Chapter 2. Atoms and Elements

Chapter 2. Atoms and Elements John Dalton Robert Millikan J.J. Thomson Ernest Rutherford The Components of Matter Elements, Compounds, Mixtures Dalton s Atomic Theory Modern Atomic Theory Atomic and Mass Number Formulas, and Masses Modified by Dr. Cheng-Yu Lai 1

Modern Atomic Theory and the Laws That Led to It The three most important laws that led to the development and acceptance of the atomic theory are as follows: The law of conservation of mass The law of definite proportions The law of multiple proportions 2014 Pearson Education, Inc.

Conservation of Mass Law of Conservation of Mass: Mass is neither created nor destroyed in chemical reactions. Aqueous solutions of mercury(ii) nitrate and potassium iodide will react to form a precipitate of mercury(ii) iodide and aqueous potassium iodide. 3.25 g + 3.32 g = 6.57 g Hg(NO3)2(aq) + 2KI(aq) HgI2(s) + 2KNO3(aq) 4.55 g + 2.02 g = 6.57 g 2012 Pearson Education, Inc. Chapter 2/3

Conservation of Mass If the product weight is more than reactant weight, what does this indicate? Error from left over in your lab rinse your beaker 2012 Pearson Education, Inc. Chapter 2/4

The Law of Definite Proportions Law of Definite Proportions: Different samples of a pure chemical substance always contain the same proportion of elements by mass. For example, the decomposition of 18.0 g of water results in 16.0 g of oxygen and 2.0 g of hydrogen, or an oxygen-to-hydrogen mass ratio of: By mass, H2O water is: 88.8 % oxygen, 11.2 % hydrogen 2012 Pearson Education, Inc. Chapter 2/5

Example 2.1 Law of Definite Proportions Two samples of carbon dioxide are decomposed into their constituent elements. One sample produces 25.6 g of oxygen and 9.60 g of carbon, and the other produces 21.6 g of oxygen and 8.10 g of carbon. Show that these results are consistent with the law of definite proportions. Solution To show this, calculate the mass ratio of one element to the other for both samples by dividing the mass of one element by the mass of the other. For convenience, divide the larger mass by the smaller one. For the first sample: For the second sample: The ratios are the same for the two samples, so these results are consistent with the law of definite proportions. No matter where is CO2 from, the mass ratio between Oxygen and Carbon is the same Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.

The Law of Multiple Proportions In 1804, John Dalton published his law of multiple proportions. When two elements (call them A and B) form two different compounds, the masses of element B that combine with element A can be expressed as a ratio of small whole numbers. An atom of A combines with either one, two, three, or more atoms of B (AB1, AB2, AB3, etc.). 2012 Pearson Education, Inc. Chapter 2/7

[2Oxygen] CO 2 [1Carbon] [1Oxygen] CO [1Carbon] CO 2 [2Oxygen / 1Carbon] 2 CO [1Oxygen / 1Carbon] 2014 Pearson Education, Inc.

Dalton s Atomic Theory (1808) John Dalton All matter is composed of extremely small particles called atoms Atoms of a given element are identical in size, mass, and other properties; atoms of different elements differ in size, mass, and other properties Atoms cannot be subdivided, created, or destroyed Atoms of different elements combine in simple whole-number ratios to form chemical compounds In chemical reactions, atoms are combined, separated, or rearranged Please memorize these 2012 Pearson Education, Inc. 9

Cathode Ray Experiment - Discovery of the Electron In 1897, J.J. Thomson used a cathode ray tube to deduce the presence of a negatively charged particle. Cathode ray tubes pass electricity through a gas that is contained at a very low pressure. J. J. Thomson measured the charge-to-mass ratio ( e/m) of the cathode ray particles by deflecting them using electric and magnetic fields, as shown in the figure. The value he measured was 1.76 108 coulombs (C) per gram. 2012 Pearson Education, Inc. 10

Millikan s Oil Drop Experiment: The Charge of the Electron American physicist Robert Millikan (1868 1953), performed his now famous oil drop experiment in which he deduced the charge of a single electron. The measured charge on any drop was always a whole-number multiple of 1.96 10 19, the fundamental charge of a single electron. 2014 Pearson Education, Inc.

