# 7.17: Concentrations: Molarity Conversions

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Learning Objectives
• Apply a molarity conversion factor to convert between a molar quantity of solute and the volume of a solution.
• Apply multiple conversion factors to solve problems that involve complex molar relationships.

As stated previously, chemists have defined several types of concentrations, which each use a different chemically-acceptable unit, or combination of units, to indicate the amount of solute that is dissolved in a given amount of solvent.  The equation that is used to calculate the molarity of a solution was presented and applied in the previous section of this chapter.  In contrast to the percent-based concentrations that have been discussed, which, once calculated, are presented as "end-result" values, a molarity is a "hidden" conversion factor that can be applied in additional problem-solving contexts.

Because determining the amount of solute that is present in a homogeneous mixture is fundamental to the quantitative study of solutions, the corresponding concentration values are chemically-significant measurements.  Therefore, since the molarity of a solution can be applied as a conversion factor to bring about a desired unit transformation, as will be discussed in the following paragraphs, chemists defined this type of concentration as a molar standard.

## Molarity Indicator

Recall that all molar standards have a corresponding indicator word or phrase that identifies which relationship and, therefore, which type of conversion factor, must be applied to solve the problem at-hand.  As discussed in the previous section, a molarity is calculated by dividing the molar amount of solute by the volume of the solution, as a whole.  During this process, no unit cancelation occurs, because the units that are present in the numerator and denominator, "mol" and "L," respectively, do not match one another.  Therefore, the unit that is used to report the molarity of a solution must contain both of these units, which are related to one another in a fractional format, "mol/L."  Alternatively, the unit for molarity can be abbreviated as a capital "M," which should be italicized, M, when typed and underlined, M, if hand-written.  This final symbolism is the indicator that a molarity conversion factor should be developed and applied to solve a given problem.

For example, calculate how many moles of potassium metabisulfite, K2S2O5, are present in a 0.64 liter solution of 1.9 M potassium metabisulfite, K2S2O5.

The italicized capital "M" indicates that a molarity conversion factor should be developed and applied to solve this problem.

## Representing Molarities as Conversion Factors

As stated above, a capital "M," which is italicized, M, when typed and underlined, M, when hand-written, is the indicator that a molarity conversion factor should be developed and applied to solve a given problem.  This symbolism is the abbreviation for "mol/L," which is the unit that is used to report the molarity of a solution.  The value "1.9 M potassium metabisulfite, K2S2O5" can, therefore, be written as "1.9 mol/L potassium metabisulfite, K2S2O5."

The "/" in this unit is read as "per," which is the indicator word that is associated with identifying a "hidden" conversion factor.  Because the word "per" implies a ratio, which is associated with the mathematical operation of division, the word "per" is represented as a fraction bar in a conversion factor.  Any number or unit read before the word "per" becomes the numerator of the proportion, and any number or unit found after the word "per" is written in the denominator.  Therefore, "1.9 mol/L" can be written as

$$\dfrac{1.9 \; {\text{mol}}}{\text{L}}$$ or $$\dfrac{\text{L}}{1.9 \; {\text{mol}}}$$

The conversion factor that is shown on the left is a direct representation of the indicated quantity, and the second conversion factor is derived by interchanging where each value is written, relative to the fraction bar.

Finally, recall that all of the numerical values that are incorporated into the equalities and conversion factors that are derived from molar standards must be associated with two units.  Because a molarity is calculated by dividing the molar amount of solute by the volume of the solution, as a whole, the chemical formula of the solute should be incorporated as the secondary unit on the molar quantity in the corresponding conversion factors, and the word "solution" should be written as the secondary unit on the volumetric value in the proportions that are being developed.  Since potassium metabisulfite, K2S2O5, is the substance that is associated with the given molarity, this chemical is the solute in the solution, and, therefore, "1.9 mol/L potassium metabisulfite, K2S2O5" can be written as

$$\dfrac{1.9 \; {\text{mol}} \; \ce{K_2S_2O_5}}{\text{L} \; \text{solution}}$$ or $$\dfrac{\text{L} \; \text{solution}}{1.9 \; {\text{mol}} \; \ce{K_2S_2O_5}}$$

Again, the conversion factor that is shown on the left is a direct representation of the given molarity, and the second conversion factor is derived by interchanging where each quantity is written, relative to the fraction bar.

