Chem 115 Chapter 1-9 Notes PDF

Chapter 1: The Basics of Chemistry  Outline  Introduction: What is Chemistry  The Properties and Transformations of Matter  The Scientific Method  The Units of Measurement  Scientific Notation  Significant Figures  Dimensional Analysis Introduction: What is Chemistry?  Chemistry is the...

Chapter 1: The Basics of Chemistry  Outline  Introduction: What is Chemistry  The Properties and Transformations of Matter  The Scientific Method  The Units of Measurement  Scientific Notation  Significant Figures  Dimensional Analysis Introduction: What is Chemistry?  Chemistry is the study of matter, its properties, and the mechanisms involved in its changes. a  Matter and Its Properties and Transformations  Matter is any substance that has mass and occupies space  The fundamental building blocks of matter are atoms, which cannot be chemically broken down into smaller components  Atoms combine to form larger molecules and compounds  The States of Matter o Matter comes in three different forms, or states, known as gases, solids and liquids  Q1.01 Pair each trait with the appropriate physical state. Premise Response 1 Solids A Atoms closely packed together 2 Liquids B Are highly dispersed 3 Gases C volume but no shape  Q1.01 Pair each trait with the appropriate physical state. Premise Response 1 Solids A Atoms closely packed together 2 Liquids C Have volume but no shape 3 Gases B Are highly dispersed Classification of Matter  Q1.02 Identify the following as either an element or a compound. [Image description: The image is of alternating sheets of purple and green spheres]. A Element B Compound  Q1.02 Identify the following as either an element or a compound. [Image description: The image is of alternating sheets of purple and green spheres]. A Element B Compound Explanation Spheres of two different colors indicate two different elements. Throughout the entire cube, a regular pattern repeats, showing the presence of a pure compound. The pattern is an alternating layer of purple and green atoms.  Q1.03 Identify the following as either an element or a compound. [Image description: The image is of stacked rows of red spheres. A Element B Compound  Q1.03 Identify the following as either an element or a compound. [Image description: The image is of stacked rows of red spheres. A Element B Compound Explanation Spheres of the same color indicate only one element. Additionally, the repeating pattern consists of layers of red spheres. Mixtures Homogeneous Heterogeneous Pure Substances Elements already consist of atomic units and cannot be broken down any further Compounds are combinations of more than one type of element and can be broken down into individual atomic units  Q1.04 Identify the following as either a homogeneous or heterogeneous mixture. [Image description: The image is of a coffee cup with swirls of white cream amidst the brown coffee. A Homogeneous mixture B Heterogeneous mixture  Q1.04 Identify the following as either a homogeneous or heterogeneous mixture. [Image description: The image is of a coffee cup with swirls of white cream amidst the brown coffee. A Homogeneous mixture B Heterogeneous mixture Explanation There are two separate layers in a cappuccino, indicating a nonuniform distribution of multiple compounds.  Q1.05 Identify the following as either a homogeneous or heterogeneous mixture. [Image description: The image is of 7 red spheres and 12 blue spheres dispersed randomly within a square.] A Homogeneous mixture B Heterogeneous mixture  Q1.05 Identify the following as either a homogeneous or heterogeneous mixture. [Image description: The image is of 7 red spheres and 12 blue spheres dispersed randomly within a square.] A Homogeneous mixture B Heterogeneous mixture Explanation Two different colors of spheres indicate atoms of two different elements. The figure shows that they are uniformly dispersed.  Q1.06 Identify each figure as either an element, compound, homogeneous mixture, or heterogeneous mixture. [Image description: Image I is of a beaker with blue spheres loosely arranged within it and slight spaces in between the spheres. Image II is of a cube consisting of alternating red and blue spheres closely packed in an organized fashion. Image III is of a box containing molecules, each composed of one red sphere and two white spheres.] Premise Response 1 I A Element 2 II B Heterogenous mixture 3 III C Homogeneous mixture D Compound  Q1.06 Identify each figure as either an element, compound, homogeneous mixture, or heterogeneous mixture. [Image description: Image I is of a beaker with blue spheres loosely arranged within it and slight spaces in between the spheres. Image II is of a cube consisting of alternating red and blue spheres closely packed in an organized fashion. Image III is of a box containing molecules, each composed of one red sphere and two white spheres.] Premise Response 1 I A Element 2 II D Compound 3 III D Compound B Heterogeneous mixture C Homogeneous mixture Explanation I)Spheres of the same color indicate only one element. II)Spheres of two different colors indicate two different elements. Throughout the entire cube, a regular pattern repeats, showing the presence of a pure compound. III)Spheres of two different colors indicate two different elements. Throughout the entire figure, exactly one red sphere is connected to two white spheres, showing the presence of a pure compound.  Physical and Chemical Changes of Matter  Physical properties of matter can be expressed or observed without a change in chemical composition. e.g. size, color, shape, texture, melting point, boiling point  Chemical properties of matter are expressed or observed only after a change in chemical composition has occurred. e.g. flammability, corrosiveness, acidity, oxidizing  Q1.07 Identify each of the following as a physical or chemical property. Premie Response 1 The coarseness of sand The tendency of fruit to easily turn brown 2 when exposed to air A Physical B Chemical  Q1.07 Identify each of the following as a physical or chemical property. Premie Response 1 The coarseness of sand A Physical The tendency of fruit to easily turn brown 2 B Chemical when exposed to air Explanation The coarseness of sand refers to its physical texture. Fruit turns brown (i.e., spoils) as the result of a chemical reaction with oxygen (an oxidation reaction). Physical and Chemical Changes of Matter Physical changes of matter retain chemical composition e.g. phase changes, mixing, size changes, dissolving Chemical changes of matter alter chemical composition e.g. combustion, rusting, spoiling  Q1.08 Identify each of the following as a physical or chemical change. Premise Response 1 Sugar mixed very well with coffee Gaseous vapor observed when handling 2 dry ice (solid CO ) A Physical B Chemical Explanation When sugar dissolves in coffee, it still remains as sugar. Its chemical composition is not altered. When dry ice gas (CO ) is released, it sublimes; that is, it changes directly from a solid to a gas.  Q1.08 Identify each of the following as a physical or chemical change. Premise Response 1 Sugar mixed very well with coffee A Physical Gaseous vapor observed when handling 2 A Physical dry ice (solid CO ) B Chemical B Chemical Explanation When sugar dissolves in coffee, it still remains as sugar. Its chemical composition is not altered. When dry ice gas (CO ) is released, it sublimes; that is, it changes directly from a solid to a gas. The Scientific Method  Q1.10 Identify each statement as either a law or a theory. Premise Response Two bodies attract each other with a force that is directly proportional to their 1 A Law masses but inversely related to their distance. Through a process known as “panspermia¸" microorganisms on 2 asteroids crash-landed on Earth 4 billion B Theory years ago¸ thus seeding the planet with the origins of life.  Q1.10 Identify each statement as either a law or a theory. Premise Response Two bodies attract each other with a force that is directly proportional to their 1 A Law masses but inversely related to their distance. Through a process known as “panspermia¸" microorganisms on 2 asteroids crash-landed on Earth 4 billion B Theory years ago¸ thus seeding the planet with the origins of life. Explanation Law: Two bodies attract each other with a force that is directly proportional to their masses but inversely related to their distance. (This statement summarizes a property of physical objects but offers no explanation). Theory: Through a process known as “panspermia,” microorganisms on asteroids crash-landed on Earth 4 billion years ago, thus seeding the planet with the origins of life. (This statement explains an observation by providing the underlying mechanism). The Units of Measurement Property English units SI units Mass pounds (lbs) kilogram (kg) Length yard (yd) Meter (m) Volume gallon (gal) Cubic Meter (m^3) Temperature Fahrenheit (°F) Kelvins (K) The Temperature Scales Prefix Multipliers  Q1.11 Rank the following in order of decreasing length. A 100 nm B 100 cm C 100 km D 100 m  Q1.11 Rank the following in order of decreasing length. C 100 km D 100 m B 100 cm A 100 nm Scientific Notation Scientific notation allows us to conveniently express very small or very large numbers. If the power > 1, the overall magnitude is >1 for positive values. 4321.1 = 4.3211x103 = 4.3211E3 If the power < 1, the overall magnitude is between 0 and 1 for positive values. 0.053244 = 5.3244x10-2 = 5.3244E-2  Q1.14 Convert 4.653x10-2 to a decimal number.  Q1.14 Convert 4.653 × 10 to a decimal number. 0.04653 Explanation The decimal is moved 2 places to the left. Significant Figures: A Measurement of Precision 2.55 inches 2.6 inches 19.9°C, 66 °F Report all measurements to 1 digit past the last digit of certainty Accuracy versus Precision ▪ Accuracy refers to the proximity of a measurement to the true value of a quantity. ▪ Precision refers to the proximity of several measurements to each other.  Q1.17 Which ruler would give the more reproducible result? A Top B Bottom  Q1.17 Which ruler would give the more reproducible result? A Top B Bottom Explanation The ruler with the greater number of increments is more reliable.  