Molarity and Normality
At the end of this chapter, the reader should be able to do the following:
1. Define the following terms: molarity, normality, mole, molar, molal, molality, gram, equivalent weight.
2. Calculate molar concentrations of solutions.
3. Calculate normal concentrations of solutions.
4. Calculate molal concentrations of solutions.
5. Interconvert molar solution concentrations with normal solution concentrations.
MOLARITY
In science, it is important that there is a standardized way of quantifying the concentrations of various solutions. A solution is a mixture of solvent and solute. The solvent is the liquid into which the solute is diluted. In the clinical laboratory, the solvent may be deionized water, ethanol, methanol, or an organic solvent. The solute may be a solid, such as dry chemicals, or a liquid. By standardizing the method in which various concentrations of solutions can be made, the concentrations of the solutions can become reproducible.
The basic solid measurement unit used in science is the gram. Many solutions made in the laboratory use gram quantities of a solid chemical in the solution. An example is 0.85% normal saline, which is used often in the hematology laboratory. To make this solution, 8.5 g of solid crystals of sodium chloride (NaCl) are dissolved into a small quantity of water in a 1.0-L volumetric flask. Once the sodium chloride is dissolved, additional water is added to the 1.0-L calibration mark. The term quantum satis (qs) is often used to describe the addition of the water to the calibration mark. For example, if you prepared the 0.85% normal saline solution, 8.5 g of NaCl would be weighed, dissolved in a small quantity of water in a 1.0-L volumetric flask, and then qs with water to the calibration mark.
The basic volume measurement unit used in science is the liter. In many solutions in the laboratory, gram quantities of a solid are diluted into a solvent to prepare a total of 1 L of solution. The quantitation of the concentration of the solution can be standardized by use of molarity (M). The molarity of a solution is comprised of three parts: the gram weight of the solute, the solute’s gram molecular weight, and the solvent quantity.
The gram weight of the solute divided by its gram molecular weight is called a mole. A molar solution refers to a solution with a particular molarity. A 1 M solution contains 1 mol of solute in 1 L of solution.
Often, in the laboratory, the concentration of the solution that is necessary for an experiment and the quantity of solution that is needed is known. What is not known and must be computed is the amount of the solute (moles) needed to derive the necessary concentration of the solution. To calculate the amount of moles needed for a solution of a particular molar concentration, the gram atomic weight, or gram molecular weight of the element or compound that is needed, must be found. These weights may be found by using a periodic table or any table that lists the gram atomic weights of the elements. The molecular weight of a compound is the sum of the atomic weights of each element in the compound. A periodic table is found on page 139. A periodic table lists each element by categorizing the elements according to atomic number, energy level, and electron orbital position. Table 5–1 lists the atomic weights of common elements used in the clinical laboratory. The atomic weights have been rounded to the hundredths. Because they have been rounded, when deciding the quantity of significant figures for a result in a calculation involving atomic weights, the preciseness of the atomic weights is ignored in the decision.
TABLE 5-1
Atomic Weights of Common Elements Used in the Clinical Laboratory
| Element Name | Element Abbreviation | Atomic Wt (rounded value) |
| Aluminum | Al | 26.98 |
| Calcium | Ca | 40.08 |
| Carbon | C | 12.01 |
| Chlorine | Cl | 35.45 |
| Copper | Cu | 63.55 |
| Hydrogen | H | 1.01 |
| Iodine | I | 126.90 |
| Iron | Fe | 55.85 |
| Lead | Pb | 207.20 |
| Lithium | Li | 6.94 |
| Magnesium | Mg | 24.30 |
| Manganese | Mn | 54.94 |
| Mercury | Hg | 200.59 |
| Molybdenum | Mo | 95.94 |
| Nickel | Ni | 58.70 |
| Nitrogen | N | 14.01 |
| Oxygen | O | 16.00 |
| Phosphorus | P | 30.97 |
| Potassium | K | 39.10 |
| Silver | Ag | 107.87 |
| Sodium | Na | 22.99 |
| Sulfur | S | 32.06 |
Example 5–1
What is the gram molecular weight of KCl?
The gram molecular weight is the sum of the atomic weights of K and Cl. Using Table 5–1, the gram atomic weight of potassium is 39.10 and the gram atomic weight of chloride is 35.45. Therefore, the gram molecular weight of KCl is as follows:
The gram molecular weight of KCl is 74.55.
What is the gram molecular weight of HCl?
The gram molecular weight of HCl is the sum of the gram atomic weights of hydrogen and chloride. Therefore, the gram molecular weight of HCl is:
The gram molecular weight of HCl is 36.46.
What is the gram molecular weight of NaOH?
