Chemistry 105 ~ Skills Summary for Laboratory Final Exam
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1 Chemistry 105 ~ Skills Summary for Laboratory Final Exam The Final Exam questions will have a variety of formats, including: true/false, multiple-choice, fill-in-the-blank, and problem-solving. Each student must work individually, and come prepared with a calculator. A Periodic Table and key mathematical formulas (% Error, % Difference, % Yield, etc.) will be provided. Good study materials include: Your Pre-Lab Assignments and Pre-Lab Quizzes (though you should realize that these questions were intended to test for a BASIC level of understanding BEFORE conducting the experiment; this means you should know a bit more now that you ve completed the experiment) The Concepts you need to know to be prepared discussions in the lab handouts The post-lab calculations on the Report Sheets The problem-solving and discussion questions on the Report Sheets For ALL calculations performed on the lab final exam, realize that your lab instructor can t possibly grade work done in your head or on your calculator, ONLY the work that you present neatly and legibly on your test paper! So to receive maximum credit on calculations, you MUST show them neatly, with correct units, and express your answers, with correct units, to the appropriate number of significant figures/decimal places. If you use one of the formulas provided on the exam (such as PV = nrt), you should show the formula FIRST, and THEN substitute the values with their units. Realize that it s quite likely that you ll be given sample experimental data, and asked to work the same sorts of calculations that you did in completing your Report Sheets. For example: Consider the Mole Relationship ( Tracking Zn +2 ) experiment. You might be given the balanced chemical equation and an initial mass of ZnCO 3, and asked to predict the mass of ZnO (or CO 2 ) that will be produced by the reaction (a stoichiometry calculation). If you re given the mass of product that was experimentally produced, you could then be asked to calculate the % Yield. If you feel that you ll need assistance in reviewing any of the skills listed below, please make arrangements to meet with your lab instructor (or other chemistry teaching assistant) in the Pastore 215 Help Office. Realize that with the exception of a few skills specific to the laboratory, these are the SAME skills needed for success in your CHM 103 lecture course. To demonstrate mastery of CHM 105 Skills, students should be able to: From the Laboratory Safety Activit:y: Interpret the chemical hazard information (color and number) indicated on National Fire Protection Association (NFPA) labels. From Lab 1: Laboratory Measurements Given a measurement, recognize the number of significant figures it contains (and the digit that contains the uncertainty). When making a measurement, identify the smallest increment on the measuring device, and then estimate the digit that contains the uncertainty. Record the measurement with units and the appropriate number of significant figures. Use rulers, graduated glassware and thermometers to make measurements. Use significant figures correctly when working ALL calculations (multiplying /dividing or adding/subtracting), and round appropriately to express the uncertainty in the result. Convert numbers between standard and scientific (exponential) notation. Use Dimensional Analysis (aka the Factor Unit Method or Factor Label Method) to solve unit conversion problems: Use unit relationship EQUALITIES as conversion factors. Set up calculations so that units CANCEL correctly and the result has the desired units. C. Graham Brittain Page 1 of 12 11/18/2008
2 Explain the difference between accuracy and precision: Accuracy is the degree of agreement between a measured value and the true value. A measurement is said to be accurate if it is very close to the true value, and inaccurate if it is not close to the true value. Precision refers to the agreement of replicate measurements of the same quantity. Given an accepted value and an experimental value, calculate the Percent Error. State the base metric unit for mass, length, volume, and temperature: The base metric unit of mass is the gram (g) The unit of length is the meter (m) The unit of volume is the liter (L) The unit of temperature is the Celsius degree ( o C) Define the most commonly used metric prefixes (mega-, kilo-, deci-, centi-, milli-, and micro-). These are related to the base unit by powers of ten. Use the metric prefixes to convert between larger to smaller units (so as to use a unit of the appropriate size for what is being measured). Explain how volume is a unit that is derived from length. State equivalencies between the more commonly used cubic length and laboratory volume units. C. Graham Brittain Page 2 of 12 11/18/2008
