Atoms, Molecules, and Ions (CHEM110) Chapter 2 PDF

Atoms, Molecules, and Ions Chapter 2 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.  Dalton’s Atomic Theory (1808) 1. Elements are composed of extremely small particles called atoms. 2. All atoms of a given element...

Atoms, Molecules, and Ions Chapter 2 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.  Dalton’s Atomic Theory (1808) 1. Elements are composed of extremely small particles called atoms. 2. All atoms of a given element are identical, having the same size, mass and chemical properties. The atoms of one element are different from the atoms of all other elements. 3. Compounds are composed of atoms of more than one element. In any compound, the ratio of the numbers of atoms of any two of the elements present is either an integer or a simple fraction. 4. A chemical reaction involves only the separation, combination, or rearrangement of atoms; it does not result in their creation or destruction. 2 Dalton’s Atomic Theory Law of Multiple Proportions 3 16 X + 8Y 8 X2Y Law of Conservation of Mass 4  Cathode Ray Tube J.J. Thomson's experiments with cathode ray tubes showed J.J. Thomson, measured mass/c that all atoms contain tiny (1906 Nobel Prize in Phy negatively charged subatomic particles or electrons. 5 Cathode Ray Tube 6  Millikan’s Experiment Measured mass of e- (1923 Nobel Prize in Physics) e- charge = -1.60 x 10-19 C Thomson’s charge/mass of e- = -1.76 x 108 C/g e- mass = 9.10 x 10-28 g 7 Types of Radioactivity (uranium compound) 8 Thomson’s Model 9  Rutherford’s Experiment (1908 Nobel Prize in Chemistry)  particle velocity ~ 1.4 x 107 m/s (~5% speed of light) 1. atoms positive charge is concentrated in the nucleus 2. proton (p) has opposite (+) charge of electron (-) 3. mass of p is 1840 x mass of e- (1.67 x 10-24 g) 10  Rutherford’s Model of the Atom atomic radius ~ 100 pm = 1 x 10-10 m nuclear radius ~ 5 x 10-3 pm = 5 x 10-15 m “If the atom is the Houston Astrodome, then the nucleus is a marble on the 50-yard line.” 11  Chadwick’s Experiment (1932) (1935 Noble Prize in Physics) H atoms: 1 p; He atoms: 2 p mass He/mass H should = 2 measured mass He/mass H = 4  + 9Be 1 n + 12C + energy neutron (n) is neutral (charge = 0) n mass ~ p mass = 1.67 x 10-24 g 12  Components of the Atom mass p ≈ mass n ≈ 1840 x mass e- 13  Atomic Number, Mass Number, and Isotopes Atomic number (Z) = number of protons in nucleus Mass number (A) = number of protons + number of neutrons = atomic number (Z) + number of neutrons Isotopes are atoms of the same element (X) with different numbers of neutrons in their nuclei Mass Number A ZX Element Symbol Atomic Number 1 2 3 1H 1H (D) 1H (T) 235 238 92 U 92 U 14 The Isotopes of Hydrogen 15 Example 2.1 Give the number of protons, neutrons, and electrons in each of the following species: (a) (b) (c) (d) carbon-14 Example 2.1 Strategy Recall that the superscript denotes the mass number (A) and the subscript denotes the atomic number (Z). Mass number is always greater than atomic number. (The only exception is H, where the mass number is equal to the atomic number.) In a case where no subscript is shown, as in parts (c) and (d), the atomic number can be deduced from the element symbol or name. To determine the number of electrons, remember that because atoms are electrically neutral, the number of electrons is equal to the number of protons. Example 2.1 Solution (a) The atomic number is 11, so there are 11 protons. The mass number is 20, so the number of neutrons is 20 − 11 = 9. The number of electrons is the same as the number of protons; that is, 11. (b) The atomic number is the same as that in (a), or 11. The mass number is 22, so the number of neutrons is 22 − 11 = 11. The number of electrons is 11. Note that the species in (a) and (b) are chemically similar isotopes of sodium. Example 2.1 (c) The atomic number of O (oxygen) is 8, so there are 8 protons. The mass number is 17, so there are 17 − 8 = 9 neutrons. There are 8 electrons. (d) Carbon-14 can also be represented as 14C. The atomic number of carbon is 6, so there are 14 − 6 = 8 neutrons. The number of electrons is 6.  