2.1 SYMBOLS
As a matter of convenience in indicating the elements, the chemist usually uses abbreviations, since they are more quickly written than names. These abbreviations are called SYMBOLS. J.J. Berzelius (1779-1848) is generally given credit for inventing the modern symbol. Berzelius proposed that all elements be given a symbol corresponding to the first letter of their names. Any repetitions would be taken care of by using in addition the second letter or a letter outstanding in the name. For hydrogen, the symbol is H; for oxygen, O; and for carbon, C. Elements such as calcium, Ca; cobalt, Co; magnesium, Mg - the first letter is always capitalized. The second letter is never capitalized for those elements with two letters. Some symbols are abbreviations of the Latin names of the elements, as Fe for iron (Latin, ferrum), Na for sodium (Latin, natrium), and Cu for copper (Latin, cuprum).
The symbol not only identifies the element, but also represents one atom of the element. To indicate more than one atom of an element a coefficient is placed in front of the symbol to designate a given number of atoms. Thus 2 Cu designates 2 atoms of copper, and 5 Fe designates 5 atoms of iron.
The following table provides a list of some of the more common elements and their symbols.
Boron (B) | Aluminum (Al) | Copper (Cu, from cuprum) |
Carbon (C) | Barium (Ba) | Iron (Fe, from ferrum) |
Fluorine (F) | Bromine (Br) | Lead (Pb, from plumbum) |
Hydrogen (H) | Calcium (Ca) | Mercury (Hg, from hydrargyrum) |
Iodine (I) | Chlorine (Cl) | Potassium (K, from kalium) |
Nitrogen (N) | Helium (He) | Silver (Ag, from argentum) |
Oxygen (O) | Magnesium (Mg) | Sodium (Na, from natrium) |
Phosphorus (P) | Platinum (Pt) | Tin (Sn, from stannum) |
Sulfur (S) | Silicon (Si) | Tungsten (W, from wolfram) |
2.2 FORMULAS
A FORMULA is a single symbol or a group of symbols which represents the composition of a substance. The symbols in a formula identify the elements present in the substance. Thus, NaCl is the formula for sodium chloride, which consists of equal numbers of atoms of the elements sodium and chlorine. Subscripts are used to indicate the relative numbers of atoms in the compound. A subscript is not used when only one atom of a given element is present. The formula for water, H_{2}O, indicates that each molecule contains two atoms of hydrogen and one atom of oxygen. Sulfuric acid is represented by the formula H_{2}SO_{4}, which indicates two atoms of hydrogen, one of sulfur, and four of oxygen. The formula for aluminum sulfate, Al_{2}(SO_{4})_{3}, specifies two atoms of aluminum, and three sulfate groups. Each sulfate group contains one atom of sulfur and four atoms of oxygen. Hence, the aluminum sulfate formula includes a total of two atoms of aluminum, three atoms of sulfur, and twelve atoms of oxygen.
The formula for a molecule of sulfur in its most common form is S_{8}, showing that each molecule of this element consists of eight atoms. The molecular formulas for elemental hydrogen and oxygen (diatomic molecules) are H_{2} and O_{2}, while that for helium (a monoatomic molecule) is He. Note that H_{2} and 2H do not mean the same thing. H_{2} represents a molecule of hydrogen consisting of two atoms of the element chemically combined. The expression 2H, on the other hand, indicates that the two hydrogen atoms are not in combination as a unit but that they are separate particles.
2.3 CHEMICAL EQUATIONS
Atoms are fundamental particles of the elements that enter into chemical changes. Substances that take part in chemical changes contain these atoms in the form of molecules or ions (atoms or groups of atoms that are electrically charged). Chemical changes involve the regrouping of atoms or ions to form new substances. The CHEMICAL EQUATION is the chemist's shorthand expression for describing a chemical change, and symbols and formulas are used to indicate the composition of the substances involved in the change. The writer of an equation must know what substances react and what substances are formed, and he must be able to represent these substances by formulas.