Mass of the Electron With this number in hand, and knowing Thomson s mass-to-charge ratio for electrons, we can deduce the mass of an electron: 9.10 10-28 g 1.96 10 19 C X 1/ (1.76 108 coulombs (C) / g) 1909 Robert Millikan determines the mass of the electron. The drop 2014oil Pearson Education, apparatus Inc. Mass of the electron is 9.109 x 10-31 kg

Conclusions from the Study of the Electron Cathode rays have identical properties regardless of the element used to produce them. All elements must contain identically charged electrons. Atoms are neutral, so there must be positive particles in the atom to balance the negative charge of the electrons Electrons have so little mass that atoms must contain other particles that account for most of the mass 2012 Pearson Education, Inc. plum-pudding model.13

Thomson s Atomic Model J.J. Thomson Thomson believed that the electrons were like plums embedded in a positively charged pudding, thus it was called the plum pudding model. 2012 Pearson Education, Inc. 14

Rutherford s Gold Foil Experiment In 1909, Ernest Rutherford (1871 1937), who had worked under Thomson and subscribed to his plum-pudding model, performed an experiment in an attempt to confirm Thomson s model. In the experiment, Rutherford directed the positively charged particles at an ultra thin sheet of gold foil. 2014 Pearson Education, Inc. Chapter 2/15

Rutherford s Gold Foil Experiment Alpha particles are helium nuclei Particles were fired at a thin sheet of gold foil Particle hits on the detecting screen (film) are recorded 2012 Pearson Education, Inc. 16

Rutherford s Gold Foil Experiment He concluded that matter must not be as uniform as it appears. It must contain large regions of empty space dotted with small regions of very dense matter. 2014 Pearson Education, Inc. Chapter 2/17

Rutherford s Findings Most of the particles passed right through A few particles were deflected VERY FEW were greatly deflected Rutherford proposed that the atom must consist mainly of empty space with the mass concentrated in a tiny central core the nucleus. Conclusions: The nucleus is small The nucleus is dense The nucleus is positively charged 18

Rutherford s nuclear theory Building on this idea, he proposed the nuclear theory of the atom, with three basic parts: 1. Most of the atom s mass and all of its positive charge are contained in a small core called a nucleus. 2. Most of the volume of the atom is empty space, throughout which tiny, negatively charged electrons are dispersed. 3. There are as many negatively charged electrons outside the nucleus as there are positively charged particles (named protons) within the nucleus (named neutrons), so that the atom is electrically neutral. Later work by Rutherford and one of his students, British scientist James Chadwick (1891 1974), demonstrated that the previously unaccounted for mass was due to Chapter neutrons, neutral particles within the nucleus. 2014 Pearson Education, Inc. 2/19

Atomic Particles are composed of the same subatomic particles: Particle Electron Proton Neutron Charge Mass (kg) -1 9.109 x 10-31 +1 1.673 x 10-27 Location Electron cloud Nucleus 0 1.675 x 10-27 Nucleus The charge of the proton is opposite in sign but equal to that of the electron. The mass of the atom is primarily in the nucleus. The mass of a neutron is similar to that of a proton. 2012 Pearson Education, Inc. 20

Atomic Numbers The sum of the number of neutrons and protons in an atom is its mass number and is represented by the symbol A A = number of protons (p) + number of neutrons (n) where X is the chemical symbol, A is the mass number, and Z is the atomic number. 2012 Pearson Education, Inc. Chapter 2/21

Atomic Number in Periodic Table 2014 Pearson Education, Inc.

