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Chapter 1 An Introduction to Chemistry “Chemistry is the science of composition, structure, Definition of chemistry properties and reactions of matter, especially atomic and molecular systems.” Broad field – includes everything in the whole universe, animate and inanimate. Agriculture – find a wheat kernel that makes best bread Astronomy – what matter is there on Mars? Animal science – find the best animal feed Geology – composition of rocks, water inclusion in lava Medicine – MRI Material science – catalysts for fuel cells Molecular biology – protein structure and folding Despite the common belief that a chemist mixes chemicals just for the love of it – and gets unimaginable results – the truth is different: a chemist tries to understand the nature and natural phenomena, composition and structure of complex systems (like biomolecules) or to synthesize new substances in order to advance science in general and contribute to the well-being of humanity. Scientific method Collect data or facts relevant to the problem. Formulate prediction (hypothesis) to explain the data Plan and test the hypothesis with additional experiments. Modify the hypothesis. laws observation hypothesis experiment theory Hypothesis – tentative explanation Theory – well-established hypothesis; explanation of general principles of certain phenomena with facts to prove it. - Einstein’s theory of relativity Scientific laws – simple statements of natural phenomena to which no exceptions are known. - Law of buoyancy: any object submerged in a fluid is acted on by a force with magnitude equal to the weight of the fluid displaced by the object. Bias- a strong preference that inhibits impartial judgment. Aristotle believed that women have less teeth than men. Matter Matter is composed of tiny particles called atoms. It can be in any of the three physical states: solid, liquid or gas. The physical state in which the matter exists depends on the nature of the matter (i.e. attractive forces in the atoms or molecules of the matter), and external factors (pressure, temperature). Physical transformations: (solid – liquid) freezing / melting, liquid-gas: vaporization / condensation, and (solid-gas) sublimation Matter is divided into pure substances, and mixtures. Pure substances are particular form of matter with definite and fixed composition, e.g. elements (Cu, Au, Al) and compounds (salt, sugar, water). Mixtures can be: homogeneous - uniform in appearance and with same properties throughout, also called: solution, or heterogeneous (two distinct phases with boundary between them). Mixtures contain two or more substances that can be separated physically (boiling, magnet, filtration, etc. Physical state and properties State Solid Liquid Gas Shape Definite Indefinite Indefinite Volume Definite Definite Indefinite Compress. V.Slight Slight High State of common materials Solids Most metals (gold, copper, zinc) Salt Sugar Sand 0 100 Temperature, oC Liquids Water Alcohol Vinegar Blood Oil Gases Air (oxygen, nitrogen,…) Acetylene Chlorine Noble gases (He, Ne …) 3 Properties and changes of matter Set of properties gives unique identity to the matter. Physical properties can be determined without destroying the substance (state, color, density, taste/odor, melting/boiling point). Chemical properties describe the ability of the substance to form new substances, either by decomposition or reaction with other substances (corrosiveness, flammability, acidity, toxicity, etc.). No two substances have identical physical and chemical properties. Physical changes are reversible changes (in size, shape, state of matter, density) when no new substances are formed. Heating of Pt wire changes its appearance from silvery metallic to glowing yellowred. Upon cooling, original appearance is restored and no new compounds are formed. Shattered glass looks different, but its (chemical) composition is the same. Chemical changes are irreversible as new compounds are formed. Upon heating, Copper wire changes its appearance like Pt, but appears black after it is cooled. Cu oxidizes to copper(II) oxide, CuO. Physical change results in a different form of the same substance, while chemical change results in a completely different substance, or substances. Chapter 2 Measurements and significant figures Measurement is a comparison of the quantity of the sample to a standard. Measurement is important – a small variation of a drug quantity may kill you! Measurement never gives an exact value. Its numerical value always contains all certain figures plus one that is estimated. Estimated figure you obtain These are called significant figures. The measuring device gives you the certain figures. by reading between the lines. Measurement of a coin Estimated using a ruler marked in figure: 5 centimeters gives the diameter to be halfway between 2 and 3 cm, or about 2.5 cm; uncertainty is in the estimated decimal. Certain figure: 2 This measurement gives the radius to be 2.5 cm. That is, two significant figures. Accuracy: closeness of the measurement to the true value. Using a ruler marked in millimeters makes the 0.5 cm a certain figure. Estimated figure: 0 Certain figures: 2, 5 This measurement gives the radius to be 2.50 cm: three significant figures. Precision: closeness to one another of a series of measurements made on the same object. Using the first ruler we got the diameter of 2.5 cm. What would be its radius? Radius = 2.5 cm / 2. Your calculator would show 1.25 cm. BUT: the second decimal place is beyond your ability to estimate. Thus, you must round the number to a single decimal place, i.e. 1.3 cm. You can keep only significant figures – the ones that are not beyond the accuracy of the measuring device, plus your estimated figure. Note that there are exact numbers: there is exactly 2 radii in a diameter! Significant Figure Rules – 1. all non-zero digits are significant. 2. zero is NOT significant if: – it is before the first nonzero digit or - at the end of a number without decimal point. Practice: 20.201 20.210 20.0002 120. # of sig. figs.: 5 5 6 3 Leading (placeholder) zeros only give the location of the decimal point and are not significant. 0.06 inch has only one sig figs. Decimal point tells you that the last zero is an estimate (uncertainty +/- 1). Without the decimal point, the number 120 can be interpreted as 120 +/- 10, thus two sig. figs. To avoid ambiguity, we use scientific notation. Scientific notation 1.2 x=2 A = 1.2 1.23 x 102 1.23 x 10-1 1.23 x 100 4.6 x 104 4.6 x 10-4 x is an exponent (or power) of 10. Exponent x is always a whole number. When x is positive, you move the decimal point x places to the right. Scientific notation A x 10x A is a number > (or =) 1 and < 10 x 102 Decimal point moves: 2 places right 1 place left doesn’t move 4 places right 4 places left When x is negative, you move the decimal point x places to the left. Standard notation 123. 0.123 1.23 46,000 0.00046 1.20 x 102 2 places left 120. 1.2 x 102 2 places left 120 no need for decimal point 100 = 1 ! no decimal point! Add 3 zeros after six Add 3 zeros before four Reverse movement of decimal point when going from standard to sci. notation When doing math Calculator often gives more digits than justified. Rounding rule – use all digits you want to retain. Then check the digit after the last one retained: - If it is 4 or less, drop it and all the digits behind. Leave the retained number unchanged. - If it is 5 or more, drop it and all the digits behind. The last digit in the retained number increases by one. To avoid accumulating errors, always round the number at the end of calculation. Sig. figs. rules for multiplication and division: The result cannot have more significant figures than there are in the measurement that has the smallest number of significant figures. 2.0 cm X 2 cm = 4 cm2 2.0 cm X 2.0 cm = 4.0 cm2 2 sig. figs. 1 sig. fig. 1 sig. fig. 2 sig. figs. 2 sig. fig. 2 sig. fig. Sig. figs. rules for addition and subtraction: The result can be no more certain than the least certain measurement in the series. 123 cm 20.5 lb + 15 lb 35.5 lb = = 14.489… + 1.006 cm 2 2.45 ft x 1.00 ft 2.45 ft + 2.6 cm = 14 lb/ft2 Calculator gives 126.606 cm Uncertainty +/- 1 127 cm Uncertainty +/- 1: Should be rounded to 36, or 2 significant figures Number with a name - Unit of measure A measurement makes no sense unless units are specified. Say, you are to get to the airport in 3. Three what? Minutes? Hours? Gallons of gas? Remembering names of different units for the same property, and the relationship between them can be cumbersome.... 12 inches in a foot, 3 foot in a yard, 22 yards in a chain, 10 chains in a furlong, 8 furlongs in a mile … 8 fl. Oz in a cup, 4 cups in a quart, 4 quarts in a gallon, 31 gallons in a barrel… … and if you make a mistake in conversion from one unit into another, you are toast! Not only in medieval times! NASA wasted $125 million for a Martian Lander that burned up in the red planet’s atmosphere because of ambiguously expressed retrorocket force. The company that built the rockets expressed the force in US Customary units (in, ft, lbs), whereas NASA assumed the data was given in metric units (meters, kilograms, Newtons). SI System Whole world (except US!) now use standardized system of units, called SI system. SI system uses an unique name for a physical property and prefixes to describe bigger or smaller units. Make the relationship between bigger / smaller units to the base unit in factor of 1000. There is even relationship between units for different physical properties! Base SI Units Physical quantity Length Mass Time Temperature Amount of subst. Name meter kilogram second Kelvin mole Abbrev. m kg s K mol Derived SI Units Physical quantity Volume Pressure Energy Electric charge Name Abbrev. cubic meter m3 Pascal Pa joule j coulomb C A cube of 0.1 m length has a volume of 1 L; filled with water, its mass is 1 kg. Because 1 m3 is a huge volume, a commonly used, non-SI units for volume are Liter (L) and milliliter (mL). 1 m3 = 1000 L. 1 L = 1000 mL 1 L = (0.1 m)3 = 1 dm3 1 L = 1.057 qt. 1 m is 3.37 inches longer than a yard. 1 kg = 2.2046 lb 1 m = 1.0936 yd SI System (continued) Prefix - milli Other commonly used, non-SI units are Pressure in: millimeters of mercury (mm Hg), atmospheres (atm) Temperature in: degrees of Celsius (oC) Physical quantity: meter 1mm = 10-3 m 1km = 103 m 1 MB = 106 B 1µm = 10-6 m 1mL = 10-3 L 1GB=109 B Prefixes Used with SI Units Greek Prefix pico- (p) nano- (n) micro- (µ) milli- (m) centi- (c) deci- (d) Greek Prefix kilo- (k) mega- (M) giga- (G) tera- (T) Meaning one-trillionth (10-12) one-billionth (10-9) one-millionth (10-6) one-thousandth (10-3) one-hundredth (10-2) one-tenth (10-1) Meaning one thousand (103) one million (106) one billion (109) one trillion (1012) Note: All prefixes for smaller units are lower case, and those for larger are upper case! Prefix kilo – is the only exception. Note: Kelvin is represented as K, not oK! K = oC + 273.15 oC = (5/9) (oF – 32) oF = (9/5) oC + 32 Density Density = mass of substance divided by the volume occupied by the mass. By definition, the density of water is exactly 1 kg / L (hence the definition of L!) Specific gravity is the ratio of densities of a substance to that of water. It is dimensionless. When immersed in water, substances with density greater than 1 kg/L (e.g. most metals) sink, and those having density less 1 (ice, most liquids, wood, etc) float. Density depends on the way atoms are packed in the crystal lattice. Volume of an irregularly shaped object can be measured by displacement of water. Density is an intensive property, i.e. does not depend on the amount of material. Other intensive properties include: color, odor, taste, specific heat etc. The opposite is extensive property, one that depends on amount present. Mass and volume are both extensive properties, but their ratio (density) is intensive. Unit Analysis Recall – density is the ratio of mass and volume. d=m/V What to do if you are given the density and the mass of mercury, and asked to calculate the volume? (Suppose m = 2.00 g and d = 13.6 g/mL) One can do it in two ways. Way #1 Use above eq. to find V as a function of the other two quantities. Recall that equation stays unchanged if a mathematical operation is applied to both sides of equal sign. solved for m m x V Multiply both sides with V: Vx d= Vxd=m V m m Vxd V = solved for V Now divide both sides with d: = d d d 2.00 g Plug in the numbers: = 0.147 mL V= g 13.6 mL 1 g g x mL g x 4 1 1 = Recall fractions: 2 Check the units: = = g g g x 1 = mL x 3 2 3 mL mL 4 HOW? What to do if I cannot make one unit disappear immediately? Try and try again! Use multiple conversion factors; keep multiplying until you reach the desired unit(s) and have all other “unwanted” units disappear. Example 1: Suppose you are given density of 2.50 g / mL, and asked to convert it to lb / gal. 2.50 g 1 lb x 1 mL 1000 mL x 436 g 3.7854 L x 1L 1 gal = 20.9 lb 1 lb = 436 g 1 gal = 3.7854 L 1 gal Example 2: How many seconds are there in (exactly) 3.5 weeks? 7 day 24 hour 3600. s x 3.5 week x x = 2,116,800. s 1 week 1 day 1 hour Homework: Chapter 2: 4 (a-d,f), 7, 9, 13, 18(b), 23, 41, 49, 55 Chapter 1: 1, 3, 8, 9, 12 1. Explain the difference between (a) A hypothesis and a theory (b) A theory and a scientific law 3. How many phases are present in the graduated cylinder? 12. Is the system that contains only one substance necessarily homogeneous? Explain. Consider Table 1.1 below and answer the following: 8. Which liquids are not mixtures? 9. Which of the gases are not pure substances? 4. State the abbreviation for each of the following units: (a) milligram, (b) kilogram, (c) meter. (d) nanometer, (f) microleter 7. How many significant figures are in each of the following numbers? (a) 0.025, (b) 22.4, (c) 0.0404, (d) 5.50 x 103 9. Round each of the following numbers to three significant digits: (a) 93.246, (b) 0.02857, (c) 4.644, (d) 34.250 13. Solve the following problems: (a) 12.62 + 1.5 + 0.25 = ? (b) (2.25 x 103) (4.80 x 104) = ? (c) (452) (6.2) / (14.3) = ? (d) (0.0394) (12.8) = ? (e) (0.4278) / (59.6) = ? (f) 10.4 + (3.75) (1.5 x 104) = ? 