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Chapter 2 — Atoms, Molecules, Ions 1 Atomic Composition • Protons – + electrical charge – mass = 1.672623 x 10-24 g – mass = 1.007 atomic mass units (u) Chapter 2 Atoms, Molecules, and Ions • Electrons – negative electrical charge – relative mass = 0.0005 u • Neutrons no electrical charge – mass = 1.009 u – Jeffrey Mack California State University, Sacramento Atomic Composition The atom is mostly empty space Atomic Number, Z All atoms of the same element have the same number of protons in the nucleus, Z 13 • protons and neutrons in the nucleus. • the number of electrons is equal to the number of protons. • electrons in space around the nucleus. Atomic number, Z Al Elemental symbol 26.981 Atomic mass (u) • extremely small. One teaspoon of water has 3 times as many atoms as the Atlantic Ocean has teaspoons of water. Atomic Weight • The atomic mass of one atom of an element is relative to one atom of another. • Example: an Oatom has approximately 16 times more mass than an Hatom. • The standard is based upon the carbon-12 isotope. • The mass of one C12 atom is 1.99265 1023 g. • The atomic mass of C12 atom is defined as exactly 12 u. • Therefore: 1 u = (the mass of one C12 atom 12) = 1.66054 1024g = 1.66054 1027 kg Mass Number, A • • • • C - 12 atom has 6 protons and 6 neutrons Mass Number (A) = # protons + # neutrons Carbon – 12 has a mass # of 12 u Carbon – 13 has a mass # of 13 u (6 + 7) and so on. • Elements with different number of neutrons are called “Isotopes”. Chapter 2 — Atoms, Molecules, Ions 2 Isotopes Isotopes For a given element “X”, an isotope is written by: Atomic number (Z) = number of protons in the nucleus. Mass number (A) = total number of nucleons in the nucleus (i.e., protons and neutrons). One nucleon has a mass of 1 amu (Atomic Mass Unit) a.k.a “Dalton” or u A Z X Isotopes have the same Z but different total number of nucleons (A). For a given element “X”, an isotope is written by: Atomic number (Z) = number of protons in the nucleus. Mass number (A) = total number of nucleons in the nucleus (i.e., protons and neutrons). A Z X Hydrogen Isotopes Shorthand notation: Carbon – 12 = 12C Carbon – 14 = 14C and so on... Isotope Composition Hydrogen has 3 isotopes 1 1 H 1 proton and 0 neutrons, protium Isotope 2 1 H 1 proton and 1 neutron, deuterium Sulfur-32 3 1 1 proton and 2 neutrons, tritium H Masses of Isotopes determined with a mass spectrometer 74 35 Br Electrons Protons Neutrons 16 16 16 35 35 39 Isotope Abundance • The mass spectrometer gives information on the mass and relative abundance of each element’s isotopes. • Each isotope is represented by a Relative Abundance. Chapter 2 — Atoms, Molecules, Ions 3 Isotope Abundance Percent Abundance • The percent abundance of each element’s isotopes are found by: Percent = Abundance Atomic Weight (Mass) Atomic mass æ % abundance of isotope 1 ö =ç ÷ ? mass of isotope 1 100 è ø æ % abundance of isotope 2 ö +ç ÷ ? mass of isotope 2 + ... 100 è ø Example: Chlorine has two isotopes: Cl-35 75.53 % Isotope Problem The average weighted atomic mass is determined by the following mathematical expression: 100 total number of atoms of all isotopes of that element Isotope Problem 35 17 The atomic masses on the periodic table are “weighted averages” of the all of an element’s individual isotope masses. number of atoms of an individual isotope Cl- 37 Cl 37 17 & 34.96885 u 24.47 % Cl 36.96590 u Isotope Problem The average weighted atomic mass is determined by the following mathematical expression: Average mass of a Cl atom mass of a Cl-35 atom fraction that are Cl-35 mass of a Cl-37 atom fraction that are Cl-37 Chapter 2 — Atoms, Molecules, Ions 4 Isotope Problem Isotope Problem The average weighted atomic mass is determined by the following mathematical expression: The average weighted atomic mass is determined by the following mathematical expression: Average mass of a Cl atom Average mass of a Cl atom mass of a Cl-35 atom m Cl (u) = m Cl-35 fraction that are Cl-35 mass of a Cl-37 atom fraction that are Cl-37 abundance abundance + m Cl-37 of Cl-35 of Cl-37 mass of a Cl-35 atom m Cl (u) = m Cl-35 fraction that are Cl-35 mass of a Cl-37 atom fraction that are Cl-37 abundance abundance + m Cl-37 of Cl-35 of Cl-37 35.45 u = 34.96885 u 0.7553 + 36.96590 u 0.2447 (4 sig. fig) Isotope Problem The average weighted atomic mass is determined by the following mathematical expression: Average mass of a Cl atom mass of a Cl-35 atom m Cl (u) = m Cl-35 fraction that are Cl-35 mass of a Cl-37 atom fraction that are Cl-37 Isotope Problem: Copper has two stable isotopes, Cu-63 and