## Calculations

In order to calculate how many moles of potassium metabisulfite, K2S2O5, are present in a 0.640 liter solution of 1.9 M potassium metabisulfite, K2S2O5, a conversion factor based on the given molarity must be applied.  Because the quantity "1.9 M potassium metabisulfite, K2S2O5" should be expressed as a conversion factor, it should not be selected as the "given" quantity in this problem.  Therefore, "0.64 liter solution" is the given quantity, and, consequently, the first molarity conversion factor that is shown above must be utilized to completely eliminate the given unit, "liter solution."  Finally, recall that "hidden" conversion factors are not exact quantities, and therefore, the number of significant figures that they contain must be considered when applying the correct number of significant figures to a calculated quantity, as shown below.

$${\text {0.640}}$$ $${\cancel{\rm{L} \; \rm{solution}}} \times$$ $$\dfrac{1.9 \; \rm{mol} \; \rm{K_2S_2O_5}}{\cancel{\rm{L} \; \rm{solution}}}$$ = $${\text {1.216}}$$ $${\rm{mol} \; \rm{K_2S_2O_5}}$$ ≈ $${\text {1.2}}$$ $${\rm{mol} \; \rm{K_2S_2O_5}}$$

Exercise $$\PageIndex{1}$$

Calculate how many milliliters of a 0.48 M copper (II) bromide solution contain 2.22 x 1023 molecules of copper (II) bromide.

Indicator Information, Equality Patterns, & Conversions
The italicized capital "M" indicates that a molarity conversion factor should be developed and applied to solve this problem.  Because this symbolism is the abbreviation for "mol/L," which is the unit that is used to report the molarity of a solution, the value "0.48 M copper (II) bromide" can be written as "0.48 mol/L copper (II) bromide."  The "/" in this unit is read as "per," which is the indicator word that is associated with identifying a "hidden" conversion factor.  Because the word "per" implies a ratio, the word "per" is represented as a fraction bar in a conversion factor, any number or unit read before the word "per" becomes the numerator of the proportion, and any number or unit found after the word "per" is written in the denominator.  Finally, all of the numerical values that are incorporated into conversion factors that are derived from molar standards must be associated with two units.  Because a molarity is calculated by dividing the molar amount of solute by the volume of the solution, as a whole, the chemical formula of the solute should be incorporated as the secondary unit on the molar quantity in the corresponding conversion factors, and the word "solution" should be written as the secondary unit on the volumetric value in the proportions that are being developed.  Since copper (II) bromide, CuBr2, is the substance that is associated with the given molarity, this chemical is the solute in the solution, and, therefore, "0.48 mol/L copper (II) bromide" can be written as

$$\dfrac{0.48 \; {\text{mol}} \; \ce{CuBr_2}}{\text{L} \; \text{solution}}$$ or $$\dfrac{\text{L} \; \text{solution}}{0.48 \; {\text{mol}} \; \ce{CuBr_2}}$$

Additionally, the phrase "molecules of copper (II) bromide" indicates that the Avogadro's number equality that is shown below will also be required to solve the given problem.

1 mol CuBr2 = 6.02 × 1023 CuBr2 molecules

Dimensional Analysis
Because the quantity "0.48 M copper (II) bromide" should be expressed as a conversion factor, it should not be selected as the "given" quantity in this problem.  Therefore, "2.22 x 1023 molecules of copper (II) bromide" is the given quantity, and, consequently, a conversion factor based on the Avogadro's number equality must be applied first, in order to completely eliminate the given unit, "molecules of copper (II) bromide."  However, the unit that results upon the cancelation of "molecules of copper (II) bromide" is "mol CuBr2," which is not the desired final unit.  Therefore, the second molarity conversion factor that is shown above must be utilized to completely eliminate the intermediate unit, "mol CuBr2."