18. Solve equation for the variable x: (b) 8.9 g/mL = 40.90 g / x 23. After you have worked out at a gym on a stationary bike for 45 min, the distance gauge indicates you have traveled 15.2 miles. What was your rate in km/hr? 41. A textbook is 27 cm long, 21 cm wide and 4.4 cm thick. What is the volume in (a) cubic centimeters, (b) liters, (c) cubic inches? 49. The normal boiling point of ethanol is 173.3 oF. What is this temperature in oC? 55. Concentrated hydrochloric acid has a density of 1.19 g/mL. Calculate the mass of 250 mL of it. Chapter 3 Elements and Compounds All known substances on Earth and probably the universe are formed by combinations of more than 100 elements. Most substances are compounds. Compounds can be decomposed into two or more simpler substances. Water can be decomposed into hydrogen and oxygen. Table salt can be decomposed into sodium and chlorine. An element is a fundamental or elementary substance that cannot be broken down into simpler substances by chemical means. Each element has a number. Beginning with hydrogen as 1, the elements are numbered in order of increasing complexity. An atom is the smallest particle of an element that can exist. It is the smallest unit of an element that can enter into a chemical reaction. A molecule is a combination of two or more atoms. If a molecule contains only atoms of one kind, it is an element. Otherwise, it is a compound. Carbon Symbols of the elements: Oxygen One or two letters. First letter always capitalized. Barium Sodium Elements are not distributed equally by nature. C O Ba Na In the universe, the most abundant element is hydrogen (91%) and the second most abundant element is helium (8.75%). Oxygen is the most abundant element in the crust of the earth (49.2%), followed by silicon. Oxygen is the most abundant element in the human body (65%), followed by carbon. A number of symbols appear to have no connection with the element. Most symbols start with the same letter as the element. Hydrogen is in a group of its own leftmost transition area rightmost nonmetals Metals, Nonmetals and Metalloids Metals Most elements are metals. Physical properties Good conductors of heat and electricity. Have high luster. High melting point and density. Metals are solid at room temperature. Mercury is the only exception. Malleable (hammered into sheets); Ductile (drawn into wires). Mix with other metals to form alloys Brass = copper + zinc Bronze = copper + tin Steel = Iron + carbon Chemical properties Little tendency to combine with other metals Readily combine with nonmetals (especially oxygen) to form compounds. In nature, minerals are formed by combinations of the more reactive metals with other elements. Some less reactive metals are found in nature in free state (elemental form). only seven elements Metalloids Physical properties Intermediate Be, Si, Ge, As, Sb, Te, and radioactive Po. Nonmetals Physical properties opposite of metals Lack luster. Poor conductors. Low melting point. Low density. Can be found in all three phases (solid, liquid, gas) at room temperature. 17 elements are nonmetals, including rare and radioactive (At) and six noble gases. Remaining ten nonmetals make almost all compounds. C, P, S, Se, I: solids Br: liquid H, O, N, F, Cl, noble gases: gases Periodic Table The periodic table was designed by Dimitri Mendelev in 1869 In the table each element’s symbol is placed inside of a box Above the symbol of the element is its atomic number. Elements are arranged in order of increasing atomic number Elements with similar chemical properties are organized in columns called families or groups. 7 N He These elements are known as the noble gases. They are non-reactive. They are found in pure (elemental) form. Ne Noble metals can also be found in elemental form. Most substances around us are mixtures or compounds. Ar Air is a mixture of nitrogen, oxygen, argon and traces of other gases (e.g. water vapor and carbon dioxide). Kr H, O, N etc. cannot exist in nature as single atoms. They are always paired (H2, O2, N2). These elements exist as diatomic molecules. Xe Rn Composition of Elements and Compounds Hydrogen and oxygen are elements. Elements cannot be decomposed into simpler substances. Free, or elemental hydrogen (H2) is found in volcanoes; It can be prepared in the laboratory. Elemental oxygen (O2) is found in air (21 vol%); It can be prepared in the laboratory. Water is a compound; its formula is H2O. It is made of two hydrogen atoms and one oxygen atom. Water contains NO free hydrogen or free oxygen. H2O A compound is a distinct substance that contains two or more elements combined in a definite proportion by weight. H2 O2 Compounds can be decomposed chemically into simpler substances – that is, into simpler compounds or elements. Atoms of the elements that constitute a compound are always present in simple whole number ratios. They are never present as fractional parts. Physical and chemical properties of a compound differ from those of the parent elements. Types of Compounds Compounds can be molecular or ionic. A molecule is the smallest uncharged individual unit of a compound formed by the union of two or more atoms. An ion is a positively or negatively charged atom or group of atoms. Cation is positively charged ion. Anion is negatively charged ion. Ionic compounds are held together by attractive forces between positively and negatively charged ions. Table salt, sodium chloride is a colorless crystalline ionic substance, 39.3% sodium and 60.7% chlorine by mass. The solid NaCl does not conduct electricity; molten NaCl, and solution of salt in H2O do. Compounds can be classified as molecular or ionic. Molecular compounds are held together by covalent bonds. Ionic compounds are held together by attractive forces between their positive and negative charges. Chemical Formulas Serve as abbreviations of the names of compounds. Tell which elements the compound is composed of and how many atoms of each element are present in a formula unit. Show the symbols of the atoms of the elements present in a compound. CaCl (1) 2 Show the ratio of the atoms of the elements present in a compound. calcium chloride calcium chlorine Ca calcium Cl chlorine 2 Cl 1 Ca Rules for Writing Chemical Formulas Formulas do not necessarily represent the arrangement of atoms. When a formula contains one atom of an element, the symbol of that element represents the one atom. The number one (1) is not used as a subscript. When the formula contains more than one atom of an element, the number of atoms is indicated by a subscript written to the right of the symbol of that atom. When the formula contains more than one of a group of atoms that occurs as a unit, parentheses are placed around the group, and the number of units of the group is indicated by a subscript placed to the right of the parentheses. There is one phosphorus atom in a phosphate group Ba3(PO4)2 There are three barium atoms Indicates two phosphate (PO4)3- groups There are four oxygen atoms in a phosphate group There are three barium, two phosphorus and eight oxygen atoms in the compound! H2O NaCl Indicates the element sodium (one atom) Indicates the element chlorine (one atom) H3PO4 Indicates the element hydrogen Indicates 3 H atoms Indicates the element oxygen Indicates 4 O atoms Indicates the element phosphorus (1 atom) Chapter 4 Properties of Matter Properties of a substance are characteristic of the substance. Each substance has a set of properties that are characteristic of that substance and give it a unique identity. Color Taste / Odor Physical state Melting point Boiling point Physical properties are the inherent characteristics of a substance that are determined without changing its composition. Chemical properties describe the ability of a substance to form new substances, either by reaction with other substances or by decomposition. Physical properties of chlorine gas (Cl2): 2.4 x heavier than air yellowish-green color disagreeable odor melting point: -101 oC boiling point: -34.6 oC Chemical properties of chlorine (Cl2): Doesn’t burn disinfectant Bleaching agent supports combustion Produces table salt with sodium Physical and Chemical Changes Physical changes are changes in physical properties (such as size, shape, and density) or changes in the state of matter without an accompanying change in composition. No new substances are formed. tearing of paper change of ice into water change of water into steam heating platinum wire Chemical change is a change in which new substances are formed that have different properties and composition from the original material. The formation of copper(II) oxide from copper and oxygen is a chemical change. The newly-formed substance has set of properties that are different from those of copper. Water decomposes when electrical energy passes through it. The products, elemental hydrogen and oxygen, have different properties from those of water. Chemical Reaction Water decomposes into hydrogen and oxygen when electrolyzed. Chemical symbols can be used to express chemical reactions reactant 2H2O yields products 2H2 + O2 Conservation of Mass No change is observed in the total mass of the substances involved in a chemical change. sodium + 46.0 g sulfur → 32.1 g 78.1 g reactant mass reactants sodium sulfide 78.1 g → 78.1 g product = mass products Energy Kinetic energy matter possesses due to its motion. Mechanical Chemical Electrical Heat Nuclear Radiant Potential energy could also thought as the energy stored in an object. Chemical energy is one form of potential energy. The heat released when gasoline burns is associated with a decrease in its chemical potential energy. The new substances formed by burning have less chemical potential energy than the gasoline and oxygen. Increasing potential energy Potential energy is the energy that an object possesses due to its relative position. Bouncing ball. Running man. Height Energy is the capacity to do work. Potential energy rises with the height increase Energy in Chemical Reactions In all chemical changes, matter either absorbs or releases energy. Examples of chemical processes leading to energy release Type of Energy Energy Source Electrical Storage batteries Light A lightstick. Fuel combustion. Heat and Light Combustion of fuels. Body Chemical changes within body cells. Examples of chemical processes leading to energy absorption Type of Energy Chemical Change Electrical Electroplating of metals. Decomposition of water into hydrogen and oxygen Light Photosynthesis in green plants. Conservation of Energy An energy transformation occurs whenever a chemical change occurs. If energy is given off in a chemical change, the products will have less chemical potential energy than the reactants. If energy is absorbed during a chemical change, the products will have more chemical potential energy than the reactants. Energy can be neither created nor destroyed, Law of Conservation of Energy though it can be transformed from one form of energy to another form of energy. higher potential energy Electrolysis of Water lower potential energy Burning of hydrogen in air Heat: Quantitative Measurement Heat is a form of energy associated with small particles of matter. Temperature is a measure of the intensity of heat, or of how hot or cold a system is. The SI unit for heat energy is the joule (pronounced “jool”). Another unit is the calorie. 1 calorie = 4.184 Joules (exactly) This amount of heat energy will raise the temperature of 1 gram of water by 1 oC. Two calories is needed to raise 1 gram of water by 2 oC. Twice as much heat energy is required to raise the temperature of 200 g of water by 10 oC as compared to 100 g of water. specific Heat = heat ( q = C )( x mass m 100 Ag water 200 Bg water 30oC 20 30oC 20 4184 J 8368 J temperature change )( x ) ∆t Specific Heat The specific heat of a substance is the quantity of heat required to change the temperature of 1 g of that substance by 1 oC. The units of specific heat in calories are: calories gram oCelcius cal g oC The units of specific heat in joules are: Joules gram oCelcius J g oC It takes about 5 times more energy to warm 1 g of water by 1 oC as to warm the same amount of aluminum by 1oC. Quantitative Calculation of Heat The relation of mass, specific heat, temperature change I can be found if (∆t), and quantity of heat lost both sides were or gained is expressed by the divided by m x ∆t general equation: ( )( specific heat of substance Problem 1: Calculate the specific heat of a solid in J/goC and if 1638 J raise the temperature of 125 g of the solid from 25.0oC to 52.6oC. ) mass of ∆t = heat substance heat = 1638 J mass = 125 g ∆t = 52.6 oC – 25.0 oC = 27.6 oC spec.heat = ? heat specific heat = mass x ∆t = 1638 J 125 g x 27.6 oC = 0.475 J g oC Heat = specific heat Q C = x mass of water m x x x change in temp. ∆t Problem 2: How much heat (in J) it takes to warm a 2.000 lb block of iron from 50 oC to 75 oC? J 453.6 g m = 2000 lb x 25 oC = 1.02 x 104 J Q = m C ∆t = 2.000 lb x x 0.449 g oC 1 lb ∆t = 25 oC From Table 4.3: C = 0.449 J/goC Problem 3: What is the final temperature of a 20.0 g block of iron after it has absorbed 100.0 J of heat, if its initial temperature was 25.0 oC? Hint: solve for ∆t, then add it to the initial temp! ∆t = Q mC 100.0 J = J 20.0 g x 0.449 g oC Calorie Measurement Place the food (e.g. Twinkie) into a vessel (“bomb”), surround it with water and measure the water temperature as you burn the food. = 11.1 oC Final temp. = 25.0 oC + 11.1 oC = 36.1 oC Advanced A sample of a metal with a mass of 212 g is heated to 125.0 oC and then problem: dropped into 375 g of water at 24.0 oC. If the final temperature of the water is 34.2 oC, what is the specific heat of the metal? When the metal enters the water, it begins to cool, losing heat to the water. At the same time, the temperature of the water rises. This process continues until the temperature of the metal and the temperature of the water are equal (34.2 oC), at which point no net flow of heat occurs. Heat gained by water = heat lost by metal mW = 375 g ∆tW = 34.2 – 24.0 oC Strategy: ∆tM = 125.0 – 34.2 oC 1. Calculate the heat (qW) gained by the water. mM = 212 g 2. Calculate the final temperature of the metal. spec.heat o 3. Calculate the specific heat of the metal. of water = 4.184 J/g C qW = qM temperature gained by water 1. ∆tW = 34.2 oC – 24.0 oC = 10.2 3. oC heat gained by water J x 10.2 oC = 1.60 x 104 J qW = 375 g x 4.184 g oC qM = qW = 1.60 x 104 J qM spec.heat = m x ∆t 1.60 x 104 J spec.heat = 125 g x 90.8 oC temperature lost by metal 2. ∆tM = 125.0 oC – 34.2 oC = 90.8 oC spec.heat = 0.831 J g oC Homework, chapter 3 (paired exerc., p.62): 5. Write the formula for each compound: a) Zinc oxide (1 atom Zn, 1 atom O) b) Potassium chlorate (1 atom K, 1 atom Cl, 3 atoms O) c) Sodium hydroxide (1 atom Na, 1 atom O, 1 atom H) d) Ethyl alcohol (2 atoms C, 6 atoms H, 1 atom O) 9. How any atoms are represented in each formula? a) KF; b) CaCO3; c) K2Cr2O7; d) NaC2H3O2; e) (NH4)2C2O4 19. Classify each material as an element, compound or mixture: a) xenon; b) sugar; c) crude oil; d) nitric acid 23. What percent of the first 36 elements on the periodic table are metals? Homework, chapter 4 (paired exerc., p.77): 7. State whether ech of the following represents a chemical or physical change: a) A steak is cooked until well done; b) Students firepolish the end of a glass rod to smoothen jagged edges; c) Chlorine bleach removes coffee stains from white lab coat; d) When two clear solutions are mixed together, the solution becomes yellow and cloudy; 13. Indicate with the plus sign (+) any of these processes that requires energy, and a negative sign (-) any that release energy: a) melting ice; b) relaxing a taut rubber band; c) a rocket launching; d) striking a match. 