Exam 5 Study Guide

Percent by Mass

Definition: The percent by mass of a solution is defined as the mass of solute divided by the total mass of the solution, multiplied by 100. This unit expresses the concentration of a solute as a percentage of the total mass of the solution.

Formula: (mass of solute (g) / mass of solution (g)) x 100

Calculation Insight: Keep in mind that the mass of the solution is calculated as the sum of the mass of the solute and the mass of the solvent. It is commonly used in laboratories to describe concentrations of solutions used in various experiments.

Symbol: % m/m

Molarity (M)

Definition: Molarity is defined as the number of moles of solute divided by the volume of solution in liters. It is one of the most widely used units of concentration in chemistry.

Formula: moles of solute / volume of solution (L)

Important Note: To convert grams of solute into moles, use the molar mass of the substance. Moreover, it is essential to convert the volume from milliliters (mL) to liters (L) for the calculation to be accurate. Molarity is particularly important for stoichiometric calculations in chemical reactions.

Symbol: M

Molality (m)

Definition: Molality is the number of moles of solute per kilogram of solvent. Unlike molarity, molality considers only the mass of the solvent, making it useful in situations where temperature changes might affect volume.

Formula: moles of solute / kg of solvent

Calculation Note: When calculating molality, it is critical to focus solely on the mass of the solvent, as it directly influences the properties of the solution, such as boiling point and freezing point.

Symbol: m

Mole Fraction

Definition: The mole fraction is defined as the ratio of the moles of a specific component to the total moles of all components in the mixture. It provides a way to express concentrations of various substances in a mixture and is particularly useful in thermodynamics and colligative properties.

Formula: moles A / moles A + moles B + moles C

Key Point: Mole fraction is dimensionless quantity, making it useful for calculations involving gas mixtures, and does not change with temperature or pressure challenges.

Symbol: XA

Percent by Mass Example

Problem: What is the percent by mass of a solution containing 10 grams of sodium chloride and 125 g of water?

Calculation Steps:

Mass of Solute = 10g (NaCl)

Mass of Solvent = 10g + 125g = 135g

Percent by Mass: 10.0/135 x 100 = 7.41%

Molarity Example

Problem: Calculate the molarity of a solution containing 10 grams of sodium chloride in 120 mL of solution.

Molar Mass Conversion: molar mass of NaCl = 58.5 g/mol

Calculate moles of NaCl: 10g x 1mol/58.5g/mol

Volume Conversion: convert 120 mL to L = 0.120 L

Molarity Calculation: 10g x (1 mol/58.5 g/mol) x (1/0.120 L) = 1.37 M NaCl(aq)

Molality Example

Problem: Calculate the molality of an iodine solution in methylene chloride with 5.00 grams of iodine in 30.0 grams of methylene chloride.

Identification: Solute is iodine, solvent is methylene chloride.

Molar Mass Conversion: 5g of iodine to moles: Molar mass of I2 = 254g/mol

Convert solvent mass: 30g of CH2Cl2 to kg: 30g = 0.03 kg

Molality calculation: 5.00g x (1mol I2/254g I2) x (1/0.030 kg CH2Cl2) = 0.656 m

Mole Fraction Example

Iodine and Methylene Chloride

Given:

• Iodine: 1.50 grams

• Methylene Chloride (CH2Cl2): 56.0 grams

1. Calculate Moles of Iodine:

Molar mass of iodine (I2): 254g

Moles of Iodine = 1.50g/254g/mol = 0.00591 mol

2. Calculate Moles of Methylene Chloride:

Molar mass of methylene chloride: approximately 84.93 g/mol

Moles of Methylene Chloride = 56.0g/84.93g/mol = 0.6588 mol

3. Calculate Mole Fraction of Iodine:

Total moles = moles of iodine + moles of methylene chloride

Total moles = 0.00591 + 0.6588 = 0.6647 moles

Mole Fraction of Iodine = 0.00591/0.6647 = 0.00886

4. Calculate Mole Fraction of Methylene Chloride:

Utilizing the previously calculated mole fraction of iodine

Mole Fraction of Methylene Chloride = 1 – Mole Fraction of Iodine = 1 – 0.00886 = 0.9911

Converting Percent by Mass to Molarity

Given a 10% glucose (C6H12O6) solution: 10.0g of glucose in 100g of solution.

Goal: Convert percent by mass to molarity, a vital process in solution preparation in laboratory settings.