Cu-65. If the masses are 62.929599 u and 64.927793 u respectively, what are the relative % abundances of the two isotopes? The weighted average atomic mass is reported as 63.546 u. abundance abundance + m Cl-37 of Cl-35 of Cl-37 35.45 u = 34.96885 u 0.7553 + 36.96590 u 0.2447 (4 sig. fig) This is the value that is reported on the periodic table. Note that: 0.7553 + 0.2447 = 1.0000 (100%) Isotope Problem: Copper has two stable isotopes, Cu-63 and Cu-65. If the masses are 62.929599 u and 64.927793 u respectively, what are the relative % abundances of the two isotopes? The weighted average atomic mass is reported as 63.546 u. Isotope Problem: Copper has two stable isotopes, Cu-63 and Cu-65. If the masses are 62.929599 u and 64.927793 u respectively, what are the relative % abundances of the two isotopes? The weighted average atomic mass is reported as 63.546 u. What is given: What is given: There are only 2 isotopes The masses of each isotope the average weighted mass (from the periodic table) What is asked: Determine the relative abundance of each isotope (fraction that occurs in nature) There are only 2 isotopes The masses of each isotope the average weighted mass (from the periodic table) Chapter 2 — Atoms, Molecules, Ions Isotope Problem: Copper has two stable isotopes, Cu-63 and Cu-65. If the masses are 62.929599 u and 64.927793 u respectively, what are the relative % abundances of the two isotopes? The weighted average atomic mass is reported as 63.546 u. 5 Isotope Problem: Copper has two stable isotopes, Cu-63 and Cu-65. If the masses are 62.929599 u and 64.927793 u respectively, what are the relative % abundances of the two isotopes? The weighted average atomic mass is reported as 63.546 u. Solution: Recall that the sum of the fractions that represent the two isotopes must equal 1 exactly. Since these quantities are unknown, let’s assign variables! Isotope Problem: Copper has two stable isotopes, Cu-63 and Cu-65. If the masses are 62.929599 u and 64.927793 u respectively, what are the relative % abundances of the two isotopes? The weighted average atomic mass is reported as 63.546 u. Solution: Recall that the sum of the fractions that represent the two isotopes must equal 1 exactly. Since these quantities are unknown, let’s assign variables! The reported average must equal the weighted sum of the two isotopes. Average Mass =Mass Cu-63 × X+ Mass Cu-65 × Y Let X equal the fraction that is Cu-63 Let Y equal the fraction that is Cu-65 Such that: X + Y = 1 The reported average must equal the weighted sum of the two isotopes. fraction that is copper-63 Average Mass =Mass Cu-63 × X+ Mass Cu-65 × Y The reported average must equal the weighted sum of the two isotopes. fraction that is copper-63 fraction that is copper-65 Average Mass =Mass Cu-63 × X+ Mass Cu-65 × Y Chapter 2 — Atoms, Molecules, Ions The reported average must equal the weighted sum of the two isotopes. fraction that is copper-63 fraction that is copper-65 Average Mass =Mass Cu-63 × X+ Mass Cu-65 × Y 6 Isotope Problem: Copper has two stable isotopes, Cu-63 and Cu-65. If the masses are 62.929599 u and 64.927793 u respectively, what are the relative % abundances of the two isotopes? The weighted average atomic mass is reported as 63.546 u. 63.546 u = 62.929599 u × X + 64.927793 u × Y Plugging in the numbers: 63.546 u = 62.929599 u × X + 64.927793 u × Y Isotope Problem: Copper has two stable isotopes, Cu-63 and Cu-65. If the masses are 62.929599 u and 64.927793 u respectively, what are the relative % abundances of the two isotopes? The weighted average atomic mass is reported as 63.546 u. 63.546 u = 62.929599 u × X + 64.927793 u × Y Isotope Problem: Copper has two stable isotopes, Cu-63 and Cu-65. If the masses are 62.929599 u and 64.927793 u respectively, what are the relative % abundances of the two isotopes? The weighted average atomic mass is reported as 63.546 u. 63.546 u = 62.929599 u × X + Since X + Y = 1 64.927793 u × Y Since X + Y = 1 We can write the equation in terms of one variable, then solve. Isotope Problem: Copper has two stable isotopes, Cu-63 and Cu-65. If the masses are 62.929599 u and 64.927793 u respectively, what are the relative % abundances of the two isotopes? The weighted average atomic mass is reported as 63.546 u. 63.546 u = 62.929599 u × X + 64.927793 u × Y Since X + Y = 1 We can write the equation in terms of one variable, then solve. 