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Date Created: 09/26/16
Exam One Chem 135 Study Guide Hayat Jabbour Chapter 4 SI-prefixes Factor Name Symbol Factor Name Symbol 10 18 exa E 10-1 deci d 10 15 peta P 10-2 centi c 10 12 tera T 10-3 milli m 9 -6 10 giga G 10 micro µ 10 6 mega M 10-9 nano n 3 -12 10 kilo k 10 pico p 10 2 hecto h 10-15 femto f -18 10 atto a Common unit conversions: Table 1.1 SI units: mass - kilogram (kg)length - meter (m)time - second (s) temperature - Kelvin (K)amount - mole (mol) charge - Coulomb electric current - ampere (A) 1 L=1000mL=1000cm 3 1 L=1.057 qt 1 gal=3.785 1 mL= 1 cm 3 Sig Fig Rules: 1. All nonzero digits are significant 2. Interior zeroes are significant 3. Leading zeroes are not significant 4. Trailing zeroes after a decimal point are always significant 5. Trailing zeroes before a decimal point and after a nonzero are always significant 6. Trailing zeroes before an implied decimal point should be avoided by using scientific notation. Density=mass/volume Avogadro’s number: 6.022x10 23 c, the speed of light in a vacuum, 2.998 x 108 m/s Molar mass: mass in grams of one mole of atoms of an element, numerically equivalent to the atomic mass of the element in amu. Structure of atom: protons and neutrons in nucleus, electrons on outside orbitals Isotopes: One of two or more atoms of the same element with the same number of protons but different numbers of neutrons and consequently different masses. Elements that form ions with predictable charges: Electromagnetic Spectrum: Electromagnetic radiation equations: c = λν ν= c/λ E = hν E=hc/λ deBroglie wavelength of a particle λ = h/mv Energy levels for electrons in a hydrogen atom from the Bohr model: E n -2.18x10 J (1/n ) 2 (n=1,2,3,…) If you're given the mass of each isotope of an element, and the average atomic mass, you can calculate the percent (%) abundance of each isotope. Let "x" be one of the abundances, and the other abundance is "1-x" ... then solve the equation using basic algebra. Example #1: Silver (Atomic weight 107.868) has two naturally-occurring isotopes with isotopic weights of 106.90509 and 108.90470. What is the percentage abundance of the lighter isotope? To avoid mistakes, use "x" as the multiplier for the isotope percentage you wish to find. In this case, you want to find the percentage of the lighter isotope, so the "x" is associated with 106.90509. Since the sum of the isotopic abundance percentages is equal to 1 (100%), the formula is: 108.90470 (1 - x) + 106.90509 (x) = 107.868 Multiplying, re-arranging and condensing the above formula results in: 108.90470 - 108.90470x + 106.90509x = 107.868 - 108.90470x + 106.90509x = - 108.90470 + 107.868 - 1.9996x = - 1.0367 x = 0.5185 Therefore, the answer is 51.85 % Rules for Writing Electron Configurations 1) Aufbau's Principle = Electrons enter orbitals of the lowest energy first. An energy level (or sublevel more precisely) will fill with electrons before electrons start to fill the next energy level or energy state. All lower energy levels are full up to the last energy level; levels are not skipped or eliminated because of energy level. 2) The Pauli Exclusion Principle = An atomic orbital may describe at most two electrons, at least one for that orbital to exist. To occupy the same orbital, two electrons must have opposite spins. If two electrons occupy the same orbital they are said to be paired. Orbitals with one electron are unpaired. 3) Hund's Rule = When electrons occupy orbitals of equal energy, one electron enters each orbital until all the orbitals contain one electron, with spins parallel, before they are paired. Second electrons then add to each orbital so that their spins are paired. There are different ways to show electron arrangements. Shorthand: 1s2 2s2 2p6 1. The first number represents the energy level. 2. The letter represents the sublevel for that particular energy level. 3. The superscript number represents the number of electrons in the sublevel of that energy level. Here is the total sequence: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p 4. You can also do an extreme form of shorthand. Since Noble Gases energy levels are filled for that level, you can represent all previous levels with the symbol of that Noble Gas in brackets. Here is an example of shorthand and extreme shorthand: Sulfur is 1s2 2s2 2p6 3s2 3p4, it can also be represented by [Ne]3s2 3p4. Ionic Radii: Ionization Energy Chapter 5 Rules for Naming Ionic Compounds Containing Polyatomic Ions Polyatomic ions are ions which consist of more than one atom. For example, nitrate ion, NO 3 contains one nitrogen atom and three oxygen atoms. The atoms in a polyatomic ion are usually covalently bonded to one another, and therefore stay together as a single, charged unit. Rule 1. The cation is written first in the name; the anion is written second in the name. Rule 2. When the formula unit contains two or more of the same polyatomic ion, that ion is written in parentheses with the subscript written outside the parentheses. Note: parentheses and a subscript are not used unless more than one of a polyatomic ion is present in the formula unit (e.g., the formula unit for calcium sulfate is "Ca4O " not "Ca(4O )"). Rule 3. If the cation is a metal ion with a fixed charge, the name of the cation is the same as the (neutral) element from which it is derived (e.g., Na = "sodium"). If the cation is a metal ion with a variable charge, the charge on the cation is indicated using a Roman numeral, in parentheses, immediately following the name of the cation (e.g., Fe3+= "iron(III)"). Rule 4. If the anion is a monatomic ion, the anion is named by adding the suffix -ide to - the root of the element name (e.g., I = "iodide"). Note: Greek prefixes are not used to indicate the number of atoms, or polyatomic ions, in the formula unit for the compound (e.g.3 2a(NO ) is named "calcium nitrate" not "calciuim dinitrate"). Metal Ion Li Li+ Na Na+ K K+ Rb Rb+ Cs Cs+ Be Be2+ Mg Mg2+ Ca Ca2+ Sr Sr2+ Ba Ba2+ Al Al3+ Zn Zn2+ Sc Sc3+ Ag Ag+ Rules for Naming Molecular Compounds: 1-Remove the ending of the second element, and add "ide" just like in ionic compounds. 