Comprehensive Study Guide: Moles, Solution Concentrations, Dilutions, Percent Composition, and Empirical Formulas

Percent Composition by Mass

Core Principles and Calculation Formula

• Definition: Percent composition by mass defines the percentage of total compound mass contributed by a specific constituent element.
• General Formula:   $\text{Percentage Mass of Element} = \frac{\text{Molar Mass of Element in Compound}}{\text{Total Molar Mass of Compound}} \times 100$
• Stoichiometric Rule: The numerator must account for the total mass of the specific element in one mole of the compound by multiplying the atomic mass of the element by its subscript in the chemical formula.
• Everyday Analogy:
  • Calculating percent composition by mass follows the same logic as determining an exam percentage score.
  • For instance, if a student scores 9 out of 13 on a mathematics test, the percentage is calculated as:     $\text{Percentage Score} = \frac{9}{13} \times 100 = 69.23\%$
  • The part of interest (9) is divided by the total (13) and multiplied by 100.

Step-by-Step Worked Example: Water ($H_2O$)

• Step 1: Calculate Total Molar Mass of Water ($H_2O$)

  • Look up atomic molar masses on the periodic table:
  • Hydrogen ($H$): $1.008\,g\,mol^{-1}$
  • Oxygen ($O$): $16.00\,g\,mol^{-1}$
  • Account for two Hydrogen atoms and one Oxygen atom:     $M = (2 \times 1.008\,g\,mol^{-1}) + 16.00\,g\,mol^{-1} = 18.016\,g\,mol^{-1}$

• Step 2: Calculate Percentage Mass of Hydrogen

  • Multiply the molar mass of Hydrogen by 2 (due to the $H_2$ subscript in $H_2O$):     $\text{Percentage Mass of } H = \frac{2 \times 1.008\,g\,mol^{-1}}{18.016\,g\,mol^{-1}} \times 100 = 11.19\% \approx 11.2\%$

• Step 3: Calculate Percentage Mass of Oxygen

  • Direct Calculation Method:     $\text{Percentage Mass of } O = \frac{16.00\,g\,mol^{-1}}{18.016\,g\,mol^{-1}} \times 100 = 88.8\%$
  • Subtraction Method:     $\text{Percentage Mass of } O = 100\% - 11.2\% = 88.8\%$

Quality Control and Validation

• Common Sense Check: Sum the individual element mass percentages within a compound to ensure they total $100\%$$11.2\% + 88.8\% = 100.0\%$
• Performing this summation serves as an immediate verification check for mathematical calculations.

Solutions and Molarity

Terminology and Fundamentals

• Solution: A homogeneous mixture consisting of two or more substances where one substance is dissolved into another.
• Solute: The dissolved substance present in a smaller proportion within the mixture.
• Solvent: The primary bulk phase medium that dissolves the solute.
• Real-World Examples:
  • Cordial Drink: Water acts as the bulk solvent, while dissolved solutes include sugar, food colors, flavoring agents, and preservatives.
  • Seawater: Water acts as the solvent, containing various dissolved salt solutes.

Concentration Formula and Calculations

• Concentration (Molarity) Definition: The quantity of solute in moles per unit volume of solution in liters.
• Mathematical Formula:   $C = \frac{n}{V}$   Where:
  • $C$ = Concentration or Molarity ($mol\,L^{-1}$ or $M$)
  • $n$ = Amount of solute in moles ($mol$)
  • $V$ = Solution volume in liters ($L$)
• Rearranged Equations:
  • Finding Solute Moles: $n = C \times V$
  • Finding Solution Volume: $V = \frac{n}{C}$

Notation Standards and Unit Distinctions

• Strict Volume Requirement: In $C = \frac{n}{V}$ and $n = C \times V$, volume must be expressed in liters ($L$).   $V(L) = \frac{V(mL)}{1000\,mL\,L^{-1}}$
• Units of Concentration:
  • Units are written as $mol\,L^{-1}$ or abbreviated as Molar ($M$).
  • Both designations represent identical concentrations ($1\,M = 1\,mol\,L^{-1}$).
• Distinguishing Capital Symbol $M$:
  • $M$ inside algebraic formulas: Represents Molar Mass ($g\,mol^{-1}$) as in $n = \frac{m}{M}$.
  • $M$ following a numerical value: Represents Molarity / Molar ($mol\,L^{-1}$) as a unit.
• Square Bracket Notation: Enclosing a chemical formula in square brackets denotes "concentration of" or "molarity of".
  • Example: $[NaCl]$ translates to "the molarity of sodium chloride".

Practical Worked Examples

Example 1: Calculating Solution Concentration

• Problem: Calculate the concentration of sodium chloride ($[NaCl]$) when $1.25\,mol$ of $NaCl$ is dissolved in $3.5\,L$ of solution.
• Calculation:   $C = \frac{1.25\,mol}{3.5\,L} = 0.357\,M$
• Units: Can be reported as $0.357\,M$ or $0.357\,mol\,L^{-1}$.

