Conceptually, that was the most challenging discussion in this chapter. Let’s move on to review ways in which compounds are represented.
Representation of Compounds
There are different ways of representing compounds and their constituent atoms. We’ve already reviewed a couple of these systems (Lewis dot structures and VSEPR theory) in Chapter 3. In organic chemistry, it is common to encounter “skeletal” representations of compounds, called structural formulas, to show the various bonds between the constituent atoms of a compound. Inorganic chemistry typically represents compounds by showing the constituent atoms without representing the actual bond connectivity or atomic arrangement. For example, C6H12O6 (which you ought to recognize is the molecular formula for glucose) tells us that this particular compound consists of 6 atoms of carbon, 12 atoms of hydrogen, and 6 atoms of oxygen, but there is no indication in the molecular formula of how the different atoms are arranged or how many bonds exist between each of the atoms.
LAW OF CONSTANT COMPOSITION
The law of constant composition states that any pure sample of a given compound will contain the same elements in the identical mass ratio. For example, every sample of H2O will contain two hydrogen atoms for every one oxygen atom, or in terms of mass, for every one gram of hydrogen, there will be eight grams of oxygen. We’re sorry to say that it doesn’t matter whether the water you drink is sourced from deep springs in the Fiji Islands or flows out of a tap connected to a shallow well. The law of constant composition says that water is water.
EMPIRICAL AND MOLECULAR FORMULAS
There are two ways to express a formula—the elemental composition—for a compound. The empirical formula gives the simplest whole number ratio of the elements in the compound. The molecular formula gives the exact number of atoms of each element in the compound and is usually a multiple of the empirical formula. For example, the empirical formula for benzene is CH, while the molecular formula is C6H6. For some compounds, the empirical and molecular formulas are the same, as in the case of H2O. For the reasons previously discussed, ionic compounds, such as NaCl or CaCO3, will only have empirical formulas.
PERCENT COMPOSITION
The percent composition by mass of an element is the weight percent of a given element in a specific compound. To determine the percent composition of an element X in a compound, the following formula is used:
% Composition = (Mass of X in formula/Formula weight of compound) × 100%
You can calculate the percent composition of an element by using either the empirical or the molecular formula; just be sure to use to the appropriate mass measurement:
MCAT Expertise
Percent composition is a common way for stoichiometry to be tested on the MCAT. Practice these problems to build up speed and efficiency for Test Day.
formula weight for empirical formula or molar mass for molecular formula. Formula weight is simply the mass of the atoms in the empirical formula of a compound.
You can determine the molecular formula if both the percent compositions and molar mass of the compound are known. The following examples demonstrate such calculations.
MCAT Expertise
When there are two methods for approaching a problem, be well versed in both. Knowing multiple ways to solve a problem will help you tackle questions efficiently. help you tackle questions efficiently.
Example: What is the percent composition of chromium in K2Cr2O7?
Solution: The formula weight of K2Cr2O7 is
Example: What are the empirical and molecular formulas of a compound that contains 40.9 percent carbon, 4.58 percent hydrogen, and 54.52 percent oxygen and has a molar mass of 264 g/mol?
Method One: First, determine the number of moles of each element in the compound by assuming a 100-gram sample; this converts the percentage of each element present directly into grams of that element. Then convert grams to moles:
Next, find the simplest whole number ratio of the elements by dividing the number of moles by the smallest number obtained in the previous step.
Finally, the empirical formula is obtained by converting the numbers obtained into whole numbers (multiplying them by an integer value).
C1H1.33O1 × 3 = C3H4O3
C3H4O3 is the empirical formula. To determine the molecular formula, divide the molar mass by the formula weight. The resultant value is the number of empirical formula units in the molecular formula.
The empirical formula weight of C3H4O3 is
Method Two: When the molar mass is given, it is generally easier to find the molecular formula first. This is accomplished by multiplying the molar mass by the given percentages to find the grams of each element present in one mole of compound, then dividing by the respective atomic weights to find the mole ratio of the elements:
Thus, the molecular formula, C9H12O9, is the direct result.
The empirical formula can now be found by reducing the subscript ratio to the simplest integral values.
Key Concept
The molecular formula is either the same as the empirical formula or a multiple of it. To calculate the molecular formula, you need to know the mole ratio (this will give you the empirical formula) and the molecular weight (molecular weight ÷ empirical formula weight will give you the multiplier for the empirical formula to molecular formula conversion).
Types of Chemical Reactions
This section reviews the major classes of chemical reactions. As it has probably already become apparent to you in your inorganic and organic chemistry classes, it would be quite impossible to memorize every single individual reaction that could occur. Fortunately, there is no need to memorize any reaction, as long as you take the time now and throughout your preparation for the MCAT to learn and understand the recognizable patterns of reactivities between compounds. Some classes of compounds react in very “stereotyped” ways, or at least appear to react in stereotyped ways, because of the MCAT’s focus on a particular subset of a chemical’s reactivities. The MCAT has a tendency to typecast certain compounds, in spite of repeated protests made by the compounds’ talent agents.
COMBINATION REACTIONS