Updated March 01, 2020 By Rosann Kozlowski Reviewed by: Lana Bandoim, B.S.
A mole calculation in a solution requires using the molarity formula. The volume of the solution and the solution concentration is needed.
Molarity is the number of moles of solute per liter of solution. A solute, which can be solid, liquid or gas, is a substance that is dissolved in a solvent. The solvent is another substance that is capable of dissolving it within its intermolecular spaces. Together, the dissolved solute and the solvent make a solution.
Molarity is also considered molar concentration because it is the measure of a concentration of a solution. The formula for molarity may be expressed as:
To fully understand molarity, the mole concept must be understood. A mole (often abbreviated to mol) is a unit of measurement. It is a certain amount. If a dozen bagels were purchased, the amount, if counted, would be 12 bagels.
A mole, like the word dozen, denotes a particular amount, too. However, the amount, called Avogadro’s number, is very large: 6.022 × 1023.
If a mole of bagels were purchased, they would nearly fill the interior space of the Earth. Although a mole of anything can be counted, it is usually reserved for incredibly small items, like atoms and molecules.
One mole of any element or chemical compound is always the same number. One mole of hydrogen would mean there are 6.022 × 1023 atoms of hydrogen.
A mole of sodium chloride, NaCl, is the same amount, 6.022 × 1023. Here, however, it is 6.022 ×1023 molecules. With molarity, consider the moles of solute as finding the number of molecules in solution.
The values for concentration and liters of solution must be given or calculated.
Example problem: Sugar, or sucrose, easily dissolves in water. How many moles of sucrose are in a 0.02 M solution?
In the problem, the molar concentration, M, is given: 0.02 M. The volume is assumed to be 1 L since the definition of molarity is moles of solute per liter of solution.
Use the formula from “Molarity Definitions and Formula” (above) to solve for moles:
Rearranging to solve for moles of solute:
mol = 0.02 mol/L × 1 L = 0.02 mol of sucrose, C12H22O11
There are 0.02 moles of sucrose in a 0.02 M sucrose solution.
Commonly, a question asks for grams of solute, especially if the substance must be measured in a laboratory setting. If the question asks for how many grams of sucrose must be added to make a 0.02 M solution, these additional steps may be followed:
While the counted amount of a mole of any substance is 6.022 x 1023, the molar mass of that substance will be different. For example, sodium chloride, NaCl, will have a different mass than table sugar, sucrose, C12H22O11.
Every element has a different molar mass, commonly located under the symbol on a periodic table. For example, one mole of carbon (C) has a mass of 12.01 g/mol. The molar mass of hydrogen (H) is 1.01 g/mol, and oxygen (O) is 16.00 g/mol.
For example, the molar mass of sucrose would be calculated by adding the molar masses of the individual elements:
Add all the individual components of sucrose together:
144.12 g/mol + 22.22 g/mol + 176 g/mol = 342.34 g/mol
The molar mass of sucrose, C12H22O11, is 342.34 g/mol
Use the number of moles calculated in Step 2 and the molar mass of sucrose from Step 3 to solve for grams:
0.02 mol of C12H22O11 × 342.34 g C12H22O11 / 1 mol C12H22O11 = 6.85 g C12H22O11
6.85 grams of sucrose dissolved in water will make a 0.01 M solution.