Q1.18 Could it be recorded as 2.54 cm? A Yes B No  Q1.19 Could it be recorded as 2.54 cm? A Yes B No Explanation 2.54 cm is acceptable because it is between 2.5 and 2.6 cm.  Q1.22 What is an appropriate measurement of the liquid in the following 10 mL graduated cylinder? Enter the numerical value without units. [Image description: The figure is of a graduated cylinder partially filled with liquid. The bottom of the meniscus is between 6.6 and 6.8 mL.]  Q1.22 What is an appropriate measurement of the liquid in the following 10 mL graduated cylinder? Enter the numerical value without units. [Image description: The figure is of a graduated cylinder partially filled with liquid. The bottom of the meniscus is between 6.6 and 6.8 mL.] 6.57E0 to 6.67E0 Explanation Read to 2 decimal places for the 10 mL graduate cylinder.  Q1.23 The __________ number of digits in a measurement, the more precise the reading. A greater B fewer  Q1.23 The __________ number of digits in a measurement, the more precise the reading. A greater B fewer Identifying Significant Figures: Some Rules All nonzero digits are significant, including all digits for a value expressed in scientific notation. For example in the number 432, there are (3) sig figs. In 1.1450×105, there are (5) sig figs. A zero is significant if it is in-between nonzero digits. For example, in the number 50002, all (5) digits are significant. A zero is significant if it is a trailing zero to the right of a decimal number. For example, in the number 432.00, all (5) digits are significant. Identifying Significant Figures: Some Rules A zero is not significant if it comes at the end of a number without a decimal. For example, in the number 4130, the zero does not count and there are (3) sig figs, not (4). A zero is not significant if it starts a decimal number less than 1. For example, in 0.0004312, none of the zeroes count and there are only (4) sig figs.  Q1.24 How many sig figs are in 300.123940?  Q1.24 How many sig figs are in 300.123940? 9 Explanation The two zeros after the 3 count since they are between nonzero digits (rule 2). The last zero counts since it is a trailing zero of a decimal number (rule 3). Exact Numbers Quantities whose precision are considered infinite are known as exact numbers. Equality statements and conversion factors. 4 laps = 1 mile 2.54 cm = 1 inch Countable quantities. 32 students 9 cars  Q1.25 Carry out the following calculation with the correct number of sig figs: 431.00791 + 52.5441 431.00791 + 52.54410 483.5520.  Q1.25 Carry out the following calculation with the correct number of sig figs: 431.00791 + 52.5441 483.5520 Explanation There are four decimal places in 52.5441 that determine the number of decimal places in the answer. 483.55201 ≈ 483.5520 We round down because 1 < 5.  Q1.26 Report 0.02315621 to 3 sig figs.  Q1.26 Report 0.02315621 to 3 sig figs. 0.0232 Explanation Three sig figs would be 0.023. The digit to the immediate right of the one in 0.231 is ≥ 5, so we round up.  Q1.27 The last operation is multiplication, so we should go by: A fewest sig figs. B fewest decimal places.  Q1.27 The last operation is multiplication, so we should go by: A fewest sig figs. B fewest decimal places. Explanation In a multiplication step, the number of significant figures in the answer equals the fewest number of significant figures present. Significant Figures in Calculations For addition and subtraction: The number of decimal places in the final answer equals the fewest number of decimal places from the combined numbers. 1.34561 – 0.341 = 1.00461 ~ 1.005 (5 decimal places) (3 places) (report answer to 3 places)  Q1.28 Carry out the following calculation with the correct number of sig figs. (431.00791 + 52.5441)/45.3200130 483.5520/45.3200130  Q1.28 Carry out the following calculation with the correct number of sig figs. (431.00791 + 52.5441)/45.3200130 10.66972 Explanation The sum in the numerator is 483.55201. Because this is addition, we go by the fewest decimal places that need to be marked: 483.5520. Dividing this by the denominator, we get 10.66972355. The last step was division, which goes by the fewest sig figs present. 483.5520 has seven sig figs, whereas 45.3200130 has ten sig figs. So, the final answer is reported to seven sig figs: 10.66972355 ~ 10.66972. Significant Figures in Calculations For multiplication and division: The number of sig figs in the final answer equals the fewest number of sig figs present from the contributing factors. 3.451 × 0.00671 = 0.02315621 ~ 0.0232 (4 sf) (3 sf) (report answer to 3 sf)  Q1.29 Carry out the following calculation with the correct number of sig figs. (24.324)(3.2122) – 9.3210  Q1.29 Carry out the following calculation with the correct number of sig figs. (24.324)(3.2122) – 9.3210 68.813 Explanation Significant Figures in Calculations For combined mathematical operations: Carry all digits until the end. “Tag” estimated digits after each step, but do not round until the very end. (1.32451 – 0.341)(1.0054) 0.98351(1.0054) 0.988820954 ~ 0.989 Dimensional Analysis ▪ We use dimensional analysis to convert one quantity to another. ▪ Most commonly, dimensional analysis utilizes conversion factors (e.g., 1 in. = 2.54 cm) 1 in. 2.54 cm or 2.54 cm 1 in. Dimensional Analysis Use the form of the conversion factor that puts the sought-for unit in the numerator: desired unit Given unit  = desired unit given unit ©01 Conversion factor 2 Pear son Edu cati Dimensional Analysis: The Use of Conversion Factors Unit A → Unit B → Unit C Let us convert $2.70/gal to ¢/L First, convert gallons into liters using the equality of 1 gallon = 3.7854 Liters Then, convert dollars into cents using the equality of 100¢ = $1 Dimensional Analysis: The Use of Conversion Factors Unit A → Unit B → Unit C Convert 8.91 inches to centimeters, where 1 inch = 2.54 cm Convert 324 centimeters to inches, where 1 inch = 2.54 cm Dimensional Analysis and Prefix Multipliers Convert 321 nm to m. In the conversion factor, the prefix multiplier goes with the prefixless unit. Convert 341 dL to mL. Dimensional Analysis and Prefix Multipliers Convert 32 in3 to cm3. Sometimes, you have to make your own conversion factor. Given: 1 inch = 2.54 cm (1 in)3 = (2.54 cm)3 13 in3 = 2.543 cm3 1 in3 = 16.4 cm3  Q1.30 Your car gets 37.0 miles per gallon (mpg) on average, and gas costs $2.70/gal. How much will you spend on fuel to travel 405 miles (mi)? Enter just the numerical value without the dollar sign.  Q1.30 Your car gets 37.0 miles per gallon (mpg) on average, and gas costs $2.70/gal. How much will you spend on fuel to travel 405 miles (mi)? Enter just the numerical value without the dollar sign. 29.6 Explanation  Q1.31 Convert 0.152 kilometers (km) to meters (m). Enter just the numerical value without units.  Q1.31 Convert 0.152 kilometers (km) to meters (m). Enter just the numerical value without units. 152  Q1.32 Convert 2341 mm to nanometers. Enter just the numerical value without units.  Q1.32 Convert 2341 mm to nanometers. Enter just the numerical value without units. 2.341x109 Explanation First convert 2341 mm to meters. The conversion factor is 10 meters on top and 1 mm on the bottom. Then, convert meters to nanometers. The conversion factor is 1 nm on top and 10 meters on the bottom.  Q1.33 Convert 341 deciliters (dL) to milliliters (mL). Enter just the numerical value without units.  Q1.33 Convert 341 deciliters (dL) to milliliters (mL). Enter just the numerical value without units. 34100 Explanation  Q1.34 How many cubic meters are in 2.43 yd3 ?  Q1.34 Hide Correct Answer Show Responses How many cubic meters are in 2.43 yd ? 1.86 Explanation Density A substance’s density is the ratio of its mass to its volume. Density is a physical property that is unique to each substance and is temperature dependent. At 20°C, the density of liquid water is 0.99823 g/mL. Density Density can be used as a conversion factor to relate mass and volume. How many mL does 32.1 grams of water occupy at 20°C?  Q1.35 How many grams of water are contained in a 0.127 L sample at 20 °C?  Q1.35 How many grams of water are contained in a 0.127 L sample at 20 °C? 127 Explanation Density: Will an object float or sink in water? Substances with a density < water’s will float. Substances with a density > water’s will sink.  Q1.36 A typical water displacement experiment involves adding the object to a known volume of water. The volume of water displaced equals which of the following? A The sum of the initial and final volumes B The difference between the initial and final volumes  Q1.36 A typical water displacement experiment involves adding the object to a known volume of water. The volume of water displaced equals which of the following? A The sum of the initial and final volumes B The difference between the initial and final volumes Explanation The volume of water displaced equals the difference between the initial and final volumes. For example: if the initial volume is 5 mL and the final volume is 8 mL, the volume of water displaced is 3 mL. Density: Method of Water Displacement The volume of water that an object displaces equals the volume of the object. A 5.0 g object is placed into a graduated cylinder with water. 90.0 mL – 65.0 mL = 25.0 mL Density = 5.0 g / 25.0 mL = 0.20 g/mL  Q1.37 A 3.24 g object is immersed into a graduated cylinder filled with water. Use the following figure to determine the object’s density. Enter just the numerical value without units. [Image description: The figure is of two graduated cylinders. The graduated cylinder on the left is filled to between 64 and 65 mL with liquid. The graduated cylinder on the right has had a chunk of solid submerged in the liquid, and the total volume is now between 89 and 90 mL.]  