# 7.2: Solutes and Solvents


Learning Objectives
• Define solvent.
• Define solute.
• Identify the solvent and solute(s) that are present in a solution.

A chemical that is present in a solution can be classified as either a solute or a solvent.  The solvent is the chemical that is present in the greatest amount and, therefore, is the substance in which each of the remaining chemicals are distributed or dissolved.  A solute is a chemical that present in a lesser amount, relative to the solvent, and must be uniformly-distributed throughout the solution.  Based on these definitions, only a single chemical can be classified as a solvent in a particular solution.  Therefore, if only two chemicals are present in a solution, one substance serves as the solvent and the other acts as a solute.  If a solution contains three or more chemicals, a single chemical must be identified as the solvent, and the remaining substances would be classified as solutes.

The qualitative behaviors and quantitative measurements of a solution, which will be discussed in greater detail in the following sections of this chapter, depend on the solvent and solutes that are contained in that solution.  Therefore, in order to understand and apply any of the concepts that are related to solution chemistry, each substance that is present in a solution must first be classified as a solute or as the solvent.  The role of a particular chemical can be assigned by identifying an indicator word within a description of the overall solution or by analyzing numerical information that is given in a problem.  Based on the definitions that are provided above, the solvent is the substance in which each of the remaining chemicals are distributed or dissolved.  Therefore, the chemical that is referenced after the indicator word "in" is classified as the solvent for a given solution.

Alternatively, the solvent and solutes of a solution can be identified by comparing the relative amounts of the substances that are contained in a solution.  By definition, the solvent is the chemical that is present in the greatest amount in a solution.  In Chapter 6, the "amount" of gas that was present was required to be represented as a molar quantity.  In contrast, the amount of a chemical that is contained in a solution can be expressed using any chemically-acceptable unit, such as

• grams or liters, which can also be modified using prefixes,
• moles,
• atoms or molecules, which can also be generalized as "particles," and
• percentages.

However, in order to compare the quantitative information that is given in a problem, all numerical values must be expressed using the same unit.  Therefore, if different units are used to report the amounts in which chemicals are present, one or more conversion factors need to be developed and applied prior to comparing their values.

For example, identify the solute and the solvent in a solution that is prepared by mixing 28.47 grams of sodium chloride and 147.2 milliliters of water.

The word "in" is not present in the given statement.  Therefore, the relative amounts of sodium chloride, NaCl, and water, H2O, must be compared to determine which chemical is classified as the solvent and which substance is the solute in this solution.

While the units that are used to report the amounts of these chemicals, "grams" and "milliliters," respectively, are both chemically-acceptable, they are not consistent with one another and, therefore, their associated values cannot be compared.  As a result, one of these units must be converted to match the other.  The most straight-forward calculation involves utilizing the density of water, which is generally assumed to be present in the liquid state of matter, unless otherwise specified, to convert the given volume of water into a corresponding mass, which would typically be reported in grams.  Recall that the density of liquid water is 1.00 g/mL or 1.00 g/cm3.  While these values are equivalent to one another, the first quantity should be utilized to solve this problem, as its unit contains both the given unit, "milliliters," and the desired unit, "grams," for expressing the quantity of water that is present in this solution.  Based on the information that was presented in Section 1.15, the desired unit transformation can be achieved either by using the density equation or by applying the density of liquid water as a "hidden" conversion factor.

The quantities that are referenced above can be incorporated into the density equation, as shown below.  The "slash" that is written within the density unit implies that the unit "milliliter," which has an associated numerical value that is understood to be an unwritten "1," could be written as a denominator.  In order to solve for m, the quantities in this equation can then be cross-multiplied, and the volume unit, "mL," is canceled during the subsequent division step.  The unit that remains, "g," is the desired unit for expressing the amount of water that is present in this solution.

$$\rm{d}$$ = $$\dfrac{ \rm{m}}{\rm{V}}$$ ​​​​

$${1.00 \; \rm{g/mL}}$$ = $${\dfrac{\rm{m}} {147.2 \; \rm{mL}}}$$

$$\dfrac{1.00 \; \rm{g}}{\rm{mL}}$$ = $${\dfrac{\rm{m}} {147.2 \; \rm{mL}}}$$​​​​

