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9.6: Units of Concentration

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    86667
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    Learning Objectives
    • Calculate concentration of a solution using different units.
    • Use concentration units to calculate the amount of solute in a solution.
    • Use molarity to determine quantities in chemical reactions.

    Percent Concentrations

    There are several ways of expressing the concentration of a solution by using a percentage. The mass/mass percent (% m/m) is defined as the mass of a solute divided by the mass of a solution times 100:

    \[\mathrm{\% \:m/m = \dfrac{mass\: of\: solute}{mass\: of\: solution}\times100\%} \nonumber \]

    If you can measure the masses of the solute and the solution, determining the mass/mass percent is easy. Each mass must be expressed in the same units to determine the proper concentration.

    Example \(\PageIndex{1}\)

    A saline solution with a mass of 355 g has 36.5 g of NaCl dissolved in it. What is the mass/mass percent concentration of the solution?

    Solution

    We can substitute the quantities given in the equation for mass/mass percent:

    \(\mathrm{\%\: m/m=\dfrac{36.5\: g}{355\: g}\times100\%=10.3\%}\)

    Exercise \(\PageIndex{1}\)

    A dextrose (also called D-glucose, C6H12O6) solution with a mass of 2.00 × 102 g has 15.8 g of dextrose dissolved in it. What is the mass/mass percent concentration of the solution?

    Answer

    7.90%

    For gases and liquids, volumes are relatively easy to measure, so the concentration of a liquid or a gas solution can be expressed as a volume/volume percent (% v/v): the volume of a solute divided by the volume of a solution times 100:

    \[\mathrm{\%\: v/v = \dfrac{volume\: of\: solute}{volume\: of\: solution}\times100\%} \nonumber \]

    Again, the units of the solute and the solution must be the same. A hybrid concentration unit, mass/volume percent (% m/v), is commonly used for intravenous (IV) fluids (Figure \(\PageIndex{1}\)). It is defined as the mass in grams of a solute, divided by volume in milliliters of solution times 100:

    \[\mathrm{\%\: m/v = \dfrac{mass\: of\: solute\: (g)}{volume\: of\: solution\: (mL)}\times100\%} \nonumber \]

    220px-Iv1-07_014.jpg800px-Baxter_sodium_chloride_irrigation.JPG
    Figure \(\PageIndex{1}\): Mass/Volume Percent.The 0.9% NaCl concentration on this IV bag is mass/volume percent (left). Such solution is used for other purposes and available in bottles (right). Figures used with permission from Wikipedia

    Using Percent Concentration in Calculations

    The percent concentration can be used to produce a conversion factor between the amount of solute and the amount of solution. As such, concentrations can be useful in a variety of stoichiometry problems as discussed in Chapter 6. In many cases, it is best to use the original definition of the concentration unit; it is that definition that provides the conversion factor.

    As an example, if the given concentration is 5% v/v alcohol, this means that there are 5 mL of alcohol dissolved in every 100 mL solution.

    5 mL alcohol = 100 mL solution

    The two possible conversion factors are written as follows:

    \(\mathrm{\dfrac{5\: mL\: alcohol}{100\: mL\: solution}}\) or \(\mathrm{\dfrac{100\: mL\: solution}{5\: mL\: alcohol}}\)

    Use the first conversion factor to convert from a given amount of solution to amount of solute. The second conversion factor is used to convert from a given amount of solute to amount of solution. Given any two quantities in any percent composition, the third quantity can be calculated, as the following example illustrates.

    Example \(\PageIndex{2}\)

    A sample of 45.0% v/v solution of ethanol (C2H5OH) in water has a volume of 115 mL. What volume of ethanol solute does the sample contain?

    Solution

    A percentage concentration is simply the number of parts of solute per 100 parts of solution. Thus, the percent concentration of 45.0% v/v implies the following:

    \(\mathrm{45.0\%\: v/v \rightarrow \dfrac{45\: mL\: C_2H_5OH}{100\: mL\: solution}}\)

    That is, there are 45 mL of C2H5OH for every 100 mL of solution. We can use this fraction as a conversion factor to determine the amount of C2H5OH in 115 mL of solution:

    \(\mathrm{115\: mL\: solution\times\dfrac{45\: mL\: C_2H_5OH}{100\: mL\: solution}=51.8\: mL\: C_2H_5OH}\)

    Exercise \(\PageIndex{2}\)

    What volume of a 12.75% m/v solution of glucose (C6H12O6) in water is needed to obtain 50.0 g of C6H12O6?