Q1.37 A 3.24 g object is immersed into a graduated cylinder filled with water. Use the following figure to determine the object’s density. Enter just the numerical value without units. [Image description: The figure is of two graduated cylinders. The graduated cylinder on the left is filled to between 64 and 65 mL with liquid. The graduated cylinder on the right has had a chunk of solid submerged in the liquid, and the total volume is now between 89 and 90 mL.] 0.130 Explanation Atomic Theory of Matter The theory that atoms are the fundamental building blocks of matter reemerged in the early 19th century, championed by John Dalton.  Chapter 2: Matter at the Atomic Level Outline Introduction Early Atomic Theory J.J. Thompson The Nuclear Model of the Atom Atomic Number Ions Atomic Mass and Isotopes The Periodic Table The Mole and Avogadro’s Number Dalton’s Postulates 1. Each element is composed of extremely small particles called atoms. Dalton’s Postulates 2. All atoms of a given element are identical to one another in mass and other properties, but the atoms of one element are different from the atoms of all other elements. Dalton’s Postulates 3. Atoms of an element are not changed into atoms of a different element by chemical reactions; atoms are neither created nor destroyed in chemical reactions. Dalton’s Postulates 4. Compounds are formed when atoms of more than one element combine; a given compound always has the same relative number and kind of atoms.  Early Atomic Theory: John Dalton (1808)  Matter consists of atoms, which are the smallest identifiable unit of matter and cannot be destroyed.  All atoms of the same element are identical to each other, but are different than atoms of another element.  Atoms combine in small whole-number ratios when forming compounds. Is it possible that any of these premises might no longer be accepted? A. Yes B. No Early Atomic Theory  Law of Conservation of Matter The total mass of substances present at the end of a chemical process is the same as the mass of substances present before the process took place. Matter is neither created or destroyed in simple chemical reactions. (Nuclear reactions are a different matter where the mass does get converted to energy). Early Atomic Theory  Law of Multiple Proportions  In chemistry the law of multiple proportions is one of the basic laws and a major tool of chemical measurement (stoichiometry)  It states that when two elements combine, they do so in a ratio of small whole numbers. For example, carbon and oxygen react to form CO or CO2, but not CO1.8. and..  If two elements can form more than one compound, then the ratios of the weights of one element in the compounds to a fixed weight of the other element are small whole numbers Early Atomic Theory  Law of Multiple Proportions Water (H2O) forms when 2.02 grams of hydrogen combine with 16.00 grams of oxygen Hydrogen peroxide (H2O2) forms when 2.02 grams of hydrogen combine with 32.00 grams of oxygen Ratio of oxygen mass that can combine with a fixed mass of hydrogen is 2:1 Sulfur (S) forms multiple compounds with oxygen. When forming sulfur dioxide (SO2), 3:2 32.066 g of sulfur react with 32.00 g of oxygen. When forming sulfur trioxide (SO3), 32.066 g of sulfur reactsukl with 48.00 g of oxygen. What is the correct ratio of the two possible masses of oxygen that can react with a fixed mass of sulfur? A. 2:1 B. 1:1 C. 3:2 D. 1:2 Early Atomic Theory  Law of Definite Proportions (aka Law of Constant Composition) The same compound will always be comprised of its constituent elements in the same proportion by mass. Water always has a ratio of 1.01 grams of hydrogen for every 8.00 grams of oxygen. Water is always 11.2% hydrogen by mass. The remaining 88.8% is oxygen. In a 200 g sample of water, a chemist determines there are 22.4 grams of hydrogen and 177.6 grams of oxygen. Does this follow Proust’s law of definite proportions? A. YES B. NO  Q2.06 In a 200 g sample of water, a chemist determines there are 22.4 grams of hydrogen and 177.6 grams of oxygen. Does this follow Proust’s law of definite proportions? A. Yes Explanation Yes. We have to calculate the percent by mass of each element. This is essentially the part divided by the whole. For hydrogen: 22.4 g/200 g × 100%= 11.2%. For oxygen: 177.6 g / 200 g × 100 = 88.8% = 100% - 11.2%. Hence, the data follows the law of definite proportions. J.J. Thompson and the Electron In 1909 Millikan experimentally determined the exact charge and mass of an electron. J.J. Thompson (1897) Discovered the Negatively charged particles in all elements Conclusions drawn so far: 1. Cathode ray consists of negatively charged particles and electrons are constituents of all matter. 2. The charge on electrons is found to be 1.602 x 10 -19 coulombs. Atomic Structure ❑ Discovery of Electrons: ❑ J J Thomson ( in 1897) ❑ He gave the charge to mass ratio of electrons. ❑ He also showed that the nature of the cathode ray did not depend on the material of the cathode. ▪ Now since a negative particle was discovered it was definite that the atom would also have an equal positive charge. ▪ Then at the time the most reasonable explanation seemed to be a plum pudding model Or should we say blueberry muffin! Thompson’s Model of the Atom (aka Plum-Pudding Model) An atom consists of a spherical volume of space that is positively charged. Electrons are dispersed throughout the sphere. The amount of negative charge equals the positive charge, resulting in atoms being inherently neutral. ▪Rutherford explained the existence of a focused positively charged nucleus in the atom. Ernest Rutherford’s Gold Foil Experiment (1911) Every atom has a positively charged dense center called the nucleus. Essentially all of an atom’s mass is found within its nucleus. The source of the deflection has to be what charge? B A. Negative B. Positive  The source of the deflection has to be what charge? A.Negative A. Negative B. Positive Explanation It has to be positive because then it will repel, or deflect, other positive particles, like α particles. The Nuclear Atom ▪ Rutherford postulated a very small, dense nucleus with the electrons around the outside of the atom. ▪ Most of the volume of the atom is empty space.  The Nuclear Atom ▪ Rutherford postulated a very small, dense nucleus with the electrons around the outside of the atom. ▪ Most of the volume of the atom is empty space. The Nuclear Model of the Atom  https://nam10.safelinks.protection.outlook.com/?url=https%3A%2F%2Fwww.f acebook.com%2Freel%2F822129376126529%3Ffs%3De%26s%3DaEkTS0%26 mibextid%3DIUR5lV&data=05%7C01%7Cmridula.satyamurti%40umb.edu%7 C779f673cc3e54bcd5d2708dbb8a1e207%7Cb97188711ee94425953c1ace13 73eb38%7C0%7C0%7C638306777650307283%7CUnknown%7CTWFpbGZsb 3d8eyJWIjoiMC4wLjAwMDAiLCJQIjoiV2luMzIiLCJBTiI6Ik1haWwiLCJXVCI6Mn0 %3D%7C3000%7C%7C%7C&sdata=eom2oMs9zjiH%2BBKRgREKq%2B4d5Qo HsZqpAn1sYMQu9eM%3D&reserved=0 Radioactivity: ▪ The spontaneous emission of radiation by an atom. ▪ First observed by Henri Becquerel. ▪ Also studied by Marie and Pierre Curie. Becquerel had inherited from his father, a supply of uranium salts, which phosphoresce on exposure to light. When the salts were placed near to a photographic plate covered with opaque paper, the plate was discovered to be fogged. The phenomenon was found to be common to all the uranium salts studied and was concluded to be a property of the uranium atom. Later, Becquerel showed that the rays emitted by uranium, caused gases to ionize and that they differed from X-rays in that they could be deflected by electric or magnetic fields. For his discovery of spontaneous radioactivity Becquerel was awarded half of the Nobel Prize for Physics in 1903, the other half being given to Pierre and Marie Curie for their study of the Becquerel radiation. https://www.nobelprize.org/prizes/physics/1903/becquerel/facts/ Radioactivity ▪ Three types of radiation were discovered by Ernest Rutherford: ▪  particles ▪  particles ▪  rays ▪ Dalton Atomic theory ▪ J J Thomson Electrons and their charge to mass ratio ▪ Millikan Charge on the electron and later the mass ▪ Henry Becquerel Radioactivity ▪ Rutherford α β and γ radiation Gold leaf experiment to show the presence of nucleus and coined the word the proton ▪ James Chadwick Neutrons Subatomic Particles ▪ Protons and electrons are particles that have a charge. ▪ Protons and neutrons have essentially the same mass. ▪ The mass of an electron is so small we ignore it. Symbols of Elements Elements are symbolized by one or two letters. Atomic Number All atoms of the same element have the same number of protons: The atomic number (Z) Atomic Mass The mass of an atom in atomic mass units (amu) is the total number of protons and neutrons in the atom. Atomic Number An atom’s atomic number is the number of protons within its nucleus. Because atoms are inherently neutral, atomic number also equals the number of electrons (neutral atoms only). On the periodic table, elements are ordered by atomic number. Isotopes: ▪ Atoms of the same element with different masses. ▪ Isotopes have different numbers of neutrons. 11 12 13 14 6C 6C 6C 6C Atomic Mass and Isotopes Isotopes are atoms of the same element that differ in their number of neutrons. An isotope’s atomic mass number is the sum of its number of protons and neutrons: Atomic mass number = number of protons + number of neutrons Carbon has three naturally occurring isotopes: C-12, C-13, C-14. The number after the hyphen is the atomic mass number Another common way to represent isotopes: 12C, 13C, and 14C Atomic Mass and Isotopes Not all isotopes are created equally in nature because the same number of proton can have different number of neutrons. They are different isotopes of the same element. An isotope’s percent natural abundance is a measure of the average amount the isotope naturally occurs relative to all of the 12C 13C 14C element’s isotopes. 