Chapter 1 The Atomic Nature of Matter 1-1 Chemistry: Science of Change 1-2 The Composition of Matter 1-3 The Atomic Theory of Matter 1-4 Chemical
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1 Chapter 1 The Atomic Nature of Matter 1-1 Chemistry: Science of Change 1-2 The Composition of Matter 1-3 The Atomic Theory of Matter 1-4 Chemical Formulas and Relative Atomic Masses 1-5 The Building Blocks of the Atom 1-6 Finding Atomic Masses the Modern Way 1-7 The Mole Concept: Counting and Weighing Atoms and Molecules 1-8 Finding Empirical and Molecular Formulas the Modern Way 1-9 Volume and Density OFB Chapter 1 1
2 Atomic Theory of Matter Law of conservation of mass: Mass is neither created nor destroyed in a chemical reaction Dalton s Atomic Theory of Matter (1808): 1. All matter consists of solid and indivisible atoms 2. All atoms of a given chemical element are identical in mass and in all other properties 3. Different elements have different kinds of atoms; these atoms differ in mass from element to element 4. Atoms are indestructible and retain their identity in all chemical reactions 5. The formation of a compound from its elements occurs through the combination of atoms of unlike elements in small whole-number ratio. OFB Chapter 1 2
3 Chemical Formulas and Relative Atomic Masses Chemical Formulas display symbols for the elements and the relative number of atoms E.g., NH 3, CO 2, CH 3 CO 2 H or C 2 H 4 O 2 Molecules are groupings of two or more atoms bound closely together by strong forces that maintain them in a persistent combination OFB Chapter 1 3
4 Building Blocks of the Atom Electrons, Protons and Neutrons Electrons discovered in 1897 by Thomson Rutherford proposed that the atomic nucleus was composed of neutral particles called Neutrons and positively charged particles called protons Neutron number = N Atomic number = Z = number of Protons Atomic mass number = A A = Z + N OFB Chapter 1 4
5 hemist/alchemy.html pages/pertable_j.htm OFB Chapter 1 5
6 C Carbon Si Silicon N Nitrogen P Phosphorus Semi Metal Non Metal OFB Chapter 1 6
7 Atomic Mass C Carbon Atomic Number A = Z + N Atomic Mass = # Protons + # Neutrons For Carbon, 12 = 6 + Neutrons Neutrons = 6 Every Carbon atom has 6 electrons, 6 protons and 6 neutrons OFB Chapter 1 7
8 Mass Spectrometry and Isotopes Mass Spectrometer accelerates ions (or molecular ions) in an electric field and then separates those ions by relative mass in a magnetic field Theory and Experiment M + e (70 ev) M + + 2e M + lower mass ions OFB Chapter 1 8
9 Mass Spectrometry and Isotopes Mass Spectrometer accelerates ions (or molecular ions) in an electric field and then separates those ions by relative mass in a magnetic field Cl Chlorine Mass Spectrometer Separation of Chlorine Relative Amount Relative Mass OFB Chapter 1 9
10 Isotopes of Cl: msec EC/ECp, S-31/P msec EC/ECa/EC p, S-32/Si- 28/P sec EC,5.583 MeV S sec EC,5.492 MeV S % Stable E+5 yr B- /EC, Ar-36/S % Stable min B-,4.917 MeV Ar min B-,3.442 MeV Ar min B-,7.480 MeV Ar sec B-,5.730 MeV Ar sec B-,9.430 MeV Ar sec B-,7.950 MeV Ar sec B-/Bn,3.920 Ar- 44/Ar msec B-/Bn, Ar-45/Ar sec B-/B-n,6.900 Ar-46/Ar nsec B-/Bn, Ar-47/Ar MASS abund. Halflife Particle, Energy Decay Product(s) Isotopic Mass 35 Cl contains protons and neutrons 37 Cl contains protons and neutrons OFB Chapter 1 10
11 Atoms Avogadro s Number is the number of 12 C atoms in exactly 12 grams of carbon N 0 = X The mass, in grams, of Avogadro's number of atoms of an element is numerically equal to the relative atomic mass of that element m ass of atom C = OFB Chapter 1 11
12 Molecules Relative Molecular Mass of a molecule equals the sum of the relative atomic masses of all of the atoms making up the molecule relative molecular mass of CO 2 = mass CO 2 molecule OFB Chapter 1 12
13 Moles A mole measures the chemical amount of a substance Mole is an abbreviation of gram molecular weight One mole of a substance equals the amount that contains Avogadro's number of atoms, molecules. One mole = Molar mass (M) of that element or molecule OFB Chapter 1 13
14 Exercise 1-6 Molecules of isoamyl acetate have the formula C 7 H 14 O 2. Calculate (a) how many moles and (b) how many molecules are present in grams of isoamyl acetate. Strategy: 1. Calculate molar mass of C 7 H 14 O 2 2. Calculate the number of moles in grams 3. Using Avogadro s number to calculate the number of molecules in the number of moles of C 7 H 14 O 2 OFB Chapter 1 14