    Answer

    \(\mathrm{50.0\: g\: C_6H_12O_6\times\dfrac{100\: mL\: solution}{12.75\: g\: C_6H_12O_6}=392\: mL\: solution}\)

    Example \(\PageIndex{3}\)

    A normal saline IV solution contains 9.0 g of NaCl in every liter of solution. What is the mass/volume percent of normal saline?

    Solution

    We can use the definition of mass/volume percent, but first we have to express the volume in milliliter units:

    1 L = 1,000 mL

    Because this is an exact relationship, it does not affect the significant figures of our result.

    \(\mathrm{\%\: m/v = \dfrac{9.0\: g\: NaCl}{1,000\: mL\: solution}\times100\%=0.90\%\: m/v}\)

    Exercise \(\PageIndex{3}\)

    The chlorine bleach that you might find in your laundry room is typically composed of 27.0 g of sodium hypochlorite (NaOCl), dissolved to make 500.0 mL of solution. What is the mass/volume percent of the bleach?

    Answer

    \(\mathrm{\%\: m/v = \dfrac{27.0\: g\: NaOCl}{500.0\: mL\: solution}\times100\%=5.40\%\: m/v}\)

    Parts per Million (ppm) and Parts per Billion (ppb)

    In addition to percentage units, the units for expressing the concentration of extremely dilute solutions are parts per million (ppm) and parts per billion (ppb). Both of these units are mass based and are defined as follows:

    \[\mathrm{ppm=\dfrac{mass\: of\: solute}{mass\: of\: solution}\times1,000,000} \nonumber \]

    \[\mathrm{ppb=\dfrac{mass\: of\: solute}{mass\: of\: solution}\times1,000,000,000} \nonumber \]

    Similar to parts per million and parts per billion, related units include parts per thousand (ppth) and parts per trillion (ppt).

    Concentrations of trace elements in the body—elements that are present in extremely low concentrations but are nonetheless necessary for life—are commonly expressed in parts per million or parts per billion. Concentrations of poisons and pollutants are also described in these units. For example, cobalt is present in the body at a concentration of 21 ppb, while the State of Oregon’s Department of Agriculture limits the concentration of arsenic in fertilizers to 9 ppm.

    In aqueous solutions, 1 ppm is essentially equal to 1 mg/L, and 1 ppb is equivalent to 1 µg/L.

    Example \(\PageIndex{4}\)

    If the concentration of cobalt in a human body is 21 ppb, what mass in grams of Co is present in a body having a mass of 70.0 kg?

    Solution

    A concentration of 21 ppb means “21 g of solute per 1,000,000,000 g of solution.” Written as a conversion factor, this concentration of Co is as follows:

    \(\mathrm{21\: ppb\: Co \rightarrow \dfrac{21\: g\: Co}{1,000,000,000\: g\: solution}}\)

    We can use this as a conversion factor, but first we must convert 70.0 kg to gram units:

    \(\mathrm{70.0\: kg\times\dfrac{1,000\: g}{1\: kg}=7.00\times10^4\: g}\)

    Now we determine the amount of Co:

    \(\mathrm{7.00\times10^4\: g\: solution\times\dfrac{21\: g\: Co}{1,000,000,000\: g\: solution}=0.0015\: g\: Co}\)

    This is only 1.5 mg.

    Exercise \(\PageIndex{4}\)

    An 85 kg body contains 0.012 g of Ni. What is the concentration of Ni in parts per million?

    Answer

    0.14 ppm

    Mole/Volume Concentration: Molarity

    Another way of expressing concentration is to give the number of moles of solute per unit volume of solution. Such concentration units are useful for discussing chemical reactions in which a solute is a product or a reactant. Molar mass can then be used as a conversion factor to convert amounts in moles to amounts in grams.

    Molarity(M) is defined as the number of moles of a solute dissolved per liter of solution:

    \[\mathrm{molarity=\dfrac{number\: of\: moles\: of\: solute}{number\: of\: liters\: of\: solution}} \nonumber \]

    Molarity is abbreviated M (often referred to as “molar”), and the units are often abbreviated as mol/L. It is important to remember that “mol” in this expression refers to moles of solute and that “L” refers to liters of solution. For example, if you have 1.5 mol of NaCl dissolved in 0.500 L of solution, its molarity is therefore

    \[\mathrm{\dfrac{1.5\: mol\: NaCl}{0.500\: L\: solution}=3.0\: M\: NaCl} \nonumber \]

    which is read as “three point oh molar sodium chloride.” Sometimes (aq) is added when the solvent is water, as in “3.0 M NaCl(aq).”