The gram molecular weight of NaOH is the sum of the gram atomic weights of sodium, oxygen, and hydrogen. Therefore, the gram molecular weight of NaOH is:
Now let us move onto determining how many grams are in 1 mol of a compound.
Example 5–2
How many grams are contained in 1 mol of potassium chloride?
The formula for a mol is as follows:
The question is asking for the quantity of grams in 1 mol of KCl. The gram molecular weight of the compound potassium chloride is calculated by adding the individual atomic weights of potassium and chloride. Potassium has an atomic weight of 39.10, and chloride has an atomic weight of 35.45. The gram molecular weight of potassium chloride then is 74.55. Next, the values that are known can be substituted into the mole equation:
Solving the equation, the result is 74.55 g KCl are contained in 1 mol of KCl.
Example 5–3
What is the molarity of a solution that contains 2.50 mol of hydrochloric acid in 1.00 L of solution?
To solve this problem, use the formula for molarity. By substituting the numbers into the formula, the following equation is formed:
Therefore, this solution has a concentration of 2.50 M, or another way of stating it is to say that the solution has a molarity of 2.50.
Therefore, this solution has a molarity of 3.00.
What is the molarity of a solution that contains 4.5 mol of Na2CO3?
Using the formula for molarity and substituting in the value of 4.5 mol, the solution has a molarity of 4.5.
Many compounds used in the laboratory consist of more than one or two elements. To calculate the correct mole concentration or molarity, you must first obtain the correct molecular weight of the compound.
Example 5–4
What is the molecular weight of the compound Na3PO4?
The first step is to list the gram atomic weight (gaw) of each individual element.
Next, each element’s gram atomic weight is multiplied by how many times that element is found in the compound.
Na: There are three sodium atoms in Na3PO4, so the 22.99 gaw of Na must be multiplied by 3 to reflect the true weight of Na in this molecule. Therefore, the weight of Na in this molecule is 22.99 × 3, which equals 68.97.
P: There is only one phosphorus atom in Na3PO4, so the 30.97 gaw of P is unchanged.
O: There are four oxygen atoms in Na3PO4.
Therefore, the 16.00 gaw of O must be multiplied by 4 to reflect the true weight of oxygen in this molecule. Thus, instead of 16.00, oxygen carries a weight of 64.00 gaw in this molecule.
Finally, the gram atomic weights for each element are added to obtain the gram molecular weight of the compound.
The gram molecular weight of Na3PO4 = 68.97 + 30.97 + 64.00 = 163.94 gmw.
What is the gram molecular weight of H3PO4?
The gram molecular weight of H3PO4 is the sum of the gram atomic weights of hydrogen, phosphorus, and oxygen. There are three hydrogen atoms and four oxygen atoms in this molecule. Therefore, the gram molecular weight of H3PO4 is:
The gram molecular weight of H3PO4 is 98.00.
What is the gram molecular weight of CaCl2?
The gram molecular weight of CaCl2 is the sum of the gram atomic weights of calcium and chlorine. There is one calcium atom and two chlorine atoms in this molecule. Therefore, the gram molecular weight of CaCl2 is:
The gram molecular weight of CaCl2 is 110.98.
In this chapter you have been given examples of how to calculate the gram molecular weight of a compound, how to determine a molar quantity, and how to calculate molarity. As you learned, a solution with a molarity of 1.0 has 1.0 mol of solute in 1.0 L of solution. When making solutions you usually know the molarity that you want and the volume that you want. What needs to be calculated is how many grams of the solute you need.
Example 5–5
Consider if you wanted to make 1.00 L of a 0.200 M solution of Na3PO4. How many grams of Na3PO4 would you need?
From Example 5–4, you already know that the gram molecular weight of the compound is 163.94. By substituting the quantities that are known into the molarity formula, you can solve for the amount of grams needed:
To prepare a 0.200 M solution of Na3PO4, 32.79 g of Na3PO4 are weighed, placed into a 1.00-L volumetric flask, dissolved in some water, and then qs to the 1.00-L calibration mark with additional water.
How many grams of NaCl would you need to make a 0.300 M solution?
The gram molecular weight of NaCl is the sum of the atomic weights of sodium and chlorine. The gram molecular weight is 22.99 + 35.45 = 58.44. Substituting into the formula for molarity, the following equation is derived:
Therefore, you would weigh out 17.53 g of NaCl, dissolve with some water in a 1.0-L volumetric flask, and then qs to the 1.0-L calibration mark with additional water to make a 0.3 M solution of NaCl.
How many grams do you need to make a 1.50 M solution of NaOH?