1. Determine Moles of Glucose:

Molar mass of glucose: approximately 180 g/mol

Moles of Glucose = 10.0g / 180g/mol = 0.0556 moles

2. Convert Solution Mass to Volume:

To find the volume of the solution, we use the density: 1.04 g/mL

Volume of solution: 100g / 1.04 g/mol = 96.15 mL = 0.09615 L

3. Calculate molarity (M):

The relationship of molarity is given by the formula: M = moles of solute / volume of solution in liters

M = 0.0556 mol / 0.09615 L = 0.578 M

This indicates that there are 0.578 moles of glucose in each liter of solution.

Converting Percent by Mass to Molality

1. Start with Composition:

In a typical 10% glucose solution, this means there are 10.0g of glucose in 100g of solution, of which 90g is the solvent (water).

2. Convert Grams to Kilograms:

90g = 0.090kg

It is essential to use kilograms for calculating molality.

3. Calculate Moles of Glucose (Already Done):

0.0556 moles

4. Calculate Molality (m):

The formula for molality is given by: m = moles of solute / kilograms of solvent

m = 0.055g mol / 0.090 kg = 0.618 m

This value indicates the concentration of glucose in terms of moles of solute per kilogram of solvent

Molarity to Molality Conversion Example: Ethanol Solution

Problem Statement: Given a 14.1 M ethanol solution is water, convert this to molality.

Given:

• Molarity of ethanol solution: 14.1 M

• Density of the solution: 0.853 g/mL

Steps to Convert Molarity to Molality

1. Convert Molarity to Volume:

Choose 1000 mL of solution for ease of calculation.

Given 14.1 M, this results in 14.1 moles of ethanol in 1000 mL, which is foundational for further calculations.

2. Calculate Mass of the Solution:

Utilize the density to determine the mass of the entire solution.

Mass of solution = 1000 mL x 0.853 g/mL = 853g

This shows the total mass derived from the volume and density.

3. Calculate Mass of the Solute (Ethanol):

Determine the mass of the ethanol from its moles:

Molar mass of ethanol (C2H6OH): 46.0 g/mol

Mass of ethanol = 14.1 moles x 46.0 g/mol = 650g

Understanding the solute’s contribution to the total mass is critical for subsequent calculations.

4. Calculate mass of solvent (Water):

This involves subtracting the solute mass from the total mass.

Mass of solvent = Mass of solution – Mass of solute

= 853g – 650g = 203g (Water)

Knowing the mass of the solvent is essential for finding molality.

5. Convert Mass of Solvent to Kilograms:

To convert grams to kilograms for molality calculations:

Mass of solvent = 203g x (1kg / 1000g) = 0.203g

6. Calculate Molality

Finally, calculate the molality using the following formula:

m = 14.1 mol / 0.203 kg = 69.5 molal (m)

this final step connects the solution back to the molality definition by converting the concentration from moles per volume to moles per mass of solvent

Molality to Molarity Conversion Example: Citric Acid Solution

Problem Statement: Convert 2.331 m molality of citric acid (HC6H7O7) into molarity.

Given:

• Density of solution: 1.1346 g/mL

Steps to Convert Molality to Molarity:

1. Identify Moles of Solute:

Utilizing the molality given allows us to establish moles per kilogram of solvent.

m = 2.331 molal = 2.331 moles of solute

this is crucial because it’s calculated as moles of solute per 1 kg of solvent (water = 1000g).

2. Calculate Mass of Solute (Citric Acid):

Just like in the pervious example, we derive the mass from the moles and the molar mass.

Molar mass of citric acid: 192.14 g/mol

Mass of citric acid = 2.331 moles x 192.14 g/mol = 448g

3. Calculate Total Mass of Solution:

Understand the total mass is the sum of solute mass and solvent mass:

Mass of solution = 448g +1000g = 1448g

This is essential in determining how the solute impacts the solution overall.

4. Convert Mass to Volume Using Density:

Converting the mass of the solution to volume takes into account the density of the solution.

Volume of solution = mass/density = 1448g / 1.1346 g/mL = 1276 mL

Convert to liters:

= 1.276 L

Knowing the volume is critical for molarity calculations.

5. Calculate Molarity:

Utilizing the formula for molarity connects back to our earlier definitions.

M = moles of solute / volume of solution in L

M = 2.331 moles / 1.276 L = 1.827 M

This encapsulates the conversion from molality to molarity systematically.

Mole Fraction to Molality Conversion

Problem Statement: A solution contains a mole fraction of iodine at 0.00889 and methylene chloride at 0.9911, which indicates that this solution consists primarily of methylene chloride as the solvent and iodine as the solute.