    Before a molarity concentration can be calculated, the amount of the solute must be expressed in moles, and the volume of the solution must be expressed in liters.

    If the quantity of the solute is given in mass units, you must convert mass units to mole units before using the definition of molarity to calculate concentration. For example, what is the molar concentration of a solution of 22.4 g of HCl dissolved in 1.56 L?

    clipboard_ee770cca226ecbfff3f7e710f94e842f1.png

    Step 1: convert the mass of solute to moles using the molar mass of HCl (36.46 g/mol):

    \[22.4\cancel{gHCl}\times \dfrac{1\: mol\: HCl}{36.46\cancel{gHCl}}=0.614\, mol\; HCl \nonumber \]

    Step 2: use the definition of molarity to determine the concentration:

    \[M \: =\: \dfrac{0.614\: mol\: HCl}{1.56L\: solution}=0.394\, M HCl \nonumber \]

    Example \(\PageIndex{5}\)

    What is the molarity of an aqueous solution of 25.0 g of NaOH in 750 mL?

    Solution

    Before we substitute these quantities into the definition of molarity, we must convert them to the proper units. The mass of NaOH must be converted to moles of NaOH. The molar mass of NaOH is 40.00 g/mol:

    \(\mathrm{25.0\: g\: NaOH\times\dfrac{1\: mol\: NaOH}{40.00\: g\: NaOH}=0.625\: mol\: NaOH}\)

    Next, we convert the volume units from milliliters to liters:

    \(\mathrm{750\: mL\times\dfrac{1\: L}{1,000\: mL}=0.750\: L}\)

    Now that the quantities are expressed in the proper units, we can substitute them into the definition of molarity:

    \(\mathrm{M=\dfrac{0.625\: mol\: NaOH}{0.750\: L}=0.833\: M\: NaOH}\)

    Exercise \(\PageIndex{5}\)

    If a 350 mL cup of coffee contains 0.150 g of caffeine (C8H10N4O2), what is the molarity of this caffeine solution?

    Answer

    0.00221 M

    Using Molarity in Calculations

    The definition of molarity can also be used to calculate a needed volume of solution, given its concentration and the number of moles desired, or the number of moles of solute (and subsequently, the mass of the solute), given its concentration and volume. As in the percent concentration, molarity can also be expressed as a conversion factor.

    Molarity is defined as moles solute per liter solution. There is an understood 1 in the denominator of the conversion factor. For example, a 3.0 M solution of sucrose means that there are three moles of sucrose dissolved in every liter of solution. Mathematically, this is stated as follows:

    3.0 moles sucrose = 1 L solution

    Dividing both sides of this expression by either side, we generate two possible conversion factors:

    \(\mathrm{\dfrac{3.0\: mol\: sucrose}{1\: L\: solution}}\) or \(\mathrm{\dfrac{1\: L\: solution}{3.0\: mol\: sucrose}}\)

    The first conversion factor can be used to convert from volume (L) of solution to moles solute, and the second converts from moles of solute to volume (L) of solution.

    clipboard_e28e447425ad8aaf91fb5b219b5864def.png

    For example, suppose we are asked how many moles of sucrose are present in 0.108 L of a 3.0 M sucrose solution. The given volume (0.108 L) is multiplied by the first conversion factor to cancel the L units, and find that 0.32 moles of sucrose are present.

    \[0.108\cancel{L\, solution}\times \dfrac{3.0\, mol\, sucrose}{\cancel{1L\, solution}}=0.32\, mol\, sucrose \nonumber \]

    How many liters of 3.0 M sucrose solution are needed to obtain 4.88 mol of sucrose? In such a conversion, we multiply the given (4.88 moles sucrose) with the second conversion factor. This cancels the moles units and converts it to liters of solution.

    \[4.88\cancel{mol\, sucrose}\times \dfrac{1\, L\, solution}{\cancel{3.0\, mol\, sucrose}}=1.63\, L\, solution \nonumber \]