19. A 250.0 g metal bar requires 5.866 kJ to change its temperature form 22 oC to 100.0 oC. Calculate the specific heat of the metal. 20. A 1.00 kg of antimony absorbed 30.7 kJ thus raising its temperature from 20.0 oC to its melting point (630.0 oC). Calculate the specific heat of antimony. 31. One gram of anthracite coal gives off 7000. cal when burned. How many joules is this? How many grams of anthracite are required to raise the temperature of 4.0 L of water from 20.0 oC to 100.0 oC? Chapter 5 Early Atomic Theory and Structure Empedocle (440 BC): all matter consists of 4 “elements” earth, fire, water and air. Democritus (470-370 BC): matter is composed of indivisible particles (Greek atomos - not cuttable). Aristotle (384-322 BC): endorsed Empedoclean theory, so that it dominated until 17th century. John Dalton (1766-1844), 2000 years after Democritus, revived concept of atoms: Dalton’s rules: 1. elements are composed of indivisible particles-atoms. 2. atoms of the same element are alike in mass and size. 3. atoms of different elements are not alike – they have different masses and sizes. 4. compounds are union of two or more atoms of different elements. 5. compounds contain simple numerical ratios of atoms. 6. more than one compound can be formed of atoms of two elements. Dalton rules (cont.) Water While most of the Dalton’s theory still holds, some modifications were necessary: An atom is composed of subatomic particles, it can be decomposed and may have isotopes of different masses. Hydrogen peroxide H2O and H2O2 have the same atoms, but different number. Law of definite composition – a compound contains two or more elements in a definite proportion by mass. Law of multiple proportions – atoms of two or more elements may combine in different ratios to produce more than one compound. compound water hydrogen peroxide Copper (I) chloride Copper(II) chloride Methyl alcohol Ethyl alcohol formula H2O H2O2 CuCl CuCl2 CH4O C2H6O percent composition 11.2 % H, 88.8% O 5.9% H, 94.1% O 64.2% Cu, 35.8% Cl 47.3% Cu, 52.7% Cl 37.5% C, 12.6% H, 49.9% O 52.1%C, 13.1% H, 34.7% O Electric Charge Matter may have electric charge 1. charge may be of two types, positive and negative 2. unlike charges attract, like repel 3. can be transferred from one object to the other by contact or induction 4. the smaller the distance, the stronger the force between two charges x (F – force, k - constant, q1, q2 - charges, r - distance) F = k x q1 2q2 r G.J. Stoney (1826-1911) named the unit of electricity electron. Joseph Thompson (1856-1940) experimentally showed the existence of electron, which is: negative in charge, deflected by magnetic and electric fields, and capable of moving small paddle wheel. Michael Faraday (1791-1867) 1. Some compounds dissolved in water conduct electricity, others decompose 2. Particles of some elements are attracted to the positive or negative electrode. 3. Concluded that they are charged and called them ions (Greek-wanderer). Svante Arhenius (1859-1927) 1. Melted NaCl also conduct electricity. NaCl breaks up to Na+ and Cl- ions. 2. Ions move to oppositely charged electrode. Ion that goes towards positively charged cathode is called cation. The other goes to positive anode is called anion. Charge and Periodic Table IA H+ Li+ Na+ K+ Rb+ Cs+ IIA Invented by Dmitri Mendeleev, the Periodic Table shows recurring trends in properties of elements and their charges. VIIIA IIIA Be2+ Mg2+ Ca2+ IIIB IVB VB Cr2+ Cr3+ Sr2+ VIB VIIB Fe2+ Fe3+ ------ VIIIB ------- Al3+ IB IIB Cu2+ Cu+ Zn2+ Ag+ Cd2+ IVA VA VIA N3- O2- P3- S2- VIIA F- ClBr- I- Ba2+ Group IA: Group IIA: always 1+ always 2+ Charge = group number Elements of B groups (transition metals) form ions of different charges Group VA: Group VIA: Group VIIA: often 3usually 2- always 1Charge = 8 - group number Group VIIIA: no charge -noble gases Subatomic particles Atom is extremely small, cannot be observed by optical microscope. Atomic diameters range from 0.1 to 0.5 nm. Consider this dot on the right. If its diameter is 1 mm, 10 million H atoms could be arranged in it. Yet, there are particles smaller than H atom. They are called subatomic particles. Three subatomic particles are sufficiently long-lived to be used in experiments: electron, proton and neutron. Proton and electron have charges opposite in sign, identical in absolute value. Proton and neutron have almost equal masses. Number of other short-lived subatomic particles are detected in recent years. quarks, neutrinos, antineutrinos, positrons and antiprotons ... Their existence is still debated. The Evolution of Atomic Theory Becquerel in 1896 discovered radioactivity. Radioactive elements spontaneously emit alpha particles, beta particles and gamma rays from their nuclei. By 1907 Rutherford found that alpha particles emitted by certain radioactive elements were helium nuclei. J.J. Thomson discovered electron in 1897. Shortly after, he proposed the first atomic ‘Plum-Pudding’ model where negatively-charged electrons (‘plums’) are embedded into a positively-charged sphere (‘pudding’). + Thomson and Goldstein discovered proton in 1907, and Chadwick discovered neutron in 1932. α particles should pass through the Rutherford’s experiment showed that the plumpudding model is incorrect. sphere because it is effectively neutral. α particle velocity ~ 1.4 x 107 m/s (~5% speed of light) The Evolution of Atomic Theory (cont.) Indeed, most of them did, but a few (1 in 20,000) were deflected, some in great If Thomson’s model was correct, αangles. It was “as if you were firing 15” particles would go straight through the foil. shell at a tissue paper, and it came back and hit you”. Rutherford’s Conclusions: 1. Atom consists of a nucleus and electrons. 2. Size of an atom is about 100,000 times bigger than that of its nucleus. Size of an atom: If the nucleus is as big as a marble, the atom is as big as football field. Atoms are mostly empty space! 3. Atoms positive charge is concentrated in the nucleus. 4. Electron has negative charge, and proton has positive charge. Mass of p is 1840 x mass of e- (1.67 x 10-24 g). In order to avoid falling onto the nucleus, electron must be constantly rotating with great speed, as if it was everywhere at once. Problem: classical physics says that the Nucleus has both protons and neutrons. moving charge (electron) radiates energy. Creation of Ions An neutral atom has equal number of protons and electrons. Remember that the proton has equal, but opposite, charge as an electron When one or more electrons are lost from an atom, a positive cation is formed. When one or more electrons are added to a neutral atom, an anion is formed. The word ‘cation’ comes from the fact that positive ion (cation) moves towards negativelycharged (cathode) in an electrolyte solution. The negative ion (anion) moves towards positively-charged electrode (anode). Atomic Numbers of the Elements The atomic number of an element is equal to the number of protons in the nucleus of that element. It gives the identity of the element. Atomic number (Z) = number of protons in nucleus Mass number (A) = number of protons + number of neutrons = atomic number (Z) + number of neutrons Isotopes: atoms of the same element with different numbers of neutrons in their nuclei. Neutrons can be envisioned as glue that holds nuclei together. The mass of 12C is 12.000 g. Mass Number A 13 Element Symbol There exists 1.11 % C and Atomic Number Z traces of 14C. The mass of C atom (12.01 g) is weighted average of all atoms. Two isotopes of uranium X 235 92 238 92 U U Protons: 92 Neutrons: 235-92=143 Protons: 92 Neutrons: 238-92=146 Three isotopes of hydrogen Hydrogen (Protium) Deuterium Tritium (radioactive) Masses of Elements and Isotopes To overcome the problem in measuring extremely small masses, a system of relative atomic masses using “atomic mass units” was devised to express the masses of elements using simple numbers. A mass of exactly 12 atomic mass units (amu) was assigned to 126 C. Element masses are average numbers, because of isotopes. Mass of H atom is 1.008 amu. The mass of a single atom is too small to measure on a balance. Using a mass spectrometer, the mass of the hydrogen atom was determined = 1.673 x 10-24 g. The standard to which the masses of all other atoms are compared to was chosen to be the most abundant isotope of carbon. 1 amu is defined as exactly equal to 1/12 the mass of a carbon - 12 atom: 1 amu = 1.6606 x 10-24 g 12 6 C Masses of elements and Isotopes (cont.) Isotopes of the same element have different masses. The listed atomic mass of an element is the average relative mass of the isotopes of that element compared to the mass of carbon-12 (exactly 12.0000…amu). To calculate the atomic mass, multiply the atomic mass of each isotope by its abundance and add the results. Percent abundances are Example: calculate average atomic mass of copper. Isotope multiplied by 100! Average Isotopic Abundance atomic mass (amu) (%) mass (amu) 63 29 Cu 62.9298 69.09 65 29 Cu 64.9278 30.91 (62.9298 amu) x 0.6909 = (64.9278 amu) x 0.3091 = 63.55 + 43.48 amu 20.07 amu 63.55 amu Finding the Number of Protons, Neutrons and Electrons Element Chlorine Gold Bromine Symbol 36Cl Au Br - Z Mass# #p 17 79 56 36 197 80 35 Fill in the table below. Symbol Element Z = #p. 2. Find mass number and #n Mass# = #p + #n 3. Find charge or #electrons (charge) = #p – #e #e = #p – (charge) 35 #e charge 19 118 79 16 17 79 18 0 0 +2 -2 45 36 -1 Charge = 0 Locate Cl in periodic table Look up Atomic number (Z) for Cl 1. From Periodic table, find: Element, symbol, atomic number and #protons. At least one of those must be given, the other three are read from Periodic Table. 17 79 #n Mass # is 36 = #p + #n #n = 36 – #p = 19 4. You can get #e and charge. Use step 3 to find #p and from that identify element. See if you can solve the other two examples. The solutions are: 135Ba2+, and 32S2-. 56 16 HW, Chapter 5 (p.100): 12, 15, 23, 27, 37 Chp. 10 (p.220): 13, 19, 33, 39 Chapter 10 The Modern Atomic Theory and Periodic Table Rutherford model, based on classical (Newtonian) physics could not explain periodicity of the elements, nor how electrons stay in orbit around nucleus. A new model was needed. Light moves with speed of 300,000 km/s. It is characterized with wavelength λ. When passed through a prism, a familiar set of colors are seen. Spectrum is continuous as colors blend. Visible spectrum is just a tiny fraction of the electromagnetic radiation. Bohr’s Theory Niels Bohr in 1913 postulated that an electron inside an atom can possess only certain values of energy. Those “packs” of energies are called quanta (pl. of quantum). Atom is in the lowest energy when its electrons are closest to the nucleus. When atom absorbs quantum of energy, electrons are moved to higher orbits. Electrons can only have allowed energies, not the energies between the allowed ones. It is like a person on a ladder who can stand on a 1st, 2nd, 3rd rung, but not between the rungs. Allowed orbits are called shells. They are characterized by the principal quantum number, n that goes from n = 1 (shell closest to the nucleus), to infinity. The maximum number of electrons a shell can hold increases with n as the radius of the shell increases. Li and Na have a single electron in the valence shell. Periodicity Elements have similar chemical properties because their valence shell configurations F and Cl have 7 electrons are similar. in the valence shell. The number of electrons in the valence shell is equal to the Roman-numeral group number for the representative elements. The dependence of chemical on valence shell occupancy holds for each group of elements. Helium is an exception. It behaves as a noble gas although it has only 2 ein the valence shell. Group IA has one electron, Group IIA has two… The lowest reactivity is found in the group VIIIA (noble gases). Elements of other groups can attain the valence shell of group VIIIA by producing compounds. Line Spectra Ground state s: 2 e- p: 6 ed: 10 e- f: 14 e- Excited state New, refined model includes subshells that are very close in energy and size. Every subshell is designated with a letter (s, p, d, f). s subhell can hold two electrons. Each subshell after it holds four electrons more than the previous. Transition n=3 n=2 produces 13.1 eV – 11.2 eV = 1.9 eV, i.e. red color. Similarly, n=4 n=2 produces 13.8 – 11.3 = 2.6, green color; other colors in the visible spectrum are also found. Bohr’s model correctly predicted all transitions of H atom; transition into n=1 produces a line in UV, into n=2 (visible) and into n≥3 (infrared). But the model failed to predict spectrum for any atom with more than one e-. n s p d f 1 2 2 2 6 3 2 6 10 4 2 6 10 14 Each ‘box’ contains 2 electrons with paired spins. Filling of Subshells Subshells exist for any n > 1. They increase in energy: s < p < d < f. Subshells may cross each other; the crossing is more pronounced when spacing between shells reduces as the principle quantum number increases. 