1. Chapter 2 Atoms and Molecules

2. Learning Objectives
Use symbols for chemical elements to write formulas for chemical compounds
Identify the characteristics of protons, neutrons, and electrons
Use the concepts of atomic number and mass number to determine the number of subatomic particles in isotopes and to write correct symbols for isotopes
Use atomic weights of the elements to calculate molecular weights of compounds
Use isotope percent abundances and masses to calculate atomic weights of elements

3. Learning Objectives (continued)
Use the mole concept to obtain relationships between number of moles, number of grams, and number of atoms for elements, and use those relationships to obtain factors for use in factor-unit calculations
Use the mole concept and molecular formulas to obtain relationships between number of moles, number of grams, and number of atoms or molecules for compounds, and use those relationships to obtain factors for use in factor-unit calculations

4. Symbols and Formulas
A unique name and symbol is used to represent each element
Based on elemental properties or are derived from names of famous scientists, places, astronomical bodies, or mythological characters
Elemental symbols: Based on the name of the element and consist of one capital letter or a capital letter followed by a lowercase letter

5. Table 2.1 - The Chemical Elements and Their Symbols

6. Table 2.1 - The Chemical Elements and Their Symbols (continued)

7. Elements in the Human Body

8. Compound Formula
Consists of the symbols of the atoms found in the molecule
Each elemental symbol represents one atom of the element
If more than one atom is present in the compound, then a subscript follows the elemental symbol

9. Table 2.2 - Examples of Compound Formulas

10. Example 2.1 - Writing Compound Formulas
Write formulas for the following compounds:
Nitrogen dioxide: One nitrogen (N) atom and two oxygen (O) atoms
Sulfuric acid: Two hydrogen (H) atoms, one sulfur (S) atom, and four oxygen (O) atoms

11. Example Solution
The single N atom will not have a subscript because ones are understood and never written, and the two O atoms will be represented by writing a subscript 2
The molecular formula is NO2
Using similar reasoning, the H atom will have a subscript 2, the S atom will have no subscript, and the O atom will have a subscript 4
The molecular formula is H2SO4

12. 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 - CS2

13. Atomic Structure Atoms are made up of three subatomic particles
Protons, neutrons, and electrons
Protons and neutrons
Tightly bound together to form the central portion of an atom called the nucleus
Electrons
Located outside the nucleus
Move rapidly throughout a relatively large volume of space surrounding the nucleus
Electrons move rapidly around a massive nucleus. This figure
is not drawn to scale. For a nucleus of the size shown, the closest electrons would be at least 80 m away.

14. Table 2.3 - Characteristics of the Fundamental Subatomic Particles

15. 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 is formed by the yellow and blue spheres?
Answer - The nucleus

16. Atomic and Mass Numbers
Atomic number of an atom
Equal to the number of protons in the nucleus and the number of electrons in an atom
Symbolically represented by Z
Mass number of an atom
Equal to the sum of the number of protons and neutrons in the nucleus of an atom
Symbolically represented by A

17. Atomic and Mass Numbers Application
Based on the information given below, what is the atomic number of fluorine?
Answer - The atomic number of fluorine is 9
In 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

18. Atomic and Mass Numbers Application (continued)
Based on the information given below, 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

19. Isotopes
Atoms that have the same number of protons in the nucleus but different numbers of neutrons
Same atomic number but different mass numbers
All isotopes of the same element have the:
Same number of electrons outside the nucleus
Same number of protons in the nucleus

20. Ways to Represent Isotopes
Represented by the following symbol:
Z is the atomic number
A is the mass number
E is the elemental symbol
Example -
This symbol represents an isotope of hydrogen that contains 11 protons in the nucleus
Represented by the elemental name, which is followed by the mass number
Example - Hydrogen-2 is an isotope of hydrogen

21. Join In (8)
Which of the following is the isotope symbol for an atom with 46 protons, 46 electrons, and 56 neutrons?
Correct Answer Pd

22. Example 2.2 - Using the Periodic Table
Use the periodic table to answer the following questions about isotopes:
What are the mass number, atomic number, and isotope symbol for an atom that contains 7 protons and 8 neutrons?
How many neutrons are contained in an atom of nickel-60?
How many protons and how many neutrons are contained in an atom with a mass number of 26 and the symbol Mg?