63.546 u = 62.929599 u × X+ 64.927793 u × (1−X) Isotope Problem: Copper has two stable isotopes, Cu-63 and Cu-65. If the masses are 62.929599 u and 64.927793 u respectively, what are the relative % abundances of the two isotopes? The weighted average atomic mass is reported as 63.546 u. Solving for X: X = 0.69500 Chapter 2 — Atoms, Molecules, Ions Isotope Problem: Copper has two stable isotopes, Cu-63 and Cu-65. If the masses are 62.929599 u and 64.927793 u respectively, what are the relative % abundances of the two isotopes? The weighted average atomic mass is reported as 63.546 u. Solving for X: 7 Isotope Problem: Copper has two stable isotopes, Cu-63 and Cu-65. If the masses are 62.929599 u and 64.927793 u respectively, what are the relative % abundances of the two isotopes? The weighted average atomic mass is reported as 63.546 u. Solving for X: X = 0.69500 Y must therefore be: Y = 0.30500 X = 0.69500 recall, Y=1X Y must therefore be: Y = 0.30500 Isotope Problem: Copper has two stable isotopes, Cu-63 and Cu-65. If the masses are 62.929599 u and 64.927793 u respectively, what are the relative % abundances of the two isotopes? The weighted average atomic mass is reported as 63.546 u. Isotopes • For more information on Elements, Isotopes and Isotope abundance go to: http://www.webelements.com Solving for X: X = 0.69500 recall, Y=1X Y must therefore be: Y = 0.30500 This means that the % abundances for the two isotopes are: 69.500 % for Cu-63 and 30.500 % for Cu-65 The Periodic Table • Dmitri Mendeleev (18341907) developed the modern periodic table. • Argued that elemental properties are periodic functions of atomic weights. • We now know that element properties are periodic functions of the ATOMIC NUMBERS. • Here you can explore more about the periodic table by looking at interesting elements such as boron, silicon, chromium and antimony. • The web site will provide you with practical applications and usage of elements. Features of the Periodic Table Periods & Groups Chapter 2 — Atoms, Molecules, Ions Regions in the Periodic Table 8 Group 1A: Alkali Metals Li, Na, K, Rb, Cs sodium metal Group 2A: Alkaline Earth Metals Be, Mg, Ca, Sr, Ba, Ra • The second group in the periodic table, Group 2A, is also composed entirely of metals that occur naturally only in compounds. • Except for beryllium (Be), these elements react with water to produce basic solutions. • Most of these oxides (such as lime, CaO) form alkaline solutions; hence, they are known as the alkaline earth metals. • Magnesium (Mg) and calcium (Ca) are the seventh and fifth most abundant elements in the earth’s crust, respectively Magnesium Magnesium oxide Gems & Minerals • Sapphire: Al2O3 with Fe3+ or Ti3+ impurity gives blue whereas V3+ gives violet. • Ruby: Al2O3 with Cr3+ impurity • Elements in the leftmost column, Group 1A, are known as the alkali metals (except H). • All the alkali metals are solids at room temperature and all are reactive with water to produce hydrogen and basic solutions. • These metals are only found in nature combined in compounds (such as NaCl), never as free elements. Group 3A: B, Al, Ga, In, Tl • Group 3A (aluminum, gallium, indium, and thallium) are metals, whereas boron (B) is a metalloid. • Aluminum (Al) is the most abundant metal in the earth’s crust at 8.2% by mass. It is exceeded in abundance only by the nonmetal oxygen and metalloid silicon. Gallium is one of the few metals that can be liquid close to room temp. Group 4A: C, Si, Ge, Sn, Pb • Beginning with Group 4A, however, the groups contain more and more nonmetals. • In Group 4A there is a nonmetal, carbon, two metalloids, silicon and germanium, and two metals, tin and lead • Because of the change from nonmetallic to metallic behavior, more variation occurs in the properties of the elements of this group than in most others. • Nonetheless, there are similarities. For example, these elements form compounds with analogous formulas such as CO2, SiO2, GeO2, and PbO2. Diamond Quartz, SiO2 Chapter 2 — Atoms, Molecules, Ions Carbon 9 Group 5A: N, P, As, Sb, Bi • One interesting aspect nonmetals (like carbon) is that an element of this type can often exist in several different and distinct forms, called allotropes. • Each allotrope has its own physical and chemical properties. Polyatomic Elements • Nitrogen in Group 5A occurs naturally in the form of the diatomic molecule (N2) It makes up about three-fourths of earth’s atmosphere. • Phosphorus is essential to life. It is an important constituent in bones, teeth, and DNA. The element glows in the dark if it is in the air. Its name is based on Greek words meaning “lightbearing”. Phosphorous also has several allotropes, the most important being white & red phosphorus. Group 6A: O, S, Se, Te, Po • In addition to nitrogen there are several elements that exist in nature as diatomic or polyatomic molecules. • When one speaks of