Noble Gas 20 Halogen The Modern Periodic Table Group Period Alkali Earth Metal Alkali Metal A molecule is an aggregate of two or more atoms in a definite arrangement held together by chemical forces. H2 H2O NH3 CH4 A diatomic molecule contains only two atoms: H2, N2, O2, Br2, HCl, CO diatomic elements A polyatomic molecule contains more than two atoms: O3, H2O, NH3, CH4 21 An ion is an atom, or group of atoms, that has a net positive or negative charge. cation – ion with a positive charge If a neutral atom loses one or more electrons it becomes a cation. 11 protons 11 protons Na 11 electrons Na + 10 electrons anion – ion with a negative charge If a neutral atom gains one or more electrons it becomes an anion. 17 protons 17 protons Cl 17 electrons Cl - 18 electrons 22 A monatomic ion contains only one atom: Na+, Cl-, Ca2+, O2-, Al3+, N3- A polyatomic ion contains more than one atom: OH-, CN-, NH4+, NO3- 23 Common Ions Shown on the Periodic Table 24 Formulas and Models 25 A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance. An empirical formula shows the simplest whole-number ratio of the atoms in a substance. molecular empirical H2O H2O C6H12O6 CH2O O3 O N2H4 NH2 26 Example 2.3 Write the empirical formulas for the following molecules: (a)biborane (B2H6), which is used in rocket propellants (b)glucose (C6H12O6), a substance known as blood sugar (c)nitrous oxide (N2O), a gas that is used as an anesthetic gas (“laughing gas”) and as an aerosol propellant for whipped creams. Example 2.3 Strategy Recall that to write the empirical formula, the subscripts in the molecular formula must be converted to the smallest possible whole numbers. Example 2.3 Solution (a)There are two boron atoms and six hydrogen atoms in diborane. Dividing the subscripts by 2, we obtain the empirical formula BH3. (b) In glucose there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. Dividing the subscripts by 6, we obtain the empirical formula CH2O. Note that if we had divided the subscripts by 3, we would have obtained the formula C 2H4O2. Although the ratio of carbon to hydrogen to oxygen atoms in C 2H4O2 is the same as that in C6H12O6 (1:2:1), C2H4O2 is not the simplest formula because its subscripts are not in the smallest whole-number ratio. Example 2.3 (c) Because the subscripts in N2O are already the smallest possible whole numbers, the empirical formula for nitrous oxide is the same as its molecular formula. Ionic compounds consist of a combination of cations and anions. The formula is usually the same as the empirical formula. The sum of the charges on the cation(s) and anion(s) in each formula unit must equal zero. The ionic compound NaCl 31 The most reactive metals (green) and the most reactive nonmetals (blue) combine to form ionic compounds. 32 Formulas of Ionic Compounds 2 x +3 = +6 3 x -2 = -6 Al2O3 Al3+ O2- 1 x +2 = +2 2 x -1 = -2 CaBr2 Ca2+ Br- 2 x +1 = +2 1 x -2 = -2 Na2CO3 Na+ CO32- 33 Example 2.4 Write the formula of magnesium nitride, containing the Mg2+ and N3− ions. When magnesium burns in air, it forms both magnesium oxide and magnesium nitride. Example 2.4 Strategy Our guide for writing formulas for ionic compounds is electrical neutrality; that is, the total charge on the cation(s) must be equal to the total charge on the anion(s). Because the charges on the Mg2+ and N3− ions are not equal, we know the formula cannot be MgN. Instead, we write the formula as MgxNy, where x and y are subscripts to be determined. Example 2.4 Solution To satisfy electrical neutrality, the following relationship must hold: (+2)x + (−3)y = 0 Solving, we obtain x/y = 3/2. Setting x = 3 and y = 2, we write Check The subscripts are reduced to the smallest whole- number ratio of the atoms because the chemical formula of an ionic compound is usually its empirical formula.  