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. In preceding sections, we focused on the composition of substances: samples of matter that contain only one type of element or compound. However, mixtures—samples of matter containing two or more substances physically combined—are more commonly encountered in nature than are pure substances. Similar to a pure substance, the relative composition of a mixture plays an important role in determining its properties. The relative amount of oxygen in a planet’s atmosphere determines its ability to sustain aerobic life. The relative amounts of iron, carbon, nickel, and other elements in steel (a mixture known as an “alloy”) determine its physical strength and resistance to corrosion. The relative amount of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological effect. The relative amount of sugar in a beverage determines its sweetness (see Figure 1). In this section, we will describe one of the most common ways in which the relative compositions of mixtures may be quantified. We have previously defined solutions as homogeneous mixtures, meaning that the composition of the mixture (and therefore its properties) is uniform throughout its entire volume. Solutions occur frequently in nature and have also been implemented in many forms of manmade technology. We will explore a more thorough treatment of solution properties in the chapter on solutions and colloids, but here we will introduce some of the basic properties of solutions. The relative amount of a given solution component is known as its concentration. Often, though not always, a solution contains one component with a concentration that is significantly greater than that of all other components. This component is called the solvent and may be viewed as the medium in which the other components are dispersed, or dissolved. Solutions in which water is the solvent are, of course, very common on our planet. A solution in which water is the solvent is called an aqueous solution. A solute is a component of a solution that is typically present at a much lower concentration than the solvent. Solute concentrations are often described with qualitative terms such as dilute (of relatively low concentration) and concentrated (of relatively high concentration). Concentrations may be quantitatively assessed using a wide variety of measurement units, each convenient for particular applications. Molarity (M) is a useful concentration unit for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter (1 L) of the solution: [latex]M = \frac{\text{mol solute}}{\text{L solution}}[/latex]
Calculating Molar Concentrations Solution [latex]M = \frac{\text{mol solute}}{\text{L solution}} = \frac{0.133 \;\text{mol}}{355 \;\text{mL} \times \frac{1 \;\text{L}}{1000 \;\text{mL}}} = 0.375 \; M[/latex] Check Your Learning
Deriving Moles and Volumes from Molar Concentrations Solution
[latex]M = \frac{\text{mol solute}}{\text{L solution}}[/latex] [latex]\text{mol solute} = 0.375 \;\frac{\text{mol sugar}}{\text{L}} \times (10 \;\text{mL} \times \frac{1 \text{L}}{1000 \;\text{mL}}) = 0.004 \;\text{mol sugar}[/latex] Check Your Learning
Calculating Molar Concentrations from the Mass of Solute Solution [latex]M = \frac{\text{mol solute}}{\text{L solution}} = \frac{25.2 \;\text{g CH}_3\text{CO}_2\text{H} \times \frac{1 \;\text{mol CH}_2\text{CO}_2\text{H}}{60.052 \;\text{g CH}_2\text{CO}_2\text{H}}}{0.500 \;\text{L solution}} = 0.839 \;M[/latex] [latex]\begin{array}{r @{{}={}} l} M & \frac{\text{mol solute}}{\text{L solution}} = 0.839\;M \\[1em] M & \frac{0.839 \;\text{mol solute}}{1.00 \;\text{L solution}} \end{array}[/latex] Check Your Learning
Determining the Mass of Solute in a Given Volume of Solution Solution [latex]M = \;\frac{\text{mol solute}}{\text{L solution}}[/latex] [latex]\text{mol solute} = M \times \text{L solution}[/latex] [latex]\text{mol solute} = 5.30 \;\frac{\text{mol NaCl}}{\text{L}} \times 0.250 \;\text{L} = 1.325 \;\text{mol NaCl}[/latex] Finally, this molar amount is used to derive the mass of NaCl: [latex]1.325 \;\text{mol NaCl} \times \frac{58.44 \;\text{g NaCl}}{\text{mol NaCl}} = 77.4 \;\text{g NaCl}[/latex] Check Your Learning When performing calculations stepwise, as in Example 4, it is important to refrain from rounding any intermediate calculation results, which can lead to rounding errors in the final result. In Example 4, the molar amount of NaCl computed in the first step, 1.325 mol, would be properly rounded to 1.32 mol if it were to be reported; however, although the last digit (5) is not significant, it must be retained as a guard digit in the intermediate calculation. If we had not retained this guard digit, the final calculation for the mass of NaCl would have been 77.1 g, a difference of 0.3 g. In addition to retaining a guard digit for intermediate calculations, we can also avoid rounding errors by performing computations in a single step (see Example 5). This eliminates intermediate steps so that only the final result is rounded.