3 Explain the relationships between the Celsius and Fahrenheit temperature scales. Given the appropriate formulas, convert a temperature from one temperature scale to another. = = Distinguish between extensive and intensive properties: Extensive properties depend on the quantity of a substance. Mass and volume are examples of extensive properties. Intensive properties are those that are inherent to (and characteristic of) a particular substance. Thus they are independent of the amount of substance that is being measured, and they can be used to help identify an unknown substance. Density, freezing/melting point, and boiling point are examples of intensive properties. Explain density: the mass per unit volume of a material. Given any two of the three quantities (mass, volume, density), calculate the third. Use the density unit relationship between mass and volume as a conversion factor in working dimensional analysis problems. From Lab 2: Densities of Solids and Liquids Given mass and volume measurements of different samples of the same substance, prepare a plot of the data from which the density of the substance can be determined. From a plot of mass versus volume, graphically determine the density of a substance. (Draw the best fit straight line, and determine the slope of that line.) From Lab 3: Separation of a Mixture into Pure Substances ~ Paper Chromatography of Metal Cations Fully describe the Classification System for Matter: Any material is either a pure substance or a mixture of pure substances. A pure substance cannot be separated into other kinds of matter by a physical change. The two types of pure substances are elements and compounds. Elements are the basic building blocks of matter; they cannot be decomposed by a chemical reaction. Compounds are made from two or more elements by a chemical reaction; thus they are a chemical combination of elements. Compounds have constant composition; for a particular compound, the elements always combine in a fixed proportion. Mixtures are a physical combination of pure substances (elements and compounds). They can be separated by a physical change, by leveraging the differences in the physical properties of the components. They also have variable composition: elements and/or compounds can be combined in any proportion to make a mixture. Heterogeneous mixtures have multiple phases. Homogeneous mixtures have just a single phase (they are consistent throughout). Homogeneous mixtures are also called solutions. C. Graham Brittain Page 3 of 12 11/18/2008
4 The Classification System for Matter: Explain how substances can be separated from one another by paper chromatography, on the basis of differences in their affinities for the mobile and stationary phase. Given a paper chromatogram, determine the Retention Factor (R f ) value of the separated substances. Explain what a larger R f value (moves high up the paper with the mobile phase) indicates about a substance. Explain what a lower R f value (leaves the mobile phase and adheres to the stationary phase) indicates about a substance. From Lab 4: Structure, Geometry, and Polarity of Molecules Explain the electron sharing (covalent bonding) that occurs between nonmetals in molecular (covalently-bonded) compounds. Explain the covalence of the nonmetal atoms. (Covalence is the number of electrons needed to achieve a noble gas valence electron configuration, and is thus the number of covalent bonds an atom will typically form.) Given the chemical formula of a simple molecule (one to three central atoms), draw a Lewis structure for the molecule. Explain the information contained in the chemical formula and in the Lewis structure for molecules (covalent compounds): The chemical formula expresses the composition of the molecule (exactly how many of each type of atom), but doesn t indicate how the atoms are connected. Lewis structures indicate the connectivity of the atoms, but do NOT clearly represent the molecular geometry about the central atoms. C. Graham Brittain Page 4 of 12 11/18/2008
5 Use electron group repulsion theory to the molecular geometry of the central atoms in a covalently-bonded compound: Count the groups of electrons around a central atom and determine the electron-group geometry. From the number of bonding and nonbonding electron groups, determine the molecular geometry of central atoms. Explain what it means for a covalent bond to be polar. State the trends in electronegativity (Table 4.4). Use electronegativity values to evaluate the polarity of covalent bonds. Use the molecular geometry of a central atom to determine whether bond dipoles are canceled or enhanced by the geometry. From Lab 5: Determining the Chemical Formula of an Ionic Compound State Avogadro s number: 1 mole of any substance = 6.02 x particles (where particles refers to atoms, ions, or molecules, depending on the nature of the substance). Write an Avogadro s statement for any element or compound: 1 mole of an element contains 6.02 x atoms of that element, and has a mass in grams equal to the atomic weight of the element. 1 mole of a molecular element (such as O 2, H 2, etc.) contains 6.02 x molecules of that element, and has a mass in grams equal to the formula weight (molecular weight) of the molecule. 1 mole of an ionic compound contains 6.02 x formula units of that compound, and has a mass in grams equal to the formula weight of the compound. 1 mole of a covalent (molecular) compound contains 6.02 x molecules of that compound, and has a mass in grams equal to the formula weight (molecular weight) of the compound. NOTE: The mass in grams of 1 mole of ANY substance is commonly referred to as the Molar Mass. Use the Avogadro s unit relationships (expressed in the Avogadro s Statement ) to convert between numbers of items, grams, and moles for any element or compound. Given sample data (the experimental masses of each of the elements in a compound): Use the molar masses of each element to convert from grams to moles. Determine the mole ratio of the elements in the compound. Determine the chemical formula of the compound. Determine the experimental mass percentage of each element in the compound. Given the chemical formula of a compound, calculate the theoretical mass percentage of each element. C. Graham Brittain Page 5 of 12 11/18/2008