Chemical Nomenclature Ionic Compounds – Often a metal + nonmetal – Anion (nonmetal), add “-ide” to element name BaCl2 barium chloride K2O potassium oxide Mg(OH)2 magnesium hydroxide KNO3 potassium nitrate 37  Transition metal ionic compounds – indicate charge on metal with Roman numerals FeCl2 2 Cl- -2 so Fe is +2 iron(II) chloride FeCl3 3 Cl- -3 so Fe is +3 iron(III) chloride Cr2S3 3 S-2 -6 so Cr is +3 (6/2) chromium(III) sulfide 38 39 40 Example 2.5 Name the following compounds: (a)Fe(NO3)2 (b)Na2HPO4 (c)(NH4)2SO3 Example 2.5 Strategy Our reference for the names of cations and anions is Table 2.3. Keep in mind that if a metal can form cations of different charges (see Figure 2.10), need to use the Stock system. Example 2.5 Solution (a)The nitrate ion (NO3−) bears one negative charge, so the iron ion must have two positive charges. Because iron forms both Fe+ and Fe2+ ions, we need to use the Stock system and call the compound iron(II) nitrate. (b)The cation is Na+ and the anion is HPO42− (hydrogen phosphate). Because sodium only forms one type of ion (Na+), there is no need to use sodium(I) in the name. The compound is sodium hydrogen phosphate. (c) The cation is NH4+ (ammonium ion) and the anion is SO32− (sulfite ion). The compound is ammonium sulfite. Example 2.6 Write chemical formulas for the following compounds: (a)mercury(I) nitrite (b)cesium sulfide (c)calcium phosphate Example 2.6 Strategy We refer to Table 2.3 for the formulas of cations and anions. Recall that the Roman numerals in the Stock system provide useful information about the charges of the cation. Example 2.6 Solution (a)The Roman numeral shows that the mercury ion bears a +1 charge. According to Table 2.3, however, the mercury(I) ion is diatomic (that is, ) and the nitrite ion is. Therefore, the formula is Hg2(NO2)2. (b)Each sulfide ion bears two negative charges, and each cesium ion bears one positive charge (cesium is in Group 1A, as is sodium). Therefore, the formula is Cs2S. Example 2.6 (c) Each calcium ion (Ca2+) bears two positive charges, and each phosphate ion ( ) bears three negative charges. To make the sum of the charges equal zero, we must adjust the numbers of cations and anions: 3(+2) + 2(−3) = 0 Thus, the formula is Ca3(PO4)2.  Molecular compounds − Nonmetals or nonmetals + metalloids − Common names − H2O, NH3, CH4 − Element furthest to the left in a period and closest to the bottom of a group on periodic table is placed first in formula − If more than one compound can be formed from the same elements, use prefixes to indicate number of each kind of atom − Last element name ends in -ide 48  Molecular Compounds HI hydrogen iodide NF3 nitrogen trifluoride SO2 sulfur dioxide N2Cl4 dinitrogen tetrachloride NO2 nitrogen dioxide N2O dinitrogen monoxide 49 Example 2.7 Name the following molecular compounds: (a)SiCl4 (b)P4O10 Example 2.7 Strategy We refer to Table 2.4 for prefixes. In (a) there is only one Si atom so we do not use the prefix “mono.” Solution (a)Because there are four chlorine atoms present, the compound is silicon tetrachloride. (b)There are four phosphorus atoms and ten oxygen atoms present, so the compound is tetraphosphorus decoxide. Note that the “a” is omitted in “deca.” Example 2.8 Write chemical formulas for the following molecular compounds: (a)carbon disulfide (b) disilicon hexabromide Example 2.8 Strategy Here we need to convert prefixes to numbers of atoms (see Table 2.4). Because there is no prefix for carbon in (a), it means that there is only one carbon atom present. Solution (a)Because there are two sulfur atoms and one carbon atom present, the formula is CS2. (b) There are two silicon atoms and six bromine atoms present, so the formula is Si2Br6.