This statement of the decomposition of water by electrolysis, "When water is decomposed by an electric current, hydrogen and oxygen are formed", can be expressed as
H_{2}O H_{2} + O_{2} (1)
The formulas for the reactants are written to the left and the formulas for the products are written to the right of the arrow. The arrow is read as "gives", "produces", "yields", or "forms". The + sign on the right side of the expression is read as "and"; it does not imply mathematical addition. When the + sign appears between the formulas for two reactants on the left side of an equation, it implies "reacts with".
As it now stands expression (1) does not conform with the Law of Conservation of Matter. Two atoms of oxygen in the molecule O_{2} could not be formed from one molecule of water containing but one oxygen atom. The equation is balanced by introducing the proper number (coefficient) before each formula. Two molecules of water will each furnish one oxygen atom for the production of one diatomic oxygen molecule. Two water molecules will then supply four hydrogen atoms for the production of two diatomic molecules of hydrogen. The balanced equation is
2 H_{2}O 2 H_{2} + O_{2} (2)
It is important to remember that the subscripts in a formula cannot be changed in order to make an equation balance. Substances have definite atomic compositions that cannot be changed by merely changing the subscripts in their formulas.
Because atoms are neither created nor destroyed in any reaction, a chemical equation must have an equal number of atoms of each element on each side of the arrow. When this condition is met, the equation is said to be balanced. For example, on the right side of the following equation there are two molecules of H_{2}O, each containing two atoms of hydrogen and one atom of oxygen.
2H_{2} + O_{2} 2H_{2}O
Thus, 2H_{2}O (read "two molecules of H_{2}O") contains 2 X 2 = 4 H atoms and 2 X 1 = 2 O atoms. Because there are also 4 H atoms and 2 O atoms on the left side of the equation, the equation is balanced.
Before we can write the chemical equation for a reaction, we must determine by experiment those substances that are reactants and those that are products. Once we know the chemical formulas of the reactants and products in the reaction, we can write the unbalanced chemical equation. We then balance the equation by determining the coefficients that provide equal numbers of each type of atom on each side of the equation. Generally, whole-number coefficients are used.
In balancing equations, it is important to understand the difference between a coefficient in front of a formula and a subscript in a formula. Changing a subscript in a formula--from H_{2}O to H_{2}O_{2}, for example--changes the identity of the chemical. The substance H_{2}O_{2}, hydrogen peroxide, is quite different from water. SUBSCRIPTS SHOULD NEVER BE CHANGED IN BALANCING AN EQUATION. In contrast, placing a coefficient in front of a formula changes only the amount and not the identity of the substance; 2H_{2}O means two molecules of water, 3H_{2}O means three molecules of water, and so on.
To illustrate the process of balancing equations, consider the reaction that occurs when methane, CH_{4}, the principal component of natural gas, burns in air to produce carbon dioxide gas, CO_{2}, and water vapor, H_{2}O. Both of these products contain oxygen atoms that come from O_{2} in the air. We say that combustion in air is "supported by oxygen," meaning that oxygen is a reactant. The unbalanced equation is:
CH_{4} + O_{2} CO_{2} + H_{2}O (unbalanced)
It is usually best to balance first those elements that occur in only one substance on each side of the equation. In our example both C and H appear in only one reactant and, separately, in one product each, so we begin by focusing attention on CH_{4}. Let's consider first carbon and then hydrogen.
Notice that one molecule of CH_{4} contains the same number of C atoms (one) as does one molecule of CO_{2}. Therefore, the coefficient for these substances must be the same, and we choose them both to be 1 as we start the balancing process. However, the reactant CH_{4} contains more H atoms (four) then does the product H_{2}O (two). If we place a coefficient 2 in front of H_{2}O, there will be four H atoms on each side of the equation:
CH_{4} + O_{2} CO_{2} + 2H_{2}O (unbalanced)
At this stage the products have more total O atoms (four--two from CO_{2} and two from 2H_{2}O) than the reactants (two). If we place a coefficient 2 in front of O_{2}, we complete the balancing by making the number of O atoms equal on both sides of the equation:
CH_{4} + 2O_{2} CO_{2} + 2H_{2}O (balanced)
For most purposes a balanced equation should contain the smallest possible whole-number coefficients. The process is largely trial and error. We balance each kind of atom in succession, adjusting coefficients as necessary. This approach works for most chemical equations.