Mole Fraction Interpretation: Mole fractions are dimensionless and represent the ratio of moles of a component to the total moles of all components in the solution. Here, 0.00889 corresponds to 0.00889 moles of iodine, while the remainder, represented by the mole fraction of methylene chloride, indicates its proportion in the solution

Finding Moles of Methylene Chloride: To ascertain the amount of methylene chloride in the solution, we observe that the mole fraction of methylene chloride is 0.9911. This implies optimal solvation, where the larger mole fraction identifies the solvent predominating over the solute, leading to 0.9911 moles of methylene chloride.

Molar Mass Conversion: by using the molar mass of methylene chloride (85 g/mol), we can convert the moles of methylene chloride to grams:

• Grams of methylene chloride: 0.9911 moles x 85 g/mol = 84.24 gram

To consider practical applications, we convert grams of kilograms (as required in the molality formula). The conversion yields:

• 84.24g = 0.08424 kg

Calculating Molality: With the determined quantities, we can compute the molality of the solution:

• Molality (M): m + moles of iodine / kg of methylene chloride = 0.00889 mol / 0.08424 kg = 0.106 molal

This means that the solution’s molality is approximately 0.106 molal, illustrating the concentration of iodine in the methylene chloride solvent. This derivation is significant for understanding the solution behavior in contexts like chemical reactions and physical property modifications

Boiling Point Definition

Defines the temperature at which a liquid’s vapor pressure is equal to the atmospheric pressure around it, allowing the liquid to boil. The normal boiling point, under standard pressure, is significant for determining the stability of a substance in different environments.

Freezing Point Definition

It is the temperature at which a liquid becomes a solid. Similar to the boiling point, it helps define the conditions under which substance exist as solids or liquids.

Molecular Weight

The sum of the atomic weights of all the atoms in a molecule of the substance.

Boiling Point Elevation

Adding a solute to a solvent results in an increase in the boiling point. The boiling point elevation is determined by the number of solute particles present in the solution. (Delta Tb = Kbm)

Freezing Point Depression

The freezing point of a solution is lower than that of the pure solvent. Adding solute disrupts the formation of the solid lattice structure of ice. (Delta Tf = Kfm)

Van’t Hoff Factor (i)

Representing the number of particles produced in solution. (Delta Tb = Kbmi) (Delta Tf = Kfmi)

Example Problem: Boiling Point Elevation of a Glucose Solution

Problem Statement: Calculate the boiling point of a 2.50 molal glucose (C6H12O6) aqueous solution.

Given:

• Kb of water = 0.512oC/molal

• Molality (m) = 2.50 molal

Calculation:

Delta Tb = Kbm = 0.512 x 2.50 = 1.28oC

Final Boiling Point: 100 + 1.28 = 101.28oC

Example Problem: Freezing Point Depression of Benzoic Acid in Benzene

Problem Statement: Calculate the freezing point of a solution containing 8.50 grams of benzoic acid (C6H5COOH) in 75.0 grams of benzene (C6H6).

Given:

• Kf for benzene = 5.065oC/molal

• Benzoic acid (non-electrolyte)

Molar Mass Calculation:

Molar mass of benzoic acid (C6H5COOH) = 122.12 g/mol

Moles = 8.50g / 122.12g/mol = 0.0697 mol

Convert mass of benzene to kg: 75.0g = 0.075 kg

Calculate Molality (m):

m = 0.0697mol / 0.075kg = 0.929 molal

freezing Point Depression Calculation:

Delta Tf = Kfm = 5.065 x 0.929 = 4.71oC

Pure Benzene freezing Point: 5.455oC

Final Freezing Point: 5.455 – 4.71 = 0.745oC

Liquids

Definite volume, indefinite shape

Solids

Fixed shape, fixed volume

Gases

Indefinite shape, indefinite volume

Dipole-Dipole Forces

Definition: Attractive forces that occur between polar molecules, characterized by regions of partial positive and partial negative charges due to uneven electron distribution.

Mechanism:

• Polar molecules align themselves so that the positive end of one molecules is attracted to the negative end of another, creating a network of attractions that affect the substance’s boiling and melting points.

• The strength of these forces depends on the polarity of the molecules involved; more polar molecules experience stronger dipole-dipole interactions.

Example; In hydrogen chloride (HCl), chlorine’s higher electronegativity compared to hydrogen results in a covalent bond where H+ (partial positive) is attracted to Cl- (partial negative).

