General Chemistry I, Unit I: Study Guide

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1 General Chemistry I, Unit I: Study Guide General Chemistry I Unit I 1 CDS Chapter 1: Atomic Molecular Theory Law of Conservation of Mass the total mass of all products of a chemical reaction is equal to the total mass of all reactants of the reaction The total mass of the elements that make up the compound must equal the mass of the starting compound Example: copper carbonate made of copper, carbon and oxygen o A 100.0g sample of copper carbonate contains 51.5g copper, 38.8g oxygen and 9.7g carbon o The sum of these constituent masses is 100g, the same as the mass of copper carbonate o Regardless of the specific sample of copper carbonate, or the mass of the sample, the fraction of the mass of the sample which is copper (or oxygen or carbon) is always the same (51.5%, 38.8% and 9.7% respectively) Law of Definite Proportions When two or more elements combine to form a compound, their masses in that compound are in a fixed and definite ratio The Law of Definite proportions does not prove anything about atoms especially not how they combine to make compounds The mass relationships between a fixed amount of nitrogen and oxygen in three different oxides shows that there are a few specific masses of oxygen which will combined with the fixed nitrogen, and those specific masses are integer multiples of each other o This reveals that oxygen exists as fixed units of mass known as an atom Law of Multiple Proportions When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in simple integer ratio o If we fix the mass of one element, the masses of the other element in the three compounds are always in simple integer ratio Atomic Molecular Theory o Each element is composed of very small, identical particles called atoms o All atoms of a single element have the same characteristic mass o The number and masses of these atoms do not change during a chemical transformation o Each compound consists of identical molecules, which are small, identical particles formed of atoms combined in simple whole number ratios CDS Chapter 2: Atomic Masses and Molecular Formulas Assumptions: knowledge of the Atomic Molecular Theory, Law of Conservation of Mass, Law of Definite Proportions, Law of Multiple Proportions 1 L oxygen + 2 L hydrogen gas = 2 L water vapor o The volume of gases is not conserved here the combined volume of the reactants is 3 L, but the product is 2 L o However, the volumes of the gases are in simple integer ratios o Arbitrary amounts of hydrogen and oxygen cannot be combined because they combine in a fixed ratio by mass Law of Combining Volumes When gases combine during a chemical reaction at fixed temperature and pressure, the volumes of the reacting gases and products are in simple integer ratios From Atomic Molecular Theory: atoms and molecules react in simple integer ratios o A possible explanation for the integer volume ratios is that these are the same integer ratios as the particles react in o Equal volumes of the two gases must contain equal numbers of particles, regardless of whether they are hydrogen or oxygen o Basis for this conclusion: uniqueness of integers Occam s Razor

2 General Chemistry I Unit I 2 Avogadro s Law Equal volumes of gas contain equal numbers of particles, if the volumes are measured at the same temperature and pressure Problem: 1 L of hydrogen + 1 L of chlorine = 2 L of hydrogen chloride o This is the same as saying 1 hydrogen atom + 1 chlorine atom = 2 hydrogen chloride atoms o This violates AMT unless hydrogen is diatomic The rest is observationbased read over to review CDS Chapter 3: Structure of an Atom Assumptions: Atomic Molecular Theory, measurements of relative atomic masses, properties of an element and its atoms o Electricity consists of individual charged particles called electrons assigned a negative charge o Atoms in elemental form are neutral (without charge) so atoms must contain a positive charge that equals the charge of the electrons there has to be an integer number of negative and ositive charges Rutherford s Gold Foil Experiment o Alpha particles positively charged, much more massive than electron o Beam of alpha particles will be fired at the gold foil o Observations and conclusions: Observation The greatest number of the alpha particles pass directly through the gold foil without any deflection. A much smaller number of alpha particles experience small deflections in paths. Conclusion The gold foil consists mostly of empty space. Since these particles are deflected, whatever they came close to must have been more massive than they are they came close to the nucleus. A very small fraction of alpha particles are deflected back in the direction of the beam they came from. A tiny fraction of the alpha particles must encounter something other than empty space in order to rebound, an alpha particle must hit something much more massive than itself. o Simple atom model: each gold atom must be mostly empty space and most of the mass of the gold atom must be concentrated into a very small fraction of the volume of the atom this concentrated mass is the nucleus. o If the positive charges of the atom are concentrated in the nucleus, then the negative charges in the atom (the electrons) must be in the much larger space outside the nucleus. Xray emissions o Since the positive charge is an integer, there must be particles of positive charge in the nucleus known as protons o Use spectroscopy to identify that a sample contains a particular compound or element just by looking at frequencies of light emitted o There is a parabolic relationship between atomic number and xray frequency o This shows that the atomic number is a physical property of the atom because it is an integer, it must be counting something within the atom the only thing it could be describing is the number of protons CDS Chapter 4: Electron Shell Model of an Atom Elements with similar atomic numbers are very different both physically and chemically Assumptions: AMT, relative atomic masses, structure of an atom, atomic numbers