Atomic Number Atomic number (Z) of an element is the number of protons in the nucleus of each atom of that element. Element Carbon Phosphorus Gold # of protons Atomic # (Z) 6 6 15 15 79 2012 Pearson Education, Inc. 79 23

Mass Number Mass number is the number of protons and neutrons in the nucleus of an isotope. Mass # = p+ + n0 Nuclide p+ n0 e- Mass # Oxygen - 18 8 10 8 18 Arsenic - 75 Phosphorus - 31 33 42 33 75 15 16 15 31 How many protons, electrons, and neutrons are present in an atom of? + 0 - Nuclide Cr p 2012 Pearson Education, Inc. 24 n 28 e 24 Mass # 52 24

Isotopes: Varied Number of Neutrons carbon-12 mass number 12 6C 6 protons 6 electrons 6 neutrons 14 6C 6 protons 6 electrons 8 neutrons atomic number carbon-14 mass number atomic number Atoms with the same number of protons but a different 2012 Pearson Education, Inc. 25 number of neutrons are called isotopes.

Atomic Masses The mass of 1 atom of carbon-12 is defined to be 12 amu. Atomic mass is the average of all the naturally isotopes of that element. 2012 Pearson Education, Inc. Chapter 2/26

Atomic Masses Isotope Symbol Composition of the nucleus % in nature amu Carbon12 12C 6 protons 6 neutrons 98.89% 12.0000 Carbon13 13C 6 protons 7 neutrons 1.11% 13.0034 Carbon14 14C 6 protons 8 neutrons <0.01% 14.0032 mass of carbon = (12 amu)(0.9889) + (13.0034 amu)(0.0111) + (14.0032 amu)(0.0001) = 11.87 amu + 0.144 amu + 0.001amu 2012 Pearson Education, Inc. = 12.01 amu 2012 Pearson Education, Inc. 1 Carbon Atomic Mass = 12.011 amu 27

Example 2.5 Atomic Mass Copper has two naturally occurring isotopes: Cu-63 with a mass of 62.9396 amu and a natural abundance of 69.17%, and Cu-65 with a mass of 64.9278 amu and a natural abundance of 30.83%. Calculate the atomic mass of copper. Solution Convert the percent natural abundances into decimal form by dividing by 100. Calculate the atomic mass using the equation given in the text. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.

Modern Atomic Theory # 1 Several changes have been made to Dalton s theory. Dalton said: Atoms of a given element are identical in size, mass, and other properties; atoms of different elements differ in size, mass, and other properties Modern theory states: Atoms of an element have a characteristic average mass which is unique to that element. 2012 Pearson Education, Inc. Chapter 2/29

Modern Atomic Theory #2 Dalton said: Atoms cannot be subdivided, created, or destroyed Modern theory states: Atoms cannot be subdivided, created, or destroyed in ordinary chemical reactions. However, these changes CAN occur in nuclear reactions 2012 Pearson Education, Inc. Chapter 2/30

Moles and Formula Mass 2012 Pearson Education, Inc. 31

Converting between Number of Moles and Number of Atoms 1 dozen = 12 1 mole = 6.022 x 1023 For example, 1 mol of marbles corresponds to 6.02214 1023 marbles. 1 mol of sand grains corresponds to 6.02214 1023 sand grains. One mole of anything is 6.02214 1023 units of that thing. 1 mole C = 6.022 x 1023 C atoms The conversion factors take the following forms: 2012 Pearson Education, Inc. 32

Atomic Masses and the Mole Avogadro s Number (NA): One mole of any substance contains 6.022 x 1023 formula units. Molar Mass: The mass in grams of one mole of any element. It is numerically equivalent to its atomic mass. Amadeo Avogadro 6.022 x 1023 is called Avogadro s Number in honor 2012Amadeo Pearson Education, Inc. Chapter 2/33 of the Italian chemist Avogadro (1776-1855).

