Comprehensive Revision Guide to Analytical Chemistry: Gravimetry, Titration, Spectroscopy, and Gas Stoichiometry

Master Analytical Chemistry Framework and Calculation Toolkit

Core Methodological Unity Across Analytical Techniques:
- Analytical chemistry investigations rely on a shared foundational skill set: chemical substance identification, stoichiometric mole ratios, concentration calculations, experimental error evaluation, and application of standard governing formulas.
- Every technique bifurcates into theoretical rationale (principles, reactions, apparatus, and source of error) and quantitative application (data interpretation, unit conversion, and multi-step stoichiometric calculation).
Fundamental Calculation Toolkit Formulas:
- Moles from Solid Mass: \n n = \frac{m}{M}\n where $n$ is the amount in moles ( $\mathrm{mol}$ ), $m$ is the mass in grams ( $\mathrm{g}$ ), and $M$ is the molar mass in grams per mole ( $\mathrm{g\,mol^{-1}}$ ).
- Moles from Solution Concentration: \n n = cV\n where $c$ is molar concentration ( $\mathrm{mol\,L^{-1}}$ ) and $V$ is the solution volume in litres ( $\mathrm{L}$ ).
- Molar Concentration: \n c = \frac{n}{V}\n
- Ideal Gas Equation: \n PV = nRT \implies P = \frac{nRT}{V}\n where $P$ is pressure in kilopascals ( $\mathrm{kPa}$ ), $V$ is volume in litres ( $\mathrm{L}$ ), $n$ is gas amount in moles ( $\mathrm{mol}$ ), $R$ is the universal gas constant ( $8.314\,\mathrm{L\,kPa\,mol^{-1}\,K^{-1}}$ ), and $T$ is absolute temperature in Kelvin ( $\mathrm{K}$ ).
- Dilution Factor ( $\text{DF}$ ) and Original Concentration: \n \text{DF} = \frac{V_{\text{final}}}{V_{\text{original aliquot}}}\n \n c_{\text{original}} = c_{\text{diluted}} \times \text{DF}\n
- Percentage Yield: \n \%\text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\n
Essential Unit Conversions:
- Volume conversion: $V(\mathrm{L}) = \frac{V(\mathrm{mL})}{1000}$ .
- Concentration conversion from molarity to mass concentration: $\mathrm{g\,L^{-1}} = \mathrm{mol\,L^{-1}} \times M$ .
- Parts per million (ppm) in dilute aqueous solutions: $1\,\mathrm{ppm} \approx 1\,\mathrm{mg\,L^{-1}}$ , meaning $\mathrm{ppm} = \mathrm{g\,L^{-1}} \times 1000$ .
- Mass per volume percentage ( $\%\,m/v$ ): grams of solute per $100\,\mathrm{mL}$ of solution (equivalent to $\frac{\mathrm{g\,L^{-1}}}{10}$ ).

Gravimetric Analysis: Theory, Error Analysis, and Worked Stoichiometry

Theoretical Foundation:
- Gravimetric analysis determines the quantity of an analyte by precipitating it from solution as an insoluble solid and accurately measuring the solid's dry mass.
- Standard target investigation: Quantification of sulfate ions ( $\mathrm{SO_4^{2-}}$ ) in fertiliser.
- Precipitation reaction: \n \mathrm{Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)}\n
- A precipitate is defined as an insoluble solid formed when substances react in solution.
Critical Experimental Terminology:
- Analyte: The target chemical species being measured ( $\mathrm{SO_4^{2-}}$ ).
- Precipitating Reagent: The chemical solution introduced to cause solid precipitation ( $\mathrm{Ba(NO_3)_2}$ , supplying $\mathrm{Ba^{2+}}$ ).
- Precipitate: The insoluble compound collected, dried, and weighed ( $\mathrm{BaSO_4}$ , an insoluble white solid).