98.93% 1.07% trace 98.93% of any sample of carbon will be C-12, while 1.07% of the same sample will be C-13. Average Mass ▪ The unit for atomic mass is amu or just u ▪ Because in the real world we use large amounts of atoms and molecules, we use average masses in calculations. ▪ Average mass is calculated from the isotopes of an element weighted by their relative abundances. Atomic Mass and Isotopes An element’s atomic mass is a weighted average of all of the masses of its naturally occurring isotopes. atomic mass= ∑(isotope fractional abundance)(exact isotope mass) Naturally occurring chlorine is 75.78% 35Cl, which has an atomic mass of 34.969 amu, and 24.22% 37Cl, which has an atomic mass of 36.966 amu. Calculate the average atomic mass (that is, the atomic weight) of chlorine Naturally occurring chlorine is 75.78% 35Cl, which has an atomic mass of 34.969 amu, and 24.22% 37Cl, which has an atomic mass of 36.966 amu. Calculate the average atomic mass (that is, the atomic weight) of chlorine 75.78% = 0.7578 and 24.22% = 0.2422, Average atomic mass = (0.7578)(34.969 amu) + (0.2422)(36.966 amu) = 26.50 amu + 8.953 amu = 35.45 amu Three isotopes of silicon occur in nature: 28Si (92.23%), which has an atomic mass of 27.97693 amu; 29Si (4.68%), which has an atomic mass of 28.97649 amu; and 30Si (3.09%), which has an atomic mass of 29.97377 amu. Calculate the atomic weight of silicon. Naturally occurring chlorine is 75.78% 35Cl, which has an atomic mass of 34.969 amu, and 24.22% 37Cl, which has an atomic mass of 36.966 amu. Calculate the average atomic mass (that is, the atomic weight) of chlorine 75.78% = 0.7578 and 24.22% = 0.2422, Average atomic mass = (0.7578)(34.969 amu) + (0.2422)(36.966 amu) = 26.50 amu + 8.953 amu = 35.45 amu Three isotopes of silicon occur in nature: 28Si (92.23%), which has an atomic mass of 27.97693 amu; 29Si (4.68%), which has an atomic mass of 28.97649 amu; and 30Si (3.09%), which has an atomic mass of 29.97377 amu. Calculate the atomic weight of silicon. Answer: 28.09 amu Isobars Isobars are nuclides of different elements (different Z) with the same mass number (A). Isobars have nearly the same mass. 40S, 40Cl, 40Ar, 40K and 40Ca Modern Periodic Table ▪ A systematic catalog of elements. ▪ Elements are arranged in order of atomic number.  Periodicity When one looks at the chemical properties of elements, one notices a repeating pattern of reactivities.  Periodic Table ▪ The rows on the periodic chart are called periods. ▪ Columns are called groups. ▪ Elements in the same group have similar chemical properties.  Periodic Table ▪ The rows on the periodic chart are called periods. ▪ Columns are called groups. ▪ Elements in the same group have similar chemical properties. Introduction to the Periodic Table  Periodic Table Nonmetals are on the right side of the periodic table (with the exception of H).  Periodic Table Metalloids border the stair-step line (with the exception of Al and Po). Periodic Table ▪ Metals are on the left side of the chart ▪ Note that there are more metals than nonmetals Octet rule  Q2.12 Sodium one electron to become a +1 cation. A loses B gains  Q2.12 Sodium one electron to become a +1 cation. A loses Explanation Let’s go ahead and see why this makes sense. Sodium is element no. 11. A neutral sodium atom has 11 protons and 11 electrons. If the number of protons changed, then it would no longer be sodium. By losing one electron, the overall charge becomes (+11) + (−10) = +1, that is Na → Na+1 + e-1 Consider magnesium (Mg). How many protons (p) and electrons (e) are in a Mg+2 cation? A 12 p, 12 e B 12 p, 10 e C 10 p, 12 e D 10 p, 10 e  Q2.13 Consider magnesium (Mg). How many protons (p) and electrons (e) are in a Mg^{2+} cation? A 12 p, 12 e B 12 p, 10 e C 10 p, 12 e D 10 p, 10 e Explanation Magnesium is element no. 12. A neutral magnesium atom has 12 p and 12 e. Magnesium typically forms a +2 cation. By losing two electrons, magnesium adopts an overall charge of (+12) + (−10) = +2, that is Mg → Mg^{2+} + 2 e^-. Hence, atoms can lose electrons to become cations.  Q2.14 How many protons and electrons are in an O-2 anion ? A 6 p, 8 e B 6 p, 6 e C 8 p, 6 e D 8 p, 8 e E 8 p, 10 e  Q2.14 How many protons and electrons are in an O-2. anion? A. 6 p, 8 e B 6 p, 6 e C 8 p, 6 e D 8 p, 8 e E 8 p, 10 e Explanation Atoms gain electrons to form anions. By gaining two electrons, oxygen achieves an overall charge of (+8) + (−10) = −2, that is O + 2 e^-→ O^{2-}.  Q2.15 How many protons and electrons are in an N3- anion? A 7 p, 7 e B 7 p, 10 e C 10 p, 10 e D 10 p, 7 e  How many protons and electrons are in an N3-anion? A 7 p, 7 e B 7 p, 10 e C 10 p, 10 e D 10 p, 7 e Explanation By gaining three electrons, nitrogen achieves an overall charge of (+7) + (−10) = −3, that is N + 3 e^- → N^{3-}. The following ions represent an isoelectronic series (i.e., all have the same number of electrons). Rank the ions in order of increasing electronic charge with the least electronic charge at the top and the most at bottom. A Nitrogen (N) with 10 e B Fluorine (F) with 10 e C Oxygen (O) with 10 e  The following ions represent an isoelectronic series (i.e., all have the same number of electrons). Rank the ions in order of increasing electronic charge with the least electronic charge at the top and the most at bottom. B Fluorine (F) with 10 e C Oxygen (O) with 10 e A Nitrogen (N) with 10 e Explanation Fluorine has 9 p and 9 e. F^- has 9 p and 10 e. Oxygen has 8 p and 8 e. O^{2-} has 8 p and 10 e. Nitrogen has 7 p and 7 e. N^{3-} has 7 p and 10 e. So the order of increasing electronic charge is F with 10 e, O with 10 e, N with 10 e. Bromine (Br, atomic mass 79.90 amu) has two naturally occurring isotopes79Br and 81Br. 79Br has an exact mass of 78.9183371 amu and is 50.69% abundant. Calculate the exact isotope mass of 81Br. Bromine (Br, atomic mass 79.90 amu) has two naturally occurring isotopes79Br and 81Br. 79Br has an exact mass of 78.9183371 amu and is 50.69% abundant. Calculate the exact isotope mass of 81Br. Atomic mass= Σ(isotope fractional abundance)(isotope mass). 79.90 amu =(0.5069)(78.9183371 amu)+ (1-0.5069) (isotope mass of 81Br) 79.90 – (40.0037051) = 0.4931 (isotope mass of 81Br) 39.8962949/0.4931 = 80.909136 isotope mass of 81Br 81Br = 80.91 amu. Metals versus Nonmetals Metals are characterized by the following: Malleable: can be formed into thin sheets Ductile: can be pulled into thin wires Lustrous: have a shiny appearance Heat and electricity conductors Relatively large densities Relatively high melting and boiling points All are solids at room temperature except mercury, Hg Tend to form cations Very few colors: grey, silver, gold/copper Metals versus Nonmetals Nonmetals tend to have the following traits: Brittle: deform easily, break upon impact, not malleable or ductile Dull appearance Insulators: poor conductors of heat and electricity Relatively lower densities, melting and boiling points than metals Various physical states at room temperature Tend to form anions Variety of colors Metalloids Label each property of silicon from the premise columns as metallic or nonmetallic. Premise 1 Lustrous 2 Brittle 3 Good conductor Response 4 High melting point E Metallic G Nonmetallic  Q2.21 Label each property of silicon from the premise columns as metallic or nonmetallic. Premise Response 1 Lustrous E Metallic 2 Brittle G Nonmetallic 3 Good conductor E Metallic 4 High melting point E Metallic Explanation Silicon is known as a metalloid, meaning it possesses intermediate characteristics of metals and nonmetals. Groups These five groups are known by their names. Group IA: The Alkali Metals Soft, dull metals (grey-white) All react violently with water Group IIA: The Alkaline Earth Metals Harder than Group IA React less violently with water than Group IA https://www.youtube.com/watch?v=M8Eovgaa2QE https://www.youtube.com/watch?v=eD9BiYvsT4k  Q2.25 Identify the column containing the halogens.  Q2.25 Hide Correct Answer Show Responses Identify the column containing the halogens. Explanation The halogens are located in the 7 A or the 17th column. Group VIIA: The Halogens Naturally occur as diatomic compounds All are very reactive Group VIIIA: The Noble Gases Naturally occur as colorless, monatomic gases All are relatively inert, or unreactive  Diatomic Molecules These seven elements occur naturally as molecules containing two atoms.  __________ are found uncombined, as monatomic species in nature. A) Noble gases B) Halogens C) Alkali metals D) Alkaline earth metals E) Chalcogens Allotropes ▪ Many elements like C, O and S exist in more than one form in nature. Oxygen : dioxygen O2 - colorless ozone O3 - blue Carbon: ▪ Diamond - an extremely hard, transparent crystal, with the carbon atoms arranged in a tetrahedral lattice. A poor electrical conductor. An excellent thermal conductor. ▪ Graphite - a soft, black, flaky solid, a moderate electrical conductor. The C atoms are bonded in flat hexagonal lattices, which are then layered in sheets. ▪ amorphous carbon ▪ fullerenes including "buckyballs", such as C60, and carbon nanotubes Phosphorus: White phosphorus - crystalline solid P4 Red phosphorus - polymeric solid Sulfur ▪Plastic (amorphous) sulfur - polymeric solid ▪Rhombic sulfur - large crystals composed of S8 molecules ▪Monoclinic sulfur - fine needle-like crystals ▪Other ring molecules such as S7 and S12 Chemical Formulas H2O, CO2, CO, CH4, H2O2,O2 ▪ The subscript to the right of the symbol of an element tells the number of atoms of that element in one molecule of the compound. Molecular Compounds Molecular compounds are composed of molecules and almost always contain only nonmetals. Types of Formulas ▪ Molecular formulas give the exact number of atoms of each element in a compound. ▪ Empirical formulas give the lowest whole-number ratio of atoms of each element in a compound. Types of Formulas ▪ Structural formulas show the order in which atoms are bonded. ▪ Perspective drawings also show the three- dimensional array of atoms in a compound. Ions ▪ When atoms lose or gain electrons, they become ions. ▪ Cations are positive and are formed by elements on the left side of the periodic chart. ▪ Anions are negative and are formed by elements on the right side of the periodic chart. Ions Ions are charged versions of atoms. Atoms lose electrons to become positively charged ions, called cations. Na → Na+ + e- 11 e- →10 e- Atoms gain electrons to become negatively charged ions, called anions. Mg → Mg2+ + 2e- 12 e- → 10 e- The Periodic Table and Ionic Charge IA VIIIA IIA IIIA VIA VA VIA VIIA Beryllium ▪ The chemistry of Beryllium is dominated by its tendency to lose two electrons to form Be2+. As this ion is so small it is highly polarizing to the neighboring atom, to the extent that its compounds are mostly covalent.  