oxygen in its natural state, one means O2. • A single oxygen atom is referred to as “atomic oxygen” and so on. • Oxygen, which constitutes about 20% of earth’s atmosphere and which combines readily with most other elements, is at the top of Group 6A. • Most of the energy that powers life on earth is derived from reactions in which oxygen combines with other substances. • Sulfur, selenium, and tellurium are often referred to collectively as chalcogens (from the Greek word, khalkos, for copper) because most copper ores contain these elements. One elemental form of S has a ring structure. Group 7A: Halogens Group 8A: Noble Gases F, Cl, Br, I, At He, Ne, Ar, Kr, Xe, Rn • The Group 7A elements, fluorine, chlorine, bromine, iodine, and radioactive astatine are nonmetals. • All exist as diatomic molecules. • These elements are among the most reactive of all elements. All combine violently with alkali metals to form salts such as table salt, NaCl. • Collectively they are know as the halogens, from the Greek “hals”, meaning “salt,” & genes, “forming.” Br2 gas Chapter 2 — Atoms, Molecules, Ions Group 8A: Noble Gases He, Ne, Ar, Kr, Xe, Rn • The Group 8A elements, helium, neon, argon, krypton, xenon, and radioactive radon are collectively know as the “Noble” or “Inert” gases. • The are the least reactive elements of the period. • All are gases, and none is abundant on earth or in earth’s atmosphere. • It was first thought that the Noble gases would not combine chemically to form stable compounds. However in 1962, a xenon & fluorine compound was first prepared, (XeF4) opening the way to the preparation of a number of other such compounds. Transition Elements • Stretching between Groups 2A and 3A is a series of elements called the transition elements. • These fill the B-groups (1B through 8B) in the fourth through the seventh periods in the center of the periodic table. • All are metals and 13 of them are in the top 30 elements in terms of abundance in the earth’s crust • Most occur naturally in combination with other elements, but a few, Cu, Ag, Au & Pt are much less reactive where they can be found in nature as pure elements. • Two rows at the bottom of the table accommodate the lanthanides [the series of elements between the elements lanthanum (Z = 57) and hafnium (Z = 72) and the actinides the series of elements between actinium (Z = 89) and rutherfordium (Z = 104). Compounds & Molecules • COMPOUNDS are a combination of 2 or more elements in definite ratios by mass. • The character of each element is lost when forming a compound. • MOLECULES are the smallest unit of a compound that retains the characteristics of the compound. (non-metal combined with a non-metal) 10 Transition Elements Lanthanides and actinides Iron in air gives iron(III) oxide Molecules, Ions, and Their Compounds Elements combine to form compounds. These new species have different chemical and physical properties than the original elements. Compounds • A compound is a distinct substance that contains two or more elements combined in a definite proportion by weight. • Atoms of the elements that constitute a compound are always present in simple whole number ratios. • They are never present as fractional parts. Examples: Never: AB A2B A½B AB2 Chapter 2 — Atoms, Molecules, Ions Molecular Formulas 11 Molecular Modeling Structural formula of glycine: • • • • Molecular Formula: #’s & types of each atom. Condensed Formula: indicates structure Structural Formula: Shows the connections (bonds) Molecular Model: Give 3-D perspective Molecular Modeling Ball & stick H H O H N C C O H H Space-filling Ions & Ionic Compounds Various ways to represent a molecule. Ionic Compounds • Ionic compounds (metals & nonmetals) constitute another major class of compounds. • They consist of ions, atoms or groups of atoms that bear a positive or negative electric charge. • Many familiar compounds are composed of ions. Table salt, or sodium chloride (NaCl) is one example. • These are generically referred to as salts. Ions & Ionic Compounds • IONS are atoms or groups of atoms with a formal positive or negative charge. • Removing electrons from an atom produces a CATION with a positive charge. • Adding a electrons to an atom gives an ANION with a negative charge. NaCl Chapter 2 — Atoms, Molecules, Ions Ions & Ionic Compounds Predicting Ion Charges 12 Forming Cations & Anions A CATION forms when an atom loses one or more electrons. An ANION forms when an atom gains one or more electrons Mg Mg2+ + 2 e- F + e- F- Charges on Common Ions -4 -3 -2 -1 +1 +2 +3 In general • metals (Mg) lose electrons forming cations • nonmetals (F) gain electrons