3 General Chemistry I Unit I 3 o Coulomb s Law (edit 8/17/15: you will need to know this for the rest of general chemistry and possibly for the rest of your god damn life) algebraic expression which relates the strength of the interaction between two charged particles to the sizes of the charges on the particles and the distance between them. The strength of the interaction between particles is either the force that one particles exerts on the other or the potential energy that exists when two particles interact. General Chemistry likes to focus on the potential energy, symbolized by V: V = (q 1)(q 2 ) r When V is a large negative number, the potential energy is very low and the charges are strongly attracted to one another it requires a lot of work to pull them apart. V will be a large negative number when The charges are large r is small Elements with very different atomic numbers can have similar chemical properties o Fluorine and chlorine are both gases and exist as diatomic molecules in nature both are highly reactive, combine with hydrogen to form acids, combine with metals to form solid salts but atomic numbers are different o Alkali metals lithium, sodium, potassium soft metals with low melting points form salts with chlorine w/ similar molecular formulas Immediately before each of these elements in the list are three elements that are also very similar to each other noble gases Immediately after each of these elements are three elements w/ similar properties alkaline earth metals Halogens o Groups of elements appear periodically in the rankings of the elements by atomic number. Periodic Law The chemical and physical properties of the elements are periodic functions of the atomic number. The attraction of an electron at distance r from the nucleus is given by the potential energy in Coulomb s Law: V(r) = (+Ze)( e) r o An electron close to the nucleus is more strongly attracted to the nucleus (more negative potential energy) o Electron in an atom with a large atomic number (Z) would be more strongly attracted to its nucleus than an electron in another atom with a smaller atomic number o The ionization energy is the amount of energy required to remove an electron from an atom. Large negative V(r) would require a large ionization energy

4 Observations of ionization energy (IE) graph: General Chemistry I Unit I 4 o Ionization energy increases and then abruptly decreases over and over with increasing atomic number. o The top of each peak is a noble gas. o The bottom of each valley is an alkali metal. o Between each alkali metal and valley, ionization energy increases almost linearly. o In each period, the IE increases as one moves from alkali metal to noble gas. o The sharp drops in ionization energy after each noble gas are due to a large increase in r (the distance between the valence electrons and the nucleus). Conclusions: o In each period of the Periodic Table, from an alkali metal to a noble gas, the outermost electrons are all about the same distance from the nucleus. o Each additional electron is added to this consistent layer electron shell o At the end of the period, the shell is full and unable to accommodate more electrons the next electron is in a new shell, much farther from nucleus. o Alkali metals have a single electron in the outermost shell that s why they re similar. o Because this outermost shell is so important, it is the valence shell and contains valence electrons. Successive electrons can be removed ionization energies for each one o The second ionization energy is the energy needed to remove an electron from the positive ion to form an ion with a +2 charge o Observations from table 4.2: The first electron in Na (sodium) is easy to remove, but the second one is harder. The first electron is in a valence shell, far from nucleus. Therefore, the r value in Coulomb s Law is large, so the electron s potential energy is small.

5 General Chemistry I Unit I 5 The second electron is closer in and more strongly attracted; its r value is small, so its potential energy has a large negative value. Conclusion: Sodium (Na) only has one valence electron. The first and second electrons in Mg are relatively easy to remove, third one is harder. Mg has two valence electrons.

6 What the CDS doesn t cover Maingroup cations and anions (2.12) General Chemistry I Unit I 6 Common Polyatomic Ions (2.12) Percent yield Limiting reactants / excess reactant Formula Name Formula Name Cation Singly charged anions (cont.) + NH 4 Ammonium NO 2 Nitrite Singly charged anions NO 3 Nitrate CH 3CO 2 Acetate Doubly charged anions CN Cyanide 2 CO 3 Carbonate ClO Hypochlorite 2 CrO 4 Chromate ClO 2 Chlorite 2 CrO 7 Dichromate ClO 3 Chlorate 2 O 2 Peroxide ClO 4 Perchlorate 2 HPO 4 Hydrogen phosphate H 2PO 4 Dihydrogen phosphate 2 SO 3 Sulfite HCO 3 Hydrogen carbonate 2 SO 4 Sulfate HSO 4 Hydrogen sulfate S 2O 2 3 Thiosulfate OH Hydroxide Triply charged anion MnO 4 Permanganate 3 PO 4 Phosphate Percent yield = Actual yield of product Theoretical yield of product 100% 1. Convert the masses of reactant to moles. 2. Compare these mole values to the balanced chemical equation. For example, if your reactants were 2 mol H 2O and 1 mol CO 2, and you found that your samples contained 0.5 mol H 2O and 0.3 mol CO 2, you should know that H 2O is the limiting reactant because the chemical equation says the amount of H 2O has to be two times that of CO 2 for a full reaction to take place.

