General Chemistry
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2-6. REACTING QUANTITIES

It has already been emphasized that all reactions occur on an atom-to-atom level. This presents a small problem to us, since we cannot hold an atom in our hand, or count out a specific number of atoms to put into a reaction. How then do we measure amounts of material that will react together? Chemists long ago solved this problem by learning how to count particles indirectly. They did this by measuring samples of the chemicals in particular ratios by their weights. To understand the means of doing this, we need to expand our concept of atomic weight to compounds in the form of the formula (or molecular) weight.

Milligram Formula (Milligram Molecular) Weight. When atoms combine to form compounds, the atomic nuclei are not affected. There is no net loss of weight. Regardless of whether the particle formed is a molecule or an ion group, it will have a formula and a formula weight. The formula weight of a compound is the sum of the atomic weights of all the atoms that appear in its chemical formula. Consider, for example, carbon dioxide:

• Atoms: C + O + O = CO₂ (molecule)
• Atomic weights: 12 + 16 + 16 = 44 (formula weight)

While we have arrived at a formula weight which is in terms of atomic mass units, it is much more useful to express it in terms of milligrams. This is known as the milligram formula weight. For the example above, CO₂, the milligram formula weight is 44 mg. This is a quantity that we can measure and see, and thus can easily work with. It also represents a reacting unit of the compound.

Molarity. A molar solution, or a one molar (1 M) solution, consists of one-gram molecular weight (GMW) of solute dissolved in enough water to make 1 liter of finished solution. Molarity, then, is the number of GMWs dissolved in enough water to make a finished solution of 1000 ml. Molar solutions may have as a solute a solid, a liquid, or a gas. Later in this subcourse, we will use the concept of molarity to explain the measurement of acidity, called the pH.

• Calculating the gram molecular weight. One-gram molecular weight of a substance is its molecular weight expressed in grams. Thus, a GMW of NaOH would be 40 grams, where the atomic weights are as follows: Na = 23, O = 16, and H = 1. Thus, .5 GMW of NaOH would be 20 grams, and so forth. A mole is one-gram molecular weight of a substance. Thus, a mole of NaOH is 40 grams of NaOH; a half-mole (.5 mole) is 20 grams; two moles of NaOH are 80 grams, and so on.
• Calculating the molarity of a solution. To find the molarity of a solution, we divide the number of gram molecular weights of solute by the number of liters of total solution. The formula may be written:

Since many problems are stated in terms of the weight of solute and require you to determine the number of gram molecular weights (moles), the following formula will be of benefit:

Example. What is the molarity of a solution containing 29.25 grams of sodium chloride in 500 ml. of total solution?

Step 1. Find the number of GMWs.

Step 2.      Find the molarity.

c. Milligram Equivalent Weight (Milliequivalent Weight). Sometimes we are interested in more than just the weight ratios of reacting compounds. Since the valence of an element is a measure of that element's combining power, the valences in a compound should be indicative of their reactivity. Therefore, chemists have modified the milligram formula weight to include the positive or negative valence of a compound. This value is called the milligram equivalent weight and is defined as the milligram molecular weight divided by the total positive or negative valence. Consider, for example, sodium hydroxide:

1. Milligram molecular weight = 40 mg
2. Total positive valence = 1
3. Milligram equivalent weight = 40 mg/1 = 40 mg

Another example is potassium phosphate (K3PO4):

1. Milligram molecular weight = 212 mg
2. Total positive valence = 3
3. Milligram equivalent weight = 212 mg / 3 = 70.7 mg

In a reaction, one milliequivalent (mEq) weight of one compound will react with one milliequivalent weight of another. If we are reacting two compounds, then, we can determine how much of each compound should be used to obtain a desired amount of product.

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