Precipitation of barium sulfate in a beaker

Sequential Experimental Procedure and Rationales:
- 1. Addition of Excess Precipitating Reagent ( $\mathrm{Ba(NO_3)_2}$ ): Ensures a stoichiometric excess of $\mathrm{Ba^{2+}}$ is present so that all sulfate ions react and precipitate completely.
- 2. Testing the Filtrate: Adding a few drops of barium nitrate to a separate aliquot of the filtered liquid. If no secondary cloudiness or precipitate appears, it confirms complete precipitation of the analyte.
- 3. Washing the Precipitate: Rinsing the collected precipitate with deionised water removes unreacted dissolved ions (such as spectator $\mathrm{NO_3^-}$ and excess $\mathrm{Ba^{2+}}$ ) and soluble impurities adhering to the solid surfaces.
- 4. Drying to Constant Mass: Heating, cooling in a desiccator, and weighing repeatedly until successive mass readings are unchanged within experimental tolerance. This ensures all residual water is vaporised and does not inflate the measured mass.
Systematic Error Analysis:
- Error: Precipitate not fully dried:
- Measured Mass: Too high (due to retained water mass).
- Calculated Sulfate: Too high.
- Error: Precipitate lost during mechanical transfer or filtration:
- Measured Mass: Too low.
- Calculated Sulfate: Too low.
- Error: Incomplete precipitation (insufficient $\mathrm{Ba^{2+}}$ added):
- Measured Mass: Too low.
- Calculated Sulfate: Too low.
- Error: Soluble impurities remain trapped within the solid:
- Measured Mass: Too high.
- Calculated Sulfate: Too high.
- Governing Rule: Calculated analyte quantity directly tracks the measured precipitate mass because mole calculations assume the solid consists solely of pure, dry $\mathrm{BaSO_4}$ .
Step-by-Step Worked Stoichiometric Calculation:
- Raw Laboratory Data:
- Mass of filter paper = $1.223\,\mathrm{g}$
- Mass of filter paper + precipitate = $2.041\,\mathrm{g}$
- Volume of fertiliser solution aliquot = $25.00\,\mathrm{mL} = 0.02500\,\mathrm{L}$
- Volume of barium nitrate added = $50.00\,\mathrm{mL} = 0.05000\,\mathrm{L}$
- Concentration of barium nitrate solution = $0.200\,\mathrm{mol\,L^{-1}}$
- Step 1: Determine Pure Precipitate Mass: \n m(\mathrm{BaSO_4}) = 2.041\,\mathrm{g} - 1.223\,\mathrm{g} = 0.818\,\mathrm{g}\n
- Step 2: Convert Mass to Moles: \n M(\mathrm{BaSO_4}) = 137.3 + 32.1 + 4(16.0) = 233.4\,\mathrm{g\,mol^{-1}}\n \n n(\mathrm{BaSO_4}) = \frac{m}{M} = \frac{0.818\,\mathrm{g}}{233.4\,\mathrm{g\,mol^{-1}}} = 3.50 \times 10^{-3}\,\mathrm{mol}\n
- Step 3: Apply Reaction Mole Ratio:
- Balanced precipitation equation reveals a $1:1$ ratio between $\mathrm{SO_4^{2-}}$ and $\mathrm{BaSO_4}$ . \n n(\mathrm{SO_4^{2-}}) = 3.50 \times 10^{-3}\,\mathrm{mol}\n
- Step 4: Calculate Molar Concentration of Sulfate in Original Aliquot: \n c(\mathrm{SO_4^{2-}}) = \frac{n}{V} = \frac{3.50 \times 10^{-3}\,\mathrm{mol}}{0.02500\,\mathrm{L}} = 0.140\,\mathrm{mol\,L^{-1}}\n
- Step 5: Convert Concentration to $\mathrm{g\,L^{-1}}$ and ppm: \n M(\mathrm{SO_4^{2-}}) = 32.1 + 4(16.0) = 96.1\,\mathrm{g\,mol^{-1}}\n \n c = 0.140\,\mathrm{mol\,L^{-1}} \times 96.1\,\mathrm{g\,mol^{-1}} = 13.5\,\mathrm{g\,L^{-1}}\n \n c(\mathrm{ppm}) = 13.5\,\mathrm{g\,L^{-1}} \times 1000 = 13\,500\,\mathrm{ppm}\n
- Step 6: Quantify Excess Reagent: \n n(\mathrm{Ba^{2+}})_{\text{initial}} = cV = 0.200\,\mathrm{mol\,L^{-1}} \times 0.05000\,\mathrm{L} = 0.0100\,\mathrm{mol}\n \n n(\mathrm{Ba^{2+}})_{\text{reacted}} = n(\mathrm{SO_4^{2-}}) = 0.00350\,\mathrm{mol}\n \n n(\mathrm{Ba^{2+}})_{\text{excess}} = 0.0100\,\mathrm{mol} - 0.00350\,\mathrm{mol} = 0.00650\,\mathrm{mol}\n \n \%\text{ unreacted } \mathrm{Ba^{2+}} = \left( \frac{0.00650\,\mathrm{mol}}{0.0100\,\mathrm{mol}} \right) \times 100 = 65.0\%\n