($${1.00 \; \rm{g}}$$)($${147.2 \; \rm{mL}}$$) = ($${\rm{m}}$$)($${\rm{mL}}$$)

$$\rm{m}$$ = $${\dfrac{({1.00 \; \rm{g}}) ({147.2 \; \cancel{\rm{mL}}})} {({\cancel{\rm{mL}}})}}$$

$$\rm{m}$$ = $${147.2 \; \rm{g}}$$

Alternatively, recall that the units for density, "g/mL" and "g/cm3," which are read as "grams per milliliter" and "grams per cubic centimeter," respectively, each contain the indicator word "per."  Therefore, the density of water can be used as a "hidden" conversion factor, as shown below.

$${147.2 \; \cancel{\rm{mL}}} \times$$ $$\dfrac{1.00 \; \rm{g}}{\cancel{\rm{mL}}}$$ = $${147.2 \; \rm{g}}$$

Regardless of the method that is used, the resultant calculated quantity of water, 147.2 grams, can now be compared to the given amount of sodium chloride, 28.47 grams, in order to determine which chemical is classified as the solvent and which substance is the solute in this solution.  Water, H2O, is the solvent in this solution, as it is the chemical that is present in the greatest amount, 147.2 g, and sodium chloride, NaCl, is a solute, as it is present in a lesser quantity, 28.47 g.  Alternatively, because a solution can only contain one solvent, by definition, after identifying water, H2O, as the solvent in this solution, sodium chloride, NaCl, can be classified as a solute "by default."

Exercise $$\PageIndex{1}$$

Identify the solute(s) and the solvent in a solution that is...

1. 85% copper, 4% lead, 6% tin, and 5% zinc, by mass.
2. prepared by dissolving molecular iodine in ethanol, C2H5OH.
3. prepared by mixing 4.0 grams of helium, 4.7 x 1025 molecules of molecular nitrogen, and 21 moles of molecular oxygen.
The word "in" is not present in the given statement.  Therefore, the relative amounts of copper, Cu, lead, Pb, tin, Sn, and zinc, Zn, must be utilized to determine which chemical is classified as the solvent and which substances are solutes in this solution.

Since the amounts of these chemicals are all reported using the same unit, "%," their values can be directly compared.  Copper, Cu, is classified as the solvent in this solution, as it is the chemical that is present in the greatest amount, 85%, and lead, Pb, tin, Sn, and zinc, Zn, are all solutes, as they are present in lesser quantities, 4%, 6%, and 5%, respectively.  Alternatively, because a solution can only contain one solvent, by definition, after identifying copper, Cu, as the solvent in this solution, lead, Pb, tin, Sn, and zinc, Zn, can be classified as solutes "by default."
The word "in" is present in the given statement.  Therefore, the chemical that is referenced after the indicator word "in," ethanol, C2H5OH, is classified as the solvent in this solution, and the remaining substance, molecular iodine, I2, is a solute, "by default."
The word "in" is not present in the given statement.  Therefore, the relative amounts of helium, He, molecular nitrogen, N2, and molecular oxygen, O2, must be compared to determine which chemical is classified as the solvent and which substances are solutes in this solution.

While the units that are used to report the amounts of these chemicals, "grams," "molecules," and "moles," respectively, are all chemically-acceptable, they are not consistent with one another and, therefore, their associated values cannot be compared.  As a result, two of these units must be converted to match the third.  The most straight-forward calculations involve utilizing atomic weight and Avogadro's number conversion factors to change the given mass of helium, He, and particle count of molecular nitrogen, N2, respectively, into molar quantities.

The mass unit "grams" indicates that a mass-based equality should be developed and applied to convert the given mass of helium, He, into a molar quantity.  Specifically, since helium, He, is an element, an atomic weight equality for this chemical should be developed by equating the atomic weight of the element, which should be reported to the hundredths place, to 1 mol of the element, and utilizing an elemental symbol as the secondary unit on both sides of the resultant equality, as shown below.