The physical state of each chemical in a chemical equation is often indicated parenthetically. We use the symbols (g), (l), (s), and (aq) for gas, liquid, solid, and aqueous (water) solution, respectively. Thus, the balanced equation above can be written:
CH_{4} (g) + 2O_{2} (g) CO_{2} (g) + 2H_{2}O (g)
Often the conditions under which the reactions proceed appear above or below the arrow between the two sides of the equation. For example, temperature or pressure at which the reaction occurs could be indicated. Sometimes a substance is introduced into the reaction which causes the reaction to occur but does not itself enter into the reaction and therefore will be placed above or below the arrow as well.
2.4 CHEMICAL REACTIONS
There are many different kinds of chemical reactions. The study of new reactions and the processes in which they occur is an on-going area of research for many chemists. To simplify this study and research, these reactions can be grouped into three basic types.
SYNTHESIS REACTIONS: When two or more elements or compounds combine to form one substance, the reaction is a synthesis reaction. This is also sometimes referred to as a Combination Reaction.
A + B C
The formation of water is an example.
2H_{2} + O_{2} 2H_{2}O
Another synthesis reaction that occurs slowly under normal atmospheric conditions is the formation of iron oxide (Fe_{2}O_{3}) from the combination of iron and oxygen. This, of course, is what happens when iron rusts. The equation is shown below. What coefficients must you use to balance the equation?
Fe + O_{2} Fe_{2}O_{3}
Other examples are:
C + O_{2} CO_{2}
N_{2} + 3H_{2} 2 NH_{3}
CaO + H_{2}O Ca(OH)_{2}
2Mg + O_{2} 2MgO
DECOMPOSITION REACTIONS: When one substance is broken down into two or more substances, a decomposition reaction has occurred. For example, mercury (II) oxide can be decomposed into the elements mercury and oxygen gas by heating.
C A + B
2HgO Hg + O_{2}
The result of a combination reaction can be reversed; in other words, a compound can be decomposed into the components from which it was formed. This type of reaction is called a decomposition reaction. Many decomposition reactions occur via oxidation-reduction as illustrated below.
Note: Chlorine is reduced, while oxygen is oxidized.
But many other decomposition reactions do not involve a corresponding oxidation and reduction of the substances as shown below.
Note, that in this example of chemical decomposition, the oxidation states of the elements involved remain constant.
Other examples are:
PbCO_{3 } PbO + CO_{2}
2NaN_{3} 2Na + 3N_{2}
REPLACEMENT REACTIONS: Another type of reaction is a replacement reaction. During a replacement reaction, one substance in a compound is replaced with another substance. These reactions are further divided into single and double replacement reactions. Single replacement reactions involves substitution of one element in a compound for another. In double replacement reactions, two compounds exchange partners.
If a piece of copper is placed in a colorless silver nitrate solution, pieces of silver will be deposited on the copper and the solution will gradually turn blue. The blue solution is copper (II) nitrate. The equation for this reaction is:
Cu + 2AgNO_{3} Cu(NO_{3})_{2} + 2Ag
This is an example of a single replacement reaction. If you look at the equation, you can see that copper has replaced silver in the compound.
Note: Copper is oxidized; silver is reduced.
The element which replaces that which is in a compound is always oxidized. The element being displaced, is always reduced. This is illustrated by the displacement of hydrogen gas by metallic iron in the example below:
The oxidation of iron is represented by:
Note that the net charge on both sides of the arrow must always be equal to each other.
The reduction of hydrogen is represented by:
Note: In both oxidation and reduction, the net charge of both sides of the arrow must always be equal.
Below is an example of a double replacement reaction. In this reaction, potassium (K) and lead (Pb) are changing places in the compounds.
