SAT Subject Test Chemistry
REVIEW OF MAJOR TOPICS
CHEMICAL FORMULAS: THEIR MEANING AND USE
As you have seen, a chemical formula is an indication of the makeup of a compound in terms of the kinds of atoms and their relative numbers. It also has some quantitative applications. By using the atomic masses assigned to the elements, we can find the formula mass of a compound. If we are sure that the formula represents the actual makeup of one molecule of the substance, the term molecular mass may be used as well. In some cases the formula represents an ionic lattice and no discrete molecule exists, as in the case of table salt, NaCl, or the formula merely represents the simplest ratio of the combined substances and not a molecule of the substance. For example, CH2 is the simplest ratio of carbon and hydrogen united to form the actual compound ethylene, C2H4. This simplest ratio formula is called the empirical formula, and the actual formula is the true formula . The formula mass is determined by multiplying the atomic mass units (as a whole number) by the subscript for that element and then adding these values for all the elements in the formula. For example:
Ca(OH)2 (one calcium amu + two hydrogen and two oxygen amu = formula mass).
1Ca (amu = 40) = 40
2O (amu = 16) = 32
2H (amu = 1) = 2.0
Formula mass Ca(OH)2 = 74 amu (or µ)
In Chapter 6, the concept of a mole is introduced. If you have 6.02 × 1023 atoms of an element, then the atomic mass units can be expressed in grams, and then the formula mass can be called the molar mass. Another example is Fe2O3.
2Fe (amu = 56) = 112
3O (amu = 16) = 48.0
Formula mass Fe2O3 = 160. amu
Know how to compute the percentage composition of an element in a compound.
It is sometimes useful to know what percent of the total weight of a compound is made up of a particular element. This is called finding the percentage composition .The simple formula for this is:
To find the percent composition of calcium in calcium hydroxide in the example above, we set the formula up as follows:
To find the percent composition of oxygen in calcium hydroxide:
To find the percent composition of hydrogen in calcium hydroxide:
This type of question always appears on the test.
ANOTHER EXAMPLE: Find the percent compositions of Cu and H2O in the compound CuSO4·5H2O (the dot is read “with”).
First, we calculate the formula mass:
1 Cu = 64 amu
1 S = 32 amu
4 O = 64 (4 × 16) amu
5 H2O = 90. (5 × 18) amu
and then find the percentages:
When you are given the percentage of each element in a compound, you can find the empirical formula as shown with the following example:
Given that a compound is composed of 60.0% Mg and 40.0% O, find the empirical formula of the compound.
1. It is easiest to think of 100 mass units of this compound. In this case, the 100 mass units are composed of 60. amu of Mg and 40. amu of O. Because you know that 1 unit of Mg is 24 amu (from its atomic mass) and, likewise, 1 unit of O is 16, you can divide 60 by 24 to find the number of units of Mg in the compound and divide 40. by 16 to find the number of units of O in the compound.
2. Now, because we know formulas are made up of whole-number units of the elements, which are expressed as subscripts, we must manipulate these numbers to get whole numbers. This is usually accomplished by dividing these numbers by the smallest quotient. In this case they are equal, so we divide by 2.5.
3. So the empirical formula is MgO.
ANOTHER EXAMPLE: Given: Ba = 58.81%, S = 13.73%, and
O = 27.46%.
Find the empirical formula.
1. Divide each percent by the amu of the element.
2. Manipulate numbers to get small whole numbers. Try dividing them all by the smallest first. In this case, divide each result by 0.43, as shown below.
3. The formula is BaSO4.
In some cases you may be given the true formula mass of the compound. To check if your empirical formula is correct, add up the formula mass of the empirical formula and compare it with the given formula mass. If it is not the same, multiply the empirical formula by the small whole number that gives you the correct formula mass. For example, if your empirical formula is CH2 (which has a formula mass of 14) and the true formula mass is given as 28, you can see that you must double the empirical formula by doubling all the subscripts. The true formula is C2H4.