1 mol He = 4.00 g He

This equality can then be utilized to convert the given mass of helium, He, into a molar quantity, as shown below.

$${\text {4.0}} {\cancel{\rm{g} \; \rm{He}}} \times$$ $$\dfrac{1 \; \rm{mol} \; \rm{He}}{4.00 \; \cancel{\rm{g} \; \rm{He}}}$$ = $${\text {1}}$$ $${\rm{mol} \; \rm{He}}$$ ≈ $${\text {1.0}}$$ $${\rm{mol} \; \rm{He}}$$

The word "molecules" indicates that an Avogadro's number equality should be developed and applied to convert the given particle count of molecular nitrogen, N2, into a molar quantity.  Since "molecules" is an indicator word, this word is inserted as the second unit on the right-hand side of this type of equality, and the chemical formula for molecular nitrogen, N2, is incorporated into both of the secondary unit positions in this equality, as shown below.

1 mol N2 = 6.02 × 1023 N2 molecules

This equality can then be utilized to convert the given particle count of molecular nitrogen, N2, into a molar quantity, as shown below.

$${\text {4.7}} \times {\text{10}^{25}} {\cancel{\rm{molecules} \; \rm{N_2}}} \times$$ $$\dfrac{1 \; \rm{mol} \; \rm{N_2}}{6.02 \times 10^{23} \; \cancel{\rm{molecules} \; \rm{N_2}}}$$ = $${\text {78.073}}$$ $${\rm{mol} \; \rm{N_2}}$$ ≈ $${\text {78}}$$ $${\rm{mol} \; \rm{N_2}}$$

The resultant calculated quantities of helium, He, 1.0 moles, and molecular nitrogen, N2, 78 moles, can now be compared to the given amount of molecular oxygen, O2, 21 moles, in order to determine which chemical is classified as the solvent and which substances are solutes in this solution.  Molecular nitrogen, N2, is the solvent in this solution, as it is the chemical that is present in the greatest amount, 78 moles, and helium, He, and molecular oxygen, O2, are both solutes, as they are present in lesser quantities, 1.0 moles and 21 moles, respectively.  Alternatively, because a solution can only contain one solvent, by definition, after identifying molecular nitrogen, N2, as the solvent in this solution, helium, He, and molecular oxygen, O2, can be classified as solutes "by default."

The solutions that are exemplified in Exercise $$\PageIndex{1}$$ contain solid, liquid, and gaseous solutes and solvents.  As shown below in Table $$\PageIndex{1}$$, solutions can be prepared using solvents and solutes in any state-of-matter combination, and the physical form of the resultant solution corresponds to the phase of its constituent solvent.

Table $$\PageIndex{1}$$: Examples of Solutions
Solvent Phase Solute Phase Example
Solid-Phase Solutions
solid solid Brass
Solvent:  Copper $$\left( s \right)$$
Solutes:  Lead $$\left( s \right)$$, tin $$\left( s \right)$$, and zinc $$\left( s \right)$$
solid liquid Dental Fillings
Solvent:  Silver $$\left( s \right)$$
Solutes:  Mercury $$\left( l \right)$$, tin $$\left( s \right)$$, and zinc $$\left( s \right)$$
solid gas Reduction Catalysts
Solvent:  Platinum $$\left( s \right)$$
Solute:  Molecular hydrogen $$\left( g \right)$$
Liquid-Phase Solutions
liquid solid Tincture of Iodine
Solvent:  Ethanol, C2H5OH $$\left( l \right)$$
Solute:  Molecular Iodine $$\left( s \right)$$
liquid liquid Alcoholic Beverages
Solvent:  Water $$\left( l \right)$$
Solute:  Ethanol, C2H5OH $$\left( l \right)$$
liquid gas Carbonated Beverages
Solvent:  Water $$\left( l \right)$$
Solute:  Carbon dioxide $$\left( g \right)$$
Gas-Phase Solutions
gas solid Therapeutic Aromatherapy
Solvent:  Molecular nitrogen $$\left( g \right)$$
Solute:  Camphor, C10H16O $$\left( s \right)$$
gas liquid Humid Air
Solvent:  Molecular nitrogen $$\left( g \right)$$
Solute:  Water $$\left( l \right)$$
gas gas Air
Solvent:  Molecular nitrogen $$\left( g \right)$$
Solutes:  Molecular oxygen $$\left( g \right)$$, carbon dioxide $$\left( g \right)$$, and argon $$\left( g \right)$$

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