Volumetric Analysis: Acid-Base Titration Principles and Calculations

Titration Overview and Apparatus Setup:
- Volumetric titration determines the unknown concentration of a solution by reacting a precisely known volume with a standard solution of known concentration.
- Target investigation: Determination of citric acid ( $\mathrm{H_3C_6H_5O_7}$ ) in commercial fruit juice using sodium hydroxide ( $\mathrm{NaOH}$ ).
- Apparatus Allocation:
- Burette: Accurately delivers the variable volume of titrant ( $\mathrm{NaOH}$ ).
- Conical Flask: Contains the precisely measured aliquot of diluted fruit juice.
- Pipette: Delivers the fixed aliquot volume to the conical flask.
- Indicator: Produces a distinct visible colour transition signalling the titration endpoint.
Chemical Stoichiometry and Acid-Base Character:
- Balanced neutralization equation: \n \mathrm{H_3C_6H_5O_7(aq) + 3NaOH(aq) \rightarrow Na_3C_6H_5O_7(aq) + 3H_2O(l)}\n
- Crucial stoichiometric ratio: $1\,\mathrm{mol}$ of triprotic citric acid reacts with $3\,\mathrm{mol}$ of monoprotic sodium hydroxide ( $1:3$ ratio).
- Acid-Base Strengths:
- Sodium hydroxide ( $\mathrm{NaOH}$ ) is a strong base that completely dissociates in aqueous solution.
- Citric acid ( $\mathrm{H_3C_6H_5O_7}$ ) is a weak polyprotic carboxylic acid that ionises only partially.
- Equivalence point conditions: Neutralisation produces the conjugate base citrate ( $\mathrm{C_6H_5O_7^{3-}}$ ), which hydrolyses water to generate hydroxide ions ( $\mathrm{OH^-}$ ). Consequently, the equivalence point pH is basic ( $\mathrm{pH} > 7$ ).
Essential Titration Terminology:
- Aliquot: A precise, measured portion of a liquid sample drawn for analysis using a volumetric pipette.
- Titre: The volume of titrant dispensed from the burette to reach the reaction endpoint.
- Concordant Titres: Titres that agree closely with one another, typically within a maximum tolerance of $0.10\,\mathrm{mL}$ .
- Equivalence Point: The theoretical stage in a titration where stoichiometric amounts of acid and base have completely neutralised each other.
- Endpoint: The observable physical signal (colour transition of the chemical indicator) signifying the end of the titration.
Indicator Selection Criteria:
- Phenolphthalein: Transitions colour in an alkaline range ( $\mathrm{pH} \approx 8.2 - 10.0$ ). This matches the basic equivalence point of a weak acid-strong base titration, providing an accurate endpoint.
- Bromothymol Blue: Transitions near neutral pH ( $\mathrm{pH} \approx 6.0 - 7.6$ ). It changes colour prematurely before true equivalence, yielding an artificially low titre and an underestimated calculated acid concentration.
Titration Error Analysis:
- Water left in burette before filling with titrant: Dilutes the $\mathrm{NaOH}$ solution; a larger titre volume is needed to deliver the required moles of base. Using the nominal standard concentration yields a calculated acid concentration that is too high.
