one more piece of information must be given - the molar mass (also called molecular mass or formula mass). EX. - An organic compound is found to contain 92.25% carbon and 7.75% hydrogen. If the molecular mass is 78, what is the molecular formula? STEP 1: Find the empirical formula. C: 92.25 g C | 1 mole C = 7.69 moles C H: 7.75 g H | 1 mole H = 7.75 moles H | 12 g C |1gH 7.69 moles C = 1 7.75 moles H = 1 So... empirical formula is CH. 7.69 7.69 STEP 2: Find molar mass of the empirical formula. C: 1 x 12.0 = 12.0 H: 1 x 1.0 = 1.0 + MM = 13.0 STEP 3: Find "multiple" number. MM of molecular formula = multiple # 78 = 6 MM of empirical formula 13 STEP 4: Write molecular formula. Multiply "multiple" # by all subscripts in the empirical formula. So... molecular formula is C6H6. Practice Questions 1) What’s the empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29.0% oxygen? 2) If the molar mass of the compound in problem 1 is 110 grams/mole, what’s the molecular formula? 3) What’s the empirical formula of a molecule containing 18.7% lithium, 16.3% carbon, and 65.0% oxygen? 4) If the molar mass of the compound in problem 3 is 73.8 grams/mole, what’s the molecular formula?

Answer Key 1) C3H3O 2) C6H6O2 3) 4)

Li2CO3 Li2CO3 (In this case, the molecular and empirical formulas are the same, a frequent occurrence for inorganic compounds)

Station 9: Midterm Information 80 multiple choice questions (all are not necessarily accounted for in this description) UNIT 1 – MEASUREMENT & MATH  Accuracy & Precision (recognizing given lab data) – 1  Density calculations- 1  Number of SFs in a measurement- 2  Percent Error- 1  Reporting answers to problems to correct number of SFs (multiple problems) UNIT 2 – MATTER & CHANGES  Chemical vs. physical properties, changes- 4  Compounds vs. elements- 2  Location of metals, nonmetals, metalloids (properties of each also) (multiple problems)  Mixtures vs. pure substances- 1  Periods & groups (definitions, names, locations) (multiple questions) UNIT 3 – ATOMS  Determining number of protons, neutrons, electrons in a nuclide - 4  Grams  moles  atoms or molecules conversions- 4  Isotopes (definition, calculation of average atomic mass)- 1  Mass number & atomic number (multiple problems)  Results of gold foil experiment – 1  Thomson’s discovery, points of Dalton’s atomic theory- 2 UNIT 4 – ELECTRONS  Aufbau principle, Hund’s rule – 1  c = . E = h. calculations, proportionality relationships – 1  Shapes of orbitals- 1  Determine location on periodic table given an electron configuration - 5  Dot diagrams, valence electrons, HOELs- 2 + multiple problems  Electromagnetic spectrum (wavelengths, frequencies, energies) (multiple problems)  Energy calculations in the hydrogen atom – 2  Ground state vs. excited state (definitions, emission & absorption of energy)- 2  Lowest to highest energy orbitals, sublevels, energy levels- (multiple problems)  Octet rule UNIT 5 – PERIODIC TABLE  History of periodic table (Mendeleev, Moseley) – 3  Main group elements (valence e-, lose vs. gain e-, size difference between atom & ion, oxidation #)- 1 +(multiple problems)  Periodic trends (atomic radius, ionization energy, electronegativity, metallic & nonmetallic character)- 10 UNIT 6 – BONDING  Definitions of ionic and covalent bonding – types of elements involved, electronegativity differences - 4  Intermolecular forces (hydrogen bonding, dipole-dipole, London dispersion)- 1  Molecular polarity- 1  Properties of ionic and molecular substances-4  Why atoms bond together- 1  Molecular Shapes- 1

UNIT 7 – CHEMICAL FORMULAS  Empirical and molecular formulas (definitions, differentiating between the two, calculate using %)- 3  Nomenclature – writing formulas and naming compounds- 4  Oxidation numbers of elements in a compound- 6  Percent composition of element in a compound- 2 IMPORTANT INFORMATION  Cations (+ ions) are smaller than the neutral atom and anions (- ions) are larger than the neutral atom

          

Elements in the same group of similar chemical properties due to having the same valence electron configuration Be able to use the Bohr model diagram! Density = mass/volume Atomic mass is the larger number; atomic number is the smaller number on the periodic table Isotopes are the same element with different masses Metalloids have both properties of metals and nonmetals and they are located on the staircase Octet rule- all elements want 8 valence electrons, except H and He which want 2 (will fill the outer energy shell) Know the relationship between wavelength, frequency and energy The Aufbau principle states that the electron occupies the lowest available energy orbital The Law of multiple proportions states that two of the same elements can be used to form different compounds ex. H2O, H2O2, H3O S orbital has a spherical shape; p orbital has a dumbbell shape