Q2.26 Identify the column containing the alkaline earth metals. Click on the column name (represented as a roman numeral) at the top. Identify the column containing the alkaline earth metals. Click on the column name (represented as a roman numeral) at the top. Explanation The alkaline earth metals are located in the 2nd column. Which element in the fourth row prefers to form a +3 cation? A K B Ca C Ga D Se  Q2.27 Which element in the fourth row prefers to form a +3 cation? A K B Ca C Ga D Se Explanation Gallium (Ga) prefers to form a +3 cation. Avogadro’s Number On the macroscopic scale, it is not practical for us to count atom by atom. Instead, we tend to represent bulk quantities in terms of grams, kilograms, etc. Avogadro’s Number 1 mole (mol) = 6.022 × 1023 individual units Avogadro’s Number To convert from atoms or molecules to moles, divide by Avogadro’s number: To convert from moles to atoms or molecules, multiply by Avogadro’s number: Molar Mass It turns out that an element’s atomic mass (in amu) directly transfers over to the macroscopic: Molar mass The mass of exactly 1 mole of an element (in units of g/mol) To convert from grams to moles, divide by molar mass: To convert from moles to grams, multiply by molar mass: Molar Mass ▪ By definition, these are the mass of 1 mol of a substance (i.e., g/mol) ▪ The molar mass of an element is the mass number for the element that we find on the periodic table ▪ The formula weight (in amu’s) will be the same number as the molar mass (in g/mol) Molar Mass and Avogadro’s Number ▪ Examples ▪ 125 g of Fe = ? Moles of Fe ▪ 125 g Fe x 1 mole Fe = 2.238 moles Fe 55.845 g Fe = 2.24 moles Fe ▪ 125 g of Fe = ? atoms of Fe ▪ 125 g Fe x 1 mole Fe x 6.02 x 10 BBatoms of Fe 55.845 g Fe 1 mole of Fe = 1.35 x 10 24 atoms of Fe ▪ 125 g NaCl = ? Moles of NaCl 2.14 Moles of NaCl ▪ 125 g NaCl x 1 mole of NaCl ? Grams of NaCl 58.44 grams of NaCl ▪ 10.00 x 1035 atoms of Cu = ? g of Cu ▪ A horseshoe made of iron weighing 16.8002 g contains how many moles of iron atoms? ▪ A) 0.0874000moles ▪ B) 3.320214 moles ▪ C) 0.64026 moles ▪ D) 0.1320 moles ▪ E) 0.301000 moles Answer E  Chapter 3: Molecules, Compounds, and Their Composition  Outline Introduction: Importance of understanding compound formulas and names Chemical formulas Ionic and covalent compounds Naming ionic compounds Naming molecular compounds Molar mass Mass percent Empirical formula Learning Objectives  Introduction Everything is made of chemicals! (a.k.a. chemical compounds) Every chemical compound has its own name and formula There are systematic naming rules. MN - Ionic Bonds NN - Mostly Ionic Bonds MM - Covalent Cation Anion Main gp Transition Chemical Formulas Systematic listing of the elements in a compound Elements are listed from most metallic to least metallic Subscripts indicate how many of an element are in the compound CO2 H2O NH3  Octet Rule o Rule of 8 (or 2 for very small atoms) o In 1904, Richard Abegg formulated that the difference between the maximum positive and negative valences of an element is frequently eight. o This rule was used later in 1916 when Gilbert N. Lewis formulated the "octet rule" in his cubical atom theory. o The octet rule refers to the tendency of atoms to prefer to have eight electrons in the valence shell. o When atoms have fewer than eight electrons, they tend to react and form more stable compounds. o Atoms tend to gain, lose, or share electrons so that each atom has a full outermost energy level, which typically consists of 8 electrons (called an octet) ______ form ions with a 2+ charge when they react with nonmetals. A) Alkali metals B) Alkaline earth metals C) Halogens D) Noble gases E) None of these choices ______ form ions with a 2+ charge when they react with nonmetals. A) Alkali metals B) Alkaline earth metals C) Halogens D) Noble gases E) None of these choices All ionic compounds are crystalline solids with high melting points.  Ionic Compounds Ionic compounds are between Metal and nonmetal Metal loses one or more electron(s) to become positively charged cation Na → Na+ Mg → Mg2+ Nonmetal gains one or more electron(s) to become negatively charged anion Cl → Cl– S → S2–  Naming Ionic Compounds Metals cations are named as such like sodium ion, potassium ion etc. but Monoatomic nonmetal ions end with an ide oxygen → oxide sulfur → sulfide nitrogen → nitride phosphorus → phosphide We get their charge from the periodic table. NaCl is sodium chloride Na2O is sodium oxide Na2S is sodium sulfide Na3P is sodium phosphide Naming Ionic Compounds containing transition metals For metals that can take on different charges like the transition metals and the heavier metals Charge is given in Roman numerals Charge can be determined from the other elements in compound For example FeO is called iron(II) oxide Fe2O3 is iron(III) oxide HgO is mercury(II) oxide Hg2O is mercury(I) oxide Common Cations Writing Formulas o Because compounds are electrically neutral, one can determine the formula of a compound this way: The charge on the cation becomes the subscript on the anion. The charge on the anion becomes the subscript on the cation. If these subscripts are not in the lowest whole-number ratio, divide them by the greatest common factor. Let us see how we can figure out the charges on the ions that make the compound- o FeO o What is the charge on the oxide ion- refer to your periodic table, see it’s position and see how many electrons are needed to complete the octet. o o Now see how many positive ions should be on iron to neutralize the charge on oxygen. o Fe2O3 o This one is even easier- o The charges are crisscrossed! o HgO o Now figure this one out using the first example o Hg2O o Now do this one … Which of the following is a transition metal and will have more than one kind of ion. A. Na B. F C. Cu D. Al E. Cs Poly atomic ions: 1. Only two cations: NH4+ and H3O+ 2. Anions are many In most chemical reactions the polyatomic ions will move as a whole. Polyatomic ions : These can be cations or anions Naming Ionic Compounds containing polyatomic ions For example; NaC2H3O2 = sodium acetate Na2CO3 = sodium carbonate Na3PO4 = sodium phosphate CuC2H3O2 = copper(I) acetate FeCO3 = iron(II) carbonate FePO4 = iron(III) phosphate Patterns in Oxyanion Nomenclature o When there are two oxyanions involving the same element: The one with fewer oxygens ends in -ite NO − : nitrite; SO 2− : sulfite 2 3 The one with more oxygens ends in -ate NO − : nitrate; SO 2− : sulfate 3 4 indicating that a chemical compound contains a high proportio n of a specified element: peroxide ; perchloride. indicating that a chemical element is in a higher than usual state of oxidation: permanganate ; perchlorate ClO − : hypochlorite ClO2− : chlorite ClO3− : chlorate ClO4− : perchlorate Inorganic Nomenclature o Write the name of the cation. o If the anion is an element, change its ending to -ide; if the anion is a polyatomic ion, simply write the name of the polyatomic ion. If the cation can have more than one possible charge, write the charge as a Roman numeral in parentheses. This happens in the case of transition metals. The other metals will have only a fixed charge. Naming Hydrated Ionic Compounds Ionic compound prefix-hydrate Name the ionic compound using its rules Take the prefix from the table for the number of waters associated with the compound For example Mg3(PO4)2 2H2O = magnesium phosphate dihydrate CuSO4 5H2O = copper(II) sulfate pentahydrate Hydrated copper(II)sulfate is vibrant blue, while anhydrous copper (II) sulfate is a white solid Some chemicals are known by their common names o H2O Water o NH3 Ammonia o H2S Hydrogen sulfide o H2O2 Hydrogen peroxide What is the correct formula for aluminum carbonate? A) AlCO3 B) Al2(CO3)3 C) Al3(CO3)2 D) Al2CO3 E) Al3CO3 When a metal and a nonmetal react, the __________ tends to lose electrons and the __________ tends to gain electrons. A) metal, nonmetal B) nonmetal, metal C) nonmetal, nonmetal D) metal, metal E) None of the above, these elements share electrons. Covalent Compounds o Covalent compounds are formed between two nonmetals. o Covalent compounds can be solids, liquids or gases at room temperature. Identifying and Naming Covalent Compounds Each of the two elements share electrons in a covalent bond Nomenclature of Binary covalent Compounds o The less electronegative atom is usually listed first. o A prefix is used to denote the number of atoms of each element in the compound (mono- is not used on the first element listed, however.) Nomenclature of Binary Compounds If the prefix ends with a or o and the name of the element begins with a vowel, the two successive vowels are often elided into one: N2O5: dinitrogen pentoxide  Let us name CO and CO2 If we get these two names right, we will be able to name everything Now let us name- CO = carbon monoxide SO2 = sulfur dioxide SO3 = sulfur trioxide SF4 = sulfur tetrafluoride The correct name for N2O5 is __________. A) nitric oxide B) nitrogen pentoxide C) nitrogen oxide D) nitrous oxide E) dinitrogen pentoxide o Remember these are between two nonmetals The correct name for CaH2 is A) calcium hydride B) hydrocalcium C) calcium dihydride D) calcium dihydroxide E) calcium hydroxide Identifying and Naming Binary Acids H + nonmetal Hydrogen is listed first in formula Naming: hydro-anion-ic acid For example HF = hydrofluoric acid HCl = hydrochloric acid HBr = hydrobromic acid Identifying and Naming Oxoacids H + polyatomic anion Hydrogen (can have more than one) is listed first in formula Naming: -ite → -ous polyatomic ion-ous acid -ate → -ic polyatomic ion-ic acid For example H2SO3 = sulfurous acid H2SO4 = sulfuric acid HNO2 = nitrous acid HNO3 = nitric acid Other than this you need to know the names of Acetic acid, phosphoric acid, chlorous acid and hypochlorous acid for now Identifying and Naming Linear Alkanes Only carbon and hydrogen in formula: CxH2x+2 Naming: prefix-ane For example CH4 = methane C2H6 = ethane C8H18 = octane Functional Groups -COOH Carboxylic group - NH2 Amino group - OH Hydroxyl group (Understand how it is different from OH - ) Naming Compounds Summary Molar Mass The molar mass of a compound is the sum of the atomic masses of each atom in the compound. Also known as: molecular weight or molecular mass For example CO2: 12.01 g/mol C + 2 x 16.00 g/mol O = 44.01 g/mol CaCO3: 40.08 g/mol Ca + 12.01 g/mol C + 3 x 16.00 g/mol O = 100.09 g/mol C8H18: 8 x 12.01 g/mol C + 18 x 1.01 g/mol H = 114.26 g/mol Mass Percent For example: Na2CO3: 2 x 23.00 g/mol Na + 12.01 g/mol C+ 3 x 16.00 g/mol O = 106.01 g/mol Na: (2 x 23.00) / 106.01)) x100 = 43.39% C : (12.01 / 106.01) x 100 = 11.33% O N a O : (3 x 16.00) / 106.01)) x 100 = 45.28% C Finding Empirical Formulas  Empirical Formulas Finding chemical formula from mass percent (or mass) data 1. Determine the mass (in grams) of each element from mass % 2. Convert the mass of each element to moles. Use the atomic mass of each element as the conversion factor. 