- Acid left in conical flask from rinsing with acid: Introduces additional acid moles into the flask; requires a larger titre of $\mathrm{NaOH}$ . Calculated acid concentration is too high.
- Endpoint overshot: Excess $\mathrm{NaOH}$ is delivered from the burette; recorded titre is too large. Calculated acid concentration is too high.
- Non-concordant titre included in calculations: Distorts the mean titre value, reducing analytical precision and reliability.
- Conical flask rinsed with deionised water: Introduces solvent water but does not alter the absolute moles of citric acid present in the measured aliquot. The calculated acid concentration remains unchanged (an acceptable and standard laboratory practice).
Step-by-Step Worked Titration Calculation:
- Laboratory Data:
- Standard $\mathrm{NaOH}$ concentration = $0.1000\,\mathrm{mol\,L^{-1}}$
- Aliquot of diluted fruit juice = $25.00\,\mathrm{mL} = 0.02500\,\mathrm{L}$
- Dilution protocol: $20.00\,\mathrm{mL}$ original juice diluted to a total of $250.0\,\mathrm{mL}$
- Concordant titres: $18.70\,\mathrm{mL}$ , $18.80\,\mathrm{mL}$ , $18.75\,\mathrm{mL}$
- Step 1: Calculate Mean Titre and Moles of Titrant: \n V_{\text{mean}} = \frac{18.70 + 18.80 + 18.75}{3} = 18.75\,\mathrm{mL} = 0.01875\,\mathrm{L}\n \n n(\mathrm{NaOH}) = cV = 0.1000\,\mathrm{mol\,L^{-1}} \times 0.01875\,\mathrm{L} = 1.875 \times 10^{-3}\,\mathrm{mol}\n
- Step 2: Calculate Moles of Citric Acid in Aliquot: \n n(\text{citric acid}) = \frac{n(\mathrm{NaOH})}{3} = \frac{1.875 \times 10^{-3}\,\mathrm{mol}}{3} = 6.25 \times 10^{-4}\,\mathrm{mol}\n
- Step 3: Calculate Diluted Solution Concentration: \n c_{\text{diluted}} = \frac{n}{V} = \frac{6.25 \times 10^{-4}\,\mathrm{mol}}{0.02500\,\mathrm{L}} = 0.0250\,\mathrm{mol\,L^{-1}}\n
- Step 4: Undo the Dilution to Find Original Concentration: \n \text{DF} = \frac{250.0\,\mathrm{mL}}{20.00\,\mathrm{mL}} = 12.50\n \n c_{\text{original}} = 0.0250\,\mathrm{mol\,L^{-1}} \times 12.50 = 0.3125\,\mathrm{mol\,L^{-1}}\n
- Step 5: Convert Concentration to Mass Units and Total Bottle Content: \n M(\mathrm{H_3C_6H_5O_7}) = 192.1\,\mathrm{g\,mol^{-1}}\n \n c = 0.3125\,\mathrm{mol\,L^{-1}} \times 192.1\,\mathrm{g\,mol^{-1}} = 60.0\,\mathrm{g\,L^{-1}}\n \n c(\mathrm{ppm}) = 60.0\,\mathrm{g\,L^{-1}} \times 1000 = 60\,000\,\mathrm{ppm}\n
- Total mass of citric acid in a $500\,\mathrm{mL}$ ( $0.500\,\mathrm{L}$ ) bottle: \n m = cV = 60.0\,\mathrm{g\,L^{-1}} \times 0.500\,\mathrm{L} = 30.0\,\mathrm{g}\n

UV-Visible Spectroscopy: Principles, Calibration, and Analysis

Instrumentation Principles:
- UV-visible spectroscopy quantifies light absorption by molecules across the ultraviolet and visible regions of the electromagnetic spectrum.
- Relationship between sample colour and absorbance:
- Observed solution colour represents transmitted or reflected light wavelengths.
- Riboflavin (vitamin $\mathrm{B_2}$ ) appears intensely yellow because it transmits yellow light while strongly absorbing complementary blue-violet light.

