Introductory Concepts
General Chemistry
Basic Definitions and Relationships Used in Chemistry
An atom is the smallest neutral component of an element that has all the chemical properties of the element. Atoms are composed of subatomic particles: protons, neutrons, and electrons. For all practical purposes, you can think of the atom as having a very small core, the nucleus, which contains both protons and neutrons. Electons form a cloud around the nucleus. The space occupied by electrons (electron cloud) is massive compared to the nucleus. The atomic number of an atom is the number of protons that the atom contains.
In many substances, groups of atoms are joined together by chemical bonds to form molecules. You can express the composition of a molecule using its molecular formula, by writing the symbols of the atoms it contains, with numeric subscripts showing how many of that kind of atom are present in the molecule.
A mole is the amount of a substance tha contains Avogadro's number of particles of that substance. Avogadro's number is about 6.02x1023. Thus, a mole of atoms contains 6.02x1023 atoms.
The atomic weight of an element is an average of the weights of all the isotopes of the element. Isotopes are atoms of the same element that differ in mass but not in atomic number (atoms with the same number of protons but different number of neutrons). Stoichiomestry is the study of the molar relationship between atoms and compounds.
- Atomic and Molecular Weight
In chemical calculations, you often need to know the mass of an element. That is accomplished by finding the atomic weight of the element on the periodic table. The atomic weight is the larger of the two numbers for each element. The smaller number is usually the atomic number. To calculate molecular weight, you add the individual atomic weights of the elements comprising the compound.
Example: What is the molecular weight of C6H6?
Look up the atomic weights of both carbon and hydrogen (C = 12 and H = 1). Multiply each atomic weight by the number of each of the corrsponding atoms, then add the values together. In this case the molecular weight is (6 x 12) + (6 x 1) = 78 g/mol.
|
One common mistake in the calculation of molecular weight is not multiplying through by the subscripts in a molecular formula. For example, a molecule with a formula of Al2(SO4)3 would actually be thought of as Al2S3O12 for calculating molecular weight.
- Empiric Formula Versus Molecular Formula
Chemical formulas are derived empirically, which means that experiments must be done to determine the actual mass of the elements in a compound or their percentage by weight. The empiric formula of a compound is the smallest possible integer ratio of the different kinds of atoms present in a compound.
The molecular formula is an integral multiple of the empiric formula. It expresses the actual number of atoms joined by chemical bonds to form a molecule. For example, the molecular formula for benzene is C6H6. Notice the 1:1 ratio of carbon and hydrogen, making the empiric formula CH.
The molecular formula can be determined if the molecular weight and the empiric formula of a compound are known. The ratio of the molecular weight of the compound to the molecular weight of the empiric formula provides the integral multiple from which the molecular formula can be determined.
Example: Glucose has an empiric formula CH2O and a molecular weight of 180g/mol. What is its molecular formula?
The empiric weight of glucose is 12 + 1 + 1 + 16 = 30. 180/30 = 6 of the empiric unit in the molecule. The molecular formula is C6H12O6.
|
How do you determine the molecular weight of a compound experimentally? Usually from studies of the compound in its gaseous state. If a compound is weighed and then vaporized at constant temperature and volume, the gaseous molecular weight can be determined.
- Description of Composition by Percent Mass
Many analytic methods do not give an empiric formula directly. Instead, a percent mass is provided for each of the elements in the sample. The empric formula can then be determined from the composition by percent mass. To determine the empiric formula from percent composition, follow the following steps:
- Assume 100 g of the compound.
- Find the number of moles of each element in the compound.
- Divide each of these number of moles by the smallest value of step 2.
- Multiply all by the smallest factor that provides a whole number.
- Values derived from step 4 are subscripts of the empiric formula.
Example: A hydrocarbon was determined to contain 20% hydrogen by mass. Determine its empiric formula.
First, realize that a hydrocarbon contains only carbon and hydrogen, so the actual percent composition is 20% hydrogen, and 80% carbon. Then, follow these steps:
- Assume 100 g of the hydrocarbon. Because 20% of 100g is 20g, and 80% of 100g is 80g, there are 20g of hydrogen, and 80g of carbon.
- Convert each of the masses to moles: (80g)(1mol/12g C) = 6.67 mol and (20g)(1mol/1g H) = 20 mol.