3. Divide all of the moles by the smallest number of moles.  Empirical Formulas 4. If all the numbers of moles of each are within 0.15 of a whole number, then round them to the nearest whole number. If any of the numbers of moles is not within 0.15 of a whole number, then multiply all the numbers of moles by an integer to make them all whole numbers. The number of moles, as a whole number, is used as the subscript in the empirical formula. Calculating Empirical Formulas The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA. Calculating Empirical Formulas Assuming 100.00 g of para-aminobenzoic acid, C: 61.31 g x 1 mol = 5.105 mol C 12.01 g 1 mol H: 5.14 g x = 5.09 mol H 1.01 g 1 mol N: 10.21 g x = 0.7288 mol N 14.01 g 1 mol O: 23.33 g x = 1.456 mol O 16.00 g  Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: C: 5.105 mol = 7.005  7 0.7288 mol H: 5.09 mol = 6.984  7 0.7288 mol N: 0.7288 mol= 1.000 0.7288 mol O: 1.458 mol = 2.001  2 0.7288 mol  Calculating Empirical Formulas These are the subscripts for the empirical formula: C7H7NO2 Molecular Formula Molecular formula is the empirical formula multiplied by an integer Molar mass of compound divided by molar mass of empirical formula = integer Empirical formula: CH2O Molecular formula: CH2O Molecular formula: C6H12O6 integer = 6 integer = 1 When a metal and a nonmetal react, the __________ tends to lose electrons and the __________ tends to gain electrons. A) metal, nonmetal B) nonmetal, metal C) nonmetal, nonmetal D) metal, metal E) None of the above, these elements share electrons.  Summary Ionic compounds are formed from a metal and a nonmetal bonding through an ionic bond. Covalent compounds are formed from a nonmetal and a nonmetal bonding through a covalent bond. Binary ionic compounds (no transition metals) are named as: metal nonmetal-ide Binary ionic compounds (with transition metals) are named as: metal(charge) nonmetal- ide Binary compounds (with polyatomic ions) are named as: metal polyatomic ion or metal(charge) Polyatomic ion Ionic compounds (with waters of hydration) are named as: ionic compound prefix-hydrate Summary Binary covalent compounds are named as: prefix-first element prefix-second element- ide Binary acids are named as: hydro-anion-ic acid Oxoacids are named as: polyatomic ion-ous acid or polyatomic ion-ic acid Linear alkanes are named as: prefix-ane Molar mass is the sum of the atomic mass of each element (multiplied by its subscript) Mass percent is: (mass part / mass whole) x 100 The empirical formula of a compound is the smallest whole number ratio of its elements, and is determined experimentally. The chemical formula of a compound is determined from the empirical formula and the molar mass of the compound.  Chapter 4: Chemical Reactions and Stoichiometry Introduction In any given physical or chemical change, matter is neither destroyed nor created. Mass of products formed must equal the mass of reactants entering the chemical reaction Chemical reaction is defined as a process where the bonds of reactant molecules break and new bonds form to generate the product, without changing any elements. Introduction Chemical reactions are represented by chemical equations, which show reactants on the left side and products on the right side. The first step in stoichiometry is to balance the chemical equation and determine the stoichiometric coefficients of both the reactants and products. Stoichiometry is a mathematical representation of the chemical reaction, which is the relationship between the various amounts of species in the reaction. Stoichiometry Antoine Lavoisier discovered that during a chemical reaction the total mass of all compounds remains the same. He was the first to introduce the concept of stoichiometry, which relates amounts of compounds involved in a chemical reaction. Stoichiometry is based on the law of conservation of mass, which states that the total mass of reactants must equal the total mass of the products. Chemical Equations The number and types of reactant atoms has to equal the number and types of product atoms. For example, consider the following generic balanced chemical reaction between reactant molecules A and B to form product molecules C and D: The lower case letters a, b, c, and d are known as coefficients. Chemical Equations We can apply these principles to the chemical reaction between methane and oxygen: One molecule of CH4 reacts with 2 molecules of O2 to produce 1 molecule of CO2 and 2 molecules of H2O. To count the total number of individual atoms, coefficients are multiplied by the subscripts. Stoichiometry Consider the following unbalanced chemical equation: C2H6 + O2 → CO2 + H2O. There are 2 carbons on the reactant side and 1 carbon on the product side. Hence, a coefficient of 2 is required for CO2. Counting H atoms, we get 6 on the reactant side and 2 on the product side. A coefficient of 3 is needed for H2O. Stoichiometry Now that the carbon and hydrogen atoms are balanced, we can examine oxygen. There are 2 O atoms on the reactant side and a total of 7 oxygen atoms on the product side. Now how do we proceed- We can only balance with whole numbers, and we need 7 O atoms. Oxygen is O2 C2H6 + 7O2  2CO2 + 3 H2O Now this messes up everything else as we now have 14 O atoms. Now we try to fix this problem 2 C2H6 + 7O2  4CO2 + 6 H2O Chemical Stoichiometry We often use stoichiometry in our daily lives without knowing it. Manufacturing or assembling a bicycle is a very good example of stoichiometry. One bicycle is made from two wheels, one frame, one handlebar, two pedals, and one seat. Chemical Stoichiometry The above numbers are equivalent to stoichiometric coefficients and are referred to as the number of moles of the reactants and products. If one component is known, then we can use these stoichiometric coefficients to determine the amounts needed of the other components to assemble Chemical Stoichiometry The number of bicycles that can be assembled can also be determined using a stoichiometric relationship: Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products Stoichiometric Calculations From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant) Mass-to-Mass Conversions aA + bB → cC + dD a, b, c, and d are the number of moles of A, B, C, and D If the mass of A is given, then we can determine mathematically the mass of B needed to react with A: How many grams of water are produced in the oxidation of 1.00 g of glucose, C6H12O6? C6H12O6(s) + O2(g) → CO2(g) + H2O(l) How many grams of water are produced in the oxidation of 1.00 g of glucose, C6H12O6? C6H12O6(s) + O2(g) → CO2(g) + H2O(l) C6H12O6(s) + 6 O2(g)→6 CO2(g) + 6 H2O(l) Stoichiometric Calculations C6H12O6 + 6 O2 6 CO2 + 6 H2O Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6… use the coefficients to find the moles of H2O… and then turn the moles of water to grams o The decomposition of KClO3 is commonly used to prepare small amounts of O2 in the laboratory: KClO3(s) → KCl (s) + O2(g). How many grams of O2 can be prepared from 4.50 g of KClO3? o The decomposition of KClO3 is commonly used to prepare small amounts of O2 in the laboratory: KClO3(s) → KCl (s) + O2(g). How many grams of O2 can be prepared from 4.50 g of KClO3? 2 KClO3(s) → 2 KCl (s) + 3 O2(g). Molar masses: KClO3= K + Cl + (O x 3) = O2 = 2 x O = Solid lithium hydroxide is used in space vehicles to remove the carbon dioxide exhaled by astronauts. The lithium hydroxide reacts with gaseous carbon dioxide to form solid lithium carbonate and liquid water. How many grams of carbon dioxide can be absorbed by 1.00 g of lithium hydroxide? Solid lithium hydroxide is used in space vehicles to remove the carbon dioxide exhaled by astronauts. The lithium hydroxide reacts with gaseous carbon dioxide to form solid lithium carbonate and liquid water. How many grams of carbon dioxide can be absorbed by 1.00 g of lithium hydroxide 2 LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l) Grams LiOH → moles LiOH → moles CO2 → grams CO 2 Molar mass of LiOH= 6.94 + 16.00 + 1.01 = 23.95 g/mol CO2=12.01 + 2(16.00) = 44.01 g/mol. 2 LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l) Limiting Reactants To Make 24 cookies you need o 2 3/4 cups all-purpose flour o 1 teaspoon baking soda o 1/2 teaspoon baking powder o 1 cup butter, softened o 1 1/2 cups white sugar o 1 egg o 1 teaspoon vanilla extract How many cookies can you make with 1 cup sugar and unlimited amount of other ingredients A. As many as you want B. 10 cookies C.16 cookies D. 24 cookies How Many Cookies Can I Make? o You can make cookies until you run out of one of the ingredients How Many Cookies Can I Make? o In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount Suppose if we take 5 g of H2 and 3.5 g of O2 Can you tell which component is limiting without converting it to moles?  