Yellow riboflavin solution in a microcentrifuge tube alongside its chemical structure

Wavelength Selection and Beer-Lambert Mechanics:
- Analyses select the wavelength of maximum absorption ( $\lambda_{\max}$ ) for the analyte to maximize measurement sensitivity.
- Concentration and absorption dynamics:
- A higher concentration solution contains more absorbing molecules within the optical path length.
- Increased absorption results in less light reaching the detector, yielding a higher absorbance ( $A$ ).
- Absorbance is directly proportional to solute concentration ( $A \propto c$ ) within a linear concentration range.

Schematic diagram of a UV-Vis spectrophotometer and visible spectrum wavelengths

Spectrophotometer Configuration and Spectral Domains:
- Instrumental Beam Path:
- Wide band light source $\rightarrow$ Monochromator $\rightarrow$ Beam splitter ( $50:50\,\%$ ) $\rightarrow$ Sample cuvette and Reference (blank) cuvette $\rightarrow$ Detector.
- Visible Light Spectrum Wavelength Bands:
- Violet: $400 - 420\,\mathrm{nm}$
- Indigo: $420 - 440\,\mathrm{nm}$
- Blue: $440 - 490\,\mathrm{nm}$
- Green: $490 - 570\,\mathrm{nm}$
- Yellow: $570 - 585\,\mathrm{nm}$
- Orange: $585 - 620\,\mathrm{nm}$
- Red: $620 - 780\,\mathrm{nm}$
Calibration Curves and Experimental Controls:
- Calibration Curve Structure:
- X-axis: Concentration of analyte (in $\mathrm{ppm}$ , $\mathrm{mg\,L^{-1}}$ , or $\mathrm{mol\,L^{-1}}$ ).
- Y-axis: Absorbance (dimensionless unit, $A$ ).
- Reference Cell (Blank): Filled with solvent only; measures and subtracts background absorbance caused by the solvent and the cuvette walls.
- Operational Rules:
- Standards and unknown samples must be analyzed at the exact same wavelength.
- Cuvettes must be optically identical, clean, and completely free of fingerprints, air bubbles, and scratches.
- The sample's absorbance must sit securely within the bounds of the standard calibration curve.
- Ban on Extrapolation: Never extrapolate beyond the tested standard range; non-linear physical and chemical effects make extrapolated values invalid. If a sample is too concentrated, it must be diluted and remeasured.
Step-by-Step Worked Spectroscopic Calculation:
- Calibration Data Table:
- Standard 1: $10\,\mathrm{ppm} \rightarrow 0.12$
- Standard 2: $20\,\mathrm{ppm} \rightarrow 0.25$
- Standard 3: $30\,\mathrm{ppm} \rightarrow 0.37$
- Standard 4: $40\,\mathrm{ppm} \rightarrow 0.50$
- Standard 5: $50\,\mathrm{ppm} \rightarrow 0.62$
- Sample Analysis:
- Measured absorbance of unknown diluted riboflavin = $0.44$
- Dilution procedure: $5.00\,\mathrm{mL}$ diluted to $250.0\,\mathrm{mL}$
- Molar mass of riboflavin = $376.4\,\mathrm{g\,mol^{-1}}$
- Step 1: Determine Diluted Concentration via Calibration Interpolation:
- An absorbance of $0.44$ lies between $0.37$ ( $30\,\mathrm{ppm}$ ) and $0.50$ ( $40\,\mathrm{ppm}$ ). \n c_{\text{diluted}} = 35.4\,\mathrm{ppm}\n
- Step 2: Calculate Dilution Factor and Original Concentration: \n \text{DF} = \frac{250.0\,\mathrm{mL}}{5.00\,\mathrm{mL}} = 50.0\n \n c_{\text{original}} = 35.4\,\mathrm{ppm} \times 50.0 = 1770\,\mathrm{ppm}\n
- Step 3: Convert to Mass Concentration ( $\mathrm{g\,L^{-1}}$ ) and Molarity ( $\mathrm{mol\,L^{-1}}$ ): \n 1770\,\mathrm{ppm} \approx 1770\,\mathrm{mg\,L^{-1}} = \frac{1770}{1000}\,\mathrm{g\,L^{-1}} = 1.77\,\mathrm{g\,L^{-1}}\n \n c = \frac{1.77\,\mathrm{g\,L^{-1}}}{376.4\,\mathrm{g\,mol^{-1}}} = 4.70 \times 10^{-3}\,\mathrm{mol\,L^{-1}}\n
- Step 4: Convert to Percentage Mass/Volume and Total Bottle Content: \n \%\,m/v = \frac{1.77\,\mathrm{g}}{1000\,\mathrm{mL}} \times 100\,\mathrm{mL} = 0.177\%\,m/v\n
- Mass of riboflavin in a $600\,\mathrm{mL}$ ( $0.600\,\mathrm{L}$ ) commercial bottle: \n m = cV = 1.77\,\mathrm{g\,L^{-1}} \times 0.600\,\mathrm{L} = 1.06\,\mathrm{g}\n