35Br This crossing is the reason for the socalled ‘potassium problem’. Present the electron configuration of 19K. 1s2 2s2 2p6 3s2 3p6 4s1 Valence electron Orbital diagram is slightly more complicated. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 Rearrange: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5 Valence electrons Subshell fill up S No need to memorize crossings. Just follow the periodic table. Move He next to H to get four blocks, each with different width (s:2, p:6, d:10, f:14). Electron configuration of 16S: Try 26Fe 1s2 2s2 2p6 3s2 3p6 4s2 3d6 1st period: 1s2. 2nd period: 2s2 2p6. 3rd period: 3s2 3p4. Valence electrons 2 8 + 6 16 1s2 2s2 2p6 3s2 3p6 4s2 3d10 Gd: 64 4p6 5s2 4d10 5p6 6s2 5d1 4f 7. When a transition metal produces cation, first to go are its electrons in the s orbital of the valence shell ! Valence electron jumps from Na to Cl The best way for Mg to achieve octet is to lose two electrons from its 3rd shell. Theoretically, Mg could also gain 6 e-, but the 6 e- excess is too much for Mg nucleus to hold on to. The best way for F to achieve octet is to gain one e-. Each e- from Mg 3rd shell jumps into 2nd shell of an F atom. Mg atom becomes Mg2+ cation. Each F atom becomes F- anion. All three ions have 8 valence e-. Compound Formation NaCl MgF2 Neutral atoms: 1 valence e- in Na 7 valence e- in Cl Note the size of atoms! Ions: no electrons in n = 3 shell of Na. Na+ has 8e- in n=2 shell. Cl- has 8e- in n=3 shell. Both ions satisfy octet rule. Note the size of ions! Atomic Size Understanding of atomic size with Bohr’s model. One consequence: shells get larger as n increases. A combination of same valence shell and increased nuclear charge shrinks the atom in a group. Quantum Mechanical Model Electron has wavelike characteristics, it can tunnel to appear in ‘forbidden’ places. Uncertainty principle states that it is impossible to know at any given time where the electron is, or where it is going. Instead, etravels around nucleus in an electron cloud or orbital. The QM model also explains why e- stays in orbit around the nucleus: An e- orbital is allowed only if the electron wave closes in on itself. Allowed orbit must close in on itself. Nonallowed orbit does not. Electron cloud exhibits wavelike motion, so that different orbitals have different shapes. It has a particular shape and energy determined by the principle quantum number n. Chapter 8 Chemical Reactions In mid 1980s chemists combined Y2O3, CuO and BaCO3 and obtained a new superconducting ceramic material, YBa2Cu3O7 that conducted electricity without resistance when cooled below liquid nitrogen temperature. Magnetic lines cannot penetrate a superconductor, so a magnet floats above it. Experimental trains based on magnetic levitation (maglev) are in construction. In a chemical reaction one or more substances (reactants) are converted into new substances (products) with different physical and chemical properties. Reactants Products Which reaction is chemical ? H2O(l) H2O(g) CO2 + O2 O2 + CO2 CO2 + H2O H2CO3 Gas molecules travel rapidly at room temperature (hundreds of miles per hour). Energy associated with motion (kinetic energy) is absorbed by colliding molecules. H2 I2 H – H+ I – I 2 atoms H 2 HI Energy is required to break + 2 atoms I H–I a chemical bond. 2 molecules HI Balanced equation H – I 2 atoms H, 2 atoms I Balanced equation must have smallest possible whole numbers as coefficients. ∆ 2 Al(s) + Fe2O3(s)→ 2 Fe(s)+ Al2O3(s) coefficient coefficient A chemical equation is a shorthand expression for a chemical change or reaction. A chemical equation must be balanced. Coefficients (whole numbers) are placed in front of substances to balance the equation and to indicate the number of units (atoms, molecules, moles, or ions) of each substance that are reacting. A chemical equation uses the chemical symbols and formulas of the reactants and products and other symbolic terms to represent a chemical reaction. Symbol + means plus. Symbol hν ν means light. Symbol means yields. Symbol means gas formation. Symbol means precipitate. Symbol ∆ means heat. Conditions required to carry out the reaction may be placed above or below the arrow. The physical state of a substance is indicated by symbols such as (s) for solid, (l) for liquid, (g) for gas and (aq) for aqueous solution. Balancing chemical equations H2 + I2 ? HI Not balanced H2 + I2 2 HI Balanced H2 + I2 Impossible! Never change subscripts in the molecular formula! Always adjust coefficients, i.e. numbers in front of the molecules. Practice N2 + 3 H2 2 NH3 Balanced N: 1 2 N: 2 H: 2 6 H: 3 6 Try some more… CH4 + 2 O2 CO2 + 2 H2O C: 1 C: 1 H: 4 H: 2 4 O: 2 4 O: 3 4 And more… C8H18 + 25 O 2 2 H2I2 H2I2 does not exist. If it did, it would be a compound different from HI. Step1: Count atoms on each side. Step 2: Start balancing metal first. Continue with any nonmetal other than O or H. Step 3: When all atoms other than O and H are balanced, start balancing either O or H. if there is either O2 or H2 on the left side, balance it last. Step 1: count atoms. Step 2: No metal atoms. Continue with C. Step 3: Balance H next. Finish with O. 8 CO2 + 9 H2O 2 C8H18 + 25 O2 16 CO2 + 18 H2O And more. C6H12O6 + 6 O2 6 CO2 + 6 H2O 2 HCl + Na2CO3 2NaCl + CO2 + H2O A Balanced Equation Shows: Mole is a number like a dozen, only much bigger. It is used to translate number of particles into grams. Formulas Number of molecules Number of atoms Number of moles Molar masses Sum of masses Must be balanced 44.09 g + 160.0 g 204.1 g ? = 132.0 g + 72.08 g 204.1 g Types of Reactions Chemists try to classify a reaction by understanding what it is doing. Ca2+(aq) + CO32-(aq) CaCO3(s) Remember that aq, s, l, g indicate physical state. The reaction is used to soften ‘hard’ water (i.e. water containing Ca2+ ions). There are other reactions serving the same purpose. Single-displacement reactions One element replaces another element in a compound. A displaces B. 2 Al(s) + Fe2O3(s) Al2O3(s) + 2 Fe(l) Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g) A + A C B C + B Double-displacement reactions PCl3(l) + 3 AgF(s) PF3(g) + 3 AgCl(s) Two compounds exchanged partners. A B + C D A D + C B Decomposition reaction One compound breaks into two or more simpler substances. 2 HgO(s) A B Heat 2 Hg(l) + O2(g) A + B H2(g) + I2(g) 2 HI(g) 2 Na(s) + Cl2(g) 2 NaCl(s) Combination reaction Two or more simple compounds 4 Ni(s) + 4 Al(s) + S8(s) 2 Ni2Al2S4(s) combine to give one or more A + B A B complex substances. Single displacement reactions Occur if the reactant in elemental form is more reactive than the element in the compound. Metal Activity Series 2 Al(s) + Fe2O3(s) Al2O3(s) + 2 Fe(l) The reaction occurs ONLY if the position of the free element in the activity series is above the element in compounds. Mg(s) + PbS(s) → MgS(s) + Pb(s) Ag(s) + CuCl2(s) → no reaction Halogen Activity Series Cl2(g) + CaBr2(s) → CaCl2(aq) + Br2(aq) I2(g) + CaBr2(s) → no reaction Double displacement reactions Accompany the following processes: release of heat Reaction goes only formation of water when one of the formation of a precipitate products leaves release of gas bubbles aqueous solution Halogens F2 Cl2 Br2 I2 increasing activity Metals K Ca Na Mg Al Zn Fe Ni Sn Pb H Cu Ag Hg AB + CD → AD + CB Double Displacement Reactions (cont.d) 1. Acid / base neutralization, or acid + metal oxide, produces water and heat HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) 2HNO3(aq) + CuO(s) → Cu(NO3)2(aq) + H2O(l) 2. Formation of insoluble precipitate AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) 3. Formation of a gas Indirect gas formation NH4Cl(aq) + NaOH(aq) → NaCl(aq) + NH4OH(aq) Heat in Chemical Reactions NH4OH(aq) → NH3(g) + H2O(l) Energy changes accompany H2(g) + Cl2(g) → 2HCl(g) + 185 kJ (exothermic) chemical reactions. Energy is Energy released 2 mol 1 mol 1 mol either released to, or absorbed from surroundings. per mole H2 or Cl2 NOT per mole HCl Since the amounts of N2(g) + O2(g) + 185 kJ → 2NO(g) (endothermic) substances are expressed in 2 mol Energy absorbed 1 mol 1 mol moles, the heat of the reaction 6CO2 + 6H2O + 2519 kJ → C6H12O6 + 6O2 is expressed per mole of 2519 kJ of energy is absorbed per 6 moles CO2, H2O, O2! either reactant or product. Energy of Activation A certain amount of energy is always required for a reaction to occur. It is called the energy of activation, Ea. The reaction will occur only if the activation energy is supplied. The activation energy can take . the form of a spark or a flame. CH4 + 2O2 → CO2 + 2H2O + 890 kJ Although the reaction releases energy, it needs Ea to start. Exothermic Reaction: Combined energy of products is lower than that of reactants. The difference is released. Endothermic Reaction: Combined energy of products is larger than that of reactants. The difference is absorbed. N2(g) + O2(g) + 185 kJ → 2NO(g) Chapter 6 Nomenclature of inorganic compounds There are ~ 11 million chemical compounds. Some of the arbitrary (common) names you may recognize (water = hydrogen oxide H2O, laughing gas = dinitrogen monoxide N2O, quicksilver = mercury Hg). One cannot memorize arbitrary names of all of them. Chemical nomenclature is the system of names for compounds. Systematization of names was devised by IUPAC (International Union of Pure and Applied Chemistry) in 1921 and is constantly updated Elements: Exceptions: Chemical formula of an element is just the symbol of the element. Some non-metals cannot exist as single atoms at normal temperatures. They form polyatomic molecules. Chemical formula of element hydrogen is not H, but H2. Same is true for N2, O2, F2, Cl2, Br2, I2, S8 and P4. Ions: Charged particles produced by adding or removing electrons from neutral atoms are called cations and anions. K K+ + eMg Mg2+ + 2eAl Al3+ + 3e- F + e- FO + 2e- O2N + 3e- N3- The bond between metals and non-metals is usually ionic. Metals give away their electrons and become positively charged cations. Nonmetals accept them and become anions. The ionic bond is formed as a result of attraction between oppositely charged ions. Charge and Periodic Table IA H+ Li+ Na+ K+ Rb+ Cs+ IIA Invented by Dmitri Mendeleev, the Periodic Table shows recurring trends in properties of elements and their charges. VIIIA IIIA Be2+ Mg2+ Ca2+ IIIB IVB VB Cr2+ Cr3+ Sr2+ VIB VIIB Fe2+ Fe3+ ------ VIIIB ------- Al3+ IB IIB Cu2+ Cu+ Zn2+ Ag+ Cd2+ IVA VA VIA N3- O2- P3- S2- VIIA F- ClBr- I- Ba2+ Group IA: Group IIA: always 1+ always 2+ Charge = group number Elements of B groups (transition metals) form ions of different charges Group VA: Group VIA: Group VIIA: often 3usually 2- always 1Charge = 8 - group number Group VIIIA: no charge -noble gases Ionic compound form three-dimensional ordered network of cations and anions called ionic lattice. There are no bonds. Electronegativity and the Polar Covalent Bond Electrons are rarely shared equally between atoms. Electronegativity (EN) is numerical rating of an atoms ability to attract to itself the shared electrons in a covalent bond. Generally, electronegativity of metals is low, and that of nonmetals is high. The difference in electronegativity determines the nature of the bond. ∆EN = 0, covalent; ∆EN = 1.0, polar covalent (23% ionic); ∆EN = 1.9, polar covalent (60% ionic); ∆EN > 1.9, ionic. Naming of a molecule follows the order of increasing electronegativity. Naming Chemical Compounds While some compounds have common names, today millions of compounds are known. Hence the need to name each compound by its chemical formula. 1. metal + nonmetal (ionic) and 2. nonmetal + nonmetal (covalent) Ionic: Less electronegative atom (usually metal) goes first; its name is unchanged; nonmetal ion keeps the stem, adds suffix –ide. If the metal can form two classes of compounds, the metal charge is given by the Roman numerals. Covalent: Because of many possibilities when two nonmetals combine, prefixes are used to tell how many atoms of each element are present in the molecular formula. Lesser electronegative element’s name stays unchanged, the other gets suffix –ide. If there is only one atom of lesser electronegative element, prefix mono– is omitted. Examples: Mon(o)- 1 NaCl Sodium chloride FeCl2 Iron(II) chloride Di2 MgS Magnesium sulfide FeCl3 Iron(III) chloride Tri3 Li3N Lithium nitride CuO Copper(II) oxide Tetr(a)- 4 Alluminum oxide Al2O3 Cu2O Copper(I) oxide Pent(a)- 5 SnO2 Hexa6 Tin(IV) oxide CoCl3 Cobalt(III) chloride Binary compounds (made of two elements): PCl3 Phosphorus trichloride P4O6 Tetraphosphorus hexaoxide NO Nitrogen monoxide NO2 N2O4 N2O5 Nitrogen dioxide Dinitrogen tetroxide Dinitrogen pentoxide HeptaOctaNonaDeca- 7 8 9 10 Common names are used. Polyatomic Ions Ammonium is the only cation. Mostly composed of a nonmetal plus one or more oxygen atoms. Exceptions: some transition metal anions and CN-. Most oxyanions exist in series that differ only in the number of oxygen atoms. If there are two members in such a series, the ion with fewer oxygens gets the -ite ending, the one with more oxygens gets the -ate ending. If there are more oxyanions in a series, the prefix hypo- is used denotes the ion with fewest oxygens, while the prefix perdenotes the oxyanions with the most oxygens. Naming: metal ion goes first, using the rules for binary compounds, followed by the polyatomic anion. ClOClO2ClO3ClO4- hypochlorite chlorite chlorate perchlorate NaNO2 NaNO3 K2CO3 Sodium nitrite Sodium nitrate Potassium carbonate The molecule is neutral: the Cu(NO3)2 Copper(II) nitrate sum of positive charges is Mg (PO ) Magnesium 3 4 2 equal to the sum of phosphate negative charges. Nomenclature of Acids A compound that dissociates in water to produce hydrogen ions (H+) is an acid. If the acid does not contain oxygen, it is named by adding prefix hydro- and the suffix –ic acid. The above names only apply for aqueous solution, i.e. when acid can dissociate. In gaseous phase compounds are named according to the