Modern Chemistry Chapter 3 Atoms: The Building Blocks of Matter law of conservation of mass- mass is neither created nor destroyed during ordinary chemical reactions or physical changes e.g. 20 g A + 20 g B  40 g AB

law of definite proportions- a chemical compound contains the same elements in exactly the same proportions by mass regardless of the size of the sample or the source of the compound e.g. If 10 grams of A combine with 20 grams of B to form compound AB, how many grams of B will be necessary to combine with 20 grams of A to form AB? Answer 40: grams of B.

law of multiple proportions- if two or more different compounds are composed of the same two elements, then the ratio of the masses of the second element combined with a certain mass of the first element is always a ratio of small whole numbers e.g. H2O & H2O2 or NO2 & N2O5

Dalton’s Atomic Theory In 1808, an English school teacher named John Dalton proposed an explanation for the law of conservation of mass, the law of definite proportions, and the law of multiple proportions.

Dalton’s Atomic Theory: 1- All matter is composed of extremely small particles called atoms. 2- Atoms of a given element are identical in size, mass and other properties. 3- Atoms cannot be subdivided, created or destroyed. 4- Atoms of different elements combine in simple whole-number ratios to form chemical compounds. 5- In chemical reactions, atoms are combined, separated, or rearranged.

Modern Atomic Theory Dalton’s theory was a good one, but it has since been modified. Atoms are divisible into even smaller particles. Atoms of a given element can have different masses. Atoms can be destroyed. Do Section Review #3 on page 71.

Section review #3 page 71 IF each compound contains 1.0 g oxygen and the three samples contain: compound A: K = 1.22 g / 1.22 = 1.0 compound B: K = 2.44 g / 1.22 = 2.0 compound C: K = 4.89 g / 1.22 = 4.0 1:2:4 ratios of potassium  multiple proportions

Chapter 3 Section 2 The Structure of the Atom atom- the smallest particle of an element that retains the chemical properties of that element

atomic nucleus- the small, densely packed, positively charged central portion of the atom that contains nearly all of its mass but nearly none of its volume neutron- The neutral particle of the nucleus of an atom. proton- The positively charged particle of the nucleus of an atom. electron cloud- The large area surrounding the nucleus of an atom in which the electrons are located. electron- the negatively charged particles of an atom

The Discovery of Electrons In the late 1800’s, electric current was passed through cathode ray tubes. It was discovered that the cathode ray was attracted to the positive pole of a magnet and repelled by the negative pole. This led to the discovery of electrons. In 1909, Robert Millikan measured the negative charge of the electron. From this, it was found that the mass of an electron is 9.109 x 10-31 kg. The mass of an electron is 1/1837th the mass of the simplest hydrogen atom. The negative charges of the electrons equal the positive charges of an atom (protons).

The Discovery of the Atomic Nucleus In 1911, Ernest Rutherford, Hans Geiger, & Ernest Marsden bombarded a thin foil of gold with positively charged particles called alpha particles. They were surprised to find that a few of the particles (1 in 8000) were reflected from the foil straight back toward their source. They reasoned that this must mean that there was a small positively charged portion of each gold atom.

Composition of the Nucleus protons- are positively charged particles in the nucleus (their + charge = - charge of an electron) mass = 1.673 x 10-27 kg or 1836/1837 the mass of a protium atom neutrons- neutral particles of the nucleus mass = 1.675 x 10-27 kg = mass of 1 electron + 1 proton

Composition of the Nucleus How can the numerous positively charged protons exist packed into the nucleus without flying apart due to their like charges repelling one another? nuclear forces- are short range forces (proton-proton, proton-neutron, & neutron-neutron) that hold the nuclear particles together.

The Size of Atoms Atomic radius ranges between 40 & 270 pm (10-12 m) Discuss how small a picometer would be. Nuclear radius = 0.001 pm. The size of the nucleus to the entire atom would be about the same as if you placed a dime at the center of the football stadium.