Example 2: Calculating Solute Moles

• Problem: Calculate the number of moles of hydrochloric acid ($HCl$) in $135\,mL$ of a $1.98\,M$ $HCl$ solution.
• Step 1: Convert Volume to Liters:   $V = \frac{135\,mL}{1000\,mL\,L^{-1}} = 0.135\,L$
• Step 2: Calculate Moles ($n$):   $n = C \times V = 1.98\,M \times 0.135\,L = 0.267\,mol$

Multi-Step Quantitative Calculations (The Chemical Tool Belt)

Interconnecting Conversion Formulas

• The mole ($n$) serves as the central bridge between physical quantities:   $\text{Mass } (m) \xleftrightarrow{n = \frac{m}{M}} \text{Moles } (n) \xleftrightarrow{n = C \times V} \text{Concentration } (C) \text{ \& Volume } (V)$$\text{Particle Count } (N) \xleftrightarrow{n = \frac{N}{N_A}} \text{Moles } (n)$

Problem-Solving Strategy

1. Deconstruct Question Statements: Isolate given numerical values and identify required target variables, ignoring extraneous text.
2. Convert Given Quantities to Moles ($n$): Convert initial given values (mass or concentration and volume) into moles.
3. Convert Moles to Target Variable: Use the calculated mole value to solve for the target output quantity.

Comprehensive Multi-Step Examples

Example 1: Mass to Concentration Calculation

• Problem: Determine the molarity of an $85\,mL$ ethanol solution containing $1.77\,g$ of ethanol ($C_2H_6O$).
• Step 1: Calculate Molar Mass of Ethanol ($C_2H_6O$):
  • Carbon: $2 \times 12.011\,g\,mol^{-1} = 24.022\,g\,mol^{-1}$
  • Hydrogen: $6 \times 1.008\,g\,mol^{-1} = 6.048\,g\,mol^{-1}$
  • Oxygen: $1 \times 15.999\,g\,mol^{-1} = 15.999\,g\,mol^{-1}$
  • Total Molar Mass ($M$) = $24.022 + 6.048 + 15.999 = 46.068\,g\,mol^{-1}$
• Step 2: Calculate Moles of Ethanol ($n$):   $n = \frac{m}{M} = \frac{1.77\,g}{46.068\,g\,mol^{-1}} = 0.03842\,mol$
• Step 3: Convert Volume to Liters:   $V = \frac{85\,mL}{1000\,mL\,L^{-1}} = 0.085\,L$
• Step 4: Calculate Concentration ($C$):   $C = \frac{n}{V} = \frac{0.03842\,mol}{0.085\,L} = 0.452\,M$
• Significant Figures and Units Verification: Input values ($1.77\,g$ and $85\,mL$) contain 3 significant figures; final answer must be stated as $0.452\,M$ with explicit molarity units.

Example 2: Concentration and Volume to Mass Calculation

• Problem: Calculate the mass of potassium iodide ($KI$) required to prepare $500\,mL$ of a $2.8\,M$ solution.
• Step 1: Convert Volume to Liters:   $V = \frac{500\,mL}{1000\,mL\,L^{-1}} = 0.5\,L$
• Step 2: Calculate Required Moles ($n$):   $n = C \times V = 2.8\,M \times 0.5\,L = 1.4\,mol$
• Step 3: Calculate Molar Mass of Potassium Iodide ($KI$):
  • Potassium ($K$) = $39.10\,g\,mol^{-1}$
  • Iodine ($I$) = $126.9\,g\,mol^{-1}$
  • Total Molar Mass ($M$) = $39.10 + 126.9 = 166.0\,g\,mol^{-1}$
• Step 4: Calculate Required Mass ($m$):   $m = n \times M = 1.4\,mol \times 166.0\,g\,mol^{-1} = 232\,g$

Dilutions

Principles and Purpose of Dilution

• Definition: Dilution is the process of adding solvent (water) to a solution to lower its concentration.
• Applications in Analytical Chemistry:
  • Diluting stock samples whose concentrations are too high for direct analytical measurement.
  • Back-calculating concentrated stock properties from measured diluted samples.
• Fundamental Physical Invariant:
  • Adding water increases volume ($V \uparrow$) and decreases concentration ($C \downarrow$).
  • The absolute number of moles of solute remains constant throughout the dilution process ($n_1 = n_2$).

The Dilution Equation

• Because solute moles remain constant ($n = C \times V$):   $C_1 \times V_1 = C_2 \times V_2$   Where:
  • $C_1$ = Initial concentration
  • $V_1$ = Initial volume taken
  • $C_2$ = Final concentration
  • $V_2$ = Final total volume

Volume Unit Rules for Dilutions

• Algebraic Cancellation: Volume units do not require conversion to liters in $C_1 V_1 = C_2 V_2$, provided both $V_1$ and $V_2$ use identical units (e.g., both in $mL$):   $C_2 = \frac{C_1 \times V_1}{V_2}$
• Warning and Best Practice:
  • While $C_1 V_1 = C_2 V_2$ allows any consistent volume unit, single-state mole calculations ($n = C \times V$) strictly require volume in liters ($L$).
  • Safety Net: Converting all volume quantities to liters ($L$) across all equations prevents unit conversion errors.