Limestone Decomposition, Gas Laws, and Environmental Impact

Thermal Decomposition Reaction:
- Thermal decomposition occurs when heat energy breaks chemical bonds, reducing a compound into simpler component substances.
- Reaction equation: \n \mathrm{CaCO_3(s) \rightarrow CaO(s) + CO_2(g)}\n
- Stoichiometric molar ratio: \n \mathrm{CaCO_3 : CaO : CO_2} = 1:1:1\n
- Ambient atmospheric gases present inside the chamber prior to heating (such as $O_2$ ) do not participate chemically in the reaction and are omitted from stoichiometric calculations.
Ideal Gas Behavior vs. Real Gases:
- Ideal Gas Model: A theoretical model where gas particles possess negligible volume, exert zero intermolecular attractions on each other, and collide with perfectly elastic impacts.
- Conditions Favoring Ideal Behavior in Real Gases:
- High Temperatures: Particles have large kinetic energies, rendering attractive intermolecular forces insignificant.
- Low Pressures: Gas particles are spaced extremely far apart, making particle volume negligible relative to the total container volume.
Gas Laws and Standard Laboratory Conditions:
- Standard Laboratory Conditions (SLC): Temperature of $25^\circ\mathrm{C}$ ( $298\,\mathrm{K}$ ), pressure of $100\,\mathrm{kPa}$ , and a standard molar volume ( $V_m$ ) of $24.8\,\mathrm{L\,mol^{-1}}$ .
- Boyle's Law: For a fixed amount of gas at constant temperature, pressure is inversely proportional to volume: \n P_1V_1 = P_2V_2\n
Limiting and Excess Reagents:
- Limiting Reagent: The reactant completely consumed first in a chemical reaction; it places an absolute stoichiometric ceiling on the maximum theoretical yield of products.
- Excess Reagent: The reactant remaining in the reaction vessel after the limiting reagent has been fully converted.
- Reaction with hydrochloric acid: \n \mathrm{CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)}\n
- Molar requirements: One mole of $\mathrm{CaCO_3}$ requires two moles of $\mathrm{HCl}$ ( $1:2$ ratio). Identifying the limiting reagent requires dividing available moles by stoichiometric coefficients, not merely comparing raw mole numbers.
Environmental Chemistry of Carbon Dioxide:
- Carbon dioxide is a greenhouse gas that absorbs outgoing thermal infrared radiation emitted by Earth's surface and re-radiates it in all directions, trapping heat within the troposphere.
- Natural Greenhouse Effect: The natural atmospheric mechanism that maintains global temperatures within a habitable range.
- Enhanced Greenhouse Effect: Additional atmospheric warming driven by anthropogenic emissions of greenhouse gases ( $\mathrm{CO_2}$ , $\mathrm{CH_4}$ , etc.).
- Industrial calcination of limestone ( $\mathrm{CaCO_3} \rightarrow \mathrm{CaO} + \mathrm{CO_2}$ ) in cement manufacture is a major direct industrial contributor to global $\mathrm{CO_2}$ emissions.
Step-by-Step Worked Stoichiometric and Gas Law Calculations:
- Decomposition Stoichiometry:
- Given: $25.0\,\mathrm{g}$ of $\mathrm{CaCO_3}$ \n M(\mathrm{CaCO_3}) = 40.1 + 12.0 + 3(16.0) = 100.1\,\mathrm{g\,mol^{-1}}\n \n n(\mathrm{CaCO_3}) = \frac{25.0\,\mathrm{g}}{100.1\,\mathrm{g\,mol^{-1}}} = 0.250\,\mathrm{mol}\n
- Due to the $1:1:1$ ratio, decomposition produces $0.250\,\mathrm{mol}$ of $\mathrm{CO_2}$ and $0.250\,\mathrm{mol}$ of $\mathrm{CaO}$ . \n m(\mathrm{CO_2}) = nM = 0.250\,\mathrm{mol} \times 44.0\,\mathrm{g\,mol^{-1}} = 11.0\,\mathrm{g}\n \n m(\mathrm{CaO}) = nM = 0.250\,\mathrm{mol} \times 56.1\,\mathrm{g\,mol^{-1}} = 14.0\,\mathrm{g}\n
- Volume of $\mathrm{CO_2}$ gas generated at SLC: \n V = nV_m = 0.250\,\mathrm{mol} \times 24.8\,\mathrm{L\,mol^{-1}} = 6.20\,\mathrm{L}\n
- Chamber Pressure Dynamics:
- Initial chamber condition: $0.400\,\mathrm{mol}$ of unreactive $O_2$ gas in a $12.0\,\mathrm{L}$ rigid vessel at $298\,\mathrm{K}$ . \n P_{\text{initial}} = \frac{nRT}{V} = \frac{0.400\,\mathrm{mol} \times 8.314\,\mathrm{L\,kPa\,mol^{-1}\,K^{-1}} \times 298\,\mathrm{K}}{12.0\,\mathrm{L}} = 82.7\,\mathrm{kPa}\n
- Post-decomposition total gas amount: \n n_{\text{total}} = n(O_2) + n(\mathrm{CO_2}) = 0.400\,\mathrm{mol} + 0.250\,\mathrm{mol} = 0.650\,\mathrm{mol}\n \n P_{\text{final}} = \frac{0.650\,\mathrm{mol} \times 8.314\,\mathrm{L\,kPa\,mol^{-1}\,K^{-1}} \times 298\,\mathrm{K}}{12.0\,\mathrm{L}} = 134\,\mathrm{kPa}\n
- Expansion of vessel to $18.0\,\mathrm{L}$ at constant temperature (Boyle's Law): \n P_2 = \frac{P_1V_1}{V_2} = \frac{134\,\mathrm{kPa} \times 12.0\,\mathrm{L}}{18.0\,\mathrm{L}} = 89.3\,\mathrm{kPa}\n
- Limiting Reagent Determination:
- Reactants: $25.0\,\mathrm{g}$ of $\mathrm{CaCO_3}$ ( $0.250\,\mathrm{mol}$ ) mixed with $150.0\,\mathrm{mL}$ ( $0.1500\,\mathrm{L}$ ) of $2.00\,\mathrm{mol\,L^{-1}}\,\mathrm{HCl}$ . \n n(\mathrm{HCl}) = cV = 2.00\,\mathrm{mol\,L^{-1}} \times 0.1500\,\mathrm{L} = 0.300\,\mathrm{mol}\n
- Required $\mathrm{HCl}$ to fully consume $0.250\,\mathrm{mol}\,\mathrm{CaCO_3}$ : \n n(\mathrm{HCl})_{\text{required}} = 0.250\,\mathrm{mol} \times 2 = 0.500\,\mathrm{mol}\n
- Since available $\mathrm{HCl}$ ( $0.300\,\mathrm{mol}$ ) is less than required ( $0.500\,\mathrm{mol}$ ):
  - Limiting Reagent: $\mathrm{HCl}$
  - Excess Reagent: $\mathrm{CaCO_3}$
- Reacted and remaining $\mathrm{CaCO_3}$ : \n n(\mathrm{CaCO_3})_{\text{reacted}} = \frac{0.300\,\mathrm{mol}}{2} = 0.150\,\mathrm{mol}\n \n n(\mathrm{CaCO_3})_{\text{excess}} = 0.250\,\mathrm{mol} - 0.150\,\mathrm{mol} = 0.100\,\mathrm{mol}\n
- Doubling the initial mass of $\mathrm{CaCO_3}$ has zero effect on product yield because $\mathrm{HCl}$ remains the limiting reagent.
- Percentage Yield Calculation:
- Theoretical mass of $\mathrm{CaO} = 14.0\,\mathrm{g}$ . Actual collected mass = $9.20\,\mathrm{g}$ . \n \%\text{ yield} = \frac{9.20\,\mathrm{g}}{14.0\,\mathrm{g}} \times 100 = 65.7\%\n
- Factors causing yield to fall below $100\%$ include incomplete decomposition, product loss during transfer, and measurement inaccuracies.