rules for nonacidic compounds. If the acid contains oxygen(s) it is called oxyacid. The rules follow the polyatomic naming. If the anion name ends in –ate, the acid is named by changing the suffix to –ic. If the anion name ends in –ite, the acid is named by changing the suffix from –ite to –ous. HF(aq) hydrofluoric acid HCl(aq) hydrochloric acid H2S(aq) hydrosulfuric acid HCl(g) hydrogen chloride Note that the salt of these acids have usual names of binary compounds. NaCl sodium chloride Examples no hydro- prefix indicates oxyacid containing sulfur possible existence of an –ous acid indicates hydrogen sulfuric acid sulfite sulfurous acid nitrite nitroous acid 2− 3 SO H 2SO3 − NO2 HNO 2 This is the way the formula is written. contains contains contains hydrogen sulfur oxygen H 2SO 4 sulfate sulfuric acid nitrate nitric acid KMnO4 K + MnO 4 SO 2− 4 H 2SO4 − NO3 HNO3 The ions are what is actually present. Polyatomic compound Acid (begins with H) Other compounds w/polyatomic ions Anion containing oxygen 1. Name cation(s) 2. Name anion Check ending on anion -ite Writing formula of a compound -ate Cation Anion name ends with –ous acid Anion name ends with –ic acid K+ H+ Zn2+ Al3+ Homework: Chpt. 8 (p.172): 3, 9, 13, 21 Chpt. 6 (p.122): 3, 6, 8, 15, 17, 25, 29 ClKCl HCl ZnCl2 AlCl3 Anions SO42O2PO43K2SO4 K2O K3PO4 H2SO4 HCl H3PO4 ZnSO4 ZnO Zn3(PO4)2 Al2(SO4)3 Al2O3 AlPO4 Chapter 7 Quantitative Composition of Compounds Making new chemicals is much like following a recipe from a cook book... 1 cup flour + 2 eggs + ½ tsp baking powder → 5 pancakes … except you don’t get to lick the spoon! What if you want to make more (or less)? Suppose you have plenty of flour and baking powder, but only 8 eggs. How many pancakes can you make? You can solve it using conversion factor: 8 eggs x 5 pancakes 2 eggs = 20 pancakes 5 pancakes 2 eggs Solve it in your head: 2 eggs makes 5 pancakes, so four times more eggs makes 20 (5x4) pancakes. Practice using the following mouthwashing, diet-buster recipe: 3 blocks cream cheese + 5 eggs + 1 cup sugar = 1 cheese cake. How many cheese cakes can we make out of 15 eggs? 1 cake 15 eggs x 5 eggs = 3 cheese cakes How much sugar do we need for 5 cheese cakes? (5) Suppose you want to ‘whip’ a batch of hydrogen iodide, following the balanced chemical equation: H2 + I2 2 HI How much H2 and I2 should you use to make 10 g of HI? A common mistake is that H2 and I2 react in one-to-one mass ratio so: 5 g H2 + 5 g I2 10 g HI The coefficients balancing the equation refer to number of atoms, not masses. Introducing the mole. The mole is like a dozen, but much, much more. The mole is Avogadro’s Number of items. 1 mole = 602,214,179,000,000,000,000,000 or 6.022 x 1023. We need the mole because the mass of an atom is too small to be measured on a balance. Remember: 1 amu = 1.6 x 10-24 g. 1 mole of anything: donuts, pancakes, is always 6.022 x atoms, molecules, ions… 1023 of that thing. 1 mole of soft drink cans is enough to cover the surface of the earth to a depth of over 200 miles. If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole. The mole translates between the number of atoms (or molecules, ions) and grams of atoms (molecules, ions). It is defined as the mass of Avogadro’s number of atoms 126C, which, in turn, weights exactly 12 g. A mole of atoms weighs the same number of grams as the atomic mass. One mole of H atoms weighs 1.0079 g. One mole of C atoms weighs 12.011 g. H2 Amadeo Avogadro The mole Atomic mass refers to: the sum of protons and neutrons in a single atom, weighted average mass of all isotopes of an element and also to the number of grams in one mole of atoms. + I2 = 2 HI 1 molecule 2 H atoms 1 molecule 2 I atoms 12 molecules 12 molecules 6.022 x 1023 molecules 1 mole 2.0158 g 6.022 x 1023 molecules 1 mole 253.81 g 2 molecule 2 x (1 atom H, 1 atom I) 24 molecules 1.204 x 1024 molecules 2 mole 255.8258 g or any number of molecules 1 mole of H2 weighs 2 x 1.0079 g = 2.0158 g Conversion factors: 1 mole 6.022 x 1023 species 1 mole molar mass Mole - mass - atoms conversions Q1: How many atoms in 0.5 mole Au? 6.022 x 1023 atoms Au 0.5 mole Au x 1 mole Au = 3.011 x 1023 atoms Au Q1a: How many moles in 7.12 x 1024 atoms of Cu? 1 mole Cu 7.12 x 1024 atoms Cu x 6.022 x 1023 atoms Cu = 11.8 mol Cu Q2: What is the mass of 0.5 mol Au? 196.967 g Au 0.5 mole Au x 1 mole Au = 98.4835 g Au Q3: How many atoms in 15.00 g Au? 6.022 x 1023 atoms Au 1 mole Au x = 4.59 x 1022 atoms Au 15.00 g Au x 1 mole Au 196.987 g Au Percent Composition Percent composition is % mass that each element in a molecule contributes to the total molar mass of the compound. Assume that you have one mole of the compound. What is the % composition of CH2O? Total mass = 12.01 g + 2.016 g + 16.00 g = 30.026 g 12.01 g %C = 30.026 g x 100 %C = 40.00 % %H = 6.71 % Practice: What is the % composition of glucose? Check your answer: it is the same as in CH2O! Types of Formulas + %O = 53.29 % 100.00 % CH2O is the empirical formula for glucose, C6H12O6 Empirical Formula: the formula of a compound that expresses the smallest whole number ratio of the atoms present. Formulas describe the relative number of atoms (or moles) of each element in a formula unit. It’s always a whole number ratio. If we can determine the relative number of moles of each element in a compound, we can determine a formula for the compound. Molecular Formula: the formula that states the actual number of each kind of atom found in one molecule of the compound. 1 molecule of C9H8O4 = 9 atoms of C, 8 atoms of H and 4 atoms of O. 1 mole of C9H8O4 = 9 mol of C, 8 mol of H and 4 mol of O atoms. Dr. Ent burned 0.5 g of the sample and obtained the total of over 1 g of products. How is that possible? Oxygen from air is a reactant! From the mass of the products (water and carbon dioxide) we determine the number of moles of C, H, and O, and from them obtain the empirical formula of the compound. 1. Determine the mass in grams of each element present, if necessary. Remember, % means “out of 100”. 2. Convert grams of CO2 and H2O (or C and H) into moles of C and H atoms. 3. Convert moles of C into grams of C. Do the same for H. 4. Add masses for C and H and subtract the sum from the mass of the sample to obtain mass of O. Convert the mass into moles of O. 5. Divide all number of moles with the smallest to obtain the subscripts of the empirical formula. 6. Divide the molar mass of the compound by the molar mass of the empirical formula. To find the molecular formula, multiply all subscripts in the empirical formula by this product. Note: steps 3, 4 apply only for finding formulas from combustion analysis. Combustion analysis shown 0.300 g H2O and 0.733 g CO2 from 0.500 g of sample. Find the empirical and molecular formula if the molar mass of the compound is 180.15 g/mol. 2 mol H atms 1 mol H2O x 0.3 g H2O x 18.01 g H 1 mol H2O 2O 2. = 0.0333 mol H at. 1 mol C atms 1 mol CO2 x 0.733 g CO2 x 44.01 g CO 1 mol CO2 2 = 0.0166 mol C at. 0.0333 mol H x 3. 1.008 g H 1 mol H = 0.0336 g H 12.01 g C = 0.199 g C 1 mol C g O = 0.5 – (0.0336 + 0.199) = 0.267 g O 1 mol O at. 0.267 g O x 16.00 g O = 0.0169 mol O at. Empirical Molar mass H: 0.0333 / 0.0166 = 2 formula emp. form. C: 0.0166 / 0.0166 = 1 CH2O 30.026 O: 0.0169 / 0.0166 ~ 1 0.0166 mol C x 4. 5. 6. Molar mass sample Molar mass emp. formula There is 6 CH2O units in the compound. = 180.15 =6 30.03 Molecular formula: C6H12O6. Find the empirical and molecular formulas if the % composition is 40.0% C, 6.70% H, 53.3% O, and the molar mass of the compound is 180.155 g/mol. ank h T ,D u o y t! n r. E 1. Assume that you have 100.00 g sample; the mass of each element is equal to the % composition. 40.0 g C, 6.70 g H, 53.3 g O. 2. 1 mol C 1 mol H 1 mol O 40.0 g C x 12.01 g C 6.70 g H x1.008 g H 53.3 g O x 16.00 g O = 3.33 mol O = 3.33 mol C = 6.65 mol H 1. Determine the mass in grams of each element present, if necessary. Remember, % means “out of 100”. 2. Convert grams of CO2 and H2O (or C and H) into moles of C and H atoms. 3. Convert moles of C into grams of C. Do the same for H. 4. Add masses for C and H and subtract the sum from the mass of the sample to obtain mass of O. Convert the mass into moles of O. 5. Divide all number of moles with the smallest to obtain the subscripts of the empirical formula. 6. Divide the molar mass of the compound by the molar mass of the empirical formula. To find the molecular formula, multiply all subscripts in the empirical formula by this product. Skip steps 3 and 4, they apply for combustion analysis only. 5. C: 3.33 / 3.33 = 1 H: 6.65 / 3.33 = 2 O: 3.33 / 3.33 = 1 Empirical formula CH2O. Emp. Formula mass = 30.026 Molar mass sample 180.155 = =6 Molar mass emp. formula 30.026 Thus, there are 6 (CH2O) units. Molecular formula: C6H12O6. 6. Practice (answer in parenthesis): 1. A compound has an empirical formula of NO2. The colorless liquid used in rocket engines has a molar mass of 92.0 g mole-1. What is the molecular formula of this substance? (N2O4) 2. A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34 g O. Determine an empirical formula for this substance. (NO2) 3. Percent composition of a compound is found to be 43.2% K, 39.1% Cl, and some O. Find the empirical formula. If the molar mass of the compound is 90.550 g mol-1, find the molecular formula. (KClO) Chapter 9 Calculations from Chemical Equations The molar mass of an element is its atomic mass in grams. It contains 6.022 x 1023 atoms (Avogadro’s number) of the element. The molar mass of a compound is the sum of the atomic masses of all its atoms. For instance: molar mass of NaCl is 22.99 + 35.45 = 65.44 g For calculations of mole-mass-number_of_particle relationships: Remember me? Conversions go through moles. 1. Use balanced equation. ∆ 2 Al + Fe2O3 Al2O3 + 2 Fe 2 mol 1 mol 1 mol 2 mol 2. The coefficient in front of a formula represents the number of moles of the reactant or product. To quantitatively convert from one quantity to another we introduce mole ratio: 1 mol Fe2O3 2 mol Al 1 mol Fe2O3 1 mol Al2O3 Mole ratio is found from the coefficients of the balanced equation. moles of desired substance Mole ratio = moles of starting substance Which conversion factor will be used depends on starting and desired substance A mole of a compound weighs the sum of all atoms in the compound. Mole – Mole Conversions - Molecules Example 1:How many moles of NaCl result from the complete reaction of 3.4 mol of Cl2? Assume that there is more than enough Na. 2 Na(s) + Cl2(g) → 2 NaCl(s) desired substance 3.4 moles Cl2 x 1 mole 2 moles 2 moles NaCl = 6.8 moles NaCl 1 mole Cl2 starting substance The following examples refer to the equation: Ca5(PO4)3F(s) + 5H2SO4(aq)→ 3H3PO4(aq) + HF(aq) + 5CaSO4(s) 1 mole 5 moles 3 moles 1 mole 5 moles Example 2: Calculate the number of moles of phosphoric acid (H3PO4) formed by the reaction of 10 moles of sulfuric acid (H2SO4) on phosphate rock: 10 moles H2SO4 x 3 moles H3PO4 5 moles H2SO4 = 6 moles H3PO4 Example 3: Calculate the number of moles of Ca5(PO4)3F needed to produce 6 moles of H3PO4. 6 moles H3PO4 x 1 mole Ca5(PO4)3F 3 moles H3PO4 = 2 moles Ca5(PO4)3F Mass – Mole conversion Example 4: Calculate the number of moles of H2SO4 necessary to yield 784 g of H3PO4. g Molar mass of H3PO4 = 97.994 mole Ca5(PO4)3F(s) + 5H2SO4 → 3H3PO4 + HF + 5CaSO4 1 mole 5 moles 3 moles 1 mole 5 moles 1. Convert the starting 1 mole H3PO4 784 g H3PO4 x substance into moles. = 8.00 moles H3PO4. 97.994 g H3PO4 2. Convert moles of starting substance into 5 moles H2SO4 8.00 moles H PO x = 13.3 moles H2SO4. 3 4 moles of desired 3 moles H3PO4 substance. 3. Convert moles of desired substance into the units specified in the problem. done. Ex. 5: Calculate the mass of phosphate rock, Ca5(PO4)3F needed to yield 200. g of HF. Molar masses: Ca5(PO4)3F = 504.31 g/mol; HF = 20.008 g/mol Step 1, 200. g HF x 1 mole HF = 10.0 moles HF x 1 mole Ca5(PO4)3F = 10.0 moles 1 mole HF 20.008 g HF Ca5(PO4)3F Step 2 504.3 g ph.r. = 5.00 kg Ca (PO ) F. Step 3: 5 4 3 10.0 moles Ca5(PO4)3F x 1 mole ph.r Step_by_step: Mass – mass conversion Ex. 6: Calculate the number of grams of H2SO4 necessary to yield 392 g of H3PO4. Ca5(PO4)3F(s) + 5H2SO4 → 3H3PO4 + HF + 5CaSO4 1 mole 5 moles 3 moles 1 mole 5 moles Molar mass H3PO4 = 97.994 g mole g Molar mass H2SO4 = 98.086 mole 1. Convert the starting substance into moles. 1 mole H3PO4 392 g H3PO4 x = 4.00 moles 97.994 g H3PO4 5 moles H2SO4 2. Convert moles of starting substance 4.00 moles H3PO4 x 3 moles H3PO4 into moles of desired substance. = 6.67 moles 3. Convert moles of desired substance into the units specified in the problem. 6.67 moles H2SO4 x 392 g H3PO4 x 98.086 g = 654 g H2SO4. 1 mole H2SO4 Combined steps: 1 mole H3PO4 x 5 moles H2SO4 x 98.086 g = 654 g H2SO4. 97.994 g H3PO4 3 moles H3PO4 1 mole H2SO4 Example 7: Find the mass of glucose that can be synthesized from 58.5 g of CO2, assuming that there is more than enough water to react with all the CO2. Molar masses are 44.01 g (CO2) and 180.16 (glucose). sunlight 6 CO 2(g) + 6 H 2O(l) → 6 O 2(g) + C6 H12 O 6(aq) 58.5 g CO2 x 1 mole CO2 1 mole glucose 180.16 g glucose = 39.9 g glucose x x 44.01 g CO2 6 moles CO2 1 mole glucose Conversion – General Case Mass to moles of starting compound Step 1 Moles of starting compound to moles of desired compound Step 2 Moles of desired comp. to units desired. Step 3 Mass – mass: All 3 steps Example 8: Calculate the mass of NH3 formed by the reaction of 112 grams of H2. N2 + 3H2 → 2NH3 grams H2 → moles H2 → moles NH3 → grams NH3 Molar masses: H2: 2.016 g/mol; NH3: 17.034 g/mol 1 mole H2 2 moles NH3 17.034 g NH3 x x = 1420 g NH3 = 1.42 kg NH3. 2.016 g H2 3 moles H2 1 mole NH3 Starting Step 1 compound result Step 2 Step 3 Moles – moles: Step 2 only Example 9: Calculate the moles of NH3 formed by the reaction of 1.5 moles of H2. 2 moles NH 1.50 moles of H2 x 3 moles H 3 = 1.00 mole NH3. 2 112 g H2 x Starting compound Step 2 result Moles – mass: Step 2 and Step 3 only Example 10: Calculate the mass of NH3 formed by the reaction of 1.50 moles of H2. 17.034 g NH 2 moles NH 1.50 moles of H2 x 3 moles H 3 x 1 mole NH 3 = 17.0 g NH3. 3 2 Starting result compound Step 2 Step 3 Conversion – General Case (cont’d) Mass – moles: Step 1 and Step 2 only Example 11: Calculate the moles of NH3 formed by the reaction of 150. g H2. 