7 General Chemistry I Unit I 7 3. Use the limiting reactant and basic stoichiometry to find the amount of product in grams. 4. The mass of excess reactant can be found by converting only the amount of excess reactant required by the chemical equation to grams and subtracting that from the initial mass of the excess reactant. (In the example above, convert only 0.25 mol of CO 2 to grams, since the amount of H 2O reacting must be twice that of the amount of CO 2.) Molarity Diluting solution Solution stoichiometry Molarity = Moles of solute Liters of solution = mol L Moles of solute (constant) = Molarity Volume = M i V i = M f V f The same as normal stoichiometry, just including molarity. Titration (3.9) Titration is a procedure for determining the concentration of a solution by allowing a measured volume of that solution to react with a second solution of another substance (the standard solution) whose concentration is known. Percent composition / empirical formulas (3.10) Combustion analysis Percent composition can be used to find the empirical formula of a compound. 1. Convert given masses to percentages of the mass of the initial sample. 2. Use these percent masses to make a hypothetical sample of 100 grams the percent composition values now become the masses of the constituent elements in this 100g sample. 3. Convert the grams to moles. 4. Divide each molar value by the smallest molar value to find the mole ratio and empirical formula. For combustion involving only carbon, hydrogen and oxygen: 1. Find the molar amounts of carbon and hydrogen from the provided masses of CO 2 and H 2O. Example: g CO 2 1 mol CO 2 1 mol C = mol C g CO 2 1 mol CO 2 2. Find the number of grams of carbon and hydrogen in the sample using the molar masses you just found.

8 General Chemistry I Unit I 8 3. Subtract the masses of carbon and hydrogen from the mass of the starting sample. This will give you the missing mass of oxygen in the compound. 4. Convert the mass of oxygen to moles. Now that you have mole values for all three of the constituent elements, you can divide each of the mole values by the smallest of the three mole values to find the C:H:O ratio and the empirical formula. For combustion involving carbon, hydrogen, oxygen and another element: 1. Determine the mass of each element. 2. Convert each of the masses to composition percentages. 3. Determine the mass of each element present in a 100g sample of the molecule you are combusting. 4. Find the molar ratios to form the empirical formula. 5. Find the molecular formula using molar mass information provided. Net ionic equations An ionic equation explicitly shows all of the ions involved in a chemical reaction. It can be derived from the molecular equation. A net ionic equation is an ionic equation without the spectator ions, which are ions that do not undergo any change during the course of the reaction (they are simply there to balance the charge). Ionic equation: Pb 2+ (aq) + 2 NO 3 (aq) + 2 K + (aq) + 2 I (aq) 2 K + (aq) + 2 NO 3 (aq) + PbI 2 (s) Net ionic equation: Pb 2+ (aq) + 2 I (aq) PbI 2 (s) Oxidation numbers (4.6) An atom in its elemental state has an oxidation number of 0. An atom in a monatomic ion has an oxidation number identical to its charge. An atom in a polyatomic ion or in a molecular compound usually has the same oxidation number it would have it were a monatomic ion. o In general, the farther left an element is in the periodic table, the more probable that it will be cationlike. Metals usually have positive oxidation numbers. The opposite is true for nonmetals. o Hydrogen can be either +1 or 1. o Oxygen is usually 2. o Halogens are usually 1. The sum of the oxidation numbers is 0 for a neutral compound and is equal to the net charge for a polyatomic ion. Redox reactions (4.64.7) Oxidation is the loss of one or more electrons by a substance (element, compound, or ion), and a reduction is the gain of one or more electrons by a substance. An oxidationreduction (redox) reaction is any process in which electrons are transferred from one substance to another. Identifying redox reactions: Reducing agents o Causes reduction o Loses one or more electrons o Undergoes oxidation o Oxidation number of atom increases Oxidizing agent o Causes oxidation o Gains one or more electrons o Undergoes reduction o Oxidation number of atom decreases

9 Halfreaction method for balancing redox reactions General Chemistry I Unit I 9 Redox stoichiometry (4.10) It s the same as titration. Really, check for yourself if you don t believe me. Bitch. Atomic radii (5.14) An atom s radius is half the distance between the nuclei of two identical atoms when they are bonded together. A comparison of atomic radius versus atomic number shows a periodic riseandfall pattern. o Atomic radii increase going down a group of the periodic table but decrease going across a row from left to right. o The increase going down a group occurs because successively larger valenceshell orbitals are occupied. o The decrease in radius from left to right across the table occurs because of an increase in effective nuclear charge caused by the increasing number of protons in the nucleus (increasing nuclear charge). Valence shell electrons are o Strongly shielded by electrons in the inner shells, which are closer to the nucleus o Less strongly shielded by other electrons in the same shell o Only weakly shielded by other electrons in the same subshell, which are the same distance from the nucleus Because electrons in the same shell are at approximately the same distance from the nucleus, they are relatively ineffective at shielding one another. At the same time, the nuclear charge Z increases across a row. Therefore, the effective nuclear charge for the valence shell electrons increases across the period, drawing all the valence shell electrons closer to the nucleus and shrinking the atomic radii.