Midterm 2 Study Guide
Midterm 2 Study Guide CHEM 1127Q 001
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This 10 page Study Guide was uploaded by Caitrín Hall on Monday March 7, 2016. The Study Guide belongs to CHEM 1127Q 001 at University of Connecticut taught by Fatma Selampinar (TC), Joseph Depasquale (PI) in Spring 2016. Since its upload, it has received 256 views. For similar materials see General Chemistry in Chemistry at University of Connecticut.
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Date Created: 03/07/16
Chapter 3: Composition of Substances and Solutions 3.1 Formula Mass and the Mole Concept Formula mass – sum of average atomic masses of all atoms in the substance’s formula Covalent formulas represent # and types of atoms in a molecule; formula mass = molecular mass Ionic compounds contain cations and anions but do not represent the composition of a discrete molecule; formula mass is not molecular mass Measured in amu The Mole – the amount of substance containing the sam12number of discrete entities as the numbers of atoms in a sample of pure C weighing exactly 12 g An amount unit like pair or dozen Avogadro’s number (N ) = 6.02 x 10 23entities composing a mole A Molar mass – mass in grams of 1 mole of that substance (g/mol) o Same numerical value of atomic/molecular mass of a substance 3.2 Determining Empirical and Molecular Formulas Percent Composition – percentage by mass of each element in the compound To find % comp. by each element, divide the experimentally derived mass of each element by the overall mass of the compound, then convert to a percentage o Ex) 12.04 g of a compound containing 7.34 g carbon, 1.85 g hydrogen, and 2.85 g nitrogen. What is the percent composition? To determine % comp. from formula mass, consider 1 mol of given compound and use its molar mass to calculate the percentage of each of its elements o Ex) find the % comp. of C 9 8 4 To determine empirical formula, use the given masses to find moles of each element, divide each by the lesser number of moles, multiply ratio (if necessary) to get the smallest possible whole number subscripts o Ex) find empirical formula of a compound with composition of 27.29% C and 72.71% O Derivation of Molecular Formulas Compare compound’s molecular or molar mass to its empirical formula mass Molar mass/empirical formula mass = n formula units/molecule Multiply each subscript of the empirical formula by n (A B ) = A B x y n nx nx 3.3 Molarity Solutions Concentration – relative amount of a given solution component Solvent – medium in which other components are dissolved; has a significantly greater concentration that that of other components Aqueous solution – a solution in which water is the solvent Solute – component of a solution present at a lower concentration than solvent o Dilute – relatively low concentration o Concentrated – relatively high concentration Molarity (M) – number of moles of solute in exactly 1 liter of solvent M = mol solute/L solution o Moles and volumes can be determined from molar concentrations o Molar concentrations can be determined from mass of solute o Mass of solute in given volume of solution can be determined from molarity Dilution of Solutions Dilution – the process whereby the concentration of a solution is lessened by the addition of solvent Common means of preparing solutions of desired concentration According to molarity, the molar amount of solute is equal to the solution’s molarity and volume in liters n = ML M1 1=M L 2 2 More general dilution equation: C1V =1 V 2 2 *C = concentration & V = volume* o Rearrange equation to find concentration of a diluted solution (solve for 2 ); volume of a diluted solution (solve fo2 V ); volume of a concentrated solution needed for dilution (solve for 1 ) 3.4 Other Units for Solution Concentrations Mass percentage – the ratio of the component’s mass to the solution’s mass (mass of component/mass of solution) x 100% Percent mass %mass, percent weight %weight, weight/weight percent (w/w)% Calculation of percent by mass o Ex) A 5.0-g sample of spinal fluid contains 3.75 mg (0.00375 g) of glucose. What is the percent by mass of glucose in spinal fluid? Divide mass of glucose by mass of sample Calculations using Mass Percentage o “Concentrated” hydrochloric acid is an aqueous solution of 37.2% HCl that is commonly used as a laboratory reagent. The density of this solution is 1.19 g/mL. What mass of HCl is contained in 0.500 L of this solution? Use solution density given to find solution’s volume