1 mole H 2 moles NH N2 + 3H2 → 2NH3 150. g H2 x 2.016 g H2 x 3 moles H 3 = 49.6 g NH3. 2 2 Starting Step 1 Step 2 result compound Mass – particles: All 3 steps Example 12: Calculate the # molecules of NH3 formed by the reaction of 150. g H2. 1 mole H2 2 moles NH3 6.022 x 1023 molecules NH3 = 2.23 x 1025 112 g H2 x x x 2.016 g H2 3 moles H2 1 mole NH3 molecules NH3. Starting Step 1 Step 3 result compound Step 2 Limiting Reactant and Yield Calculations The amount of the product(s) depends on the reactant that is used up during the reaction, i.e. limiting reactant. One bicycle needs 1 frame, 1 seat and 2 wheels, therefore not more than 3 bicycles can be made. The number of seats is the limiting part (reactant); one frame and two wheels are parts in excess; 3 bicycles is the yield. Limiting Reactant and yield Calculations (cont’d) Example 13: How many moles of Fe3O4 can be obtained by reacting 16.8 g Fe with 10.0 g H2O? Which substance is the limiting reactant? Which substance is in excess? How much of the reactant in excess remains unreacted? Strategy: 1. Write and balance equation. 2. Calculate the number of moles of product for each reactant; 3. The reactant that gives the least moles of (the same!) product is the limiting reactant. 4. Find the amount of reactant in excess needed to react with the limiting reactant. Subtract this amount from the starting quantity to obtain the amount in excess. 5. Find the yield from the limiting reactant. Balanced equation: 3 Fe (s) + 4 H2O (g) Fe3O4 (s) + 4 H2 (g) Yield 1 mol Fe x 1 mol Fe3O4 yield from Fe: 16.8 g Fe x = 0.100 mol Fe O . 3 4 55.85 g Fe 3 mol Fe limiting reactant Least moles Fe3O4? 1 mol H2O 1 mol Fe3O4 From H2O: 10.0 g H2O x 18.02 g H O x 4 mol H O = 0.139 mol Fe3O4. 2 2 Reacted H2O 16.8 g Fe x Excess: 10.0 g – 7.01 g = 2.99 g H2O. 1 mol Fe x 55.85 g Fe 18.02 g H2O 4 mol H2O = 7.01 g H2O. x 1 mol H2O 3 mol Fe Answer: Yield is 0.100 mol Fe3O4, Fe is the limiting reactant, 2.99 g H2O is in excess. Percent Yield Calculations done so far assumed that the reaction gives maximum (100%) yield. Many reactions (especially organic) do not give the 100% yield, due to: side reactions, reversible reactions, product losses due to human factor. Theoretical yield: Amount calculated from the chemical equation. Actual yield: Amount obtained experimentally. Actual yield Percent yield: x 100 % Theor. yield Strategy: Find limiting reactant. Calculate theoretical yield. Calculate percent yield. Example 14: If 65.0 g CCl4 was prepared by CS2 + 3 Cl2 CCl4 + S2Cl2 reacting 100. g CS2 and 100. g of Cl2, calculate the percent yield. Molar masses: CS2: 76.15; Cl2: 70.90; CCl4: 153.81 g/mol 1 mol CS2 1 mol CCl4 100. g CS2 x x = 1.31 mol CCl4. 76.15 g CS2 1 mol CS2 1 mol Cl2 1 mol CCl4 x = 0.470 mol CCl4. 70.90 g Cl2 3 mol CS2 Limiting reactant 153.81 g CCl4 0.470 mol CCl4 x = 72.3 g CCl4. 1 mol CCl4 Theoretical yield 65.0 g CCl4 65.0 g CCl4 HW, Chp. 7: 1, 5, 15, 26, 33 x 100 % = 89.9 % Actual yield 72.3 g CCl4 Chp. 9: 3, 7, 13, 15, 23, 29 Percent yield 100. g Cl2 x Chemical Bonds: Formation of Compounds Chapter 11 1st shell 2nd shell s d subshell p d f This presentation has 8 groups. Periods in the Periodic System of elements are called shells; subshells (s, p, d, f…) contain orbitals (s:1, p: 3, d: 5, f: 7), and each orbital can be occupied by 2 electrons with opposite spins. Elements are classified in groups having similar chemical characteristics. Noble gases have completely filled p orbital with 8 e-. 4s orbital is populated before 3d. Before we continue with filling 4p, 10 e- must go into 3d. We put them in a loop since we are stalled filling 4th shell. Atomic properties Ionization energy … is the energy required to remove an e- from the atom. and increases bottom to top, up a group. It increases left to right across a period… That is the behavior exactly opposite to that of atomic radii! Ionization Energy vs. Atomic Number Atomic radii increase right to left across the period, and top to bottom down the group. Chemical Bonding A molecule is a collection of atoms bound together. It is considered as an element if all atoms are of the same type (e.g. H2), or a compound if it is made of different atoms. A bond between two atoms of hydrogen will occur spontaneously if the atoms are within a close proximity, i.e. when attractive forces between the nucleus of one atom and the eof the other overcome repulsive forces among their e-, thus pulling the atoms together. 74 pm In a covalent bond between two hydrogen atoms, each atom apparently has 2 electrons in its shell. That gives each H atom the electronic configuration of the closest noble gas, He. Electrons are shared between two atoms. Covalent bond H:H can be ..H H presented as H–H any of the following ways: Remember that the line stands for a pair of e-! 1 pm = 10-12 m The bond that is formed is called covalent bond. (Co-means partner, valent refers to valence electrons). It always releases energy. Filled-shell e- (or core e-) are almost never involved in the bond as they are too close to their own nucleus. 2 He atoms repel each other and will never form a bond. Molecules and the Octet Rule Elements ‘want’ to have the electronic configuration identical to that of noble gas (8e-). When form a molecule, atoms achieve the octet (8e-) by sharing e- with other atoms. Hydrogen is an exception, as it only needs one more electron to fill its 1s orbital. How many electrons an element needs to satisfy the octet rule can be found by observing the Roman numerals in the periodic table. C: 8eH: 2e- All atoms in the molecule have the electronic configuration identical to that of the closest noble gas. Core electrons Valence electrons Nonmetals usually gain electrons, metals loose them. O: 8eH: 2e- Step 1: Find the total number of Drawing Dot Diagrams valence e- by adding up the group numbers of all atoms. For ions, CO2 CO adjust the dot count accordingly Group IVA (14) Group IVA Group VIA Group VIA (16) (subtract e- for cation, add for anion). 4e- + 2 x 6e4 e- + 6 eStep 2: Unless told otherwise, Total: 16e assume that the first nonTotal: 10eO–C–O hydrogen atom in the formula of C–O the group is the central atom. 16e – 4e = 12eConnect atoms with single bonds. in lone pairs The central atom must form multiple bonds, hence H can never be the central atom. Step 3: Subtract 2e- for each bond from total #e- to get #e- in lone pairs. Step 4: Put in the remaining electrons, two at a time, as lone pairs. Start with the terminal atoms, and continue with the central atom if there are any electron pairs left. Step 5: Check that each atom has octet satisfied (doublet for H). If not, move electron pair(s) from the adjacent atom to form multiple bonds. Practice with CO32-, SO3, etc. 10e – 2e = 8ein lone pairs .. : C – .O. : .. .. :O . . – C – .O. : or .. .. :O . . – C – .O. : or .. .. :O . . – C – .O. : .. .. O ..=C=O .. .. : .O. – C Ξ O : :CΞO: .. :OΞC–O . .: Equivalent structures, or resonance forms. .. : C – .O. : C2H4 CO324e + (3 x 6e) + 2e = 24eO O C total O 24e – (3 x 2e) = 18ein lone pairs .. :O .. Which C is the central atom? 2x4e + 2x1e + 2x7e = 24etotal BOTH ! 2x4e+4x1e=12e- total H H C cannot H H 12e – (5x2e) = 2e pull C C C C H H .. O: .. H H :.O. : Octet rule not satisfied for C C C H H .. :O .. C C :.O. : O: .. C2H2Cl2 in lone pairs H H :C C H H Octet rule not satisfied for C electrons from Cl : Cl . . : : Cl . .: H .. : Cl : H C C C C H :Cl ..: :Cl . .: H .. :Cl : H has no e- to give in for double bonds 2Two more resonance structures for CO32ion are possible. SO3 is identical to this; try it yourself. HW chapter 11: 1, 3, 13, 29, 37 The bond between metals and non-metals is usually ionic. Metals give away their e- and become positively charged (cations). Nonmetals accept them and become anions. The ionic bond is formed as a result of attraction between oppositely charged ions. The compound is called ionic compound, and the three-dimensional ordered network of the ions is called ionic lattice. Electronegativity and the Polar Covalent Bond The difference in EN defines the bond. ∆EN = 0, covalent; ∆EN = 1.0, polar covalent (23% ionic); ∆EN = 1.9, polar covalent (60% ionic); ∆EN > 1.9, ionic. Electrons are rarely shared equally between atoms. Electronegativity (EN) is numerical rating of an atoms ability to attract to itself the shared electrons in a covalent bond. Generally, electronegativity of metals is low, and that of nonmetals is high. The least electronegative atom (except H!) is the central atom in dot structures. Polar covalent bond is a covalent bond in which e- are shared unequally (large ∆EN). A partial negative charge (δ-) occurs on the more EN atom. A partial positive charge (δ+) occurs on the less EN atom. Ionic and covalent are two extremes at the ends of a continuum bonding types. The Shape of the Molecules Valence Shell Electron Pair Repulsion (VSEPR) theory is the model mostly used to predict molecular shape. Electron pairs on the central atom repel one another. The two dimensional dot structure of methane, CH4. gives the angles between electron pairs of 90o. But the dot structure angles are arbitrary. Molecules are three dimensional, and the electron pairs would be further away if the third dimension is considered. In fact, the shape of methane molecule is tetrahedral; the bond angles between electron pairs is 109.5o. H | H–C–H | H Four electron pairs around an atom assume tetrahedral arrangement. When there are not enough electrons for single bonds the molecule forms multiple bonds and the structure differs. VSEPR theory treats each multiple bond as a single electron group, because it occupies roughly the same region of space. The number of electron groups around an atom is called the atom’s steric number (SN). Dot structures of formaldehyde and acetylene are arbitrarily shown with angles of 90o. Their true geometry has bond angles of 120o and 180o, respectively. :O: || H–C–H H H | | CΞC formaldehyde acetylene Dot structures :O: || C H H–CΞC–H H True geometry If the central atom has a lone pair of electrons, that electron pair is included in the molecular shape. The dot diagram of ammonia presents the atom in a plane. The steric number on nitrogen is 4 (3 bonding pairs and a lone pair). The e- pairs on the N assume tetrahedral arrangement. .. H–N–H | Electrons in the lone H pair occupy more VSEPR arrangements of electron groups around an atom having no lone pair electrons O=C=O We describe the shape of the molecule, not that of its electrons. Lone pair(s) of electrons are therefore ignored. The molecule of NH3 has a pyramidal shape. space than the bonding pairs. They squeeze H-N-H bond angle to approx. 107o. Rule of thumb: each lone pair of e- on a period-2 atom compresses the remaining bond angles around that atom by ~2o. Using VSEPR 1. Draw a dot diagram. 2. Count the number of epairs around the central atom, including lone pairs (i.e. the steric number, SN). A multiple bond counts as a single e- group. 3. Find the best arrangement of the electrons using SN. 4. Pretending the lone epairs are invisible, describe the resulting shape of the molecule. Practice on H2O and O3. H-O-H bond angle ~105o. bent H | : O: O-O-O bond angle 120o. SN=4 bent H O || O: O SN=3 Chapter 13 Water and Properties of Liquids Liquids have intermediate properties between solids and gases. Liquids are almost incompressible, have definite volume and assume the shape of the container. Densities of liquids are usually lower than that of their solids. Water is an exception. Evaporation or vaporization is the escape of molecules from liquid into gaseous state. During evaporation, liquid that stays behind is cooler. The opposite process is condensation. Sublimation is the escape of molecules directly from solid into gas, bypassing liquid state. Vapor pressure is the pressure exerted by a gas at evaporation equilibrium with its liquid, so that: liquid gas Vapor pressure depends only on condensation temperature, not on the amount of liquid. Open container completely evaporates. Closed container reaches equilibrium between liquid and gas. Vapor Pressure Measurement 1 atm = 760 torr 20 oC 20 oC a. b. a. The system is evacuated. Manometer attached to the flask shows equal pressure in both legs. b. Water is added. Liquid evaporates. Manometer shows increase in pressure. 