Molarity and Normality

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.

Additional Examples

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:

1.01 + 35.45 = 36.46.

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:

22.99 + 16.00 + 1.01 = 40.00.

The gram molecular weight of NaOH is 40.00.

Now let us move onto determining how many grams are in 1 mol of a compound.

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.

NOTE: The problem calls for the solution to have a final volume of 250 mL. Molarity is based on a liter quantity (1000 mL), so the volume must be expressed in the formula in terms of liters.

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.

Additional Examples

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:

Solving for X:

Therefore, 25.0 g of NaOH dissolved in 250 mL of water will make a 2.50 M solution.

Additional Examples

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?

Molarity=(gramsofsolutegrammolecularweightofsolute)1.00Lofsolution

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.

Additional Examples

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:

1mMsolution=1mmolofsolute1Lofsolution

1mmolofsolute=numberofmilligramsofsolutegrammolecularweightofsolute

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:

Additional Examples

How many milligrams are needed to make 150 mL of a 10.0 mM solution of NaCl?

By substituting the values of this problem into the millimolar formula, the answer of 87.66 mg is derived.

What is the millimolar concentration of a solution consisting of 25 mg of NaOH in 500 mL of water?

The millimolar concentration of this solution is 1.25 mM.

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