forming anions Transition Metal Cations By losing or gaining electrons, an atom has same number of electrons as the nearest Noble gas atom. Nonmetal Ions Names NONMETAL + n e- Xnwhere n = 8 - Group no. • These elements may have more than one charge states. • Oxidation Number (charge) found from the formula. Group 4A C4-,carbide Example: FeCl2 chlorine forms a -1 charge so Fe = +2 Group 5A N3-, nitride Group 6A O2-, FeCl3 chlorine forms a -1 charge so Fe = +3 +3 + 3 (-1) = 0 oxide F-, fluoride S2-, sulfide Cl-, chloride +2 + 2 (-1) = 0 Example: Group 7A Br-, bromide Name derived by adding “ide” suffix. I-, iodide Chapter 2 — Atoms, Molecules, Ions 13 Metal Ions Names Metal Ions Names Main group elements All metal cations are named after there representative element followed by the word “ion”. sodium barium Na Na+ sodium ion Ba Ba2+ Barium ion Al3+ aluminum ion aluminum Al Transition elements Since the transition elements have multiple charge states, each must be identified uniquely. Cu Cu+ copper (I) ion Cu Cu2+ copper (II) ion Polyatomic Ions • Polyatomic anions are groups of atoms (molecules) with a net charge. • You must MEMORIZE the names and formulas in Table 2.4 • This requires practice in the form of time! • Flash cards often help. Have your friends quiz you! Remember: chem – is – try Note: many O containing anions have names ending in –ate (or -ite). Polyatomic Anions Names Two polyatomic anions are named like simple ions with the suffix (ending) -ide. Example: OH- hydroxide ion CN- cyanide ion Polyatomic Ions NH4+ ammonium ion Most other complex ions end in -ate or -ite. SO 24- sulfate ion SO32- sulfite ion One of the few common polyatomic cations Chapter 2 — Atoms, Molecules, Ions Compounds Formed From Ions CATION + ANION COMPOUND Na+ + Cl NaCl All ionic compounds are neutral in charge. The individual charges of the cations and anions must sum to zero. 14 Ionic Compounds NH4+ Cl- ammonium chloride, NH4Cl Naming Ionic Compounds Metal of fixed charge with a complex ion Cation Anion Formula Naming Ionic Compounds Al2(SO4)3 Name spoken as: Al - two, sulfate, taken three times K+ OH Ca2+ OH Ca(OH)2 Calcium hydroxide Na+ SO24- Na2SO4 Sodium sulfate Al3+ 24 Al2(SO4)3 Aluminum sulfate SO KOH Potassium hydroxide Naming Ionic Compounds Metal of variable charge with a complex ion Cation Anion Formula Name Pb2+ SO24- PbSO4 lead (II) sulfate Pb4+ SO24- Pb(SO4)2 lead (IV) sulfate Fe3+ NO3- Fe(NO3)3 Iron (III) nitrate Fe2+ NO2- Fe(NO2)2 Iron (II) nitrite or Al -two, parenthesis, sulfate, three Properties of Ionic Compounds Forming NaCl from Na(s) and Cl2(g) • A metal atom can transfer an electron to a nonmetal atom. • The resulting cation and anion are attracted to each other by electrostatic forces. Chapter 2 — Atoms, Molecules, Ions Electrostatic Forces COULOMB’S LAW 15 Electrostatic Forces • The oppositely charged ions in ionic compounds are attracted to one another by ELECTROSTATIC FORCES. • These forces are governed by COULOMB’S LAW. • As ion charges increase, the attractive forces between oppositely charged ions increases. • As the distance between ions increase, the attractive forces decreases. Electrostatic Forces Affect of Coulomb’s Law COULOMB’S LAW NaCl, Na+ and Cl-, m.p. 804 oC The bond distance in an ionic compound is measures as a function of the sum of the individual ionic radii. Molecular Compounds Compounds without Ions MgO, Mg2+ and O2m.p. 2800 oC MgO with the greater charge and smaller bond distance has the higher melting point. Naming Molecular Compounds When non-metals combine, they form molecules. They may do so in multiple forms: CO “carbon monoxide” CO2 Carbon dioxide CH4 methane BCl3 boron trichloride CO2 “carbon dioxide” Because of this we need to specify the number of each atom by way of a prefix. 1 mono 6 hexa 2 di 7 hepta 3 tri 8 octa 4 tetra 9 nona 5 penta 10 deca Chapter 2 — Atoms, Molecules, Ions Examples: Formula Naming Molecular Compounds Name: BCl3 boron trichloride SO3 sulfur trioxide NO nitrogen monoxide we don’t write: 16 CO2 Carbon dioxide nitrogen monooxide or mononitrogen monoxide CH4 methane N2O4 BCl3 boron trichloride dinitrogen tetraoxide Counting Atoms Mg burns in air (O2) to produce white magnesium oxide, MgO. All are formed from two or more nonmetals. Ionic compounds generally involve a metal and nonmetal (NaCl) Counting Atoms: The Mole Chemistry is a quantitative science—we need a “counting unit” the: MOLE How can we figure out how much oxide is produced from a given mass of Mg? Counting Atoms: The Mole Avogadro’s Number and the Mole • The concept of a mole is defined so that we may equate the amount of matter (mass) to the number