The gram molecular weight of NaOH is 40.00. Substituting into the formula for molarity, the result is 60.0 g of NaOH that must be added to a 1.0-L volumetric flask, dissolved with water if in powder form, and then qs to the 1.0-L calibration mark.
Sometimes the amount of a solution needed is not 1.00 L but a different quantity, such as 100.0 mL or 2.5 L. Often, calculations for solutions are modified to result in a quantity that is compatible with the type of laboratory glassware available. Because most solutes are measured by use of volumetric flasks, it makes little sense to calculate a quantity of solute that is not easily measured in the laboratory. For example, the stock of volumetric flasks in Laboratory A consists of a few 250-mL volumetric flasks, two 1-L flasks, and one 2-L flask. It would be very difficult to accurately measure 800.0 mL of solute with the supplies available. Therefore, many calculations that deal with molarity use 50.0-mL, 100.0-mL, 250.0-mL, 500.0-mL, and 1.0-L quantities in their equations because these are the most commonly used sizes of volumetric flasks.
Example 5–6
If 250 mL of a 1.00 M solution of NaOH are needed, how many grams of NaOH are necessary to make this solution?
The first step is to calculate the molecular weight of NaOH.
From Table 5–1: Na = 22.99 gaw, O=16.00 gaw, H = 1.01 gaw.
The gram molecular weight of NaOH = 22.99 + 16.00 + 1.01 = 40.00 gmw.
Next, substitute into the molarity formula the items that are known.
Next, using algebra, solve for X:
The 1.00 M solution is prepared by dissolving 10.0 g of NaOH in a 250-mL volumetric flask with a small quantity of solvent (usually deionized water) and qs to the mark with solvent.
How many grams of HCl are needed to make 500 mL of a 1.00 M solution?
From earlier examples you know that the gram molecular weight of HCl is 36.46. Substituting into the molarity formula the values for this problem, 18.23 g of HCl will be needed. Remember, the molarity formula is based on a 1.0 L volume, so 500 mL is 0.500 L. The answer is derived from multiplying 1.00 (the molar quantity) by 0.500, then multiplying that answer (0.500) by the molecular weight of 36.46.
How many grams of NaCl are needed to make 2.0 L of a 1.00 M solution?
From earlier examples you know that the gram molecular weight of NaCl is 58.44. Substituting into the molarity formula the values for this problem, the answer is 116.88 g of NaCl.
Putting It All Together
In the preceding examples, the molarity of the solution was always 1.00. Let us see some examples when the total volume is not 1.0 L and the molarity of the solution is not 1.00.
Example 5–7
In Example 5–6 the amount of grams needed to make 250 mL of a 1.00 M solution of NaOH was determined. How many grams would be needed to make 250 mL of a 2.5 M solution?
From Example 5–6 you know that the gram molecular weight of NaOH is 40.00. Substituting the new values into the molarity formula yields the following formula:
Therefore, 25.0 g of NaOH dissolved in 250 mL of water will make a 2.50 M solution.
How many grams are needed to make 300 mL of a 3.20 M solution of HCl?
The gram molecular weight of HCl is 36.46. Substituting the values of this problem into the molarity formula yields a result of 35.0 g.
How many grams of CaCl2 are needed to make 1500.0 mL of a 0.5250 M solution?
The gram molecular weight of CaCl2 is 110.98. Substituting the values of this problem into the molarity formula yields a result of 87.397 g of CaCl2.
Sometimes, in the laboratory, a solution may be prepared but the molarity of that solution may need to be confirmed. To confirm the molarity of a given solution, simply substitute into the basic molarity equation the items that are known and solve for the unknown molarity.
Example 5–8
You weigh out 25.0 g of NaCl and dissolve them into 1.00 L of water. What is the molarity of the solution that you just made?
Therefore, by dissolving 25.0 g of NaCl in a 1.00-L volumetric flask with water and qs with water, a 0.428 M solution of NaCl is prepared.
You weigh out 40.0 g of KCl and dissolve them into 500 mL of water. What is the molarity of the solution that you just made?
Using the molarity formula, the following equation is derived:
Therefore, by dissolving 40.0 g of KCl in 500 mL of water, you will prepare a 1.073 M solution.
You weigh out 2.50 g of CuSO4 and dissolve them into 0.100 L of water. What is the molarity of the solution that you just made?
The gram molecular weight of CuSO4 is 159.61. The molarity of this solution is 0.157 M.
Another term that is frequently used is millimolar. The millimolarity of a solution is calculated by the following formula:
Example 5–9
A solution contains 2.75 g of HCl in 1 L of solution. What is the millimolar concentration of this solution?
