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2. Spencer L. Seager Michael R. Slabaugh www.cengage.com/chemistry/seager Jennifer P. Harris Chapter 2: Atoms and Molecules

3. SYMBOLS & FORMULAS A unique symbol is used to represent each element. Formulas are used to represent compounds. ELEMENTAL SYMBOLS A symbol is assigned to each element. The symbol is based on the name of the element and consists of one capital letter or a capital letter followed by a lower case letter. Some symbols are based on the Latin or German name of the element.

4. CHEMICAL ELEMENTS & THEIR SYMBOLS

5. CHEMICAL ELEMENTS & THEIR SYMBOLS (continued)

6. Elements from Group 7A chlorine bromine iodine

7. COMPOUND FORMULAS A compound formula consists of the symbols of the elements found in the compound. Each elemental symbol represents one atom of the element. If more than one atom is represented, a subscript following the elemental symbol is used.

8. COMPOUND FORMULAS EXAMPLES Carbon monoxide, CO one atom of C one atom of O Water, H 2 O two atoms of H one atom of O Ammonia, NH 3 one atom of N 3 atoms of H

9. COMPOUND FORMULAS PRACTICE The clear liquid is carbon disulfide. It is composed of carbon (left) and sulfur (right). If carbon disulfide contains one atom of carbon for every two atoms of sulfur, what is the chemical formula for carbon disulfide? Answer: CS 2

10. ATOMIC STRUCTURE Atoms are made up of three subatomic particles, protons, neutrons, and electrons. The protons and neutrons are tightly bound together to form the central portion of an atom called the nucleus. The electrons are located outside of the nucleus and thought to move very rapidly throughout a relatively large volume of space surrounding the small but very heavy nucleus.

11. SUBATOMIC PARTICLES Protons are located in the nucleus of an atom. They carry a +1 electrical charge and have a mass of 1 atomic mass unit (u). Neutrons are located in the nucleus of an atom. They carry no electrical charge and have a mass of 1 atomic mass unit (u). Electrons are located outside the nucleus of an atom. They carry a -1 electrical charge and have a mass of 1/1836 atomic mass unit (u). They move rapidly around the heavy nucleus.

12. SUBATOMIC PARTICLE CHARACTERISTICS

13. ATOMIC STRUCTURE REVIEW Which subatomic particles are represented by the pink spheres? Answer: electrons Which subatomic particles are represented by the yellow and blue spheres? Answer: protons and neutrons What structure do the yellow and blue spheres form? Answer: the nucleus

14. ATOMIC & MASS NUMBERS ATOMIC NUMBER OF AN ATOM The atomic number of an atom is equal to the number of protons in the nucleus of the atom. Atomic numbers are represented by the symbol Z. MASS NUMBER OF AN ATOM The mass number of an atom is equal to the sum of the number of protons & neutrons in the nucleus of the atom. Mass numbers are represented by the symbol A.

15. ATOMIC & MASS NUMBERS APPLICATION Based on the information given above, what is the atomic number of fluorine? Answer: The atomic number of fluorine is 9. On the periodic table, the atomic number is written as a whole number above the symbol F. In the written description, fluorine is said to have 9 protons (the atomic number is the number of protons). In the symbol, the number 9 is written in the atomic number or Z (lower left) position.

16. ATOMIC & MASS NUMBERS APPLICATION Based on the information given above, what is the mass number of fluorine? Answer: The mass number of fluorine is 19. In the written description, fluorine is said to have 9 protons and 10 neutrons (the mass number is the sum of the numbers of protons and neutrons). In the symbol, the number 19 in written in the mass number or A (upper left) position. Note: The periodic table does not show the mass number for an individual atom. It lists an average mass number for a collection of atoms!

17. ISOTOPES Isotopes are atoms that have the same number of protons in the nucleus but different numbers of neutrons. That is, they have the same atomic number but different mass numbers. Because they have the same number of protons in the nucleus, all isotopes of the same element have the same number of electrons outside the nucleus.

18. ISOTOPE SYMBOLS Isotopes are represented by the symbol, where Z is the atomic number, A is the mass number, and E is the elemental symbol. An example of an isotope symbol is. This symbol represents an isotope of nickel that contains 28 protons and 32 neutrons in the nucleus. Isotopes are also represented by the notation: Name-A, where Name is the name of the element and A is the mass number of the isotope. An example of this isotope notation is magnesium-26. This represents an isotope of magnesium that has a mass number of 26. 60 28 Ni

19. RELATIVE MASSES The extremely small size of atoms and molecules makes it inconvenient to use their actual masses for measurements or calculations. Relative masses are used instead. Relative masses are comparisons of actual masses to each other. For example, if an object had twice the mass of another object, their relative masses would be 2 to 1.

20. ATOMIC MASS UNIT (u) An atomic mass unit is a unit used to express the relative masses of atoms. One atomic mass unit is equal to 1/12 the mass of a carbon-12 atom. A carbon-12 atom has a relative mass of 12 u. An atom with a mass equal to 1/12 the mass of a carbon-12 atom would have a relative mass of 1 u. An atom with a mass equal to twice the mass of a carbon-12 atom would have a relative mass of 24 u.