23. Example Solution (a)
The mass number, A, equals the sum of the number of protons and the number of neutrons
A = = 15
The atomic number, Z, equals the number of protons
Z = 7
According to the periodic table, the element with an atomic number of 7 is nitrogen, with the symbol N
The isotope symbol is

24. Example Solution (b)
According to the periodic table, nickel has the symbol Ni, and an atomic number, Z, of 28
The mass number, 60, is equal to the sum of number of protons and the number of neutrons
The number of protons is equal to the atomic number, 28
Therefore, the number of neutrons is 60 – 28 = 32
The atom contains 32 neutrons

25. Example Solution (c)
According to the periodic table, the element with the symbol Mg is magnesium, which has an atomic number of 12
Therefore, the atom contains 12 protons
Since A, the number of protons plus neutrons is equal to 26, the number of neutrons is 26 – 12, or 14
The atom contains 14 neutrons

26. Relative Masses
Numbers that are given beneath the symbol and name for each element in the periodic table
Provide simple means of comparing the masses of atoms

27. Atomic Mass Unit (u) Used to express the relative masses of atoms
1 u = 1/12 the mass of a carbon-12 atom
One carbon-12 atom has a relative mass of 12 u
Atom with a mass equal to 1/12 the mass of a carbon-12 atom would have a relative mass of 1 u
Atom with a mass equal to twice the mass of a carbon-12 atom would have a relative mass of 24 u

28. Atomic Weight
Relative mass of an average atom of an element expressed in atomic mass units
Example - According to the periodic table, the atomic weight of N is 14.0 u and Si is 28.1 u
It also means that two N atoms have a total mass very close to the mass of a single Si atom

29. Molecular Weight
Relative mass of a molecule expressed in atomic mass units
Calculated by adding together the atomic weights of the atoms in the molecule
Example - The formula for a molecule of water is H2O
This means one molecule of water contains two atoms of hydrogen, H, and one atom of oxygen, O
Molecular weight of water is then the sum of two atomic weights of H and one atomic weight of O

30. Join In (9) What is the molecular weight of SO3? 48.07 amu 48.07 g
Correct Answer
80.07 amu

31. Example 2.4 - Atomic Weights and Molecular Weights
Use atomic weights from the periodic table to determine the molecular weight of urea, CH4N2O, the chemical form in which much nitrogenous body waste is excreted in the urine

32. Example Solution
According to the formula given, a urea molecule contains one carbon atom, C, four hydrogen atoms, H, two nitrogen atoms, N, and one oxygen atom, O
The molecular weight is calculated as follows:
Rounded to four significant figures, the correct answer is u

33. Molecular Weight Practice
The clear liquid is carbon disulfide, CS2
It is composed of carbon (left) and sulfur (right)
What is the molecular weight for carbon disulfide?
Answer

34. Isotopes and Atomic Weights
Atomic weight of elements that occur as mixtures of isotopes is the average relative mass of the atoms in the isotope mixture
Average mass of each particle in a group of atoms is obtained by dividing the total mass of the group by the number of particles in the group
Practical way of determining the average mass of a group of isotopes is to use an imaginary sample of an element containing 100 atoms
Use the percentage of each isotope to represent the number of atoms of each isotope in the group
Total mass equals the sum of masses contributed by each isotope

35. Isotopes and Atomic Weights - Example
A specific example is shown below for the element boron that consists of 19.78% boron-10 with a mass of u and 80.22% boron-11 with a mass of u
This calculated value matches the value given in the periodic table

36. Example 2.5 - Isotope and Atomic Weight Relationships
Calculate the atomic weight of chlorine, given that the naturally occurring element consists of 75.53% chlorine-35 (mass = u) and 24.47% chlorine-37 (mass = u)

37. Example Solution
This result is slightly different from the periodic table atomic weight value of because of slight errors introduced in rounding the isotope masses to four significant figures

38. Avogadro’s Number and the Mole Concept
Avogadro’s number - Number of atoms or molecules in a specific sample of an element or compound
Mole (mol): Number of particles contained in a sample of an element or compound with a mass in grams equal to the atomic or molecular weight, respectively
1 mol = 6.022×1023
Example - 1 mol S atoms = 6.02×1023 S atoms = 32.1 g S
Following factors can be generated for use in factor-unit calculations:
1 mol S atoms = 6.02×1023 particles S atoms
6.02×1023 S atoms = 32.1 g S
1 mol S atoms = 32.1 g S

39. The Mole and Chemical Calculations
Mole concept can be used to obtain factors that are useful in chemical calculations involving both elements and compounds

40. Mole Calculation Example (1)
Calculate the number of moles of Ca contained in a g sample of Ca
Solution:
Notice 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

41. Example 2.7 - Factor-Unit Calculations for Sulfur
Determine the following using the factor-unit method of calculation and factors obtained from the preceding three relationships given for sulfur (S):
The mass in grams of 1.35 mol of S
The number of moles of S atoms in 98.6 g of S
The number of S atoms in 98.6 g of S
The mass in grams of one atom of S

42. Example Solution
Known quantity is 1.35 mol of S, and the unit of the unknown quantity is grams of S
Factor comes from the relationship 1 mol S atoms = 32.1 g S
Known quantity is 98.6 g of S, and the unit of the unknown quantity is moles of S atoms
Factor comes from the same relationship used in (a)