Converting between Mass and Amount (Number of Moles) The mass of 1 mol of atoms of an element is the ATOMIC molar mass. For example, H=1, C=12.01 found in the periodic table An element s molar mass in grams per mole is numerically equal to the element s atomic mass in atomic mass units (amu = g/mole). 1 mole H= 1 g 1 mole C = 12.01 g 2014 Pearson Education, Inc.

Conceptual Plan We now have all the tools to count the number of atoms in a sample of an element by weighing it. First, we obtain the mass of the sample. Then, we convert it to the amount in moles using the element s molar mass. Finally, we convert it to the number of atoms using Avogadro s number. The conceptual plan for these kinds of calculations takes the following form: 2014 Pearson Education, Inc.

Calculations with Moles: Converting moles to grams How many grams of lithium are in 3.50 moles of lithium? 3.50 mol Li 6.94 g Li 1 mol Li = 2012 Pearson Education, Inc. 45.1 g Li 36

Calculations with Moles: Converting grams to moles How many moles of lithium are in 18.2 grams of lithium? 18.2 g Li 1 mol Li 6.94 g Li = 2.62 mol Li 2012 Pearson Education, Inc. 37

Calculations with Moles: Using Avogadro s Number How many atoms of lithium are in 3.50 moles of lithium? 3.50 mol 6.02 x 1023 atoms 1 mol 2012 Pearson Education, Inc. = 2.07 x 1024 atoms 38

Calculations with Moles: Using Avogadro s Number How many atoms of lithium are in 18.2 g of lithium? 18.2 g Li 1 mol Li 6.94 g Li 6.022 x 1023 atoms Li 1 mol Li (18.2)(6.022 x 1023)/6.94 = 1.58 x 1024 atoms Li 2012 Pearson Education, Inc. 39

Compound Formula Mass A compound is a pure substance made of two or more elements chemically joined together. The mass of an individual molecule or formula unit also known as molecular mass or molecular weight Mass of 1 molecule of H2O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu 1. Know chemical formula 2. Count the atoms from chemical formula 3. Sum of the masses of the atoms in a single molecule or formula unit whole = sum of the parts!

Identifying Elements and Formula Subscripts Each new element is identified by a capital letter H2O There are 2 atoms of Hydrogen There are 1 atoms of Oxygen If there is not a subscript listed, it is understood to be 1. Example: NaCl There is one atom of Sodium There is one atom of Chlorine Example: H2SO4 The elements in Sulfuric Acid Hydrogen 2H Sulfur 1S Oxygen 4O

Molar Mass of Compounds The molar mass of a compound the mass in grams of 1 mol of its molecules or formula units is equivalent to its formula mass. 1 mole of H2O contains 2 moles of H and 1 mole of O: molar mass = 1 mole H2O = 2(1.01 g H) + 16.00 g O = 18.02 g so the molar mass of H2O is 18.02 g/mole Molar mass = formula mass (in g/mole)

Calculating Formula Mass Its the sum of the individual atomic masses of each atom constituting the molecular elements and compound. 1. Count atoms from chemical formula 2 Nitrogen 1 gold HCl NaHCO3 Hydrogen 1 Chlorine - 1 Sodium 1 Hydrogen 1 Carbon 1 Oxygen -3 2. Sum of the individual atomic masses 2(14.01 g) = 2 Nitrogen 1(14.01 g) = 1 gold

Calculating Formula Mass Its the sum of the individual atomic masses of each atom constituting the molecule. Calculate the formula mass of 1 mole of magnesium carbonate, MgCO3. 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g/mole Molar Mass: The mass in grams of one mole of any element. It is numerically equivalent to its atomic mass.

Calculating Percentage Composition Calculate the percentage composition of magnesium carbonate, MgCO3. From previous slide: 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g 24.31 Mg 100 28.83% 84.32 12.01 C 100 14.24% 84.32 48.00 O 100 56.93% 84.32 100.00

Periodic Table The periodic table can also be divided into main-group elements, whose properties tend to be largely predictable based on their position in the periodic table. transition elements or transition metals, whose properties tend to be less predictable based simply on their position in the periodic table. 2014 Pearson Education, Inc.