21. ATOMIC WEIGHT The atomic weight of an element is the relative mass of an average atom of the element expressed in atomic mass units. Atomic weights are the numbers given at the bottom of the box containing the symbol of each element in the periodic table. According to the periodic table, the atomic weight of nitrogen atoms (N) is 14.0 u, and that of silicon atoms (Si) is 28.1 u. This means that silicon atoms are very close to twice as massive as nitrogen atoms. Put another way, it means that two nitrogen atoms have a total mass very close to the mass of a single silicon atom.

22. MOLECULAR WEIGHT The relative mass of a molecule in atomic mass units is called the molecular weight of the molecule. Because molecules are made up of atoms, the molecular weight of a molecule is obtained by adding together the atomic weights of all the atoms in the molecule. The formula for a molecule of water is H 2 O. This means one molecule of water contains two atoms of hydrogen, H, and one atom of oxygen, O. The molecular weight of water is then the sum of two atomic weights of H and one atomic weight of O: MW = 2(at. wt. H) + 1(at. wt. O) MW = 2(1.01 u) + 1(16.00 u) = 18.02 u

23. MOLECULAR WEIGHT PRACTICE The clear liquid is carbon disulfide, CS 2. It is composed of carbon (left) and sulfur (right). What is the molecular weight for carbon disulfide? Answer: MW = 1(atomic weight C) + 2(atomic weight S) 12.01 u + 2(32.07 u) = 76.15 u

24. ISOTOPES & ATOMIC WEIGHTS Many elements occur naturally as a mixture of several isotopes. The atomic weight of elements that occur as mixtures of isotopes is the average mass of the atoms in the isotope mixture. The average mass of a group of atoms is obtained by dividing the total mass of the group by the number of atoms in the group. A practical way of determining the average mass of a group of isotopes is to assume the group consists of 100 atoms and use the percentage of each isotope to represent the number of atoms of each isotope present in the group.

25. ISOTOPES & ATOMIC WEIGHTS (continued) The use of percentages and the mass of each isotope leads to the following equation for calculating atomic weights of elements that occur naturally as a mixture of isotopes. According to this equation, the atomic weight of an element is calculated by multiplying the percentage of each isotope in the element by the mass of the isotope, then adding the resulting products together and dividing the resulting mass by 100.

26. ISOTOPES & ATOMIC WEIGHTS EXAMPLE A specific example of the use of the equation is shown below for the element boron that consists of 19.78% boron-10 with a mass of 10.01 u and 80.22% boron-11 with a mass of 11.01u. This calculated value is seen to agree with the value given in the periodic table.

27. THE MOLE CONCEPT THE MOLE CONCEPT APPLIED TO ELEMENTS The number of atoms in one mole of any element is called Avogadro's number and is equal to 6.022x10 23. A one-mole sample of any element will contain the same number of atoms as a one-mole sample of any other element. One mole of any element is a sample of the element with a mass in grams that is numerically equal to the atomic weight of the element. EXAMPLES OF THE MOLE CONCEPT 1 mole Na = 22.99 g Na = 6.022x10 23 Na atoms 1 mole Ca = 40.08 g Ca = 6.022x10 23 Ca atoms 1 mole S = 32.07 g S = 6.022x10 23 S atoms

28. THE MOLE CONCEPT (continued) THE MOLE CONCEPT APPLIED TO COMPOUNDS The number of molecules in one mole of any compound is called Avogadro's number and is numerically equal to 6.022x10 23. A one-mole sample of any compound will contain the same number of molecules as a one-mole sample of any other compound. One mole of any compound is a sample of the compound with a mass in grams equal to the molecular weight of the compound. EXAMPLES OF THE MOLE CONCEPT 1 mole H 2 O = 18.02 g H 2 O = 6.022x10 23 H 2 O molecules 1 mole CO 2 = 44.01 g CO 2 = 6.022x10 23 CO 2 molecules 1 mole NH 3 = 17.03 g NH 3 = 6.022x10 23 NH 3 molecules

29. THE MOLE CONCEPT (continued) THE MOLE AND CHEMICAL CALCULATIONS The mole concept can be used to obtain factors that are useful in chemical calculations involving both elements and compounds. One mole quantities of six metals; top row (left to right): Cu beads (63.5 g), Al foil (27.0 g), and Pb shot (207.2 g); bottom row (left to right): S powder (32.1 g), Cr chunks (52.0 g), and Mg shavings (24.4 g). One mole quantities of four compounds: H 2 O (18.0 g); small beaker NaCl (58.4 g); large beaker aspirin, C 9 H 8 O 4, (180.2 g); green (NiCl 2 · 6H 2 O) (237.7 g).