of particles (mole). • The Standard is based upon the C -12 isotope. • The mass of one C - 12 atom is 1.99265 10-23 g. • The atomic mass of C - 12 is defined as exactly 12 u. • Therefore: 1u = (the mass of one 12C atom 12) = 1.66054 10-24g = 1.66054 10-27 kg 1 mole is the amount of substance that contains as many particles (atoms, molecules) as there are in 12.0 g of 12C. 518 g of Pb, 2.50 mol Counting Atoms: The Mole Avogadro’s Number and the Mole • Since one mole of C -12 has a mass of 12g (exactly), 12g of 12C contains 6.022142 1023 C -12 atoms. • But carbon exists as 3 isotopes: C -12, C -13 & C -14 • The average atomic mass of carbon is 12.011 u. • From this we conclude that 12.011g of carbon contains 6.022142 1023 C-atoms Is this a valid assumption? Yes, since NA is so large, the statistics hold. Chapter 2 — Atoms, Molecules, Ions 17 Particles and the Mole “Avogadro’s Number” Named for Amedeo Avogadro 1776-1856 Molar Mass Since we can equate mass (how much matter) with moles (how many particles) we now have a conversion factor that relates the two. mols 6.0221415 1023 molar mass (g/mol) = grams The Molar Mass of a substance is the amount of matter that contains one-mole or 6.022 1023 particles. aka: Avogadro's number (NA) There is Avogadro’s number of particles in a mole of any substance. One-Mole Amounts The atomic masses on the Periodic Table also represent the molar masses of each element in grams per mole (g/mol) Grams, Moles, & Molar Mass The molar mass of an atom is a conversion factor that relates mass (grams) to moles and vice versa. (how much matter) (number of atoms) g mol Avogadro's number (NA) relates moles numbers of individual particles: 1 mol Mass/Moles Conversions 6.022 1023 particles 6.022 1023 particles mol Problem: How many magnesium atoms are there in 150.0 g of magnesium? Chapter 2 — Atoms, Molecules, Ions 18 Problem: How many magnesium atoms are there in 150.0 g of magnesium? Problem: How many magnesium atoms are there in 150.0 g of magnesium? Solution: Use the molar mass of Mg from the periodic table and Avogadro's number. Solution: Use the molar mass of Mg from the periodic table and Avogadro's number. 150.0g Mg Problem: How many magnesium atoms are there in 150.0 g of magnesium? Problem: How many magnesium atoms are there in 150.0 g of magnesium? Solution: Use the molar mass of Mg from the periodic table and Avogadro's number. Solution: Use the molar mass of Mg from the periodic table and Avogadro's number. 150.0g Mg ´ 1 mol Mg 24.305g Mg 150.0g Mg ´ 1 mol Mg 6.022 x 1023 Mg atoms ´ 24.305g Mg 1 mol Mg Problem: How many magnesium atoms are there in 150.0 g of magnesium? Problem: How many magnesium atoms are there in 150.0 g of magnesium? Solution: Use the molar mass of Mg from the periodic table and Avogadro's number. Solution: Use the molar mass of Mg from the periodic table and Avogadro's number. 150.0g Mg ´ 1 mol Mg 6.022 x 1023 Mg atoms ´ 24.305g Mg 1 mol Mg 150.0g Mg ´ 1 mol Mg 6.022 x 1023 Mg atoms ´ 24.305g Mg 1 mol Mg = 3.717 ´ 1024 Mg atoms Chapter 2 — Atoms, Molecules, Ions Problem: How many magnesium atoms are there in 150.0 g of magnesium? 19 How many grams of chromium are there in 25.1g of chromium (III) acetate? Solution: Use the molar mass of Mg from the periodic table and Avogadro's number. Cr(C2H3O2)3 = 229.13 g/mol 1 mol Mg 6.022 x 1023 Mg atoms 150.0g Mg ´ ´ 24.305g Mg 1 mol Mg 25.1g Cr(C2 H3O2 )3 ´ 1 mol Cr(C2 H3O 2 )3 229.13g Cr(C2 H3O 2 )3 ´ = 3.717 ´ 1024 Mg atoms 5.70 g Cr 3 sf (4 sig figs) big number as expected! Molecular Weight & Molar Mass 1 mol Cr 52.00g Cr ´ = 1 mol Cr(C2 H3O2 )3 1mol Cr What is the molar mass of ethanol, C2H6O? Molecular weight = sum of the atomic weights of all atoms in the molecule. 1 mol contains: Molar mass = molecular weight in grams 2 mol C (12.01g C/1 mol) = 24.02 g C 6 mol H (1.01g H/1 mol) = 6.06 g H 1 mol O (16.00g O/1 mol) = 16.00 g O TOTAL = molar mass Molar Masses Note that the mass of water is included in the molar mass of a compound. = 46.08 g/mol Percent Composition • Each element in a compound my be represented by its weight (mass) percent. • This is calculated by the following: Wt.