15 Exercise 1-6 Molecules of isoamyl acetate have the formula C 7 H 14 O 2. Calculate (a) how many moles and (b) how many molecules are present in 0.250g of isoamyl acetate. Solution: 1. Calculate molar mass of C 7 H 14 O 2 2. Calculate the number of moles in grams 3. Using Avogadro s number calculate the number of molecules in n moles of C 7 H 14 O 2 OFB Chapter 1 15
16 Percentage Composition from Empirical or Molecular Formula Exercise 1-8 Tetrodotoxin, a potent poison found in the ovaries and liver of the globefish, has the empirical formula C 11 H 17 N 3 O 8. Calculate the mass percentages of the four element in this compound. Strategy: 1. Calculate molar mass of C 11 H 17 N 3 O, by finding the mass contributed by each element 2. Divide the mass for each element by the total mass of the compound. OFB Chapter 1 16
17 Exercise 1-8 Tetrodotoxin has the empirical formula C 11 H 17 N 3 O 8. Calculate the mass percentages of the four element in this compound. Solution: 1. Calculate molar mass of C 11 H 17 N 3 O 8, by finding the mass contributed by each element M M = = Divide the mass for each element by the total mass of the compound. %C %H %N %O = = = = OFB Chapter 1 17
18 Millimoles and Milligrams 1 mmol = 1 millimole =1x10-3 mol 1mg = 1 milligram =1x10-3 g OFB Chapter 1 18
19 Exercise 1-10 Moderate Heating of mg of a compound containing nickel, carbon and oxygen and no other elements drives off all of the carbon and oxygen in the form of carbon monoxide (CO) and leaves mg of metallic nickel behind. Determine the empirical formula of the compound. Strategy: 1 mmol = 1x10-3 mol 1mg = 1x10-3 g 1. Write the reaction 2. Use the Law of conservation of mass to find the amount of CO 3. Find the number of moles (or mmol) of CO and Nickel 4. Find the ratios of the moles for each substance by dividing each by the smallest one, i.e., normalize to the smallest. OFB Chapter 1 19
20 Exercise 1-10 Moderate Heating of mg of a compound containing nickel, carbon and oxygen and no other elements drives off all of the carbon and oxygen in the form of carbon monoxide (CO) and leaves mg of metallic nickel behind. Determine the empirical formula of the compound. Solution: 1. Write the reaction 2. Use the law of conservation of mass to find the amount of CO 3. Find the number of moles of CO and Nickel 4. Find the ratios of the moles for each substance by dividing each by the smallest one, i.e., normalize to the smallest. OFB Chapter 1 20
21 Exercise 1-10 Moderate Heating of mg of a compound containing nickel, carbon and oxygen and no other elements drives off all of the carbon and oxygen in the form of carbon monoxide (CO) and leaves mg of metallic nickel behind. Determine the empirical formula of the compound. Solution: 1. Write the reaction 2. Use the law of conservation of mass to find the amount of CO 3. Find the number of moles of CO and Nickel 4. Find the ratios of the moles for each substance by dividing each by the smallest one, i.e., normalize to the smallest. OFB Chapter 1 21
22 Volume and Density Exercise 1-13 The density of liquid mercury at 20 deg C is g cm -3. A chemical reaction requires mol of mercury. What volume (in cubic centimeters) of mercury should be measured out at 20 C? Strategy: 1. Use density and mass to find volume. Rearrange m d = V V = m d 2. Density is given, can find mass from the number of moles of mercury which is given 3. Solve for volume. OFB Chapter 1 22
23 Exercise 1-13 The density of liquid mercury at 20 deg C is g cm -3. A chemical reaction requires mol of mercury. What volume (in cubic centimeters) of mercury should be measured out at 20 C? Solution: 1. Use density and mass to find volume. m V = d 2. Density is given, can find mass from the number of moles of mercury which is given 0.56 moles Hg = m grams m = grams Hg Hg X 1mole Hg g Hg 3. Solve for volume. m grams V = = = 8.26 cm 3 d grams per cm 3 OFB Chapter 1 23
24 Chapter 1 The Atomic Nature of Matter Examples / Exercises (for your practice only, not to be turned in) All (1-1 thru 1-13) HW Problems (to be turned in a recitation, see due date) Textbook errors OFB Chapter 1 24
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= 16.00 amu. = 39.10 amu
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