2-When naming molecular compounds prefixes are used to dictate the number of a given element present in the compound. " mono-" indicates one, "di-" indicates two, "tri-" is three, "tetra-" is four, "penta-" is five, and "hexa-" is six, "hepta-" is seven, "octo-" is eight, "nona-" is nine, and "deca" is ten. 3-If there is only one of the first element, you can drop the prefix. For example, CO is carbon monoxide, not monocarbon monoxide. 4-If there are two vowels in a row that sound the same once the prefix is added (they "conflict"), the extra vowel on the end of the prefix is removed. For example, one oxygen would be monooxide, but instead it's monoxide. The extra o is dropped. Generally, the more electropositive atom is written first, followed by the more electronegative atom with an appropriate suffix. For example, H2O (water) can be called dihydrogen monoxide (though it's not usually). Organic molecules (molecules made of C and H along with other elements) do not follow this rule. Examples of Molecular Compound Names: SO2 is called sulfur dioxide SiI4 is called silicon tetraiodide SF6 is called sulfur hexafluoride CS2 is called carbon disulfide Empirical formulas Problem #1: A compound is found to contain 50.05 % sulfur and 49.95 % oxygen by weight. What is the empirical formula for this compound? The molecular weight for this compound is 64.07 g/mol. What is its molecular formula? Solution: 1) Assume 100 g of the compound is present. This changes the percents to grams: S ⇒ 50.05 g O ⇒ 49.95 g 2) Convert the masses to moles: S ⇒ 50.05 g / 32.066 g/mol = 1.5608 mol O ⇒ 49.95 g / 16.00 g/mol = 3.1212 mol 3) Divide by the lowest, seeking the smallest whole-number ratio: S ⇒ 1.5608 / 1.5608 = 1 O ⇒ 3.1212 / 1.5608 = 2 4) Write the empirical formula: SO 2 5) Compute the "empirical formula weight:" 32 + 16 + 16 = 64 6) Divide the molecule weight by the "EFW:" 64.07 / 64 = 1 7) Use the scaling factor computed just above to determine the molecular formula: SO 2imes 1 gives SO f2r the molecular formula Problem #2: A 1.50 g sample of hydrocarbon undergoes complete combustion to produce 4.40 g of CO a2d 2.70 g of H O.2What is the empirical formula of this compound? Solution: 1) Determine the grams of carbon in 4.40 g CO an2 the grams of hydrogen in 2.70 g H 2. carbon: 4.40 g x (12.011 g / 44.0098 g) = 1.20083 g hydrogen: 2.70 g x (2.0158 g / 18.0152 g) = 0.3021482 g 2) Convert grams of C and H to their respective amount of moles. carbon: 1.20083 g / 12.011 g/mol = 0.09998 mol hydrogen: 0.3021482 g / 1.0079 g/mol = 0.2998 mol 3) Divide each molar amount by the lowest value, seeking to modify the above molar amounts into small, whole numbers. carbon: 0.09998 mol / 0.09998 mol = 1 hydrogen: 0.2998 mol / 0.09998 mol = 2.9986 = 3 We have now arrived at the answer: the empirical formula of the substance is CH 3 Note: I did not check for the presence of oxygen. The problem said hydrocarbon, which are compounds with only C and H. Sometimes the problem will be silent about what type of compound it is, give only C and H data, but oxygen will also be in the compound. See comment in problem #3. Chapter 6&7 Resonance Electronegativity Difference Type of Bond Formed 0.0 to 0.2 nonpolar covalent (Cl2) 0.3 to 1.4 polar covalent(HCl) > 1.5 Ionic(NaCl) Finding Bond Angle 1. Write the Lewis dot structure for the molecule. Assume that you have to determine the bond angles in BF 3. B is less electronegative than F, so B becomes the central atom. If we have three F atoms, that means that we are going to use all three electrons from the B. www.vias.org This gives us three bonding pairs of electrons and 0 nonbonding pairs. Thus, the steric number, SN — the number of bonding and non-bonding electron pairs — is 3. The SN is also known as the number of ELECTRON DOMAINS. 2. Use the steric number and VSEPR theory to determine the electron domain geometry of the molecule. To get the VSEPR geometry, imagine that there is a sphere around the central atom. Place the electron pairs on the surface of the sphere so that they are as far apart as possible. This is how two to six electron domains arrange themselves on the surface of a sphere. Electron Domain Geometries Three electron domains arrange themselves evenly around the equator of the sphere to give a trigonal planar shape. 3. Use the VSEPR shape to determine the angles between the electron domains. From elementary math, we know that a circle is composed of 360 °. We divide this number by the number of electron domains and get 360°3=120°. Thus, the bond angles in BF 3are 120 °. Formal charge = [# of valence electrons] – [electrons in lone pairs + number of bonds] Sigma and Pi Counting rules: 1 single bond = 1 sigma bond 1 double bond = 1 sigma bond + 1 pi bond 1 triple bond = 1 sigma bond + 2 pi bonds
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