30. MOLE CALCULATIONS The mole-based relationships given earlier as examples for elements provide factors for solving problems. The relationships given earlier for calcium are: 1 mole Ca= 40.08 g Ca = 6.022x10 23 Ca atoms Any two of these quantities can be used to provide factors for use in solving numerical problems. Examples of two of the six possible factors are: and

31. MOLE CALCULATION EXAMPLE Calculate the number of moles of Ca contained in a 15.84 g sample of Ca. The solution to the problem is: We see in the solution that the g Ca units in the denominator of the factor cancel the g Ca units in the given quantity, leaving the correct units of mole Ca for the answer.

32. MOLE CALCULATIONS (continued) The mole concept applied earlier to molecules can be applied to the individual atoms that are contained in the molecules. An example of this for the compound CO 2 is: 1 mole CO 2 molecules = 1 mole C atoms + 2 moles O atoms 44.01 g CO 2 = 12.01 g C + 32.00 g O 6.022x10 23 CO 2 molecules = 6.022x10 23 C atoms + (2) 6.022x10 23 O atoms Any two of these nine quantities can be used to provide factors for use in solving numerical problems.

33. MOLE CALCULATION EXAMPLES Example 1: How many moles of O atoms are contained in 11.57 g of CO 2 ? Note that the factor used was obtained from two of the nine quantities given on the previous slide.

34. MORE MOLE CALCULATION EXAMPLES Example 2: How many CO 2 molecules are needed to contain 50.00 g of C? Note that the factor used was obtained from two of the nine quantities given on a previous slide.

35. MORE MOLE CALCULATION EXAMPLES Example 3: What is the mass percentage of C in CO 2 ? The mass percentage is calculated using the following equation: If a sample consisting of 1 mole of CO 2 is used, the mole- based relationships given earlier show that: 1 mole CO 2 = 44.01 g CO 2 = 12.01 g C + 32.00 g O

36. MORE MOLE CALCULATION EXAMPLES (continued) Thus, the mass of C in a specific mass of CO 2 is known, and the problem is solved as follows:

37. MORE MOLE CALCULATION EXAMPLES Example 4: What is the mass percentage of oxygen in CO 2 ? The mass percentage is calculated using the following equation: Once again, a sample consisting of 1 mole of CO 2 is used to take advantage of the mole-based relationships given earlier where: 1 mole CO 2 = 44.01g CO 2 = 12.01 g C + 32.00g O

38. MORE MOLE CALCULATION EXAMPLES (continued) Thus, the mass of O in a specific mass of CO 2 is known, and the problem is solved as follows: We see that the % C + % O = 100%, which should be the case because C and O are the only elements present in CO 2.

39. MOLE CALCULATIONS HELP

40. LABORATORY APPLICATION The mass of an empty crucible is 20.057 g. Copper sulfate pentahydrate is added to the crucible. The combined mass is 45.551 g. What is the mass of the copper sulfate pentahydrate? 45.551 g crucible + copper sulfate pentahydrate −20.057 g crucible 25.494 g copper sulfate pentahyrdate

41. LABORATORY APPLICATION (continued) The crucible is heated and allowed to cool. The mass of the crucible and anhydrous solid after heating is 38.547 g. What is the mass of the anhydrous solid? What is the mass of the water that was removed by heating? 38.547 g crucible + anhydrous copper sulfate −20.057 g crucible 18.490 g anhydrous copper sulfate 25.494 g copper sulfate pentahydrate −18.490 g anhydrous copper sulfate 7.004 g water

42. LABORATORY APPLICATION (continued) How many moles of water (H 2 O) were removed by heating? How many moles of anhydrous copper sulfate (CuSO 4 ) remained after heating? What is the ratio of moles anhydrous copper sulfate to moles of water? 0.11584 moles CuSO 4 : 0.3887 moles H 2 O 1 mole CuSO 4 : 3.355 moles H 2 O

43. LABORATORY APPLICATION (continued) The formula for a hydrate is written as CuSO 4 ·x H 2 O, where x is the number of moles of water associated with each mole of CuSO 4. Based on the calculated mole ratio, what is the experimentally determined formula for copper sulfate pentahydrate? CuSO 4 ·3.355 H 2 O The actual formula for copper sulfate pentahydrate is CuSO 4 ·5 H 2 O. Why is there a difference between the actual formula and the experimentally determined formula? Answer: The copper sulfate pentahydrate was not heated thoroughly enough. Some of the water remained in the “anhydrous” copper sulfate, which caused the number of moles of water removed by heating to be smaller than the actual value.

44. LABORATORY APPLICATION (continued) What is the mass percentage of H 2 O in CuSO 4 ·5 H 2 O? What mass of water could have been removed from the 25.494 g copper sulfate pentahydrate, if the sample had been heated completely? or

45. LABORATORY APPLICATION (continued) What percentage of the total amount of water was removed during the experiment? Had all the water been removed, what would the mass of the anhydrous copper sulfate been after heating? Would this have been the mass displayed on the balance? Why or why not? Answer: No, the mass displayed on the balance would have been 36.353 g because the anhydrous copper sulfate was in the 20.057 g crucible. 25.494 g copper sulfate pentahydrate −9.1983 g water 16.296 g anhydrous copper sulfate