Chapter 2:Atoms and Molecules : Chapter 2:Atoms and Molecules Slide 2: SYMBOLS AND FORMULAS
A unique symbol is used to represent each element.
Formulas are used to represent compounds.
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. Slide 4: chlorine bromine iodine Slide 5: 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. Slide 6: EXAMPLES OF COMPOUND FORMULAS
Carbon monoxide, CO (one atom of C and one atom of O are represented).
Water, H2O (two atoms of H and one atom of O are represented).
Ammonia, NH3 (one atom of N and 3 atoms of H are represented). Slide 7: THE STRUCTURE OF ATOMS
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. Slide 8: PROTONS
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. Slide 9: CHARACTERISTICS OF SUBATOMIC PARTICLES Slide 10: 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 and neutrons in the nucleus of the atom.
Mass numbers are represented by the symbol A. Slide 11: 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. Slide 12: SYMBOLS FOR ISOTOPES
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. Click here to play Coached Problem 60
28 Ni Slide 13: 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. Slide 14: 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. Slide 15: 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. Slide 16: 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 H2O. 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(16.0 u) = 18.02 u Slide 17: ISOTOPES AND 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. Slide 18: 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. Slide 19: 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. Click here to play Coached Problem Slide 20: 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 6.022x1023 .
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 = 6.022x1023 Na atoms
1 mole Ca = 40.08 g Ca = 6.022x1023 Ca atoms
1 mole S = 32.06 g S = 6.022x1023 S atoms Click here to play Chemistry Interactive Slide 21: 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 6.022x1023.
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. Click here to play Coached Problem Slide 22: EXAMPLES OF THE MOLE CONCEPT
1 mole H2O = 18.02 g H2O = 6.022x1023 H2O molecules
1 mole CO2 = 44.01 g CO2 = 6.022x1023 CO2 molecules
1 mole NH3 = 17.03 g NH3 = 6.022x1023 NH3 molecules
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. Slide 23: CALCULATIONS INVOLVING ELEMENTS
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 = 6.022x1023 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 Slide 24: EXAMPLE PROBLEM
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. Slide 25: Click here to play Coached Problem Click here to play Coached Problem Click here to play Coached Problem Slide 26: CALCULATIONS INVOLVING COMPOUNDS
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 CO2 is:
1 mole CO2 molecules = 1 mole C atoms + 2 moles O atoms
44.01 g CO2 = 12.01 g C + 32.00 g O
6.022x1023 CO2 molecules = 6.022x1023 C atoms +
(2) 6.022x1023 O atoms.
Any two of these nine quantities can be used to provide factors for use in solving numerical problems. Slide 27: Example 1: How many moles of O atoms are contained in 11.57 g of CO2?
Note that the factor used was obtained from two of the nine quantities given on the previous slide. Slide 28: Example 2: How many CO2 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. Slide 29: Click here to play Coached Problem Click here to play Coached Problem Click here to play Coached Problem Slide 30: Example 3: What is the mass percentage of C in CO2?
The 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=44.01 g CO2=12.01 g C + 32.00 g O. Thus, the mass of C in a specific mass of CO2 is known, and the problem is solved as follows: Slide 31: Example 4: What is the mass percentage of oxygen in CO2?
The 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 = 12.01 g C + 32.00g O Slide 32: Thus, the mass of O in a specific mass of CO2 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 CO2. Click here to play Coached Problem