2KCl + Pb(NO_{3})_{2} PbCl_{2} + 2KNO_{3}
Sometimes in a replacement reaction, one of the compounds formed is not soluble in the reaction mixture, the compound forms a solid that comes out of solution. This solid is called a precipitate.
2.5 ATOMIC WEIGHTS
It has been determined by experiment that an atom of hydrogen (the lightest element) weighs 1.7 X 10^{-24} grams. Such a method of expressing weights of atoms is somewhat inconvenient. Since the weights of the individual atoms of all the elements are of the same small order of magnitude, 10^{-24} g, it has been found more convenient to use relative weights in place of actual weights measured in grams or other units. THE RELATIVE WEIGHTS OF THE ATOMS OF THE DIFFERENT ELEMENTS ARE KNOWN AS ATOMIC WEIGHTS AND ARE PROPORTIONAL TO THE ACTUAL WEIGHTS OF THE ATOMS. For example, the atomic weight of hydrogen is 1.00797 on this arbitrary scale, that of carbon is 12.01115, and that of oxygen is 15.9994. Hence, carbon atoms weigh about twelve times as much as hydrogen atoms, and oxygen atoms are approximately sixteen times heavier than hydrogen atoms. Sulfur atoms weigh about twice as much as oxygen atoms, or 32.064 on this arbitrary scale. THE STANDARD USED IN EXPRESSING THE ATOMIC WEIGHTS IS A PARTICULAR KIND OF CARBON ATOM CALLED THE CARBON-12-ISOTOPE.
In addition to identifying the element and representing one atom of that element, a chemical symbol represents one chemical unit of weight of an element. The symbol O represents one atom of oxygen, or 15.9994 units of weight of oxygen. The unit of weight may be a gram, an ounce, a pound, etc. In quantitative work, the chemist finds it convenient to use the atomic weight of an element expressed in grams and called a GRAM-ATOMIC WEIGHT, or GRAM-ATOM. Thus, one gram-atomic weight of oxygen and of hydrogen are 15.9994 grams and 1.00797 grams respectively.
2.6 AVOGADRO'S NUMBER
It can be shown that there is the same number of atoms, 6.023 X 10^{23}, in one gram-atomic weight of any element. This number is called AVOGADRO'S NUMBER in honor of the Italian professor of physics Amedeo Avogadro (1776-1856). It is important to remember that one gram-atomic weight of any element, e.g., 15.9994 g of oxygen, 32.064 g of sulfur, or 1.00797 g of hydrogen, all contain 6.023 X 10^{23} atoms. Knowing the atomic weight of an element and Avogadro's number, one can easily calculate the weight in grams of one atom of the element.
The weight of one oxygen atom would be, therefore
15.1 grams/6.023 X 10^{23} = 2.656 X 10^{-23} grams
2.7 FORMULA WEIGHTS, MOLECULAR WEIGHTS AND MOLES
THE SUM OF THE ATOMIC WEIGHTS OF ALL THE ATOMS SHOWN BY THE FORMULA OF A SUBSTANCE IS ITS FORMULA WEIGHT. Formula weights are relative, just as are the atomic weights upon which they are based. The formula weight of sulfuric acid (H_{2}SO_{4}) is 98, (2 X 1) + (1 X 32) + (4 X 16) = 98. For most of our work it will be permissible to round off the atomic weights of the elements.
The formula weight of a substance expressed in grams is the GRAM-FORMULA WEIGHT or a MOLE of that substance. Thus, the gram-formula weight of sulfuric acid is 98 grams, whereas its formula weight is simply 98.
When the true formula for a molecule of a substance is known, the molecular weight can be calculated by adding the atomic weights of the atoms given in the formula of the substance. It is quite correct to speak of the molecular weight (or formula weight) of substances which consist of discrete molecules such as methane, CH_{4}; carbon dioxide, CO_{2}; and water, H_{2}O.
When comparing one mole of a substance to one mole of another substance it is important to note that both contain the same number of molecules. Thus, one mole of sulfuric acid (H_{2}SO_{4}), contains the same number of molecules as one mole of hydrogen gas (H_{2}), even though sulfuric acid has a gram molecular weight of 98 grams and hydrogen gas that of 2 grams.