Significance: Critical in determining the boiling points of polar substances; for instance, stronger dipole-dipole interactions in compounds like HCl lead to higher boiling points compared to nonpolar substances.

London Forces

Definition: The weakest type of intermolecular force that results from temporary dipoles formed when electrons are distributed unevenly in atoms or nonpolar molecules.

Mechanism:

• These forces arise due to fluctuations in electron density, creating instant dipoles that induce adjacent dipoles in neighboring molecules.

• Although individually weak, they become significant in larger molecules with greater surface area and molecular weight.

Importance:

• Present in all molecules (both polar and nonpolar), London dispersion forces play a crucial role in the physical properties of gases, liquids, and solids.

• Example: Larger molecules, like long-chain hydrocarbons, exhibit stronger London dispersion forces compared to smaller molecules such as methane.

Hydrogen Bonds

Definition: A strong type of intermolecular force that occurs specifically when hydrogen is covalently bonded to highly electronegative atoms like fluorine (f), oxygen (O), or nitrogen (N).

Characteristics:

• Hydrogen bonds are stronger than dipole-dipole and London forces but weaker than covalent bonds, influencing physical properties such as boiling and melting points substantially.

Mechanism:

• The hydrogen atom develops a partial positive charge, enabling it to attract lone pairs of electrons from nearby electronegative atoms.

Example: Water (H2O):Each water molecule can form hydrogen bonds with up to four other water molecules, creating a dynamic network that accounts for its high boiling point and unique surface tension.

Vapor Pressure

The strength of intermolecular forces inversely affects vapor pressure; stronger forces lead to lower vapor pressures as fewer molecules can escape the liquid phase.

Boiling Point

There is a direct correlation between the strengths of intermolecular forces and boiling points; stronger interactions result in higher boiling points.

Viscosity

Defined as a liquid’s thickness or resistance to flow, viscosity is influenced by the strength and type of intermolecular forces present in the liquid.

Clausius Clapeyron Equation

ln(P2/P1) = Delta Hvap/R (1/T1 – 1/T2)

P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively. R=8.31 J/mol(K) is the universal gas constant, a pivotal factor in calculating changes in heat and phase transition. Delta Hvap refers to the heat required for a phase change from liquid to gas, critical for understanding boiling and evaporation processes.

Example of Clausius Clapeyron Equation

Carbon Disulfide (CS2) has a normal boiling point of 46oC, which indicates that at this temperature, it transitions from a liquid to a gas when surrounded by atmospheric pressure.

Given Values:

• Normal boiling point: T1 = 46oC

• Vaporization enthalpy Delta Hvap = 26.8 kJ/mol = 26,800 J/mol, representing the amount of energy needed to convert one mole of liquid to gas as its boiling.

• Vapor pressure at T2 = 35oC, where we need to calculate P2, which helps to ascertain the vapor pressure of the liquid at a temperature lower than its boiling point .

Steps to Calculate Vapor Pressure:

1. Convert all temperatures to Kelvin:

T1 = 46 + 273 = 319K

T2 = 35 + 273 = 308K

2. Set up equation using the Clausius Clapeyron Formula:

ln(P2/P1) = Delta Hvap/R (1/T1 – 1/T2)

R is the universal gas constant, approximately 8.31 J/molK, which plays a crucial role in thermodynamic equations relating temperature and pressure in gases.

3. Substitute:

ln(P2/760) = 26800/8.31 (1/308 – 1/319)

4. Perform calculations step by step to ensure accuracy:

5. Finally, find P2 which approximately equals 530 mmHg, demonstrating the relationship between temperature and vapor pressure and its applications in various scientific contexts.

Osmosis

The phenomenon of solvent flow through a semipermeable membrane to equalize the solute concentrations on both sides of the membrane.

Exothermic

A chemical reaction or a physical change in which heat is evolved or is released from the system (q is negative).

Endothermic

A chemical reaction or a physical change in which heat is absorbed by the system (q is positive).

Condensation

The process by which a gas turns back into a liquid, typically occurring when a gas cools and loses energy.

Sublimation

Direct transition from a solid to a gas, which bypasses the liquid state altogether. This process is also an exothermic process and is often observed in substances like dry ice (solid carbon dioxide) that sublimate at room temperature.

Evaporation

This refers to the transition of a liquid into a gas, which occurs through processes such as boiling and evaporation.

Enthalpy of Vaporization

The energy required to convert a liquid into a gas at its boiling point.

Enthalpy of Fusion

This is the amount of heat energy required to change a solid into a liquid at its melting point.