2.48 mol.109 mol Suppose if we take 5 g of H2 and 3.5 g of O2 Can you tell which component is limiting without converting it to moles? No Suppose if we take 5 g of H2 and 5 g of O2 For the reaction: 2H2 + O2 2H2O Can you tell which component is limiting without converting it to moles? The Limiting Reactant The limiting reactant is the reactant that limits the amount of product that can be formed. The limiting reactant is always completely consumed during the course of the reaction. An excess reactant is defined as any of the other reactants still present after the consumption of the limiting reactant. The most important commercial process for converting N2 from the air into nitrogen- containing compounds is based on the reaction of N2 and H2 to form ammonia (NH3): N2(g) + H2(g)→ NH3(g) How many moles of NH3 can be formed from 3.0 mol of N2 and 6.0 mol of H2? The most important commercial process for converting N2 from the air into nitrogen- containing compounds is based on the reaction of N2 and H2 to form ammonia (NH3): N2(g) + H2(g)→ NH3(g) How many moles of NH3 can be formed from 3.0 mol of N2 and 6.0 mol of H2? First we need to balance our reaction: N2(g) + 3H2(g)→ 2NH3(g) 3 mol of N2 x 3 mol H2 = 9 mol of H2 1 mol of N2 6.0 mol of H2 x 1 mol N2 = 2 mol of N2 3mol of H2 Consider the reaction 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s). A mixture of 1.50 mol of Al and 3.00 mol of Cl2 is allowed to react. (a) Which is the limiting reactant? (b) How many moles of AlCl3 are formed? (c) How many moles of the excess reactant remain at the end of the reaction? Consider the reaction 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s). A mixture of 1.50 mol of Al and 3.00 mol of Cl2 is allowed to react. (a) Which is the limiting reactant? (b) How many moles of AlCl3 are formed? (c) How many moles of the excess reactant remain at the end of the reaction? The Percent Yield Chemical reactions sometimes appear to not proceed according to the stoichiometric coefficients. A reason for this is that most reactions have “side reactions,” which compete for reactants and produce undesired products. As a result, some of the reactants will be used while some will be left over when the reaction is completed. It is important to determine the efficiency of a chemical reaction. Percent Yield How much product is synthesized in ideal conditions is called the theoretical yield The amount of product actually produced by a reaction (i.e., in the lab) will usually be less than the theoretical yield and is referred to as the actual yield. chemists refer to the percent yield of the chemical reaction, which is mathematically represented as follows: Theoretical Yield o The theoretical yield is the amount of product that can be made o In other words, it’s the amount of product possible as calculated through the stoichiometry problem o This is different from the actual yield, the amount one actually produces and measures Percent Yield A comparison of the amount obtained to the amount it was possible to make Actual Yield Percent Yield = x 100 Theoretical Yield Percent Yield Suppose you mixed enough chocolate chip cookie dough to make 12 chocolate chip cookies according to the recipe. However, you end up with only 10 chocolate chip cookies. Chemical Reactions in Aqueous Solutions Chemical Reactions in Aqueous Solutions A solution is defined as any homogenous mixture that is physically and chemically the same throughout the whole system. A solution has two main components: - the solvent (present in large amounts; water in the case of aqueous solutions), - the solute, the substance that is dissolved in the solute and usually present in relatively small amounts. There can be more than one solute in a solution, but there is never more than one solvent. Chemical Reactions in Aqueous Solutions Ions of the transition elements, such as manganese (Mn), nickel (Ni), and copper (Cu), produce interesting and colorful homogenous solutions when dissolved in water Molarity To track and quantify the amount of solute added to water to make an aqueous solution, we use solution concentration to express the amount of solute present in a solution. This is known as molarity (M) and is mathematically expressed as follows: Calculate the molarity of a solution made by dissolving 23.4 g of sodium sulfate (Na2SO4) in enough water to form 125 mL of solution. moles of solute Molarity (M) = volume of solution in liters o How would you make 200.0 ml of 0.5 M NaCl solution. o First calculate how many moles of NaCl are needed. o Then convert the moles to g of NaCl M = mol mol = ML = 0.5 mol x 0.200L = 0.100 mol NaCl L L 0.100 mol NaCl x 58.442 g NaCl = 5.844 g NaCl 1 mol NaCl = 6 g NaCl o How many grams of Na2SO4 are there in 15 mL of 0.50 M Na2SO4? (b) How many milliliters of 0.50 M Na2SO4 solution are needed to provide 0.038 mol of this salt? o a) How many grams of Na2SO4 are there in 15 mL of 0.50 M Na2SO4? (b) How many milliliters of 0.50 M Na2SO4 solution are needed to provide 0.038 mol of this salt? moles of solute Molarity (M) = volume of solution in liters Solubility, Non-electrolytes and Electrolytes Ionic solids and some polar compounds, such as strong acids and strong bases, tend to have the highest solubility and form ions when dissolved in water. The result is an electrolyte solution, which is defined as any aqueous solution that conducts electricity. Electrolytes o Substances that dissociate into ions when dissolved in water. o A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so,for example, sugar Solubility, Non-electrolytes and Electrolytes Examples of strong electrolytes include ionic compounds such as sodium chloride (NaCl), strong bases such as potassium hydroxide (KOH), and strong acids like hydrochloric acid (HCl) and nitric acid (HNO3). On the other hand, when other compounds, such as weak acids and weak bases, dissolve in water, fewer ions are produced, resulting in lower conductivity These are called weak electrolytes. Solubility, Non-electrolytes and Electrolytes Both acetic acid (CH3COOH, weak acid) and ammonia (NH3, weak base) are considered examples of weak electrolytes. Although these compounds are highly soluble in water, acetic acid is a weak acid and, therefore, a weak electrolyte. Non-electrolytes are substances that dissolve in water to form neutral molecules and have essentially no effect on electrical conductivity Examples of non-electrolytes that are very soluble in water (but essentially non-conductive) are ethanol (C2H6O), ethylene glycol (C2H6O2), glucose (C6H12O6), and sucrose (C12H22O11). Solubility of Ionic Compounds How can we determine which ionic solids dissolve in water, and which do not? Strong Electrolytes Are… o Strong acids Strong Electrolytes Are… o Strong acids o Strong bases Solubility of Ionic Compounds Solubility o All alkali metal compounds are water soluble, no matter what the anion o All Ammonium compounds are soluble, no matter what the anion. Solubility NO3- (nitrate) ClO4- (perchlorate) CH3CO2- (acetate) All these anions form only soluble compounds, no matter what the cation happens to be attached to them. Types of Chemical Reactions Double Displacement Reactions A double-displacement or replacement or metathesis reaction is a type of chemical reaction where the positive ions (cations) and the negative ions (anions) of the two reactants switch places and form two new compounds or products. Both precipitation and acid–base neutralization reactions are considered examples of double-replacement reactions. Double Replacement (exchange) Reactions o Metathesis comes from a Greek word that means “to transpose” AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq) Molecular Equation The molecular equation lists the reactants and products in their molecular form. AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq) Precipitation Reactions Precipitation reactions occur when two soluble aqueous ionic solutions react to form a solid precipitate (i.e. an insoluble ionic solid in water). According to the solubility rules for these ionic solids in water, NaCl, AgNO3, and NaNO3 are completely soluble in water while AgCl is insoluble in water and forms a white precipitate. Ionic Equation o In the ionic equation all strong electrolytes (strong acids, strong bases, and soluble ionic salts) are dissociated into their ions. o This more accurately reflects the species that are found in the reaction mixture. Ag+ (aq) + NO3- (aq) + K+ (aq) + Cl- (aq)  AgCl (s) + K+ (aq) + NO3- (aq)  Net Ionic Equation o To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right. o The only things left in the equation are those things that change (i.e., react) during the course of the reaction. o Those things that didn’t change (and were deleted from the net ionic equation) are called spectator ions. Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq)  AgCl (s) + K+(aq) + NO3-(aq)  Net Ionic Equation o To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right. o The only things left in the equation are those things that change (i.e., react) during the course of the reaction. Ag+(aq) + Cl-(aq)  AgCl (s)  Writing Net Ionic Equations 1. Write a balanced molecular equation. 2. Dissociate all strong electrolytes. Do not dissociate a. The solids, b. liquids and c. gases The liquids and gases are covalently linked so they will not dissociate in water d. the weak acids are also left as is, as they dissociate a negligible amount in water. 3. Cross out anything that remains unchanged from the left side to the right side of the equation. 