It is often convenient in quantitative chemical work to use milli-moles in place of moles of a substance. A MILLIMOLE OF A SUBSTANCE IS ITS FORMULA WEIGHT EXPRESSED IN MILLIGRAMS.
EXAMPLE: How many gram-atoms (moles) of sulfur are there in 80 grams of sulfur?
gram-atoms of S (moles) | = | grams of S |
gram-atomic weight of S | ||
= | 80 grams | |
32 grams moles | ||
= | 2.5 |
2.8 PERCENTAGE COMPOSITION FROM FORMULAS
The percentage by weight of each element in a compound may easily be calculated from the formula of the compound. Simply put, to find the percentage of anything divide the partial by the total and multiply by 100.
PERCENTAGE = PARTIAL/TOTAL X 100
For example, suppose we calculate the percentage of hydrogen and that of oxygen in the compound water, H_{2}O.
% of Hydrogen in H_{2}O:
gram-atomic weight of H | = | (2 X 1) | = | 2 |
gram-formula weight of H_{2}O | = | (2 X 1) + (1 X 16) | = | 18 |
2 | X | 100 | = | 11% |
18 |
% of Oxygen in H_{2}O:
gram-atomic weight of O | = | (1 X 16) | = | 16 |
gram-formula weight of H_{2}O | = | (2 X 1) + (1 X 16) | = | 18 |
16 | X | 100 | = | 89% |
18 |
Samples of pure water, no matter what their source may be, contain the same elements, hydrogen and oxygen, united in the same proportion by weight, 11 parts of hydrogen to 89 parts of oxygen. A similar statement may be made about any pure compound. This is one of the fundamental concepts of chemistry and is known as the LAW OF DEFINITE PROPORTIONS, or the LAW OF DEFINITE COMPOSITION. This law may be stated as follows: DIFFERENT SAMPLES OF A PURE COMPOUND ALWAYS CONTAIN THE SAME ELEMENTS IN THE SAME PROPORTIONS BY WEIGHT. This law, among others, convinced John Dalton (1766-1844) of the atomistic nature of matter and led him to outline the atomic theory.
2.9 EMPIRICAL FORMULAS
When a new compound is created in the laboratory, it is usually easy to determine its percent composition experimentally. From the percent composition data, you can calculate the empirical formula of the compound. The EMPIRICAL FORMULA is the lowest whole-number ratio of the atoms of the elements in a compound. For example, a compound is composed of 40% carbon, 6.7% hydrogen and 53.3% oxygen.. If you divide each of these percentages by the atomic weight of the respective elements you will arrive at the following ratios:
40 | = | 3.33 | 6.7 | = | 6.7 | 53.3 | = | 3.33 |
12 | 1 | 16 |
When each of these numbers 3.33, 6.7 and 3.33 are reduced to the smallest whole number we arrive at 1,2,1. Thus the ratio, and therefore the empirical formula, of our compound is CH_{2}O.
If by other means we are able to determine the actual molecular weight (180) of the substance, it becomes a simple matter to determine the true molecular formula. For instance, in the example above if we divide the weight of CH_{2}O (30) into the molecular weight of the compound (180) containing carbon, hydrogen and oxygen we will see that the actual or true molecular formula has six (6) time as much of each element as the empirical formula, thus C_{6}H_{12}O_{6}.