43. Example 2.7 - Solution (continued 1)
Known quantity is 98.6 g of S, and the unit of the unknown quantity is the number of S atoms
Factor comes from the relationship 6.02×1023 S atoms = 32.1 g S
Known quantity is one S atom, and the unit of the unknown is grams of S
Factor comes from the same relationship used in (c), 6.02×1023 S atoms = 32.1 g S

44. Example 2.7 - Solution (continued 2)
Note that the factor is the inverse of the one used in (c) even though both came from the same relationship
Thus, we see that each relationship provides two factors

45. The Mole Concept Applied to Compounds
One mole of any compound is a sample of the compound with a mass in grams equal to the molecular weight of the compound
1 mol CO2 molecules = 6.02 ×1023 CO2 molecules = 44.0 g CO2
Following relationships can be used to generate factors for use in factor-unit calculations:
1 mol CO2 molecules = 6.02 ×1023 CO2 molecules
6.02 ×1023 CO2 molecules = 44.0 g CO2
1 mol CO2 molecules = 44.0 g CO2

46. Mole Calculation Example (2)
How many moles of O atoms are contained in g of CO2?
Solution
Note that the factor used was obtained from two of the six quantities given on the previous slide

47. Mole Calculation Example (3)
How many CO2 molecules are needed to contain g of C?
Solution:
Note that the factor used was obtained from two of the six quantities given on a previous slide

48. Mole Calculation Example (4)
What is the mass percentage of C in CO2?
Solution:
Mass percentage is calculated using the following equation:
If a sample consisting of 1 mole of CO2 is used, the mole-based relationships given earlier show that 1 mole CO2 = g CO2 = g C g O
Thus, the mass of C in a specific mass of CO2 is known and the problem is solved as follows:

49. Mole Calculation Example (5)
What is the mass percentage of oxygen in CO2?
Solution:
Mass percentage is calculated using the following equation:
Once again, a sample consisting of 1 mole of CO2 is used to take advantage of the mole-based relationships given earlier where:
1 mole CO2 = 44.01g CO2 = g C g O

50. Mole Calculation Example (5) (continued)
Thus, the mass of O in a specific mass of CO2 is known and the problem is solved as follows:
Notice that the % C + % O = 27.29% % = 100%, which should be the case because C and O are the only elements present in CO2

51. Example 2.8 - Factor-Unit Calculations for Carbon Dioxide
Determine the following using the factor-unit method of calculation and factors obtained from the preceding three relationships given for carbon dioxide, CO2:
The mass in grams of 1.62 mol of CO2
The number of moles of CO2 molecules in 63.9 g of CO2
The number of CO2 molecules in 63.9 g of CO2
The mass in grams of one molecule of CO2

52. Example Solution
The known quantity is 1.62 mol of CO2, and the unit of the unknown quantity is g CO2
The factor comes from the relationship 1 mol CO2 molecules = 44.0 g CO2
The known quantity is 63.9 g of CO2, and the unit of the unknown quantity is mol of CO2 molecules
The factor comes from the same relationship used in (a)

53. Example 2.8 - Solution (continued 1)
The known quantity is, again, 63.9 g of CO2, and the unit of the unknown is the number of CO2 molecules
The factor comes from the relationship 6.02×1023 CO2 molecules = 44.0 g CO2
The known quantity is one CO2 molecule, and the unit of the unknown is g CO2
The factor comes from the same relationship used in (c), but the factor is the inverse of the one used in (c)

54. The Mole and Chemical Formulas
Chemical formulas represent the numerical relationships that exist among atoms in a compound
Example - H2O represents a 2:1 fixed ratio of hydrogen atoms to oxygen atoms in a water molecule
Following relationships can be derived:
6.02×1023 H2O molecules contain 12.04×1023 H atoms and 6.02×1023 O atoms
1 mol of H2O molecules contains 2 mol of H atoms and 1 mol of O atoms
18.0 g of water contains 2.0 g of H and 16.0 g of O

55. Mole Calculations: Steps

56. Join In (14) How many molecules of Cl2 are present in a 54 g sample?
1.3×10–24
9.0×10–23
4.6×1023
3.3×1025
Correct Answer
4.6×1023

57. Join In (15)
How many moles of carbon dioxide (CO2) are in a 175 g sample?
0.160 moles
0.251 moles
3.98 moles
6.25 moles
Correct Answer
3.98 moles

58. Example 2.11 - Mass Percentage Calculations
Ammonia (NH3) and ammonium nitrate (NH4NO3) are commonly used agricultural fertilizers
Which one of the two contains the higher mass percentage of nitrogen (N)?

59. Example Solution
In each case, the mass percentage of N is given by
We will use 1 mol of each compound as a sample because the mass in grams of 1 mol of compound and the mass in grams of N in the 1 mol of compound are readily determined
One mol of NH3 weighs 17.0 g and contains 1 mol of N atoms, which weighs 14.0 g

60. Example 2.11 - Solution (continued)
Similarly, 1 mol of NH4NO3 weighs 80.0 g and contains 2 mol of N atoms, which weigh 28.0 g