Solutions and Concentration Calculations Study Guide

General Course Overview and Expectations

Chemistry 112 covers seven chapters throughout the semester. This is a lower number of chapters compared to Chemistry 111, but the material is significantly more math-based. The questions are more involved and drawn out, meaning they take longer to complete, which balances the workload over the semester.

Chapter 13 is taught out of chronological order at this university. While some universities stick to the numerical order, the decision was made here to move Chapter 13 to a later point in the semester.

Introduction to Solutions

Chapter 12 focuses entirely on solutions, which are defined as homogeneous mixtures. The prefix "homo-" means the same, indicating that a solution has a uniform consistency and composition throughout; it is perfectly and evenly mixed.

• Solute: The substance being dissolved. There can be multiple solutes in a single solution (e.g., dissolving both salt and sugar in water).
• Solvent: The substance doing the dissolving. There is typically only one solvent in a solution. In cases where it is difficult to determine which substance is dissolving which, such as in air, the solvent is defined as the substance present in the greatest amount. Conversely, the solute is the substance present in a lesser amount.

Types of Solutions

While most people visualize a solid (like salt or sugar) dissolving in a liquid (like water), solutions can exist in various phases:

• Aqueous Solutions: These are solutions where water is the solvent ($H_2O$ is the solvent). This is the most common type of solution in chemistry and biology. If a solvent is not explicitly specified for a salt or glucose solution, it is assumed to be water.
• Gaseous Solutions: Air is a prime example of a solution. It is a mixture of nitrogen, oxygen, carbon dioxide, hydrogen, argon, and other gases. In the case of air, nitrogen is the solvent because it exists in the highest concentration, while the other gases are solutes.

Concentration Concepts and Qualitative Terms

Concentration is a ratio that compares the amount of solute to the amount of solution (or solvent) it is dissolved in. Dividing two numbers in this context represents a comparison between those two quantities. Concentrations are always expressed as fractions.

• Dilute: A solution with a relatively small amount of solute per amount of solution.
• Concentrated: A solution containing a large amount of solute compared to the solution it is dissolved in.

Using terms like "dilute" or "concentrated" is qualitative. To quantify concentration, specific units must be used. For example, in a hypothetical "Kool-Aid lab," concentration could be measured in units such as $\text{packets per gallon}$. In scientific practice, universal concentration units are required.

Universal Concentration Units

Different fields of study prefer different units. Chemists often use molarity ($M$), biologists may use parts per million ($ppm$) or percent, physicists might use mole fraction ($\chi$), and certain applications require molality ($m$).

Molarity ($M$)

Molarity is the most frequently used unit. It represents the number of moles of solute per liter of solution.

$M = \frac{\text{moles of solute}}{\text{liters of solution}}$

• A mole is a specific number of particles, similar to how a "dozen" means 12. It is a quantity so large ($6.022 \times 10^{23}$) that one could not finish counting to it in a lifetime.
• The units for molarity are $\text{mol/L}$, often abbreviated as a capital $M$.
• A concentration of $2.5\,M$ is pronounced as "two and a half molar."

Mass Percent (Percent by Mass)

This unit describes the ratio of the mass of the solute to the total mass of the solution.

$\text{Mass Percent} = \left( \frac{\text{grams of solute}}{\text{grams of solution}} \right) \times 100\%$

• The grams in the numerator and denominator cancel out, leaving only a percentage as the unit.
• A $2.5\%$ solution by mass contains $2.5\,g$ of solute for every $100\,g$ of total solution.

Mole Fraction ($\chi$)

Mole fraction represents the fraction of total moles in a mixture that belongs to a specific substance. It is symbolized by the Greek letter Chi ($\chi$), which resembles a fancy "X."

$\chi = \frac{\text{moles of substance}}{\text{total moles of solution}}$

• Since the unit is $\text{moles/moles}$, the units cancel, resulting in a unitless decimal.
• A mole fraction of $0.25$ means that a quarter of the total moles in the solution are that specific substance.

Molality ($m$)

Molality is a new unit for many students and is symbolized by a lowercase $m$. It is uniquely defined using the mass of the solvent rather than the total solution.

$m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}$

• A concentration of $2.5\,m$ is pronounced as "two and a half molal."
• Temperature Independence: Unlike molarity, molality does not change with temperature. Molarity is based on liters (volume), which can expand when heated or contract when cooled. Molality is based on mass (kilograms), which remains constant regardless of temperature or even location (e.g., it remains the same on the Moon).