Section 3.3 Counting Atoms atomic number- is equal to the number of protons in the nucleus of each atom of an element mass number- is the number of protons plus neutrons in a single atom of an element.

Counting Atoms isotopes- are atoms of the same element that have different masses due to different numbers of neutrons nuclide- is a general term for a specific isotope of an element We designate isotopes using one of two different designations. Hydrogen has three isotopes; protium (m# = 1), deuterium (m# = 2), and tritium (m# = 3). hyphen notation H-1 H-2 H-3 1 2 3 nuclear symbol notation H H H 1 1 1

Isotopes Do practice problems #1, 2, & 3 on page 80.

Practice problems page 80 1- bromine-80  35 protons 35 electrons 80-35 = 45 neutrons 2- carbon-13  13 C 6 3- 15 electrons, so 15 protons & element is phosphorus. 15 + 15  phosphorus-30

Identifying Isotopes element symbol atomic # mass # p+ n0 e- calcium 40 ____ uranium 92 235 ____ uranium 238 ____ 13 27 ____ I-127 ____

Relative Atomic Masses atomic mass unit (amu)- exactly 1/12th of the mass of a carbon-12 atom average atomic mass- is the weighted average of the atomic masses of the naturally occurring isotopes of an element see table 4 on page 82

Calculating Average Atomic Mass We will be calculating the “average mass” of the science textbooks in the classroom. This is the approximate method used to determine average atomic mass of the isotopes of an element. 1- Using a bathroom scale, find the weight of the physics textbooks and count their number. What is the average weight of each book? 2- Find the weight of the chemistry textbooks and count their number. What is the average weight of each book? 3- Add the weights of the textbooks and add the numbers of books. 4- Divide the total weight by the total number of books to find the average weight. How does this compare to the average weight of each book?

Relating Mass to Numbers of Atoms mole (mol)- the amount of a substance that contains the same number of particles as there are in 12 grams of carbon-12 The concept is similar to that of a dozen.

Avogadro’s number- is equal to the number of particles in one mole of a substance and is equal to 6.022 x 1023

Molar Mass molar mass- the mass of one mole of a pure substance We can find molar mass by using the average atomic mass found on the periodic table and changing the units from amu to grams.

Mole-molar mass conversions #mole x molar mass = mass (#grams) mass (#grams) ÷ molar mass = #moles

Mole Hill #moles (mol) ÷ molar mass x molar mass (g/mol) (g/mol) mass in grams mass in grams (g) (g)

Gram to Mole Conversions We can use conversion factors to convert between grams and moles. 2.00 mol He x 4.00 g He = 8.00 g He 1 mol He 8.00 g He x 1 mol He = 2.00 mol He 4.00 g He Do practice problems #1-4 and #1-3 on page 85

Mole to Gram Conversions pg 85 2.25 mol Fe x 55.85 g/mol = 126 g Fe 0.375 mol K x 39.10 g/mol = 14.7 g K .0135 mol Na x 22.99 g/mol = 0.310 g Na 16.3 mol Ni x 58.69 g/mol = 957 g Ni

Gram to Mole Conversions pg 85 5.00 g Ca ÷ 40.08 g/mol = 0.125 mol Ca 3.60 x 10-5 g Au ÷ 196.97 g/mol = 1.83 x 10-7 mol Au 0.535 g Zn ÷ 65.39 g/mol = 8.18 x 10-3 mol Zn

Conversions Using Avogadro’s Number 3.01 x 1023 Ag atoms x 1 mole Ag atoms 6.022 x 1023 Ag atoms = 0.500 moles of Ag 1.20 x 108 atoms Cu x 1 mol Cu atoms x 63.55 g 6.022 x 1023 Cu atoms = Cu = 1.27 x 10-14 g Cu Do section review problems #2-6 on page 87.