Questions & Discussion

Question 1: Gravimetric Theory:
- Prompt: Why is a precipitate dried to constant mass?
- Response: To ensure the measured mass is no longer affected significantly by remaining water. Repeating heating, cooling, and weighing cycles proves that all volatile solvent has been driven off.
Question 2: Gravimetric Application:
- Prompt: A precipitate has a mass of $0.467\,\mathrm{g}$ and a molar mass of $233.5\,\mathrm{g\,mol^{-1}}$ . How many moles are present?
- Response: $0.00200\,\mathrm{mol}$ .
- Calculation: \n n = \frac{m}{M} = \frac{0.467\,\mathrm{g}}{233.5\,\mathrm{g\,mol^{-1}}} = 0.00200\,\mathrm{mol}\n
Question 3: Titration Theory:
- Prompt: Why is rinsing the conical flask with deionised water acceptable?
- Response: It does not change the moles of acid in the aliquot. Although adding water changes the concentration in the flask, titration stoichiometry depends entirely on the absolute number of solute moles transferred by the pipette.
Question 4: Titration Application:
- Prompt: A reaction has a $1:2$ acid-to-base mole ratio. If $0.0060\,\mathrm{mol}$ of base reacts completely, how many moles of acid reacted?
- Response: $0.0030\,\mathrm{mol}$ .
- Calculation: \n n(\text{acid}) = \frac{n(\text{base})}{2} = \frac{0.0060\,\mathrm{mol}}{2} = 0.0030\,\mathrm{mol}\n
Question 5: Spectroscopy Theory:
- Prompt: What should you do if an unknown sample's absorbance is outside the calibration range?
- Response: Dilute the sample and measure again. Extrapolating beyond the highest standard is invalid because Beer-Lambert linearity often fails at high concentrations.
Question 6: Spectroscopy Application:
- Prompt: A $4.00\,\mathrm{mL}$ sample is diluted to $100.0\,\mathrm{mL}$ . The diluted concentration is $8.0\,\mathrm{ppm}$ . What was the original concentration?
- Response: $200\,\mathrm{ppm}$ .
- Calculation: \n \text{DF} = \frac{100.0\,\mathrm{mL}}{4.00\,\mathrm{mL}} = 25.0\n \n c_{\text{original}} = 8.0\,\mathrm{ppm} \times 25.0 = 200\,\mathrm{ppm}\n
Question 7: Gas Laws Theory:
- Prompt: At constant temperature and volume, what happens to pressure if the amount of gas doubles?
- Response: Pressure doubles. According to $P = \frac{nRT}{V}$ , pressure is directly proportional to the number of gas particles.
Question 8: Limiting Reagents:
- Prompt: The reaction $A + 2B \rightarrow \text{products}$ starts with $0.40\,\mathrm{mol}\,A$ and $0.50\,\mathrm{mol}\,B$ . Which is limiting?
- Response: $B$ .
- Calculation: Consuming $0.40\,\mathrm{mol}\,A$ requires $0.40 \times 2 = 0.80\,\mathrm{mol}\,B$ . Because only $0.50\,\mathrm{mol}\,B$ is available, $B$ runs out first.