- Divide each of these by the smallest value of step 2. 6.67/6.67 = 1 and 20/6.67 = 3.
- Multiply all by the smallest factor that provides a whole number. In this case the number is 1.
- This the empric formula is CH3
|
- Mole Concept and Avogadro's Number
Although we have already defined some of these terms, we should talk about them in more detail because an understanding of the mole concept is critical to the success on the MCAT.
Elements react in certain ratios by weight. This observation led to the modern understanding of moles and stoichiometry. Because molecules and atoms are extremely small, it is not inconvenient to use quantities such as dozens or scores to define a quantity of atoms or molecules. The number of dozens of atoms needed to add up to 1 gram is an extremely large number. The mole is a quantity that makes dealing with huge numbers of molecules much easier. For example, a mole of atoms is the number of atoms of C-12 that weigh exactly 12.0g. A mole of atoms is the number of atoms of O-16 that weigh exactly 16.0g. And a mole is the number of compounds of H2O-18 that weigh exactly 18.0g. Also, as defined previously, a mole of atoms is equivalent to 6.02x1023 atoms.
Conveniently, a mole of any type of atom or compound is the number of atoms the weight of which is equal to the atomic or molecular weight. Therefore, an appropriate unit for the atomic weights on the periodic table is grams per mole (g/mol).
Example: How many moles of O2 are there in 48g of O2?
Moles O2 = (48g)(1 mol/32g) = 1.5 mol.
|
Example: How many moles in 50.0g of CH4?
Moles CH4 = (50.0g)(1 mol/16.04g) = 3.12 moles
|
Because you know that a mole of a substance contains Avogadro's number of particles of that substance, it is easy to calculate how many atoms are in a mass quantity of a compound.
Example: How many carbon and hydrogen atoms are in 25g of CH4?
[25g CH4/ (mol CH4/16g)](1 mol C/1 mol CH4)(6.02x1023)=9.41x1023C atoms
[25g CH4/ (mol CH4/16g)](4 mol H/1 mol CH4)(6.02x1023)=3.76x1024H atoms
|
- Density
Matter is something that possesses mass and occupies space or volume. Density is a relationship that describes how mass is related to volume and is given by the relationship density = m/v in which m is the mass (unit: usually grams) and v is the volume (units: liters, millimeters, or cm3).
Density is an intrinsic property of a substance, meaning that it is not dependent on the amount of matter. The volume of a substance does change with temperature, however, so density is dependent on temperature and therefore is usually reported at a given temperature.
You are probably aware of density differences in things around you. For example, you know that a block of iron weighs more than a block of wood of equal volume and that oil floats on water. You have also seen examples of density differences in the organic laboratory, when a liquid such as chloroform (CHCl3) tends to form a layer beneath a less dense aqueous layer.
- Oxidation Number
Before beginning a review of oxidation numbers, you should review the meaning of oxidation and reduction.
Oxidation: a loss of electrons or an increase in oxidation number.
Reduction: a gain of electrons or a decrease in oxidation number.
The oxidation number or oxidation state of an atom in a compound is an assigned numeric representation of the positive or negative character of the atom. In other words, it is the number of electrons that an atom appears to have gained over or lost from its normal complement when it is combined with other atoms.
You should become familiar with some of the basic rules for assigning oxidation numbers, which are summarized as follows:
- The sum of the oxidation numbers of atoms in a molecule or ion must equal the overall charge of the species (e.g., for O2, the sum of the oxidation states of the two oxygen atoms must be zero; for SO4-2, the oxidation states on the sulfur and four oxygen atoms must sum to -2).
- The oxidation number of a free, uncharged element is zero (e.g., O2, H2, Na, Cl2).
- Alkali metals, found in the first column of the periodic table (group I), have an oxidation number of +1 in compounds(e.g., Na+, K+).
- Alkaline earth metals, found in the second column of the periodic table (group II), have an oxidation number of +2 in compounds(e.g., Ca2+, Mg2+).
- Halogens, found in the column of the periodic table second from the right (group VII), usually have an oxidation number of -1 (e.g., C--, Br-).
- Oxygen has an oxidation number of -2, except in peroxides (e.g., H2O2) in which it is -1, or in compounds with flourine (e.g., OF2), in which it is +2.