4. Write the net ionic equation with the species that remain. HF(aq)+NaOH(aq)>NaF(aq)+H2O(l) HF(aq)+Na⁺(aq)+OH⁻(aq)→ Na⁺(aq)+F⁻(aq) + H2O Writing Net Ionic Equations Double Displacement Reactions Some examples of double displacement reactions: Gas-Forming Reactions There are many aqueous reactions where at least one product is a gas. The most common ones are those producing CO2(g), SO2(g), NH3(g), or H2S(g). The visible sign is the formation of bubbles in solution All but CO2(g) give off an appreciable odor When a carbonate salt reacts with an acid, CO2(g) is produced via an intermediate reaction: Gas-Forming Reactions o These double replacement reactions do not give the product expected. o The expected product decomposes to give a gaseous product (CO 2 or SO2) along with water. CaCO3 (s) + 2HCl (aq) CaCl2 (aq) + (H2CO3) CaCl2 (aq)+CO2(g) + H2O(l) NaHCO3 (aq) + HBr (aq) NaBr (aq) + (H2CO3) NaBr (aq) +CO2 (g) + H2O(l) SrSO3 (s) + 2 HI (aq) SrI2 (aq) + (H2SO3) SrI2 (aq) +SO2 (g) + H2O(l) Gas-Forming Reactions Just as in the previous examples, a gas is formed as a product of this reaction: Na2S (aq) + H2SO4 (aq)  Na2SO4 (aq) + H2S (g) This reaction gives the predicted product, but you better carry it out in the hood, or you will be very unpopular! Acid-Base Neutralization Reactions An Arrhenius acid is a substance that dissolves in water to produce hydrogen ions, H+. An Arrhenius base is a substance that dissolves in water to produce hydroxide ions, OH-. HA(aq) → H+(aq) + A−(aq) BOH(aq) → B+(aq) + OH−(aq) HCl dissociates as follows: NaOH dissociates as follows: HCl(aq) → H+(aq) + Cl−(aq) NaOH(aq) → Na+(aq) + OH−(aq) Acids: o Substances that increase the concentration of H+ when dissolved in water (Arrhenius). Monoprotic: HCl(aq)  H+(aq) + Cl–(aq) Diprotic: H2SO4(aq)  H+(aq) + HSO4–(aq) HSO4–(aq)  H+(aq) + SO42–(aq) Tripotic: H3PO4  Bases: o Substances that increase the concentration of OH− when dissolved in water (Arrhenius) Ionic base: NaOH(s) Na+(aq) + OH–(aq) Molecular bases: Ammonia and amines. These do not contain OH– ions but produce hydroxide ions when dissolved in water Ammonia - NH3(aq) + H2O(l) NH4+(aq) + OH–(aq) An amine: CH3NH2(aq) + H2O(l)  CH3NH3+(aq) +OH–(aq) o Consider solutions in which 0.1 mol of each of the following compounds is dissolved in 1 L of water: Ca(NO3)2 (calcium nitrate), C6H12O6 (glucose), CH3COONa (sodium acetate), and CH3COOH (acetic acid). Rank the solutions in order of increasing electrical conductivity, based on the fact that the greater the number of ions in solution, the greater the conductivity. o Consider solutions in which 0.1 mol of each of the following compounds is dissolved in 1 L of water: Ca(NO3)2 (calcium nitrate), C6H12O6 (glucose), CH3COONa (sodium acetate), and CH3COOH (acetic acid). Rank the solutions in order of increasing electrical conductivity, based on the fact that the greater the number of ions in solution, the greater the conductivity. o Answers: C6H12O6 (nonelectrolyte) < CH3COOH (weak electrolyte, existing mainly in the form of molecules with few ions) < CH3COONa (strong electrolyte that provides two ions, Na+ and CH3COO–) < Ca(NO3)2 (strong electrolyte that provides three ions, Ca2+ and 2 NO3 ) Acid-Base Neutralization Reactions Acid–base neutralization reactions occur when an acid reacts with a base to form a salt and water: HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l) Neutralization Reactions Generally, when solutions of an acid and a base are combined, the products are a salt and water. HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l) Neutralization Reactions When a strong acid reacts with a strong base, the net ionic equation is… HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l) H+ (aq) + Cl- (aq) + Na+ (aq) + OH-(aq)  Na+ (aq) + Cl- (aq) + H2O (l) Neutralization Reactions When a strong acid reacts with a strong base, the net ionic equation is… HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l) H+ (aq) + Cl- (aq) + Na+ (aq) + OH-(aq)  Na+ (aq) + Cl- (aq) + H2O (l) H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq)  Na+ (aq) + Cl- (aq) + H2O (l) Neutralization Reactions Observe the reaction between Milk of Magnesia, Mg(OH)2, and HCl. Mg(OH)2 is insoluble in water. When it reacts with stomach HCl it turns into soluble MgCl2 and water o If you have a weak acid in a reaction and you need to write the net ionic equation, you do not dissociate it, as weak acids are only partially dissociated.  Oxidation-Reduction Reactions o An oxidation occurs when an atom or ion loses electrons. o A reduction occurs when an atom or ion gains electrons. o i l r i g Oxidation-Reduction Reactions One cannot occur without the other. Electron Transfer Reactions or Redox Reactions Electron transfer reactions (oxidation–reduction reactions, or redox reactions) are defined as reactions where electrons move from one atom or molecule to another atom or molecule The substance that loses one or more electrons is said to be oxidized while the substance that gains one or more electrons is said to be reduced Oxidation Is Loss Reduction Is Gain Identifying Oxidizing and Reducing Agents The two main components of the electron transfer reaction are the oxidizing and reducing agents. The oxidizing agent is defined as the reactant (either an element or a compound) that gains electron(s) (i.e., is reduced). The reducing agent is the reactant (either element or compound) that loses electron(s) (i.e., is oxidized). Electron Transfer Reactions Oxidation States Oxidation States indicate the number of electrons that the atom already gained or lost when chemically bonded with another element Any pure element in the periodic table will have an oxidation state of zero. Group IA cations will have an oxidation state of +1. Group VI anions will usually have an oxidation state of −2, particularly when the element is combined with more metallic elements. Oxidation Numbers o Elements in their elemental form have an oxidation number of 0. o Fe metal = 0 o Zn metal = 0 o O2 gas = 0 o The oxidation number of a monatomic ion is the same as its charge. Oxidation Numbers Cl⁻ = -1 Fe ²⁺ +2 o Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.  Oxygen has an oxidation number of −2, except in the peroxide ion in which it has an oxidation number of −1.  Hydrogen is −1 when bonded to a metal, +1 when bonded to a nonmetal. Oxidation Numbers o Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.  Fluorine always has an oxidation number of −1.  The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions. Oxidation Numbers o The sum of the oxidation numbers in a neutral compound is 0. o The sum of the oxidation numbers in a polyatomic ion is the charge on the ion. Oxidation States What is the oxidation state of the Cr atom (transition metal) in potassium dichromate K2Cr2O7? The 2 Cr atoms have an oxidation state of +12; therefore, the oxidation state of 1 Cr atom is +6. Determine the oxidation number of sulfur in each of the following: (a) H 2S, (b) S8, (c) SCl2, (d) Na2SO3, (e) SO42–. (a) H2S: Hydrogen is −1 when bonded to a metal, +1 when bonded to a nonmetal. 2(+1) + x = 0, x = -2 (b) S8: Because this is an elemental form of sulfur, the oxidation number of S is 0 (c) SCl2: The halogens have an oxidation number of −1 when they are in binary compounds. x + 2(–1) = 0 , x = +2 (d) Na2SO3 :Sodium, an alkali metal, always has an oxidation number of +1 in its compounds Oxygen has a common oxidation state of –2 2(+1) + x + 3(-2) = 0 x =+ 4 (e) SO42– : The oxidation state of O is –2. This is not a neutral entity so the sum of the oxidation numbers is equal to the charge. x + 4(–2) = –2, x = +6 A strong electrolyte is one that __________ completely in solution. A) reacts B) dissociates C) disappears D) ionizes Identifying Oxidizing and Reducing Agents Identify the oxidizing and reducing agents in the following reaction between iron(II) oxide (FeO) and graphite (C): FeO (s) + C (s) Fe (s) + CO (g) Identifying Oxidizing and Reducing Agents Identify the oxidizing and reducing agents in the following reaction between iron(II) oxide (FeO) and graphite (C): FeO (s) + C (s) Fe (s) + CO (g) 2+ 2- 0 0 2+ 2- FeO (s) + C (s) Fe (s) + CO (g)  Identifying Oxidizing and Reducing Agents Identify the oxidizing and reducing agents in the following reaction between iron(II) oxide (FeO) and graphite (C): FeO (s) + C (s) Fe (s) + CO (g) 2+ 2- 0 0 2+ 2- FeO (s) + C (s) Fe (s) + CO (g) Fe in FeO converted into Fe(s) by gaining 2e-; therefore, FeO is the oxidizing agent (the reactant that gains e-) C(s) is converted into CO by losing 2e-; therefore, C(s) is the reducing agent (the reactant that lost e-) Single-Displacement Reaction and Activities of Metals a type of oxidation–reduction reaction where an element or ion moves out of one compound and into another: A + MX M + AX Element A has replaced cation M in compound MX to become a new compound AX and the free element M. Displacement Reactions o In displacement reactions, ions oxidize an element. o The ions, then, are reduced. Displacement Reactions In this reaction, silver ions oxidize copper metal. Cu (s) + 2 Ag+ (aq)  Cu2+ (aq) + 2 Ag (s) Single-Displacement Reaction and Activities of Metals Single-displacement reaction between solid copper wire and silver nitrate in aqueous solution. Displacement Reactions The reverse reaction, however, does not occur. Cu2+ (aq) + 2 Ag (s)  Cu (s) + 2 Ag+ (aq) x The Metals Activity Series Any Metal in the series can be oxidized by the ion of the elements below it Cu (s) + 2 Ag+ (aq)  Cu2+ (aq) + 2 Ag (s) Only the metals above hydrogen in the activity series are able to react with acids to release hydrogen gas Copper will not react with an acid to give H2 gas but Ni will Activity Series Active Metals Noble Metals Any Metal in the series can be oxidized by the ion of the elements below it Cu (s) + 2 Ag+ (aq)  Cu2+ (aq) + 2 Ag (s) Dilution Lemonade Concentrate Dilution MiVi = MfVf How many milliliters of 3.0 M H2SO4 are needed to make 450 mL of 0.10 M H2SO4? 15mL o (b) How many milliliters of 5.0 M K2Cr2O7 solution must be diluted to prepare 250 mL of 0.10 M solution? o (c) If 10.0 mL of a 10.0 M stock solution of NaOH is diluted to 250 mL, what is the concentration of the resulting solution? o Stoichiometric Calculations require the chemicals to be in molar quantities o When a gram amount is given it can directly be changed to moles. o In solutions the molarity and the volume are given and can be used to calculate the number f moles o The rest is the same as in any other problem…… Using Molarities in Stoichiometric Calculations Calculate the oxidation number of Cr in Cr2O72- A. +2 B. + 4 C. +6 D.+8 Titration The analytical technique in which one can calculate the concentration of a solute in a solution. Titration The quantity of Cl– in a municipal water supply is determined by titrating the sample with Ag+. The reaction taking place during the titration is The end point in thi