Determining the Volume of Solution Containing a Given Mass of Solute Solution [latex]\text{g solute} \times \frac{\text{mol solute}}{\text{g solute}} = \text{mol solute}[/latex] Then, use the molarity of the solution to calculate the volume of solution containing this molar amount of solute: [latex]\text{mol solute} \times \frac{\text{L solution}}{\text{mol solute}} = \text{L solution}[/latex] Combining these two steps into one yields:
[latex]\text{g solute} \times \frac{\text{mol solute}}{\text{g solute}} \times \frac{\text{L solution}}{\text{mol solute}} = \text{L solution}[/latex][latex]75.6 \;\text{g CH}_3\text{CO}_2\text{H} (\frac{\text{mol CH}_3\text{CO}_2\text{H}}{60.05 \;\text{g}}) (\frac{\text{L solution}}{0.839 \;\text{mol CH}_3\text{CO}_2\text{H}}) = 1.50 \;\text{L solution}[/latex] Check Your Learning Dilution is the process whereby the concentration of a solution is lessened by the addition of solvent. For example, we might say that a glass of iced tea becomes increasingly diluted as the ice melts. The water from the melting ice increases the volume of the solvent (water) and the overall volume of the solution (iced tea), thereby reducing the relative concentrations of the solutes that give the beverage its taste (Figure 3). Dilution is also a common means of preparing solutions of a desired concentration. By adding solvent to a measured portion of a more concentrated stock solution, we can achieve a particular concentration. For example, commercial pesticides are typically sold as solutions in which the active ingredients are far more concentrated than is appropriate for their application. Before they can be used on crops, the pesticides must be diluted. This is also a very common practice for the preparation of a number of common laboratory reagents (Figure 4). A simple mathematical relationship can be used to relate the volumes and concentrations of a solution before and after the dilution process. According to the definition of molarity, the molar amount of solute in a solution is equal to the product of the solution’s molarity and its volume in liters: [latex]n = ML[/latex] Expressions like these may be written for a solution before and after it is diluted: [latex]n_1 = M_1L_1[/latex] [latex]n_2 = M_2L_2[/latex] where the subscripts “1” and “2” refer to the solution before and after the dilution, respectively. Since the dilution process does not change the amount of solute in the solution,n1 = n2. Thus, these two equations may be set equal to one another: [latex]M_1L_1 = M_2L_2[/latex] This relation is commonly referred to as the dilution equation. Although we derived this equation using molarity as the unit of concentration and liters as the unit of volume, other units of concentration and volume may be used, so long as the units properly cancel per the factor-label method. Reflecting this versatility, the dilution equation is often written in the more general form: [latex]C_1V_1 = C_2V_2[/latex] where C and V are concentration and volume, respectively.
Use the simulation to explore the relations between solute amount, solution volume, and concentration and to confirm the dilution equation.
Determining the Concentration of a Diluted Solution Solution [latex]C_1V_1 = C_2V_2[/latex] Since the stock solution is being diluted by more than two-fold (volume is increased from 0.85 L to 1.80 L), we would expect the diluted solution’s concentration to be less than one-half 5 M. We will compare this ballpark estimate to the calculated result to check for any gross errors in computation (for example, such as an improper substitution of the given quantities). Substituting the given values for the terms on the right side of this equation yields: [latex]C_2 = \frac{0.850 \;\text{L} \times 5.00 \frac{\text{mol}}{\text{L}}}{1.80 \;\text{L}} = 2.36 \;M[/latex] This result compares well to our ballpark estimate (it’s a bit less than one-half the stock concentration, 5 M). Check Your Learning
Volume of a Diluted Solution Solution [latex]C_1V_1 = C_2V_2[/latex] Since the diluted concentration (0.12 M) is slightly more than one-fourth the original concentration (0.45 M), we would expect the volume of the diluted solution to be roughly four times the original volume, or around 44 mL. Substituting the given values and solving for the unknown volume yields: [latex]V_2 = \frac{(0.45\;M)(0.011 \;\text{L})}{0.12 \; M}[/latex] The volume of the 0.12-M solution is 0.041 L (41 mL). The result is reasonable and compares well with our rough estimate. Check Your Learning
Volume of a Concentrated Solution Needed for Dilution Solution [latex]C_1V_1 = C_2V_2[/latex] Since the concentration of the diluted solution 0.100 M is roughly one-sixteenth that of the stock solution (1.59 M), we would expect the volume of the stock solution to be about one-sixteenth that of the diluted solution, or around 0.3 liters. Substituting the given values and solving for the unknown volume yields: [latex]V_1 = \frac{(0.100\;M)(5.00 \;\text{L})}{1.59 \; M}[/latex] Thus, we would need 0.314 L of the 1.59-M solution to prepare the desired solution. This result is consistent with our rough estimate. Check Your Learning |
What concentration of solution is used by the number of moles of the substances in the solution?
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