20 oC 30 oC c. d. c. Equilibrium established. Manometer shows constant pressure difference, 17.5 torr. d. Temperature raised to 30 oC. Equilibrium reestablished. Manometer shows constant pressure difference of 31.8 torr. Vapor pressure and temperature Vapor pressure of any gas at the boiling point is equal to the atmospheric pressure. Vapor pressure of ethyl ether is the highest at any temp. TBP TBP Vapor pressure: Ether > Alc. > Water. TBP Rate of evaporation: Ether > Alc. > Water. proportional to vapor pressure. Volatility Boiling point: Ether < Alc. < Water Substances that readily evaporate are volatile. Vapor pressure of ethyl ether at 20 oC: 442.2 torr Volatile Vapor pressure of water at 20 oC: 17.5 torr Vapor pressure of mercury at 20 oC: 0.0012 torr Moderately volatile Nonvolatile Boiling Point Curves Normal Boiling Point Boiling point at standard pressure (1 atm, or 760 torr). Each point on the curve represents a vapor-liquid equilibrium at a particular temperature and pressure. At 500 torr, ethyl ether boils at ~22 oC, alcohol at ~68 oC, and water at 89 oC. Freezing or Melting Point The temperature at which the solid and liquid are in equilibrium. Changes of State Majority of substances change phases upon heating: solid liquid gas. 1 atmosphere pressure TBP ethyl ether TBP alcohol TBP water 34.6oC 78.4oC 100.0oC Heating curve for a pure substance CO2 is an exception (dry ice sublimes). A – B: solid state B – C: melting C – D: liquid state D – E: evaporation E – F: vapor state Temperature is constant during melting and boiling – all heat used to break solid (at boiling point) or liquid forces. liquid solid evaporation condensation melting freezing gas liquid Heat of Fusion and Heat of Vaporization We learned before that amount of heat Qheating = (mass) (spec.heat) (temp.change) depends on mass and temp. change. Energy (heat) needed to change 1 g Energy (heat) needed to change 1 g of a solid at its melting point into Constant of a liquid at its boiling point into liquid is heat of fusion. temperature! vapor is heat of vaporization. Qfusion = (mass) (spec.heat of fusion) Qvaporization = (mass) (spec.heat of vaporization) Example 1: How many joules is needed to change 20.0 g of ice at 0 oC to steam at 100. oC? Qheating = (mass) (spec.heat) (temp.change) Qtot = Qfusion + Qheating + Qvaporization Qfusion = (20.0 g) x (335 J/g) Qheating = (20.0 g) x (4.184 J/goC) x (100. oC) Qvaporization = (20.0 g) x (2260 J/g) Hydrogen Bond } Qtot = 60.3 kJ produces unusually high melting & boiling point Hydrogen Bonding (cont.) H bonding exists between H directly bonded to one of the three most electronegative elements (Fluorine, Oxygen, and Nitrogen), and F, O or N of another molecule. . . H bond . . H – O :. . . H – O : | | H H H bonded to O No H bond H H | .. | H–C–O–C-H | .. | H H Ethyl ether Surface Tension and Capillary Action A droplet of liquid falling forms a sphere due to attractions to other liquid molecules – surface tension. Cohesive forces within Spontaneous rise mercury liquid (left) are of liquid in a stronger than adhesive forces between Hg and narrow tube – walls of the container. capillary action. Opposite is true for H O. 2 No H bonded to F, O, or N H bonds are intermolecular forces. Hydrates Some ionic solutions retain water upon evaporation. It becomes the part of the crystalline compound – water of crystallization. The formula is written as: ionic compound, dot , # water molecules… CuSO4 5 H2O and name them by adding # (Latin) hydrate. . Copper(II) sulfate pentahydrate. Hydrates are true compounds and the water is an integral part of it. Formula mass CuSO4 5 H2O: 63.55+32.07+64.00+5x18.02 = 249.7 Percent composition of water is (5x18.02 / 249.7) x 100 = 36.08% dry CuSO4 – white Hydrate = blue . . 5 H O(s) CuSO (s) + 5 H O(g) 2 4 2 .. O δ- .. Water can be removed by intense heat: CuSO4 The reaction is reversed when water is added. Water, a Unique Liquid Water indicator H H Water covers ~75% of Earth. 97% of water is in the oceans. Only 3% is fresh water, of which 2/3 is locked up in ice polar caps. δ+ Solid form (ice) has lower density than liquid water. Water is very stable molecule, can stand temperatures up to 2000 oC. It does not conduct electricity when pure, but decomposes into H2 and O2 in solutions of ions. 2 H2 + O2 --> 2 H2O + 484 kJ Water can be formed by 2 C2H2(g) + 5 O2 4 CO2 + 2 H2O(l) + 1212 kJ Combustion, HCl(aq) + NaOH(aq) --> NaCl(aq) + 2 H2O Neutralization, Metabolic reaction C6H12O6(aq) + 6 O2 6 CO2(g) + 6 H2O(l) + 2519 kJ Water reactions with metals: Cold water reacts with Na, K, Ca: Steam reacts with Zn, Al and Fe: Reactions of water Na + H2O H2 + NaOH Fe + H2O(g) --> H2 + Fe3O4 Remind yourself of the activity series: the above six metals are the most active. Another three metals are more active than H: Pb, Sn, and Ni and react with acids only; Cu, Ag, Hg and Au are below H in the series and do not react with acids or H2O. Water also reacts with certain nonmetals. Anhydride means: without water. Most reactive: 2 F2 + 2 H2O(l) --> 4 HF(aq) + O2 Less reactive: Cl2(g) + H2O(l) HCl(aq) + HOCl(aq) To test whether a metal or nonmetal is an anhydride, try Least reactive: C(s) + H2O(g) CO(g) + H2(g) Water reacts with metal and non-metal oxides: Basic anhydride: CaO + H2O(l) Ca(OH)2(aq) Acid anhydride: CO2(g) + H2O(l) H2CO3(l) Water Purification Screening, flocculation and sedimentation, sand filtration, aeration, disinfection. Hard water contains Mg2+ and Ca2+ ions Additional water purification is done by distillation, Ca2+, Mg2+ precipitation, ion exchange and demineralization. HW (p.332): 7, 11, 17, 25 to remove H2O until all hydrogen is removed. Ca OH ∆ OH CaO + H2O H2SO4 ∆ SO3 + H2O Chapter 12 Compressibility Shape Volume Gaseous State of Matter Solid N Y Y Liquid N N Y Gas Y N N An Ideal Gas consists of point-like particles that do not attract or repel one another at all. Gas molecules move freely until they hit the wall. Unlike liquids, gases expand to fill the container. When air is removed from a can, there are no gas molecules to oppose the atmospheric pressure exerted from the outside, so the can collapses. Air pump Pressure = Force Area Pressure can be measured by the height of mercury column. When air is removed from the tube, there is no resistance to the flow of mercury from the bottom container. Mercury fills the tube until the atmospheric pressure is compensated by the weight of mercury column. We do not feel atmospheric pressure due to random motion of air. But you feel the wind! Pressure imbalance in ear: A difference in pressure across the eardrum membrane causes the membrane to be pushed out – we call it a “popped eardrum.” Gas Pressure Normal atmospheric pressure: 1 atm = 760 mm Hg = 760 torr = 29.9 in Hg. Behavior of an ideal gas can be totally described with pressure (P), volume (V), temperature (T) and the number of moles of gas (n). Pressure (P) – atmospheres Temperature (T) – Kelvin Volume (V) – liters Amount of Gas (n) – moles If different units are given, convert them into atm, K, L, moles Describing the Gas Volume is directly proportional to Pressure (P) proportional to temperature (T). temperature (Charle’s Law): V α T Faster moving gas molecules exert higher Pressure is inversely proportional to pressure on the walls of the container (Guythe volume (or proportional to 1/V). Lussac’s Law). As the volume increases, pressure PαT decreases (Boyle’s Law). T in K! P α 1/V α means proportional. Pressure is proportional to the number of moles of gas. More gas molecules results in more collision with the walls of the container. Pαn P α nT/V or P = const. x P=R nT V R = 0.082 L atm mol K nT V PV = nRT Ideal gas law R is called ideal gas constant, and is the same for any gas. Examples Example 1. What is the volume of 5.34 moles of an Ideal Gas with a pressure of 0.692 atm and a temperature of 45°C? PV = nRT V = nRT T = 45°C + 273.15K = 318K sig.figs? P (5.34 mol) x (0.082 L atm / mol K) x (318 K) V= = 201 L 0.692 atm Example 2. What is the temperature (in °C) of 0.0520 g of H2 gas enclosed in a 762 mL container, if its pressure is 24.6 inches Hg? 1 mole H2 1L 762 mL x = 0.762 L 0.0520 g H2 x = 0.0260 moles H2 2.016 g H2 1000 mL PV 1 atm T= PV = nRT 24.6 in Hg x = 0.822 atm nR 29.9 in Hg T= 0.822 atm x 0.762 L = 294 K = 21°C 0.0260 mol x (0.082 L atm / mol K) Example 3. What is the mass of O2 gas in a 3.4 L container at 25°C with a pressure of 625 mm Hg? 1 atm T = 25°C + 273K = 298K P = 625 mm Hg x = 0.822 atm 760. mm Hg 0.822 atm x 3.4 L n= = 0.11 moles O2 m = 0.11 mol x 32.0 g = 3.5 g O2 298 K x (0.082 L atm / mol K) 1 mol Dalton’s Law of Partial Pressures Ptotal = PA + PB + PC + … The total pressure of a mixture of gases is the sum of the partial pressures exerted by each of the gases in the mixture. When collecting oxygen over water (usual way), water vapor contributes to the total pressure. To determine the amount of O2, the water vapor pressure must be subtracted form the total pressure. PO2 = Ptot – PH2O Ptot = PO2 + PH2O Example 4: What is the volume of the dry oxygen if it was collected over water at 23oC and 760. torr in a 500 mL container? Water vapor pressure at 23oC is 21.2 torr. Step 2: Organize data: Step 1: determine the pressure of dry O2: P1 = 739 torr P2 =760 torr PO2 = Ptot – PH2O = 760 torr – 21.2 torr = 739 torr. V1 = 500 mL V2 = ? Step 3: Solve as a Boyle’s law: or P1V1 = P2V2 P α 1/V P = const. / V PV = const. P1V1 = const. P2V2 = const. V2 = P1V1 739 torr x 500 mL = 486 mL dry O2. = P2 760. torr Standard temperature and pressure (STP) is 1 atm and 0oC. Avogadro’s Law Equal volumes of different gases at the same T, P contain the same number of molecules. H2 + 1 molecule 1 mol 1 volume Cl2 1 molecule 1 mol 1 volume 2 HCl 2 molecules 2 mol 2 volumes Different gases at the same P, T have equal kinetic energy, and since they occupy the same volume, they must have the same number of molecules to satisfy the relationship PV = nRT. Example 5: Calculate the molar mass Experimentally determined that 1 mol of ANY gas occupies 22.4 L at STP. If the mass and volume of the gas at STP are known, one can calculate the molar mass M. Since M = mass / number of moles, M=m/n n=m/M Density of Gases Density is: mass / volume and for a gas is expressed in g/L. Recall that densities of solids and liquids are given in g/mL! Gas densities are up to 1000 times lower than the densities of their liquids. of a gas that occupies 2.00 L at STP and its mass is 3.23 g. Remember units of molar mass: g/mol. 3.23 g 22.4 L = 36.2 g/mol M= x 2.00 L 1 mol Substituting in Ideal Gas Law: m mRT PV = nRT PV = RT M= M PV Gas volume depends on the T and P. Example 6: Calculate the density of Cl2 at STP. Molar mass of Cl2 = 70.90 g/mol 70.90 g 1 mol = 3.17 g/L d = 1 mol x 22.4 L Gas Stoichiometry Volume - mol conversion using Avogadro’s law if STP, or ideal gas law if not STP conditions. Mass - moles and atoms - moles conversions as in chapter 9. 1 mol (@ STP) 22.4 L Mass (g) x 1 mol molar mass (g) Volume (L) x # atoms x Volume B Volume A Mass A Atoms A Moles A Moles B Mass B Atoms B Mol A – Mol B from coeff. of balanced equation. Mol B – Volume, Mass or atoms, use inverse conversion factors. Example 7: What volume of O2 at STP can you make from 10. g KClO3 (p.c.)? 2 KClO3 2 KCl + 3 O2 1 mol 6.022 x 1023 atoms 10. g p.c.x1 mol KClO3 x 3 mol O2 x 22.4 L = 2.7 L 122.55 g p.c. 2 mol KClO3 1 mol O2 Example 8: What volume of H2 g - mol A mol A - mol B mol - Liters B will be formed when 50.0 g of Al reacts with HCl at 30.0 oC 2 Al(s) + 6 HCl (aq) 2 AlCl3(aq) + 3 H2(g) and 700. torr? 3 mol H2 Find volume at given T,P: V = nRT/P 1 mol Al Moles H2: 50.0 g Al x x 26.98 g Al 2 mol Al L atm x (0.082 2.78 mol H = 2.78 mol H2 mol K) x 303 K 2 700. torr x (1 atm/760. torr) Not STP conditions, must use ideal gas law. = 75.1 L H2 Real Gases Most real gases behave nearly as predicted by ideal gas law. Deviations occur when molecules are crowded (high P, low T). Air Pollution oxygen Oxygen allotrope ozone is produced and decomposed in stratosphere, protecting us from damaging UV light. Simultaneous processes occur: Chlorofluorocarbons in O2 O+O UV light aerosols, refrigerators O3 O2 + O decompose ozone O2 + O O3 ozone producing ozone hole. formation of ozone decomposition of ozone Now over Antarctica but Concentration of CO2 in air has risen steadily since industrial may spread over revolution (end of populated areas. 1800s), producing green house effect and warming Earth: global warming. sunlight Homework, chp.12 (p. 301): 11, 15, 19, 27, 35, 43, 50 Solutions Chapter 14 Homogeneous mixture Solute separation A solution is a homogeneous mixture of Heterogeneous mixture two or more substances. Its composition is the same throughout, i.e. concentration of the substances is the same. A solution consists of solute (or solutes) present in lesser amount, and solvent, present in greatest amount. Solvent / solute separation A solution can be made of steps require energy. Solvent separation substances in different phases. Solvation step Solvent solute example must Gas gas air overcome gas solid dusty air separation gas liquid humid air steps for the Liquid solid sugar solution substance to Solvation liquid liquid vinegar be soluble. liquid gas soda Solid solid all alloys Grinding and mixing two or more solids will never yield a true solution. A liquid (or gaseous) phase is needed. Attractive forces release energy. Solubility of ions in water insoluble soluble all Na+, K+, NH4+ most CO32-, PO43-, all NO3-, C2H3O2- OH-, S2-. (except Na+,K+,NH4+) most SO42-, Cl-, Br-, I- Solubility and effects of P,T Solubility is the maximum mass of solute that dissolves in 100 g of solvent. Solubility of most solids increases with an increase in temperature. Solution containing the maximum mass of solute in 100 g H2O is called saturated solution; it can be calculated from the graph. Example: prepare saturated solution of NaNO3 in 250 g of water at 20 oC. 250 g H2O x 90 g NaNO3 = 225 g NaNO3 100 g H2O Soaps and detergents wash (non-polar) grease by the action of their molecules, consisting of polar head and non-polar tail. “Like dissolves like” Solubility of gases decreases with temperature, but increases with pressure. “Tadpole’s” tail is attracted to non-polar grease. Water makes hydrogen bonds with polar head. Saturated solution of NaNO3 is 90 g NaNO3 in 100 g H2O at 20 oC. If one adds 100 g NaNO3 in 100 g H2O, 90 g dissolves, 10 g remain undissolved. At a given temperature, the mass of solid that goes into solution equals to the mass that precipitates from the saturated solution. solute (undissolved) solute (dissolved) Unsaturated solution contains less solute per 100 g H2O than Supersaturated solution has more the saturated solution. Supersaturated solutions are unstable, a small disturbance will cause rapid precipitation of the excess of solute. Cooling of a saturated solution makes the excess of solute precipitate. Rate of dissolving depends on temperature, particle size, concentration and stirring. In solid state no reaction will occur: Solution as a Reaction Medium NaCl(s) + AgNO3(s) no reaction Water breaks the crystal lattice, reaction Ions are locked within the crystal lattice. proceeds. NaCl(aq) + AgNO3(aq) AgCl(s) + NaNO3(aq) Na+(aq) + Cl-(aq) + Aq+(aq) + NO3-(aq) AgCl(s) + Na+(aq) + NO3-(aq) Percent Composition % composition by Grams of solute x 100 Volume of solution mass / volume Concentration can be expressed as % composition by Volume of solute x 100 percent composition. There are 3 types. volume / volume Volume of solution Amount of solute Example 1: What is the mass % composition x 100 Concentration = Amount of solution of 273 g solution that contains 35.0 g NaCl? Grams of solute % composition by mass Grams of solute x 100 Mass % = Grams of solution x 100 Grams of solution 35.0 g NaCl = x 100 = 12.8 mass % Units must be specified to prevent ambiguity! 