% = mass (g) of indiv. element in one formula unit ´ 100 mass of one formula unit (g) • The sum of the wt. % of each element in the compound must sum to be 100%. Chapter 2 — Atoms, Molecules, Ions 20 Percent Composition Percent Composition • A pure compound always consists of the same elements combined in the same proportions by weight. • Therefore, we can express molecular composition as PERCENT BY WEIGHT • Consider some of the family of nitrogenoxygen compounds: • NO2, nitrogen dioxide and closely related, NO, nitrogen monoxide (or nitric oxide) Ethanol, C2H6O 52.13% C 13.15% H 34.72% O Structure of NO2 Percent Composition Percent Composition Consider NO2: What is the weight percent of each element in the compound? Consider NO2: What is the weight percent of each element in the compound? Solution: From the formula of the compound, find the moles and mass of each element, then use the masses and molar mass to find the mass percentages. Wt% N = 1mol N 14.01g 1mol NO2 ´ ´ ´ 100 = 30.45% N 1mol NO2 1mol N 46.01g NO2 Wt% O = 2mol O 16.00g O 1mol NO2 ´ ´ ´ 100 = 69.55%O 1mol NO2 1mol O 46.01g NO2 1 mol N ? 14.01g N 16.00g O + 1 mol O ? = 46.01 g/mol 1mol N 1mol O Describing Compound Formulas: Percent Composition Given the relative percentages of each element in a compound, 10 % X, 20 % Y, 30 % Z etc… one can find the empirical formula of the compound. The empirical formula of a compound or molecule represents the simplest ratio of each element in 1 mol of the compound or molecule. Or: 100 30.45% = 69.55% O Example: A compound is found to be 64.82 % carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? Chapter 2 — Atoms, Molecules, Ions 21 Problem Strategy Map Example: A compound is found to be 64.82 % The solution to the problem involves the following series of steps. Solution: determine X, Y & Z in (CXHYOZ) Example: A compound is found to be 64.82 % Example: A compound is found to be 64.82 % carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? Solution: determine X, Y & Z in (CXHYOZ) Solution: determine X, Y & Z in (CXHYOZ) 1. Since the percentages for each element sum to 1. Since the percentages for each element sum to 100%, if one equates % to grams (g), the sum of 100%, if one equates % to grams (g), the sum of the masses must be 100g. the masses must be 100g. (i.e. one can assume 100g of the compound) carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? (i.e. one can assume 100g of the compound) 64.82 g C 21.59 g O 13.59 g H Example: A compound is found to be 64.82 % Example: A compound is found to be 64.82 % carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? 2. Convert the grams of each element to moles. 2. Convert the grams of each element to moles. (g element X mole X etc…) (g element X mole X etc…) 1 mol C 64.82g C ´ = 5.397 mol C 12.011 g C Chapter 2 — Atoms, Molecules, Ions 22 Example: A compound is found to be 64.82 % Example: A compound is found to be 64.82 % carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? 2. Convert the grams of each element to moles. 2. Convert the grams of each element to moles. Oxygen in a compound is always monatomic… (g element X mole X etc…) 1 mol C 64.82g C ´ = 5.397 mol C 12.011 g C 21.59g O ´ 1 mol O 16.00 g O = 1.349 mol O Oxygen in a compound is always monatomic… The same applies to hydrogen! (g element X mole X etc…) 1 mol C 64.82g C ´ = 5.397 mol C 12.011 g C 21.59g O ´ 1 mol O 16.00 g O 13.59g H ´ 1 mol H = 13.48 mol H 1.0079 g H = 1.349 mol O Example: A compound is found to be 64.82 % Example: A compound is found to be 64.82 % carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? 3. Divide each of the individual moles by the smallest number of moles to gain the molar ratios for each element in the compound. 5.397 mol C 1.349 mol O 13.48 mol H These are the formula subscripts. (C XHYOZ etc…) 3. Divide each of the individual moles by the smallest number of moles to gain the molar ratios for each element in the compound. 5.397 mol C 1.349 mol O 13.48 mol H These are the formula subscripts. (C XHYOZ etc…) Subscript for C X= 5.397 = 4.001 1.349 Example: A compound is found to be 64.82 % Example: A compound is found to be 64.82 % carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? 3. Divide each of the individual moles by the smallest number of moles to gain the molar ratios for each element in the compound. 5.397 mol C 1.349 mol O 13.48 mol H These are the formula subscripts. (C XHYOZ etc…) 3. Divide each of the individual moles by the smallest number of moles to gain the molar ratios for each element in the compound. 5.397 mol C 1.349 mol O 13.48 mol H These are the formula subscripts. (C XHYOZ etc…) Subscript for C X= Subscript for H 13.48 5.397 = 4.001 Y = = 9.992 1.349 1.349 Subscript for C X= Subscript for H Subscript for O 13.48 1.349 5.397 = 4.001 Y = = 9.992 Z = = 1.000 1.349 1.349 1.349 Chapter 2 — Atoms, Molecules, Ions 23 Example: A compound is found to be 64.82 % Example: A compound is found to be 64.82 % carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? **If the ratios are fractional (0.5, 1.5 or 0.333) multiply each ratio by a whole number to get even number Rounding to the nearest whole numbers: CXHYOZ formula subscripts. X = 4.001 = 4 Y = 9.992 =10 Z = 1.000 = 1 Examples: 0.5 2 = 1 0.25 4 = 1 0.333 3 = 1 C4H10O The empirical formula is: The results of this calculation tells us only about the empirical formula of the compound. Example: A compound is found to be 64.82 % Example: A compound is found to be 64.82 % carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? 