Ions: Losing and Gaining Electrons The number of electrons in a neutral atom is equal to the number of protons in its nucleus (designated by its atomic number Z). Positively charged ions, such as Na+, are called cations. Negatively charged ions, such as F, are called anions. 2014 Pearson Education, Inc.

Ions and the Periodic Table In a chemical changes, however, atoms can lose or gain electrons and become charged particles called ions. A main-group metal tends to lose electrons, forming a cation with the same number of electrons as the nearest noble gas. A main-group nonmetal tends to gain electrons, forming an anion with the same number of electrons as the nearest noble gas. 2014 Pearson Education, Inc.

Ions and the Periodic Table 2014 Pearson Education, Inc.

Predicting Ionic Charges Group 1: Lose 1 electron to form 1+ ions H+ Li+ Na+ K+ Rb+ 33) Give the number of protons in Na+1. A) 10 B) 13 C) 9 D) 11 E) 12 Answer: D number of electrons becomes 10 Cs+

Predicting Ionic Charges Group 2: Loses 2 electrons to form 2+ ions Be2+ Mg2+ Ca2+ Sr2+ Ba2+ 30) What is the identity of element Q if the ion Q2+ contains 10 electrons? A) C B) O C) Ne D) Mg Answer: D

Predicting Ionic Charges B3+ Al3+ Ga3+ Group 13: Loses 3 electrons to form 3+ ions

Predicting Ionic Charges C22- C4- Caution! and are both called carbide Group 14: Loses 4 electrons or gains 4 electrons

Predicting Ionic Charges N3- Nitride P3- Phosphide As3- Arsenide Group 15: Gains 3 electrons to form 3- ions 32) Give the number of electrons in P-3. A) 18 B) 12 C) 19 D) 15 E) 16 Answer A and the number of protons is 15

Predicting Ionic Charges O2- Oxide S2- Sulfide Se2- Selenide Group 16: Gains 2 electrons to form 2- ions

Predicting Ionic Charges F1- Fluoride Cl1- Chloride Br1- Bromide I1- Iodide Group 17: Gains 1 electron to form 1- ions

Ions and the Periodic Table In general, the alkali metals (group 1A) have a tendency to lose one electron and form 1+ ions. The alkaline earth metals (group 2A) tend to lose two electrons and form 2+ ions. The halogens (group 7A) tend to gain one electron and form 1 ions. The oxygen family nonmetals (group 6A) tend to gain two electrons and form 2 ions. 2014 Pearson Education, Inc.

Ions and the Periodic Table For the main-group elements that form cations with predictable charge, the charge is equal to the group number. For main-group elements that form anions with predictable charge, the charge is equal to the group number minus eight. Transition elements may form various different ions with different charges. 2014 Pearson Education, Inc.

Example 2.6 Converting between Number of Moles and Number of Atoms Calculate the number of copper atoms in 2.45 mol of copper. Sort You are given the amount of copper in moles and asked to find the number of copper atoms. Given: 2.45 mol Cu Find: Cu atoms Strategize Convert between number of moles and number of atoms by using Avogadro s number as a conversion factor. Conceptual Plan Relationships Used 6.022 1023 = 1 mol (Avogadro s number) Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.

Example 2.8 The Mole Concept Converting between Mass and Number of Atoms How many copper atoms are in a copper penny with a mass of 3.10 g? (Assume that the penny is composed of pure copper.) Sort You are given the mass of copper and asked to find the number of copper atoms. Given: 3.10 g Cu Find: Cu atoms Strategize Convert between the mass of an element in grams and the number of atoms of the element by first converting to moles (using the molar mass of the element) and then to number of atoms (using Avogadro s number). Conceptual Plan Relationships Used 63.55 g Cu = 1 mol Cu (molar mass of copper) 6.022 1023 = 1 mol (Avogadro s number) Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.