273 g solution Chemists usually work with moles. Molarity Molarity is defined as the number of moles of solute dissolved in 1 L of solution… Molarity (M) = Moles of solute 1 L of solution Not # of moles of solute plus 1 L solvent! To prepare 1 L of NaCl, measure molar mass of NaCl, pour it to a 1L volumetric flask with enough water to dissolve it, and then add water to the mark. Another way to do it is to use more concentrated solution (called stock solution), by dilution equation: Mstock soln. x Vstock soln. = Mdiluted soln. x Vdiluted soln. Moles before dilution = Moles after dilution Find the number of milliliters of the stock solution and dilute it with water. Moles solute x Liters of solution = Moles solute M x Vdiluted soln. Vstock soln. = diluted soln. Liters solution Mstock soln. Dilution does not affect the number of moles (M x V)! M V Example 2a: Prepare 500. mL of 0.15 M NaCl using solid NaCl. 1 L solution 0.15 mol NaCl 500. mL x x 1000 mL 1 L solution 58.443 g NaCl x = 4.4 g NaCl 1.00 mol NaCl Example 2b: Prepare 500. mL of 0.15 M NaCl using stock solution of 0.50 M NaCl. 0.15 M NaCl x 500. mL soln.. Vstock soln. = 0.50 M NaCl = 150 mL stock solution + 350 mL water Molar mass of solute dilute conc. M1V1=M2V2 V2 = Example 3: How many grams of NaCl is in 200. mL of 1.00 M NaCl? 1 L solution 1.00 mol NaCl 200. mL x x 1000 mL 1 L solution 58.443 g NaCl x = 11.7 g NaCl 1.00 mol NaCl M1V1 M2 Example 4: Prepare 400. mL of 2.00 M NaCl using 3.00 M NaCl stock solution. 2.00 M NaCl x 400. mL soln.. Vstock soln. = 3.00 M NaCl = 267 mL stock solution + 133 mL water. Molality (m) is the number of moles of solute per kilogram of solvent. Note the difference in writing molarity (M) and molality (m) and the difference in their definitions. m = mol solute / kg solvent, therefore, molality is independent of volume. Colligative Properties of Solutions NaCl is often used to “melt” ice from streets; ethylene glycol/water mixture in car radiators works both as antifreeze and boiling point elevation agent. Freezing Point Depression, Boiling Point Elevation and vapor-pressure lowering are known as colligative properties of solutions. They depend on the number of solute particles in a solution and not on the nature of the particles. Reasoning: Vapor pressure is the tendency of water molecules to escape from the liquid surface into gas. If, say, 10 % of molecules on the surface molecules are nonvolatile, the resulting effect is that the vapor pressure is lowered. If the vapor pressure is lowered, the boiling point is elevated because liquid boils once its vapor pressure equals that of the atmospheric pressure. Also, if the vapor pressure is lowered, so is the freezing point because liquid vapor pressure curve of the solution does not intercept the solid vaporpressure at the freezing point of the pure solution, but rather below it. ∆tf = m x Kf ∆tb = m x Kb oC o = mol solute x C kg solvent kg solvent mol solute Example 5: Find molality of solution prepared by dissolving 150.0 g C6H12O6 (molar mass 180.156) in 600.0 g H2O. 150.0 g x 1 mol = 0.8326 mol 180.156 g molality = 0.8326 mol / 0.6 kg = 1.388 m Example 6: what is the freezing point of solution made by dissolving 100.0 g ethylene glycol in 200. g H2O? molar mass C2H6O2 = 62.05 g 100.0 g (1mol / 62.05 g) = 1.61 mol molality = 1.61 mol / 0.2 kg = 8.05 m Kf = 1.86 oC kg solvent/mol solute ∆tf = m x Kf = 8.05 (mol solute/kg solvent) x (1.86 oC kg solvent/mol solute) = 15.0 oC The freezing point of solution is lowered by 15 oC from that of solvent (water), so The actual freezing point is: 0 oC – 15.0 oC = -15.0 oC (5.00 oF). Osmosis … is diffusion of a liquid through a semipermeable membrane. The membrane allows solvent (usually water) to diffuse, but prevents diffusion of larger molecules. Solvent diffuses from the place with more solvent (water) to the place with less solvent. Blood cells: in blood plasma (0.15 M NaCl, 0.9% saline), they are normal. in hypertonic solution (1.6% saline) the blood cells shrink because water leaves the cell plasma. In hypotonic solution (0.2% saline) blood cells swell because water diffuses into them. Osmosis is also the reason bacteria cannot survive in sugar solutions. Plants take H2O from ground by osmosis. HW (p.365): 1, 3, 21, 27(a-c), 33 normal hypertonic hypertonicsoln. soln. hypotonic hypotonicsoln. soln. Chapter 15 Electrolytes, Acids and Bases Acid (Latin acidus - sour): sour taste; turns plant dye litmus red; dissolves metals producing H2 gas. Substances that are neither acids nor bases were called neutral substances. Electrolytes Nonelectrolytes Vinegar (HC2H3OH) Acetone (C3H6O) Hydrogen chloride (HCl) Carbon monoxide Baking soda (NaHCO3) Ethanol (C2H5OH) Table salt (NaCl) Methane (CH4) Milk of magnesia (Mg(OH)2) Sucrose (C12H22O11) Sulfuric acid (H2SO4) Turpentine (C10H22) Base: bitter taste; turns plant dye litmus blue; aqueous solutions feel slippery to touch. Electrolyte conducts electric current, non-electrolyte does not. Michael Faraday (early 1800) proposed that some mobile charged particles must exist in solution: ions. Svante Arhenius hypothesized that ions ‘Radical, ridiculous’ Ph.D. come from the dissociation of the solute. thesis, Nobel prize later. An electrolyte is a solute that dissolves in A nonelectrolyte is a solute that dissolves water and dissociates into ions, yielding in water without producing ions. The a solution that conducts electricity. solution consists of intact solute molecules. Ionic compounds are usually metal plus Water soluble molecular substances nonmetal or group of nonmetals. usually consist entirely of nonmetals. Exceptions: HX (X-halide) - polar covalent, Quick quiz: find electrolytes from the list: produce acids in H2O; ammonium salts. Al(NO3)3, (CH3)2O, (NH4)2SO4, CH3OH, HBr Weak and Strong Electrolytes HCl and HF are both hydrogen halides. A strong electrolyte is one that completely dissociates into ions upon dissolving in water. A weak electrolyte is one that only partially dissociates into ions upon dissolving in water. HCl(g) Dissolves in H2O H+(aq) + Cl-(aq) 100% dissociation All HCl molecules dissociate, no HCl molecules – light bulb shines strongly. HF(g) Dissolves in H2O H+(aq) + F-(aq) Far more undissociated HF molecules than ions – light bulb barely lights. Arhenius defined acid as an electrolyte that produces H+ ions when dissolved in water. In fact, H+ are always hydrated with a water molecule, it is more precise to say that the ions are H3O+ ions. A strong acid is a water-soluble compound that dissociates extensively to produce a large Strong acids #H3O+ Weak acids Hydrochloric (HCl) 1 1 Hydrofluoric (HF) number of H3O+ (hydronium) ions. Hydrobromic (HBr) 1 1 Hypochlorous (HClO) Depending on the maximum number Hydroiodic (HI) 1 1 Acetic (HC2H3O2) of H3O+ ions that can be produced Nitric (HNO3) 1 2 Carbonic (H2CO3) from 1 mol of acid, it is referred to as Sulfuric (H SO 2 3 Phosphoric (H3PO4) 2 4) monoprotic, diprotic or triprotic (1, 2 or 3 moles of H3O+). A di- or tri-protic acid may not be stronger than a monoprotic. Weak acid Dissociation constant, Keq, shows how many protons can be obtained from 1 mol of acid. H3PO4 + H2O H3O+ + H2PO4- Keq = 7.5 x 10-3 H2PO4- + H2O H3O+ + HPO42- Keq = 6.2 x 10-8 H3O+ + PO43- Keq = 4.2 x 10-13 Strong acid H3O+ + HSO4- Keq > 1.0 x 103 H3O+ + SO42- Keq = 1.2 x 10-2 Phosphoric acid is considered a HPO42- + H2O weak acid because even the first dissociation step is weak. The 3rd dissociation step is the weakest. H2SO4 + H2O Sulfuric acid is considered strong acid HSO4- + H2O because of the first dissociation step. The second dissociation step is weak. Bases are opposite of acids. When added in a proper amount to an acidic solution, the acidic properties are destroyed and the solution becomes neutral. Arhenius’ definition: Any electrolyte that contains a metal ion and hydroxide group and produces hydroxide (OH-) ions when dissolved in water. All bases given here are strong and completely dissociated. Bases can be monobasic, Lithium hydroxide, LiOH di- and tri- basic. Potassium hydroxide, KOH Calcium hydroxide, Ca(OH)2 Quick quiz: What do you get when you add HNO3(aq) to Ba(OH)2(aq)? Sodium hydroxide, NaOH Magnesium hydroxide, Mg(OH)2 Barium hydroxide, Ba(OH)2 HNO3(aq) + Ba(OH)2(aq) H2O(l) + Ba(NO3)2(aq) Arhenius theory cannot explain basicity of NH3 - it has no OH- ions. Brønsted and Lowry independently explained it, defining a base as a substance that removes H3O+. To accept a proton (H+) from the hydronium ion, the base must have a free electron pair. When a proton leaves H3O+ ion, H2O stays behind. Base - anything that accepts a proton. Acid - anything that donates a proton. This new theory can explain acetyleneamide reaction, and formation of ammonium chloride in gas phase. Note that acetate ion and carbonate ion come from dissociation of weak acids! When a solution is acidic neutral basic Parenthesis mean: concentration [H3O+] > = < Weak bases Ammonia, NH3 Acetate ion, C2H3O2Carbonate ion, CO32- [OH-] H2O + H2O Keq = 1.8 x 10-5 Keq = 5.6 x 10-10 Keq = 2.1 x 10-4 H3O+ + OH- [H3O+] x [OH-] K = K x [H O]2 = [H O+] x [OH-] w eq 2 3 Water also dissociates by the Keq = [H O] x [H O] 2 2 constant process called auto-ionization. pH Scale Auto-ionization of water produces equal concentrations of H3O+ and OH- ions: [H3O+] = [OH-] = 1 x 10-7 M O+] 10-7 Acid solution: [H3 >1x M. Basic solution: [OH-] > 1 x 10-7 M. When [H3O+] rises, [OH-] lowers to satisfy the ionic product of water, Kw = 1 x 10-14. Performing calculations with so small concentrations is tedious. Instead, scientists use logarithms. pH = - log [H3O+] Taking a logarithm is asking “to what power must I raise 10 to get the displayed number?” Example: log 558 = 2.75, because 102.75 = 558. Note that each one-fold decrease in pH represents tenfold increase in acidity. To convert from pH to molar H3O+ concentration, use the formula: [H3O+] = 10-(pH) Kw = Keq x [H2O]2 = [H3O+] x [OH-] Kw = [H3O] x [OH-] = [1 x 10-7] x [1 x 10-7] Kw = 1 x 10-14 Reactions of Acids and Bases acid + metal hydrogen + ionic salt 2 HCl(aq) + Ca(s) H2(g) + CaCl2(aq) acid + base salt + water HBr(aq) + KOH(aq) KBr(aq) + H2O(l) acid + metal oxide salt + water 2 HCl(aq) + Na2O(aq) 2 NaCl(aq) + H2O(l) acid + carbonate salt + H2O + CO2 2 HCl(aq) + Na2CO3(aq) 2 NaCl(aq) + H2O(l) + CO2(g) 2 KOH(aq) + 2 Al(s) + 6 H2O(l) 2 KAl(OH)4(aq) amphoteric hydroxides are capable of + 3 H2(g) reacting as a base or an acid Reactions of amphoteric hydroxides Zn(OH)2(s) + 2 HCl(aq) ZnCl2(aq) + 2 H2O(l) Zn(OH)2(s) + 2 NaOH(aq) Na2Zn(OH)4(aq) Reactions with certain metals 2 NaOH(aq) + Zn(s) + H2O(l) Na2Zn(OH)4(aq) + H2(g) Titrations Stoichiometric procedure to determine solution concentrations. Done by acid / base neutralization. Needed: 1. buret with known conc. of titrant 2. Known volume of the titrate in the Erlenmeyer flask 3. Indicator (usually phenolphtalein). Fill the buret with the titrant to 0.00 mL mark, drain it by opening the stopcock. Stop when indicator changes color and read the final volume of the titrant. Balanced eq.: NaOH(aq) + HCl(aq) H2O + NaCl(aq) moles of (known) base = moles of (unknown) acid Mbase x Vbase = Macid x Vacid 0.100 mol NaOH 1 mol HCl 1 x1 mol NaOH x MHCl = 0.02250 L NaOH x 1L NaOH 0.05000 L HCl Writing Net Ionic Equations HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) Na+(aq) + Cl-(aq) + H2O(l) spectator ions cancel H+(aq) + OH-(aq) H2O(l) Formula eq. Total ionic eq. Net ionic eq. All three equations must have balanced number of atoms and electrical charges. Strong electrolytes are written in their ionic form. Weak electrolytes, nonelectrolytes, precipitates and gases are written in their molecular form. Net ionic equation includes only substances that have undergone a chemical change. 2 AgNO3(aq) + BaCl2(aq) BaNO3(aq) + 2 AgCl(s) 2 Ag+(aq) + 2 NO3-(aq) + Ba2+(aq) + 2 Cl-(aq) Ba2+(aq) + 2 NO3-(aq) + 2 AgCl(s) AgCl precipitates, Ba2+ and NO3- are spectator ions. Ag+(aq) + Cl-(aq) AgCl(s) Try H2SO4 + Ba(OH)2; Mg + HCl, Na2CO3 + H2SO4. Colloidal Dispersions Colloids are particles that are intermediate between true solution and suspension (e.g. fine sand in water, which settles down once the shaking stops). Colloids (e.g. milk) neither settle down, nor form a true solution; the colloidal particle sizes are between true solute ions or molecules, and particles of mechanical suspension (i.e. between 10-4 and 10-7 cm) with a diameter of ~ 500 nm. Thus, colloidal particle is 1000 times larger in diameter, and about 109 times larger in volume. Ba2+ + SO42- BaSO4(s) 2 H+ H2(g) 2H+ + CO32- H2CO3(aq) H2CO3(aq) H2O(l) + CO2(g) Colloidal particles can be removed by dialysis (as in artificial kidneys). Colloidal dispersion shows Tyndal effect (light scatter). Resisting pH Changes – Buffer Solutions A buffer is a solution of a weak acid and its conjugate base. Addition of a strong acid or base to a buffer changes pH only slightly. Human blood (pH = 7.4) may not change by more than 0.4 pH units. Gastric juice has pH as low as 1.3. Only a fraction of a drop of gastric juice would change the pH of blood enough to kill you, if it were not buffered by HCO3- / CO32- system. Each acid has a conjugate base, and each base has conjugate acid. The difference is in one proton. A buffer replaces added strong acid with its weak acid, or added strong base with its weak base. OH- + HC2H3O2 C2H3O2- + H2O Added base Weak acid in buffer Conjugate weak base Neutral water H3O+ + C2H3O2- HC2H3O2 + H2O Added acid Weak base in buffer Conjugate weak acid Neutral water Weak acid conjugate has all properties of a base. Strong acid conjugate is neutral. In acetate buffer, the weak acid, HC2H3O2 goes after any added OH- ions. Simultaneously, the acid conjugate, C2H3O2- goes after any added H3O+. No buffer entirely cancels out the effect of adding strong acid or base. pH still changes, but a lot less than in water. Buffer is exhausted when either weak acid or its conjugate gets used up. HW (p.401-403) 1, 3, 5(a-c), 13, 15, 33 What to remember: -Solubility, soluble ions and effect of P,T -Define saturated soln -Define molarity and molality of a solution -Recognize electrolyte -Define strong/weak electrolyte -Define hydronium and hydroxide ions -Define acid and base -Concentration of hydronium/hydroxide ions in pure water -Define pH -Formula eq, total ionic and net ionic eq. -Define buffer Note that you eat acidic, but not basic compounds!