Practice Exercise
1. Determine the number of atoms in 3 grams of Ca.
1 Mole | = | Ca | = | 40 g | = | 6.023 X 10^{23} |
3 g | X |
X | = | (6.023 X 10^{23)}(3 g) |
(40 g) |
X | = | 4.51725 X 10^{23} |
2. Determine the number of atoms in 5 grams of Cu.
1 Mole | = | Cu | = | 63.54 g | = | 6.023 X 10^{23} |
5 g | X |
X | = | (6.023 X 10^{23)}(5 g) |
(63.54 g) |
X | = | 4.7395 X 10^{22} |
3. What is the approximate weight of 5 liters of SO_{2} at S.T.P?
1 Mole | = | SO_{2} | = | 64 g | = | 6.023 X 10^{23} | = | 22.4 L |
X |
5 L |
X | = | (64 g)(5 L) |
(22.4 L) |
X | = | 14.28 g |
4. What is the weight in grams of 16.8 liters of SO_{3}?
1 Mole | = | SO_{3} | = | 80 g | = | 6.023 X 10^{23} | = | 22.4 L |
X |
16.8 L |
X | = | (80 g)(16.8 L) |
(22.4 L) |
X | = | 60 g |
5. Find the number of molecules in 4 mL of oxygen gas (O_{2}).
1 Mole | = | O_{2} | = | 32 g | = | 6.023 X 10^{23} | = | 22.4 L | = | 22,400 mL |
X |
4 mL |
X | = | (6.023 X 10^{23)}(4 mL) |
(22,400 mL) |
X | = | 1.075 X 10^{20} |
6. What weight of N_{2} contains the same number of molecules as 48 grams of SO_{2}?
1 Mole | = | SO_{2} | = | 64 g | = | 6.023 X 10^{23} | = | 22.4 L |
48 g |
X |
X | = | (48 g)(6.023 X 10^{23}) |
(64 g) |
X | = | 4.51725 X 10^{23} |
As you can see below, the answer 4.51725 X 10^{23}, is required to complete the second phase of this problem.
1 Mole | = | N_{2} | = | 28 g | = | 6.023 X 10^{23} | = | 22.4 L |
X | 4.51725 X 10^{23} |
X | = | (28 g)(4.51725 X 10^{23}) |
(6.023 X 10^{23}) |
X | = | 21 g |
7. What weight of O_{2} contains the same number of molecules as 11 grams of CO_{2}?
1 Mole | = | CO_{2} | = | 44 g | = | 6.023 X 10^{23} | = | 22.4 L |
11 g |
X |
X | = | (11 g)(6.023 X 10^{23}) |
(44 g) |
X | = | 1.50575 X 10^{23} |
As you can see below, the answer 1.50575 X 10^{23}, is required to complete the second phase of this problem.
1 Mole | = | O_{2} | = | 32 g | = | 6.023 X 10^{23} | = | 22.4 L |
X | 1.50575 X 10^{23} |
X | = | (32 g)(1.50575 X 10^{23}) |
(6.023 X 10^{23}) |
X | = | 8 g |
8. What volume of N_{2} contains the same number of molecules as 32 grams of SO_{2}?
1 Mole | = | SO_{2} | = | 64 g | = | 6.023 X 10^{23} | = | 22.4 L |
32 g |
X |
X | = | (32 g)(6.023 X 10^{23}) |
(64 g) |
X | = | 3.0115 X 10^{23} |
As you can see below, the answer 3.0115 X 10^{23}, is required to complete the second phase of this problem.
1 Mole | = | N_{2} | = | 28 g | = | 6.023 X 10^{23} | = | 22.4 L |
3.0115 X 10^{23} | X |
X | = | (22.4 L)(3.0115 X 10^{23}) |
(6.023 X 10^{23}) |
X | = | 11.2 L |
9. What is the percentage of carbon in sodium carbonate (Na_{2}CO_{3})?
Na_{2}CO_{3} | = | Na | = | 2 | X | 23 | = | 46 |
C | = | 1 | X | 12 | = | 12 | ||
O | = | 3 | X | 16 | = | 48 | ||
106 |
Partial | X | 100 | = | % |
Total |
12 | X | 100 | = | 11.32% |
106 |
10. What is the percentage of aluminum in aluminum sulfate (Al_{2}(SO_{4})_{3})?
Al_{2}(SO_{4})_{3} | = | Al | = | 2 | X | 27 | = | 54 |
S | = | 3 | X | 32 | = | 96 | ||
O | = | 12 | X | 16 | = | 192 | ||
342 |
Partial | X | 100 | = | % |
Total |
54 | X | 100 | = | 15.79% |
342 |
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