and mass, then use the given mass percentage to calculate solute mass Volume percentage = (volume solute/volume solution) x 100% Calculations using volume percentage o Ex) Rubbing alcohol (isopropanol) is usually sold as a 70%vol aqueous solution. If the density of isopropyl alcohol is 0.785 g/mL, how many grams of isopropyl alcohol are present in a 355 mL bottle of rubbing alcohol? Given amount of solution x given volume% x given density Mass-volume percentage – ratio of a solute’s mass to the solution’s volume; (m/v) 6 Parts per million (ppm) = (mass solute/mass solution) x 10 ppm Parts per billion (ppb) = (mass solute/mass solution) x 10 ppb 9 Chapter 4: Stoichiometry of Chemical Reactions 4.1 Writing and Balancing Chemical Equations Chemical equation – symbolic representation of a chemical reaction 1. The substances undergoing reaction are called reactants, and their formulas are placed on the left side of the equation 2. The substances generated by the reaction are called products, and their formulas are placed on the right side of the equation 3. Plus signs separate individual reactant and product formulas, and an arrow separates the reactant and product sides of the equation 4. The relative numbers of reactant and product species are represented by coefficients (numbers placed immediately to the left of each formula); it is common to use the smallest possible whole-number coefficients Balancing Equations Balanced – equal numbers of atoms for each element involved in the reaction are represented on the reactant and product sides H2O ⟶ H +2O (u2balanced) 2H2O ⟶ 2H + 2 (ba2anced) Additional Information in Chemical Equations Physical states of reactants and products are indicated with a parenthetical abbreviation following the formulas; (s), (l), (g), (aq) Equations for Ionic Reactions Molecular equation – doesn’t explicitly represent the ionic species that are present in the solution CaCl (aq) + 2AgNO (aq) ⟶ Ca(NO ) (aq) + 2AgCl(s) 2 3 3 2 Complete ionic equation – explicitly represents all dissolved ions 2+ − + − 2+ − Ca (aq) + 2Cl (aq) + 2Ag (aq) + 23O (aq) ⟶ Ca (aq) + 2NO3 (aq) + 2AgCl(s) Spectator ions – presence is required to maintain charge neutrality but are not chemically nor physically changed by the process Net ionic equation – complete ionic equation MINUS spectator ions Cl (aq) + Ag (aq) ⟶ AgCl(s) 4.2 Classifying Chemical Reactions Precipitation Reactions and Solubility Rules Precipitation reaction – dissolved substances react to form one or more solid products; many involve exchange of ions between ionic compounds in aq. solution (referred to as double displacement, double replacement, or metathesis reactions) Solubility – the extent to which a substance may be dissolved in a solvent Soluble substances have relatively large solubility Insoluble substances have relatively low solubility A substance will precipitate when concentration exceeds solubility in a solution Acid-Base Reactions An acid-base reaction is one in which a hydrogen ion is transferred from one chemical species to another o Acid – any substance that will dissolve in water to yield hydronium + ions, H3O HCl(aq) + H O(aq) ⟶ Cl (aq) + H O (aq) 2 3 Strong acids completely react in water Weak acids partially react with water generate small + amount of H3O ions o Base – a substance that will dissolve in water to yield hydroxide ions, OH- Commonly compounds composed of alkali or alkaline earth metal cations combined with hydroxide Strong bases completely dissociate in water NaOH(s) ⟶ Na (aq) + OH (aq)− Weak bases react partially to yield hydroxide ions NH (aq) + H O(l) ⇌ NH + (aq) + OH (aq) 3 2 4 Neutralization reaction – acid-base reaction in which the reactants are an acid and a base, while the products are often and salt and water (neither reactant is water) acid + base ⟶ salt + water Oxidation-Reduction Reactions Oxidation = loss of electrons = increase in oxidation number Reduction = gain of electrons = decrease in oxidation number 2Na(s) + Cl (g) ⟶ 2NaCl(s) 2 It helps to separate the reaction into 2 half-reactions 2Na(s) ⟶ 2Na (s) + 2e − − − C2 (g) + 2e ⟶ 2Cl (s) o Here, the sodium is a reducing agent (reductant)—usually metals—that that is oxidized, while chlorine is an oxidizing agent (oxidant)—usually nonmetals) that is reduced Oxidation number – the charge an element’s atoms would possess if the compound was ionic o The oxidation # of an atom in elemental form is zero o The oxidation # of a monatomic ion = ion’s charge o Oxidation numbers for common nonmetals: Hydrogen: +1 when combined with nonmetals, −1 when combined with metals Oxygen: −2 in most compounds, sometimes −1, very rarely -1/2, positive values when combined with F Halogens: -1 for F, -1 for other halogens except when combined with oxygen or other halogens (then they are varying positive #s) o The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion Subclasses of redox reactions o Combustion – the reductant, or fuel, and oxidant, usually oxygen, react vigorously and produce significant amounts of heat, and often light, in the form of a flame o Single-displacement (replacement) reactions – redox reactions in which an ion in solution is displaced by the oxidation of a metal Zn(s) + 2HCl(aq) ⟶ ZnCl (aq)2+ H (g) 2 Balancing Redox Reactions via the Half-Reaction Method 1. Write the two halfreactions 2. Balance all elements except oxygen and hydrogen 3. Balance oxygen atoms by adding H O2molecules + 4. Balance hydroge atoms by adding H ions 5. Balance charge by adding electrons 6. If necessary, multiply each halfreaction’s coefficients by the smallest possible integers to yield equal numbers of electrons in each 7. Add the balanced halfreactions together and simplify by removing species that appear on both sides of the equation 8. For reactions occurring in basic media (excess hydroxide ions), carry out these additional steps: − + o Add OH ions to both sides of the equation to equal the number of H ions o On the side of the equation containing both H and OH ions, combine these ions to yield water o Remove any redundant water molecules 9. Check to see that both the number of atoms and the total charges are balanced 4.3 Reaction Stoichiometry Stoichiometry – relationships between the amounts of reactants and products of a chemical reaction Used to determine the amount of one reactant required to react with a given amount of another reactant or to yield a given amount of product Stoichiometric factor – ratio of coefficients in a balanced chemical equation, used in computations relating amounts of reactants and products Example problem: How many carbon dioxide molecules are produced when 0.75 mol of propane is combusted according to this equation? C H +5O ⟶3CO +4H O 3 8 2 2 2 4.4 Reaction Yields Limiting reactant – the reactant present in an amount lower than required by the reaction stoichiometry, thus limiting the amount of product generated Excess reactant – the reactant present in an amount greater than required by the reaction stoichiometry Percent yield – the extent to which a reaction’s theoretical yield is achieved Theoretical yield – the amount of product that may be produced by a reaction under specified conditions Actual yield – the amount of product obtained in practice Percent yield = (actual yield/theoretical yield) x 100% 4.5 Quantitative Chemical Analysis Quantitative analysis – the determination of the amount or concentration of a substance Titration Buret – device used to make incremental additions of a solution Titration analysis – quantitative chemical analysis method that involves measuring the volume of a reactant solution required to completely react with the analyte—substance whose concentration must be measured Titrant – substance whose concentration is known Equivalence point – volume of titrant solution required to react completely with the analyte; provides a stoichiometric amount of titrant for the sample’s analyte according to the titration reaction o Indicators are added to solutions to impart a change in color at or near EQ o Other indicator: distinct change in appearance (i.e. halt of bubble formation) End point – the volume of titrant actually measured; theoretically equal to EQ point Gravimetric analysis – quantitative analysis in which a sample is subjected to some treatment that causes a change in the physical state of the analyte that permits its separation from the other components Mass measurement of a component of the analysis system & known stoichiometric amounts of compounds involved calculation of analyte concentration Required change of state may be achieved by physical and chemical processes o Often the analyte is subjected to a precipitation reaction precipitate is isolated mass of precipitate is used to calculate analyte concentration Combustion analysis – gravimetric method of analysis in which the weighed sample of a compound is heated to a high temperature under a stream of oxygen gas, resulting in complete combustion to yield gaseous products of known identities
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