4. Determining the Molecular Formula 4. Determining the Molecular Formula For some compounds, the molecular formula is a multiple of the empirical formula: empirical formula n x CXHYOZ Since the molecular formula is a multiple is scaled by a factor “n”, the molecular and empirical molar masses must also scale by the same ratio. molecular formula = ( ) ( ) molar formula mass g CnXHnYOnZ n = 2,3,4… mol =n empirical formula mass g mol Example: A compound is found to be 64.82 % Example: A compound is found to be 64.82 % carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? 4. Determining the Molecular Formula 4. Determining the Molecular Formula • The empirical formula weight for C4H10O is 74.12 g/mol • In a separate experiment, the molar mass of the compound was determined to be 222.1 g/mol. • What is the molecular formula? 222.1 g / mol = 2.996 or 3 74.12 g / mol (C4H10O )´ 3 = C12H30O3 The molecular formula of the compound is: C12H30O3 Chapter 2 — Atoms, Molecules, Ions Determine the Formula of a Compound of Sn & I 24 Example: 1.056g of Sn is reacted with 1.947g of I2. After the reaction is complete, 0.601g of Sn is recovered. What is the formula for SnIx? Sn(s) + I2(s) SnIx unknown formula Example: 1.056g of Sn is reacted with 1.947g of I2. After the reaction is complete, 0.601g of Sn is recovered. What is the formula for SnIx? Example: 1.056g of Sn is reacted with 1.947g of I2. After the reaction is complete, 0.601g of Sn is recovered. What is the formula for SnIx? What do we know? • The mass of I2 reacted. • The mass of Sn reacted. (mass initial – mass left over) What do we know? • The mass of I2 reacted. • The mass of Sn reacted. (mass initial – mass left over) What does that tell us? • Moles of Sn and I2 reacted. Example: 1.056g of Sn is reacted with 1.947g of I2. After the reaction is complete, 0.601g of Sn is recovered. What is the formula for SnIx? Example: 1.056g of Sn is reacted with 1.947g of I2. After the reaction is complete, 0.601g of Sn is recovered. What is the formula for SnIx? What do we know? • The mass of I2 reacted. • The mass of Sn reacted. (mass initial – mass left over) What does that tell us? • Moles of Sn and I2 reacted. What does that yield? • The compound’s formula! mass of Sn that combined • Mass of Sn initially • Mass of Sn recovered • Mass of Sn reacted moles of Sn used: 0.455 g Sn ´ with the I2: = 1.056 g = 0.601 g = 0.455 g 1 mol Sn = 3.83 x 10-3 mol Sn 118.7 g Sn Chapter 2 — Atoms, Molecules, Ions 25 Example: 1.056g of Sn is reacted with 1.947g of I2. After the reaction is complete, 0.601g of Sn is recovered. What is the formula for SnIx? Example: 1.056g of Sn is reacted with 1.947g of I2. After the reaction is complete, 0.601g of Sn is recovered. What is the formula for SnIx? mols of I that combined with the Sn: Now find the ratio of number of moles of moles of I and Sn that combined. 1.947 g I2 ´ 1 mol I2 2 mol I ´ = 1.534 ´ 10-2 mol I 253.81 g I2 1 mol I2 1.534 x 10-2 mol I 3.83 x 10-3 mol Sn = 4.01 mol I 1.00 mol Sn Empirical formula is SnI4 Hydrates Naming Hydrates A hydrate is a substance composed of an inorganic salt and physically bound water. MXnH2O salt prefix: water n = is the ratio of moles of water to 1 mole of the salt n= Salt name + prefix hydrate mols H 2O mols MX Heating Hydrates mono, di, tri etc… BaCl22H2O barium chloride dihydrate sodium sulfate pentahydrate Hydrate Formulas When a hydrate is heated, the water is liberated. Ä ® BaCl2 (s) +2H2O(g) BaCl2 × 2H2O(s) ¾¾ • For every one mole of hydrate, n moles (in this case 2) of water are liberated. • After heating, any mass that remains is due to the salt residue. • Therefore, any mass loss in heating is due to the water loss. mass sample – mass residue = mass water lost The solution to the problem involves the following series of steps. Na2SO45H2O Chapter 2 — Atoms, Molecules, Ions Hydrate Formulas From the mass of the water lost, one can find the mols of water associated with the hydrate. Ä ® BaCl2 (s) +2H2O(g) BaCl2 × 2H2O(s) ¾¾ Knowing the mass and chemical identity of the anhydrous salt… One can determine the mols of the anhydrous salt. The ratio of mols of water to mols of salt (n) is found by: n= mols H2O from hydrate mols anhydrous salt 26