Essential Memorization and Study Requirements

Certain information is not provided on the exam reference sheet and must be memorized for quizzes and exams:

• Metric Conversions: Specifically the prefixes "kilo-" ($k$) and "milli-" ($m$).
• Polyatomic Ions: On the course Blackboard page under "Reference Documents," there is a table of polyatomic ions. Students must know the names and formulas of the items highlighted in red (approximately 5 or 6 ions).
• Concentration Formulas: The formulas for molarity ($M$), molality ($m$), mass percent, and mole fraction ($\chi$) are not on the reference sheet.
• Practice Policies: During recitation quizzes, no notes or external help are allowed. Only the designated exam reference sheet is permitted.

Problem-Solving Strategies and Rounding

Types of Concentration Problems

1. Finding Concentration: Calculating the concentration given information about the solute and solvent/solution.
2. Calculating Mass of Solute: Determining how much solute is required to achieve a specific concentration in a given amount of solvent.
3. Converting Concentration Units: Translating one concentration unit into another (e.g., converting molarity to molality). This is often considered the most difficult task of the semester.

Rounding and Significant Figures

• Internal Steps: For multi-step problems, do not round intermediate values on paper. Use the full number from the calculator to avoid compounding errors. Rounding too early will lead to an answer that is too far from the correct value.
• Grading: Homework systems typically accept an answer within a certain range rather than strictly enforcing significant figures, unless specified. Exam questions are multiple-choice, so being close to the calculated value is sufficient.
• Lab Exception: Chemistry labs are treated as separate courses and are very strict regarding significant figures; incorrect significant figures often result in a score of zero.

Calculation Examples

Example: Calculating Mass Percent

Calculate the mass percent if $3.00\,g$ of $NaCl$ is dissolved in $150\,g$ of $H_2O$.

• Solute mass: $3.00\,g$
• Solvent mass: $150\,g$
• Solution mass: $150\,g + 3.00\,g = 153\,g$
• Calculation:$\text{Mass Percent} = \left( \frac{3.00\,g}{153\,g} \right) \times 100\% = 1.96\%$

Example: Calculating Molality ($m$)

Using the same solution ($3.00\,g$ of $NaCl$ in $150\,g$ of $H_2O$), calculate the molality.

1. Find Moles of Solute ($NaCl$):     The molar mass of $NaCl$ (found on the periodic table) is $58.44\,g/mol$.     $\text{moles} = \frac{3.00\,g}{58.44\,g/mol} = 0.0513\,mol\,NaCl$

2. Find Kilograms of Solvent ($H_2O$):$150\,g = 0.150\,kg\,H_2O$

3. Calculate Molality:$m = \frac{0.0513\,mol}{0.150\,kg} = 0.342\,m$

Example: Finding Mass of Solute from Molality

Determining the mass of methanol ($CH_3OH$, molar mass $32.04\,g/mol$) needed to be added to $500\,g$ of water to create a $0.6\,m$ solution.

1. Define the Goal: We need grams of solute ($CH_3OH$).
2. Define the Concentration Unit: $0.6\,m = \frac{0.6\,mol\,CH_3OH}{1\,kg\,H_2O}$.
3. Dimensional Analysis:$\text{Mass} = 500\,g\,H_2O \times \left( \frac{1\,kg}{1000\,g} \right) \times \left( \frac{0.6\,mol\,CH_3OH}{1\,kg\,H_2O} \right) \times \left( \frac{32.04\,g}{1\,mol\,CH_3OH} \right) = 9.61\,g\,CH_3OH$

Advanced Concentration Conversions

When converting from one concentration unit (like mass percent) to another (like molality), use the following strategy:

1. Break the given unit into a numerator and denominator: For a $3\%$ hydrogen peroxide solution ($H_2O_2$), assume exactly $100\,g$ of solution. This gives you $3\,g$ of solute and $100\,g$ of solution.
2. Calculate the masses of individual components:$\text{grams solute} + \text{grams solvent} = \text{grams solution}$     In a $3\%$ solution ($100\,g$ total), if there are $3\,g$ of solute, there must be $97\,g$ of solvent.
3. Perform necessary unit conversions: Convert the $3\,g$ of solute to moles and the $97\,g$ of solvent to kilograms to find the molality.