The unit that results upon the cancelation of the intermediate unit "mol CuBr2" is "L solution," which still is not the desired final unit.  In order to convert the intermediate unit "L solution" to the desired final unit, "milliliters of solution," one of the following prefix modifier equalities must be applied as a third conversion factor.

mL = 10-3 L or mL = 0.001 L or 1,000 mL = L

Before using one of these relationships to cancel the the intermediate unit "L solution," the word "solution" must be added as a secondary unit on each side of the equal sign, as shown below.

$${2.22 \times 10^{23} \cancel{\rm{molecules} \; \rm{CuBr_2}}} \times$$ $$\dfrac{1 \; \bcancel{\rm{mol} \; \rm{CuBr_2}}}{6.02 \times 10^{23} \cancel{\rm{molecules} \; \rm{CuBr_2}}}$$ × $$\dfrac{\cancel{\rm{L} \; \rm{solution}}}{0.48 \; \bcancel{\rm{mol} \; \rm{CuBr_2}}}$$ × $$\dfrac{1,000 \; \rm{mL} \; \rm{solution}}{\cancel{\rm{L} \; \rm{solution}}}$$

= $${\text {768.2724...}}$$ $${\rm{mL} \; \rm{solution}}$$ ≈ $${\text {770}}$$ $${\rm{mL} \; \rm{solution}}$$

When using a calculator, each conversion factor should be entered in parentheses, or the "=" key should be used after each division, and any quantity that is expressed in scientific notation should be offset by an additional set of parentheses.  Finally, recall that "hidden" conversion factors are not exact quantities, and therefore, the number of significant figures that they contain must be considered when applying the correct number of significant figures to a calculated quantity, as shown above.
Exercise $$\PageIndex{2}$$

Consider the following chemical equation.

___ $$\ce{Pb} \left( s \right) +$$ ___ $$\ce{H_3PO_4} \left( aq \right) \rightarrow$$ ___ $$\ce{H_2} \left( g \right) +$$ ___ $$\ce{Pb_3(PO_4)_2} \left( s \right) + \text{q}$$

1. Balance this equation by writing coefficients in the "blanks," as necessary.
2. Classify this reaction as a combination, a decomposition, a single replacement, a double replacement, or a combustion reaction.
3. Classify this reaction as endothermic or exothermic.
4. Calculate how many liters of molecular hydrogen gas are produced if all of the solute in a 0.615 liter solution of 9.25 M phosphoric acid, H3PO4, is consumed at STP.
In order to balance a chemical equation, the quantities of each type of element and polyatomic ion that are present in the reactants and the products of the reaction must be determined.  Because heat, q, is not a chemical material, there is no need to associate a balancing coefficient with the heat that is produced in this reaction.  This reaction contains the phosphate ion, PO4–3, which is a polyatomic anion, lead, Pb, and hydrogen, H.  Because the phosphate ion is present in both a reactant and a product in the given equation, this polyatomic ion should be balanced as a single entity.  Recall that parentheses are utilized in ionic chemical formulas to offset a polyatomic ion as a unit, and the subscript that is written outside of the parentheses indicates the quantity in which that ion is present within the ionic compound.  If parentheses are not explicitly-written around the formula of a polyatomic ion, an unwritten subscript of "1" is understood to correspond to that polyatomic unit.  The quantities in which these chemicals are present in the given reaction equation are summarized in the table that is shown below.

Element or Ion Reactants Products Balanced
Pb 1 3 H 3 2 PO4–3 1 2 None of the components of this reaction are balanced, and, therefore, one or more coefficients must be written in the "blanks" above, in order to balance this reaction.  Furthermore, since lead, Pb, is represented in its atomic form, the balancing process should not be initiated using this element.

In order to balance the phosphate ion, PO4–3, a coefficient should be written in the "blank" that corresponds to this ion on the left side of the equation, as fewer phosphate ions are present on this side of the reaction arrow.  The value of this coefficient, 2, is determined by dividing the larger quantity of this element, 2, by its smaller count, 1.  Inserting this coefficient results in the chemical equation that is shown below.