To solve this problem use the millimolar formula:
NORMALITY
Another term used to quantify solutions is normality. Normality is different from molarity as normality accounts for the dynamics of the interaction of the dissolved solute with the solvent. The normality of a solution differs from the molarity because it is determined by the number of equivalent weights per liter, not the number of moles per liter. An equivalent weight is the amount of replaceable H+ or OH– ion or charge for the element or compound. This amount is usually reflected by the valence of the element or compound. The equivalent weight is calculated by dividing the gram molecular weight of an element or compound by the valence. Table 5–2 lists the valences of the most common elements used in the clinical laboratory. Some elements have more than one valence, depending on the oxidative state of that element. For elements with more than one valence, the most frequent valence is in bold.
TABLE 5-2
Valences of the Most Common Elements Used in the Clinical Laboratory
| Element Name | Element Abbreviation | Valence(s) |
| Aluminum | Al | +3 |
| Calcium | Ca | +2 |
| Carbon | C | +4, +2, −4 |
| Chlorine | Cl | +7, +5, +3, +1, −1 |
| Copper | Cu | +2, +1 |
| Hydrogen | H | +1, −1 |
| Iodine | I | +7, +5, +1, −1 |
| Iron | Fe | +3, +2 |
| Lead | Pb | +4, +2 |
| Lithium | Li | −1 |
| Magnesium | Mg | +2 |
| Manganese | Mn | +2, +4. +7 |
| Mercury | Hg | +2, +1 |
| Molybdenum | Mo | +6, +4, +3 |
| Nickel | Ni | +2 |
| Nitrogen | N | +5, +4, +3, +2, +1, −3 |
| Oxygen | O | −1, −2 |
| Phosphorus | P | +5, +3, −3 |
| Potassium | K | −1 |
| Silver | Ag | +1 |
| Sodium | Na | +1 |
| Sulfur | S | +6, +4, +2, −2 |
Modified from Masterton WL, Slowinski EJ, Stanitski CL: Chemical principles, ed 5, Philadelphia, 1981, Saunders College.
Example 5–10
What is the gram equivalent weight of HCl?
HCl has one replaceable hydrogen ion. Therefore, the equivalent weight would be as follows:
The gram equivalent weight of HCl = 36.46.
What is the gram equivalent weight of NaOH?
NaOH has one replaceable hydroxyl ion. Therefore, the gram equivalent weight of NaOH is the gram molecular weight of 40.0 divided by 1.0, or 40.0.
What is the gram equivalent weight of NaCl?
In this example, there is no hydrogen or hydroxyl group that is replaced. Instead, one atom of sodium combines with one atom of chloride to form sodium chloride. The gram equivalent weight of NaCl would be the same as its gram molecular weight, which is 58.44.
The examples above have all had a valence of 1 and the gram equivalent weight was equal to the gram molecular weight. Now let us see examples when the valence does not have a value of 1.0.
Example 5–11
What is the gram equivalent weight of CaCl2?
The gram molecular weight of CaCl2 is 110.98. There are two chloride ions that can react in solution; therefore, the valence is 2.
What is the gram equivalent weight of H2SO4?
To solve, first determine the valence. There are two hydrogen ions that can react in solution; therefore, the valence is 2. The gram equivalent weight is equal to the gram molecular weight divided by 2, or 98.08 divided by 2, yielding 49.04.
What is the gram equivalent weight of H3PO4?
The gram equivalent weight is one third of the gram molecular weight, as there are three replaceable hydrogen atoms in this compound. Therefore, the gram equivalent weight of H3PO4 is 98.0 divided by 3, or 32.67.
The normality of a solution is calculated using the following formula, which is not much different from that used to calculate molarity:
Substituting into the first formula yields the following:
To use this formula, first calculate the 
, then divide that number into the number of grams of solute. Last, divide the result of the equation by the number of liters of solution.
Example 5–12
Calculate the normality of a solution containing 98.0 g of H2SO4 in 0.250 L of solution.
To solve this problem, first calculate the gram molecular weight of the compound. From Table 5–1, the gram molecular weight is equal to 98.08. Next, substitute all of the known numbers into the formula.
To solve this equation, first divide the gram molecular weight of solute by the valence:
Next, substitute the 49.04 g equivalent weight into the equation:
Next, divide the grams of solute by the gram equivalent weight:
Last, substitute the equivalent weights into the equation and solve for normality:
What is the normality of a solution containing 65 g of HCl in 1.0 L of water?
The valence of HCl is 1.0. Using the normality formula yields the following:
What is the normality of a 1.0 L solution of 25.0 g of K2CO3?
The valence of this compound is 2; therefore, the gram equivalent weight will be the gram molecular weight divided by 2, or 69.10. Substituting this value into the formula for normality yields:
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