    Example \(\PageIndex{6}\)
    1. What volume of a 0.0753 M solution of dimethylamine [(CH3)2NH] is needed to obtain 0.450 mol of the compound?
    2. Ethylene glycol (C2H6O2) is mixed with water to make auto engine coolants. How many grams of C2H6O2 are in 5.00 L of a 6.00 M aqueous solution?
    Solution

    1. To solve for the volume, multiply the "given" (0.450 mol of dimethylamine) with the molarity conversion factor (0.0753 M). Use the proper conversion factor to cancel the unit "mol" and get the unit volume (L) of solution:

    \(\mathrm{0.450\: mol\: dimethylamine\times\dfrac{1\: L\: solution}{0.0753\: mol\: dimethylamine}=5.98\: L\: solution}\)

    2. The strategy in solving this problem is to convert the given volume (5.00 L) using the 6.00 M (conversion factor) to solve for moles of ethylene glycol, which can then be converted to grams.

    Step 1: Convert the given volume (5.00 L) to moles ethylene glycol.

    \(\mathrm{5.00\: L\: solution\times\dfrac{6.00\: mol\: C_2H_6O_2}{1\: L\: solution}=30.0\: mol\: C_2H_6O_2}\)

    Step 2: Convert 30.0 mols C2H6O2 to grams C2H6O2. Molar mass of C2H6O2= 62.08 g/mol

    \(\mathrm{30.0\: mol\: C_2H_6O_2\times\dfrac{62.08\: g\: C_2H_6O_2}{1\: mol\: C_2H_6O_2}=1,860\: g\: C_2H_6O_2}\)

    The same two-step problem can also be worked out in a single line, rather than as two separate steps, as follows:

    clipboard_e35a97b86f4b5bb4d012a423b61e91fc1.png

    \(\mathrm{5.00\: L\: solution\times\dfrac{6.00\: mol\: C_2H_6O_2}{1\: L\: solution}\times\dfrac{62.08\: g\: C_2H_6O_2}{1\: mol\: C_2H_6O_2}=1,860\: g\: C_2H_6O_2}\)

    The final answer is rounded off to 3 significant figures. Thus, there are 1,860 g of C2H6O2 in the specified amount of engine coolant.

    Note: Dimethylamine has a “fishy” odor. In fact, organic compounds called amines cause the odor of decaying fish.

    Exercise \(\PageIndex{6}\)
    1. What volume of a 0.0902 M solution of formic acid (HCOOH) is needed to obtain 0.888 mol of HCOOH?
    2. Acetic acid (HC2H3O2) is the acid in vinegar. How many grams of HC2H3O2 are in 0.565 L of a 0.955 M solution?
    Answer

    a. 9.84 L

    b. 32.4 g

    Solution Stoichiometry

    Of all the ways of expressing concentration, molarity is the one most commonly used in stoichiometry problems because it is directly related to the mole unit. Consider the following chemical equation:

    HCl(aq) + NaOH(s) → H2O(ℓ) + NaCl(aq)

    Suppose we want to know how many liters of aqueous HCl solution will react with a given mass of NaOH. A typical approach to answering this question is as follows:

    steps.jpg
    Figure \(\PageIndex{2}\): Typical approach to solving Molarity problems

    In itself, each step is a straightforward conversion. It is the combination of the steps that is a powerful quantitative tool for problem solving.

    Example \(\PageIndex{7}\)

    How many milliliters of a 2.75 M HCl solution are needed to react with 185 g of NaOH? The balanced chemical equation for this reaction is as follows:

    HCl(aq) + NaOH(s) → H2O(ℓ) + NaCl(aq)

    Solution

    We will follow the flowchart to answer this question. First, we convert the mass of NaOH to moles of NaOH using its molar mass, 40.00 g/mol:

    \(\mathrm{185\: g\: NaOH\times\dfrac{1\: mol\: NaOH}{40.00\: g\: NaOH}=4.63\: mol\: NaOH}\)

    Using the balanced chemical equation, we see that there is a one-to-one ratio of moles of HCl to moles of NaOH. We use this to determine the number of moles of HCl needed to react with the given amount of NaOH:

    \(\mathrm{4.63\: mol\: NaOH\times\dfrac{1\: mol\: HCl}{1\: mol\: NaOH}=4.63\: mol\: HCl}\)

    Finally, we use the definition of molarity to determine the volume of 2.75 M HCl needed:

    \(\mathrm{2.75\: M\: HCl=\dfrac{4.63\: mol\: HCl}{volume\: of\: HCl\: solution}}\)

    \(\mathrm{volume\: of\: HCl=\dfrac{4.63\: mol\: HCl}{2.75\: M\: HCl}=1.68\: L\times\dfrac{1,000\: mL}{1\: L}=1,680\: mL}\)

    We need 1,680 mL of 2.75 M HCl to react with the NaOH.