Worked Example: Diluting Zinc Chloride ($ZnCl_2$)

• Problem: Calculate the final concentration ($C_2$) when $20\,mL$ of $1.43\,M$ zinc chloride ($ZnCl_2$) solution is diluted up to $250\,mL$
• Given Variables:
  • $C_1 = 1.43\,M$
  • $V_1 = 20\,mL$
  • $V_2 = 250\,mL$
  • $C_2 = \text{unknown}$
• Formula Rearrangement:   $C_2 = \frac{C_1 \times V_1}{V_2}$
• Substitution and Calculation:   $C_2 = \frac{1.43\,M \times 20\,mL}{250\,mL} = 0.114\,M$
• Common Sense Check: Final concentration ($0.114\,M$) is lower than initial concentration ($1.43\,M$), confirming dilution occurred.

Empirical and Molecular Formulas

Definitions and Chemical Examples

• Empirical Formula: The simplest whole-number ratio of atoms of each element in a compound.
• Molecular Formula: The actual stoichiometric number of atoms of each element in a single molecule.
• Comparative Examples:
  • Hydrogen Peroxide:
  • Molecular Formula: $H_2O_2$ (active whitening agent in toothpastes, hair bleaches, active oxygen cleaners).
  • Empirical Formula: $HO$
  • Dicyanogen:
  • Molecular Formula: $C_2N_2$
  • Empirical Formula: $CN$
  • Octene Derivative:
  • Molecular Formula: $C_8H_{16}$
  • Empirical Formula: $CH_2$
  • Nonane:
  • Molecular Formula: $C_9H_{20}$
  • Empirical Formula: $C_9H_{20}$ (subscripts share no common divisor; empirical formula equals molecular formula).
  • Hydrazine:
  • Molecular Formula: $N_2H_4$ (rocket propellant; highly toxic compound formed when mixing ammonia and bleach cleaners).
  • Empirical Formula: $NH_2$
  • Diborane:
  • Molecular Formula: $B_2H_6$ (rocket propellant).
  • Empirical Formula: $BH_3$

Determining Empirical Formula from Percent Mass

• Example Compound: Allicin (the organosulfur compound responsible for the smell of garlic).

• Percent Composition Data:

  • Carbon ($C$): $44.4\%$
  • Hydrogen ($H$): $6.21\%$
  • Sulfur ($S$): $39.5\%$
  • Oxygen ($O$): $9.86\%$

• Step-by-Step Procedure:

  1. Assume a $100\,g$ Total Sample Mass:
  • Convert percentages directly to mass in grams:
    • $m_C = 44.4\,g$
    • $m_H = 6.21\,g$
    • $m_S = 39.5\,g$
    • $m_O = 9.86\,g$
  1. Convert Mass to Moles ($n = \frac{m}{M}$):
  • $n_C = \frac{44.4\,g}{12.01\,g\,mol^{-1}} = 3.70\,mol$
  • $n_H = \frac{6.21\,g}{1.008\,g\,mol^{-1}} = 6.16\,mol$
  • $n_S = \frac{39.5\,g}{32.06\,g\,mol^{-1}} = 1.23\,mol$
  • $n_O = \frac{9.86\,g}{16.00\,g\,mol^{-1}} = 0.616\,mol$
  1. Divide by Smallest Mole Value:
  • Smallest calculated value is $n_O = 0.616\,mol$
  • Carbon Ratio: $\frac{3.70}{0.616} = 6$
  • Hydrogen Ratio: $\frac{6.16}{0.616} = 10$
  • Sulfur Ratio: $\frac{1.23}{0.616} = 2$
  • Oxygen Ratio: $\frac{0.616}{0.616} = 1$
  1. Write Empirical Formula:
  • Empirical Formula = $C_6H_{10}S_2O$

Determining Molecular Formula from Molar Mass

• Formula Relationship: The molecular formula is an integer multiple ($k$) of the empirical formula:   $k = \frac{\text{Experimental Molar Mass of Compound}}{\text{Molar Mass of Empirical Formula}}$
• Allicin Calculation:
  • Molar mass of empirical formula $C_6H_{10}S_2O$:     $M_{\text{empirical}} = (6 \times 12.01) + (10 \times 1.008) + (2 \times 32.06) + (1 \times 16.00) = 162\,g\,mol^{-1}$
  • Measured compound molar mass = $162\,g\,mol^{-1}$
  • Multiplier factor ($k$):     $k = \frac{162\,g\,mol^{-1}}{162\,g\,mol^{-1}} = 1$
  • Molecular Formula of Allicin = $C_6H_{10}S_2O$
• Scaling Example: If experimental molar mass were $324\,g\,mol^{-1}$ ($k = 2$), the molecular formula subscripts would double to $C_{12}H_{20}S_4O_2$.