- Hydrogen has an oxidation number of +1 when bonded to a non-metal and -1 when bonded to a metal (e.g., in H2O, the hydrogen is bonded to oxygen, a non-metal, so its oxidation state is +1; in LiH, hydrogen is bonded to a metal, so its oxidation state is -1).
1. Common oxidizing and reducing agents
Oxidizing agent: The species in a redox reaction that accepts electrons. The result of this action is that the oxidizing agent is reduced and another species is oxidized. Usually nonmetal elements such as groups VI (eg. oxygen) and VII (eg. chlorine) gain electrons to obtain a noble gas configuration.
Reducing agent: The species in a redox reaction that donates electrons. The result of this action is that the reducing agent is oxidized and the other species is reduced. Usually metallic elements from groups I and II (eg. sodium or magnesium) or transitional metals (eg. silver and titanium) lose electrons to obtain a more stable electron configuration.
2. Redox titration
As in other titrations, a redox titration uses a known concentration of one reactant in a quantitative reaction with a known stoichiometry to determine the unknown concentration of a different reactant. The distinguishing feature of a redox reaction is that it involves oxidation and reduction of the reacting species. The following is a redox titration used in quantitative analysis of iron ore.
Example: Given the following reaction in which 34.6 ml of 0.11 KMnO4 was required to oxidize all the Fe2+ in a 3.52 g sample of iron ore, determine the mass percent of iron in the sample.
5Fe2+(aq) + MnO4-(aq) + 8H+ --> 5Fe3+(aq) + Mn2+(aq) + 4H2O
Calculate the number of moles of oxidizing agent:
mol KMnO4=(0.11 mol/L)(0.0346 L)=0.00381 mol
Use the stoichiometry of the reaction to calculate the number of moles of the unknown agent:
mol Fe2+=(0.00381 mol KMnO4)(5 mol Fe2+/1 mol KMnO4)
= 0.01905 mol Fe2+
Convert this value to grams:
g Fe2+=(0.01905 mol Fe2+)(55.85 g/mol Fe2+)=1.064 g Fe2+
Calculate the percentage by mass:
mass % Fe2+=(1.064 g Fe2+/3.51 g ore)(100%)= 30.2%
|
Describing Reactions by Chemical Equations
Chemical reactions can be described in words, but chemists use chemical equations as a 'shorthand' to describe these reaction.
Conventions for Writing Chemical Equations
The reactants are the substances that react with each other; they are shown on the left hand side of the equations. The products are the substances formed in the reaction; they are shown on the right hand side of the equation. A left-to-right arrow indicates that the reactants are converted to products.
Example:
Na(s) + 1/2Cl2(g) --> NaCl(s)
Reactants form products.
|
The symbols after the chemical formulas represent the physical state of the substances: (s) refers to solids, (g) refers to gases, (l) refers to liquids, and (aq) refers to aqueous solutions.
Balancing Equations, Including Oxidation-Reduction Equations
Two basic rules for balancing an equation follow:
Mass balance: The number of each type of atom on both sides of the equation must be equal.
Charge balance: The net charge on both sides of the equation must be equal.
For most equations, you should balance by inspection (trial and error). The following example shows the mass balance concept. Note that balancing is done by trial and error.
Example:
Na(s) + Cl2(g) --> NaCl(s)
This equation does not obey the mass balance condition. To balance it, make sure you have two chlorine atoms on both sides:
Na(s) + Cl2(g) --> 2NaCl(s)
This change disrupts the mass balance of the sodium atoms, so put a 2 in front of the sodium on the left.
2Na(s) + Cl2(g) --> 2NaCl(s)
|
In redox reactions, balancing by trial and error is difficult and inefficient. These equations are hard to balance because of the different numbers and different kinds of atoms on the two sides of the equations. In addition, the charges often do not balance. Although the technique outlined subsequently seem advanced, the principles involved are simply an application of the mass and charge balancing rules.
Method for Balancing Oxidation-Reduction Reaction:
1. Identify species being oxidized and reduce.
2. Determine oxidation states.
3. Balance all elements except O and H.
4. Make changes in oxidation state opposite.
5. Balance charge; use H+ in acid or OH- in base.
6. Balance hydrogen and oxygen with H2O.
|
|