    The same multi-step problem can also be worked out in a single line, rather than as separate steps, as follows:

    clipboard_e4d09e91b60ea3da4eb10e445fe634f61.png

    \(\mathrm{185\: g\: NaOH\times\dfrac{1\: mol\: NaOH}{40.00\: g\: NaOH}\times\dfrac{1\: mol\: HCl}{1\: mol\: NaOH}\times\dfrac{1\: L\: HCl\: solution}{2.75\: mol\: HCl}\times\dfrac{1000\: mL\: HCl\: solution}{1\: L\: HCl\: solution}=1,680\: mL\: HCl\: solution}\)

    Our final answer (rounded off to three significant figures) is 1,680 mL HCl solution.

    Exercise \(\PageIndex{7}\)

    How many milliliters of a 1.04 M H2SO4 solution are needed to react with 98.5 g of Ca(OH)2? The balanced chemical equation for the reaction is as follows:

    \[H_2SO_{4(aq)} + Ca(OH)_{2(s)} \rightarrow 2H_2O_{(ℓ)} + CaSO_{4(aq)} \nonumber \]

    Answer

    1,280 mL

    The generic steps for performing stoichiometry problems such as this are shown in Figure \(\PageIndex{3}\). You may want to consult this figure when working with solutions in chemical reactions. The double arrows in Figure \(\PageIndex{3}\) indicate that you can start at either end of the chart and, after a series of simple conversions, determine the quantity at the other end.

    9.3.jpg
    Figure \(\PageIndex{3}\): Diagram of Steps for Using Molarity in Stoichiometry Calculations. When using molarity in stoichiometry calculations, a specific sequence of steps usually leads you to the correct answer.
    Solutions in Our Body

    Many of the fluids found in our bodies are solutions. The solutes range from simple ionic compounds to complex proteins. Table \(\PageIndex{1}\) lists the typical concentrations of some of these solutes.

    Table \(\PageIndex{1}\): Approximate Concentrations of Various Solutes in Some Solutions in the Body*
    Solution Solute Concentration (M)
    blood plasma Na+ 0.138
    K+ 0.005
    Ca2+ 0.004
    Mg2+ 0.003
    Cl 0.110
    HCO3 0.030
    stomach acid HCl 0.10
    urine NaCl 0.15
    PO43 0.05
    NH2CONH2 (urea) 0.30
    *Note: Concentrations are approximate and can vary widely.
    Looking Closer: The Dose Makes the Poison

    Why is it that we can drink 1 qt of water when we are thirsty and not be harmed, but if we ingest 0.5 g of arsenic, we might die? There is an old saying: the dose makes the poison. This means that what may be dangerous in some amounts may not be dangerous in other amounts.

    Take arsenic, for example. Some studies show that arsenic deprivation limits the growth of animals such as chickens, goats, and pigs, suggesting that arsenic is actually an essential trace element in the diet. Humans are constantly exposed to tiny amounts of arsenic from the environment, so studies of completely arsenic-free humans are not available; if arsenic is an essential trace mineral in human diets, it is probably required on the order of 50 ppb or less. A toxic dose of arsenic corresponds to about 7,000 ppb and higher, which is over 140 times the trace amount that may be required by the body. Thus, arsenic is not poisonous in and of itself. Rather, it is the amount that is dangerous: the dose makes the poison.

    Similarly, as much as water is needed to keep us alive, too much of it is also risky to our health. Drinking too much water too fast can lead to a condition called water intoxication, which may be fatal. The danger in water intoxication is not that water itself becomes toxic. It is that the ingestion of too much water too fast dilutes sodium ions, potassium ions, and other salts in the bloodstream to concentrations that are not high enough to support brain, muscle, and heart functions. Military personnel, endurance athletes, and even desert hikers are susceptible to water intoxication if they drink water but do not replenish the salts lost in sweat. As this example shows, even the right substances in the wrong amounts can be dangerous!


    9.6: Units of Concentration is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by LibreTexts.

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