6 Explain that for an ionic compound, the chemical formula expresses the fixed, relative proportions of the elements in the compound. Remember that ionic compounds should NEVER be thought of as just a PAIR of oppositely-charged ions, as the chemical formula seems to imply. If you look closely at the illustration of a sodium chloride (NaCl) salt crystal to the right, you ll see that it is definitely NOT just pairs of Na + and Cl - ions, but an extensive, orderly, three-dimensional network of charged particles. Each Na + ion is attracted to the Cl - ions immediately above, below, to the right, to the left, in front, and behind it. And each Cl - ion is attracted to the Na + ions surrounding it. So the chemical formula for any salt does NOT describe a small grouping of ions, it describes the fixed proportion in which vast numbers of the oppositely-charged ions must combine in order to appropriately balance out the positive and negative charges. From Lab 6: A Six-Bottle Study of Ionic Compounds Given a chemical equation describing a reaction, BALANCE the chemical equation by placing numerical coefficients in front of the chemical formulas of the participants in the reaction. Realize that you CANNOT balance an equation by changing the chemical formulas (by changing the fixed proportion of the elements in a compound). Given a sentence describing a chemical reaction, translate it into a balanced chemical equation. (Realize that the first step is to write the correct chemical formula for every element or compound participating in the reaction.) Given a table of the solubilities of common ionic compounds, use it to identify the solid or gaseous products of a chemical reaction. Then indicate the physical states of all participants in the balanced chemical equation using the appropriate subscripts: (g), (l), (s), (aq). C. Graham Brittain Page 6 of 12 11/18/2008
7 From Lab 7: The Mole Relationship in Chemical Reactions Use the mole relationships in a balanced chemical equation to work stoichiometry problems. That is, for a given amount (grams, moles) of one participant in a chemical reaction, predict the amounts of other participants in the reaction as dictated by the stoichiometry of the balanced chemical equation. Use the percentage yield calculation to compare an experimental (actual) yield to a predicted (theoretical) yield. From Lab 8: Determining the Ideal Gas Constant Use the Kinetic Molecular Model to explain the properties of gases, liquids and solids: In the solid state: Particle motion is restricted to vibrations. Intermolecular attractive forces are the strongest. In the liquid state: Particles move freely around one another, but in close proximity. Intermolecular attractive forces are weaker. In the gas state: Particles move freely and are widely separated from one another. They experience elastic collisions, and intermolecular attractive forces are at a minimum. A Kinetic Molecular view of solids, liquids, and gases: C. Graham Brittain Page 7 of 12 11/18/2008
8 Use the Gas Law relationships to predict the behavior of a fixed number of moles of a gas when it is compressed, expanded, heated or cooled. Boyle s Law: For a constant number of moles of gas at a constant temperature, the VOLUME of the gas is INVERSELY proportional to its pressure. This means that as the pressure on the gas is increased at constant temperature, the volume of the gas must decrease. Or as the pressure on the gas is released, the gas can expand (increase its volume). Charles Law: For a constant number of moles of gas at a constant pressure, the VOLUME of the gas is DIRECTLY proportional to its TEMPERATURE. This means that as the gas is heated at constant pressure (its temperature is increased, the gas will expand (its volume increases). For a constant number of moles of gas at a constant volume, the PRESSURE of the gas is DIRECTLY proportional to its TEMPERATURE. This means that as the gas is heated (its temperature is increased, the pressure of the gas must increase if volume is held constant. Avogadro s Law: If two samples of gas have exactly the same volume, and exist at exactly the same temperature and pressure, they must contain the same number of moles of gas. C. Graham Brittain Page 8 of 12 11/18/2008
9 Use the Ideal Gas Law for calculations regarding the behavior of a sample of gas (n moles) at specified conditions of pressure, volume, and temperature (P, V, T). In using the Ideal Gas Law: P = pressure of the gas in atmospheres (atm) V = volume of the gas in liters (L) T = temperature of the gas in Kelvin (K) The Ideal Gas Constant, R = L atm/mol = From Lab 9: Solubilities of Ionic and Molecular Substances Explain what a solution is (a homogeneous mixture, with the components present as atoms, ions, or molecules). Give specific examples of various types of solutions (solid in liquid, gas in liquid, etc.) Explain the like dissolves like rule for solubility. Be able to predict the solubility of given solutes in specified solvents, based on an assessment of their chemical structure. Consider the thought process required to make this prediction: Is the solute ionic or molecular? IF the solute is molecular, are the solute molecules primarily polar or nonpolar? (What is the degree of polarity? Mostly hydrophilic water loving? Or mostly hydrophobic water fearing? Are there any hydrogen-bonding groups? How many? And how large is the nonpolar hydrophobic portion of the molecule?) Are the solvent molecules primarily polar or nonpolar? (Again, what is the degree of polarity?) What type(s) of intermolecular attractive forces can the solute and solvent molecules use in interacting with one another? To the right: How an IONIC compound dissolves in water: the ions dissociate and are hydrated by the water molecules. To the left: How a POLAR molecule dissolves in water: the molecules go for a swim among the water molecules and interact with them via dipole-dipole interactions or hydrogen bonding. C. Graham Brittain Page 9 of 12 11/18/2008