___ $$\ce{Pb} \left( s \right) +$$ 2 $$\ce{H_3PO_4} \left( aq \right) \rightarrow$$ ___ $$\ce{H_2} \left( g \right) +$$ ___ $$\ce{Pb_3(PO_4)_2} \left( s \right) +$$ $$\text{q}$$

As phosphoric acid, H3PO4, contains both hydrogen, H, and the phosphate ion, PO4–3, incorporating this coefficient alters the amounts of both of these chemicals on the reactant side of the equation, as indicated in the table that is shown below.  Inserting this coefficient balances the phosphate ion, PO4–3, as intended.  The quantities in which lead, Pb, and hydrogen, H, are present are still unequal.  Therefore, one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction.

Element or Ion Reactants Products Balanced
Pb 1 3 H $$\cancel{\rm{3}}$$ 6  2 PO4–3 $$\cancel{\rm{1}}$$ 2 2 In order to balance hydrogen, H, a coefficient should be written in the "blank" that corresponds to this element on the right side of the equation, as fewer hydrogens are present on this side of the reaction arrow.  The value of this coefficient, 3, is determined by dividing the larger quantity of this element, 6, by its smaller count, 2.  Inserting this coefficient results in the chemical equation that is shown below.

___ $$\ce{Pb} \left( s \right) +$$ 2 $$\ce{H_3PO_4} \left( aq \right) \rightarrow$$ 3 $$\ce{H_2} \left( g \right) +$$ ___ $$\ce{Pb_3(PO_4)_2} \left( s \right) +$$ $$\text{q}$$

As hydrogen, H, is the only element that is present in this chemical formula, incorporating this coefficient does not alter the amounts of lead, Pb, or phosphate ion, PO4–3, that are present on the product side of the equation, as indicated in the table that is shown below.  Inserting this coefficient balances hydrogen, H, as intended.  The quantities in which lead, Pb, are present are still unequal.  Therefore, one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction.

Element or Ion Reactants Products Balanced
Pb 1 3 H $$\cancel{\rm{3}}$$ 6  $$\bcancel{\rm{2}}$$ 6 PO4–3 $$\cancel{\rm{1}}$$ 2 2 In order to balance lead, Pb, a coefficient should be written in the "blank" that corresponds to this element on the left side of the equation, as fewer leads are present on this side of the reaction arrow.  The value of this coefficient, 3, is determined by dividing the larger quantity of this element, 3, by its smaller count, 1.  Inserting this coefficient results in the chemical equation that is shown below.

3 $$\ce{Pb} \left( s \right) +$$ 2 $$\ce{H_3PO_4} \left( aq \right) \rightarrow$$ 3 $$\ce{H_2} \left( g \right) +$$ ___ $$\ce{Pb_3(PO_4)_2} \left( s \right) +$$ $$\text{q}$$

As lead, Pb, is the only element that is present in this chemical formula, incorporating this coefficient does not alter the amounts of hydrogen, H, or phosphate ion, PO4–3, that are present on the reactant side of the equation, as indicated in the table that is shown below.  Inserting this coefficient balances lead, Pb, as intended.  Therefore, all of the components of this equation are now balanced.

Element or Ion Reactants Products Balanced
Pb $$\cancel{\rm{1}}$$ 3 3 H $$\cancel{\rm{3}}$$ 6  $$\bcancel{\rm{2}}$$ 6 PO4–3 $$\cancel{\rm{1}}$$ 2 2 No fractional coefficients are written in the equation that is shown above.  Furthermore, because of the unwritten "1" that is understood to occupy the fourth "blank" in this equation, these coefficients cannot be divided.  Therefore, the final equation that is presented above is chemically-correct, as written.
This reaction is classified as a single replacement because an elemental reactant, lead, Pb, and the cationic portion of a compound exchange positions.
Because the variable for heat, q, is written on the right side of the reaction arrow in the given chemical equation, heat is generated as a product.  Therefore, this reaction is exothermic.
Indicator Information, Equality Patterns, & Conversions
The italicized capital "M" indicates that a molarity conversion factor should be developed and applied to solve this problem.  As multiple chemicals are referenced in the given statement, the chemical formula for the substance that is written in closest physical proximity to the indicator "M," "phosphoric acid," H3PO4, is incorporated into the molarity conversion factor.  Because this symbolism is the abbreviation for "mol/L," which is the unit that is used to report the molarity of a solution, the value "9.25 M phosphoric acid, H3PO4" can be written as "9.25 mol/L phosphoric acid, H3PO4."  The "/" in this unit is read as "per," which is the indicator word that is associated with identifying a "hidden" conversion factor.  Because the word "per" implies a ratio, the word "per" is represented as a fraction bar in a conversion factor, any number or unit read before the word "per" becomes the numerator of the proportion, and any number or unit found after the word "per" is written in the denominator.  Finally, all of the numerical values that are incorporated into conversion factors that are derived from molar standards must be associated with two units.  Because a molarity is calculated by dividing the molar amount of solute by the volume of the solution, as a whole, the chemical formula of the solute should be incorporated as the secondary unit on the molar quantity in the corresponding conversion factors, and the word "solution" should be written as the secondary unit on the volumetric value in the proportions that are being developed.  Since phosphoric acid, H3PO4, is the substance that is associated with the given molarity, this chemical is the solute in the solution, and, therefore, "9.25 mol/L phosphoric acid, H3PO4" can be written as