Section review page 87 2a- sodium-23  11 protons, 11 electrons, 12 neutrons b- calcium-40  20 protons, 20 electrons, 20 neutrons c- copper-64  29 protons, 29 electrons, 35 neutrons d- silver-108  47 protons, 47 electrons, 61 neutrons 3a- silicon-28 b- iron-56 4- potassium  39.10 amu & 39.10 g/mol 5a- 2.00 mol x 14.0 g/mol = 28.0 g N 6a- 12.15 g / 24.3 g/mol = 0.50 mol Mg

Section Review page 87 #7- 2.06 mol Cu 222 g Ag Which has the larger mass? 2.06 mol Cu x 63.5 g/mol = 130.8 g Cu 222 g Ag > 130.8 g Cu Which beaker has the larger number of atoms? 222 g Ag ÷ 107.9 g/mol = 2.06 mol Ag SINCE both beakers have the same number of moles, they have equal numbers of atoms.

Chapter 3- Practice problems 1- U-235 92 electrons & 92 protons 235-92 = 143 neutrons 2- U-238 92 electrons & 92 protons 238-92 = 146 neutrons 3- C-14 6 electrons & 6 protons 14-6 = 8 neutrons 4- I-127 53 electrons & 53 protons 127-53 = 74 neutrons 5- K-41 19 electrons & 19 protons; 41 – 19 = 22 neutrons

#6) 4.25 mol Na x 22.99 g/mol = 97.7 g Na #7) 0.0013 mol Au x 196.97 g/mol = 0.26 g Au #8) 111.5 mol Ca x 40.08 g/mol = 4469 g Ca #9) 2.5 mol C x 12.01 g/mol = 30 g C #10) 0.025 mol Ag x 107.87 g/mol = 2.7 g Ag

#11) 100.3 g Ca ÷ 40.07 g/mol = 2.503 mol Ca #12) 72.0 g O ÷ 16.00 g/mol = 4.50 mol O #13) 0.06 g C ÷ 12.01 g/mol = 0.005 mol C #14) 5.4 g Au ÷ 196.97 g/mol = 0.027 mol Au #15) 3.449 x 1011 g He ÷ 4.00 g/mol = 8.62 x 1010 mol He

Chapter 3 Review Do problems #2, 6-11, 17-19, 21-24, & 28 on pages 89 & 90 of the textbook. Do the Math Tutor problems #1 & 2 on page 92. Do the Standardized Test Prep on page 93.

Chapter 3 vocabulary law of conservation of mass law of definite proportions- law of multiple proportions- atom- atomic nucleus- neutron- ******** electron cloud- electron- protons- neutrons- nuclear forces- atomic number- mass number-. isotopes- atomic mass unit average atomic mass- mole Avogadro’s number- molar mass-

Chemistry Chapter 3 Test 30 multiple choice Questions: definitions & uses of the Laws of Conservation of Mass, Definite Proportions, & Multiple Proportions Dalton’s Atomic Theory: its 5 points & modifications the cathode ray experiment & the discovery of electrons Rutherford’s experiment & the discovery of the atomic nucleus definitions of proton, neutron, electron, atomic nucleus (& its characteristics), nuclear forces, atomic number, mass number, isotopes, average atomic mass, mole, molar mass, & Avogadro’s number determine the number of protons, electrons, & neutrons of an element from its atomic and mass numbers mass to mole & mole to mass calculations

Honors Chemistry Chapter 3 Test 50 Multiple Choice questions definitions & implications of the Laws of Conservation of Mass, Definite Proportions, Multiple Proportions, Dalton’s Atomic Theory (and its modifications) implications of the cathode ray experiment & the discovery of the electron Rutherford’s experiment & the discovery of the atomic nucleus description of an atom, atomic nucleus, & electron cloud definitions & implications of atom, proton, neutron, electron, nuclear forces, isotopes, average atomic mass, atomic number, mass number, Avogadro’s number, mole, molar mass describe the isotopes of hydrogen (protium, deuterium, tritium) determine numbers of protons, electrons & neutrons from atomic & mass numbers mass to mole & mole to mass conversions