10 From Lab 10: Factors Affecting the Rate of a Chemical Reaction Explain how Reaction Rate is defined, as well as how it can be measured. Reaction Rate can be defined as: The change in concentration of the REACTANT as a function of time (the rate of DISAPPEARANCE of the REACTANT) The change in concentration of the PRODUCT as a function of time (the rate of FORMATION of the PRODUCT) Concentration is always expressed in MOLARITY units (M = moles/liter). So a reaction rate typically has units of: moles per liter per unit time: moles/l sec, OR moles/l min, OR moles/l hour, etc. = = = Note that the brackets [ ] mean concentration of, and that the concentration is ALWAYS expressed in MOLARITY units (moles per liter = moles/l = M). Given sample data (change in reactant concentration, change in time), calculate the reaction rate. Describe the Collision Theory that explains reaction rates: In order for reactant particles (atoms, ions, or molecules) to react, they must collide. At the instant of collision, the kinetic energy of particle motion is converted to collision energy. The collision energy must be greater than or equal to the Activation Energy (E a ) in order to have an effective (successful) collision one that has sufficient energy to break key bonds and rearrange the atoms. The Activation Energy (E a ) is the minimum energy that must be supplied by the collision in order for the reaction to occur by a particular pathway. It is the difference between the energy level of the reactants and the energy level of the transition state (the intermediate that exists at the moment of collision, when the kinetic energy, converted into collision energy, is used to break bonds and rearrange the atoms). A certain fraction of all molecules in a sample will have the necessary Activation Energy to react, and that fraction will increase with increasing temperature. The proper spatial orientation of the colliding molecules is also essential to an effective collision. Explain (in terms of the Collision Theory) the four factors that influence the rate of a chemical reaction: The ionic or molecular nature of the reactants The concentration of the reactants The temperature of the reactants The presence of a catalyst From Lab 11: An Introduction to Acids, Bases, ph, and Buffers Explain the Bronsted-Lowry definitions of an acid, a base, and an acid-base reaction: An acid is a compound which can donate a proton (hydrogen ion, H + ) A base is a compound that can accept a proton. An acid-base reaction is then simply the transfer of a proton from the acid to the base. Write chemical reactions which illustrate an acid donating a proton to produce its conjugate base, and a base accepting a proton to produce its conjugate acid. C. Graham Brittain Page 10 of 12 11/18/2008
11 Explain what the terms strong and weak mean when applied to acids (and bases): Strong acids dissociate completely in water (ionize 100%). Weak acids dissociate to an equilibrium extent, which can be described by the value of the Acid Dissociation Equilibrium Constant, K a. Given the molarity of a strong acid or strong base solution, calculate the hydronium ion concentration, and then the ph of the solution. Realize that the dissociation of ANY weak acid (HA) in water can be represented as: HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) And that the equilibrium constant expression for the dissociation of ANY weak acid can be written as: = As this equilibrium constant describes the extent of dissociation of a weak acid in water, it s known as the Acid Dissociation Constant, K a. And realize that the ionization of ANY weak base (B) in water can be represented as: B (aq) + H 2 O (l) OH - (aq) + HB + (aq) And that the equilibrium constant expression for the dissociation of ANY weak base can be written as: = As this equilibrium constant describes the extent of ionization of a weak base in water, it s known as the Base Ionization Constant, K b. Use equilibrium constants for the dissociation of weak acids (K a, acid dissociation constant) to compare the proton-donating ability of weak acids. That is, use the K a values to compare the relative amounts of hydronium ion, H 3 O +, present in each weak acid solution at equilibrium. C. Graham Brittain Page 11 of 12 11/18/2008
12 Use equilibrium constants for the ionization of weak bases (K b, base ionization constants) to compare the proton-accepting ability of weak bases. That is, use the K b values to compare the relative amounts of hydroxide ion, OH-, present in each weak base solution at equilibrium. Explain how weak acids and their conjugate bases (or weak bases and their conjugate acids) can be used to buffer the ph of a solution. Show/describe the amphiprotic nature of water (the self-ionization of water) that occurs in EVERY aqueous solution. Use the Ion Product Constant of Water to convert between hydroxide and hydronium ion concentrations in ANY aqueous solution. = = Explain the ph method of expressing the hydronium ion concentration of an aqueous solution. ph = log [H 3 O + ] Given a hydronium ion concentration, calculate the ph. Given a ph, calculate the hydronium ion concentration. That is, if ph = log [H 3 O + ], then [H 3 O + ] = 10 ph Explain the ph (and relative hydronium and hydroxide ion concentrations) of pure water, a neutral solution, an acidic solution, and a basic solution. C. Graham Brittain Page 12 of 12 11/18/2008
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