$$\dfrac{9.25 \; {\text{mol}} \; \ce{H_3PO_4}}{\text{L} \; \text{solution}}$$ or $$\dfrac{\text{L} \; \text{solution}}{9.25 \; {\text{mol}} \; \ce{H_3PO_4}}$$

Additionally, because both of the chemicals that are referenced in the problem, molecular hydrogen, H2, and phosphoric acid, H3PO4, are also present in the given reaction equation, a stoichiometric equality should also be developed and applied to solve this problem.  The chemical formulas for molecular hydrogen, H2, and phosphoric acid, H3PO4, are incorporated into the secondary unit positions on the left and right sides, respectively, of this equality.  Since the coefficient that is associated with molecular hydrogen, H2, is a "3" in the chemical equation that was balanced in Part (a), a 3 is inserted into the numerical position on the left side of this equality.  Finally, because the coefficient that corresponds to phosphoric acid, H3PO4, is a "2" in the balanced chemical equation, a 2 is inserted into the numerical position on the right side of this equality, as shown below.

3 mol H2 = 2 mol H3PO4

Finally, the phrase "at STP" indicates that an STP equality should also be developed and applied to solve this problem.  The chemical formula for the gas that is referenced in the problem, molecular hydrogen, H2, should be incorporated into both of the secondary unit positions in this equality, as shown below.

1 mol H2 = 22.4 L H2

Dimensional Analysis
Because the quantity "9.25 M phosphoric acid, H3PO4" should be expressed as a conversion factor, it should not be selected as the "given" quantity in this problem.  Therefore, "0.615 liter solution" is the given quantity, and, consequently, the first molarity conversion factor that is shown above must be utilized to completely eliminate the given unit, "liter solution."  However, the unit that results upon the cancelation of "liter solution" is "mol H3PO4," which is not the desired final unit.  Therefore, a second conversion factor, based on the stoichiometric equality, must be applied.  The unit that results upon the cancelation of the intermediate unit "mol H3PO4" is "mol H2," which still is not the desired final unit.  Therefore, the STP equality must be applied as a third conversion factor, as shown below.

$${\text {0.615}}$$ $${\cancel{\rm{L} \; \rm{solution}}} \times$$ $$\dfrac{9.25 \; \bcancel{\rm{mol} \; \rm{H_3PO_4}}}{\cancel{\rm{L} \; \rm{solution}}}$$ × $$\dfrac{3 \; \cancel{\rm{mol} \; \rm{H_2}}}{2 \; \bcancel{\rm{mol} \; \rm{H_3PO_4}}}$$ × $$\dfrac{22.4 \; \rm{L} \; \rm{H_2}}{1 \; \cancel{\rm{mol} \; \rm{H_2}}}$$ = $${\text {191.142}}$$ $${\rm{L} \; \rm{H_2}}$$ ≈ $${\text {191}}$$ $${\rm{L} \; \rm{H_2}}$$

When using a calculator, each conversion factor should be entered in parentheses, or the "=" key should be used after each division.  Finally, recall that "hidden" conversion factors are not exact quantities, and therefore, the number of significant figures that they contain must be considered when applying the correct number of significant figures to a calculated quantity, as shown above.

7.17: Concentrations: Molarity Conversions is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.