Combustion Analysis, Chemical Formulas, and Reaction Stoichiometry

Principles of Combustion Analysis

• Definition of Combustion Analysis: A laboratory technique used to determine the elemental composition of a substance (typically organic compounds containing carbon, hydrogen, and sometimes oxygen) by burning it in an excess of oxygen gas.
• Reaction Inputs: The substance being analyzed (e.g., a compound $C_xH_y$ or $C_xH_yO_z$) is heated in the presence of excess oxygen ($O_2$).
• Reaction Products: Upon complete combustion, the reaction will always produce Carbon Dioxide ($CO_2$) and Water ($H_2O$).
  • Carbon dioxide is typically produced as a gas, depending on temperature.
  • Water may initially be a liquid but often turns into a gas due to the heat of the reaction.
• Experimental Setup and Absorbers:
  • The system uses specific "absorbers" to capture the combustion products.
  • These absorbers have specific polarities and pore sizes meant to trap target molecules.
  • The First Absorber: This unit is designed to absorb produced water ($H_2O$).
  • The Second Absorber: This unit is designed to absorb produced carbon dioxide ($CO_2$).
• Data Collection via Mass Difference:
  • The mass of each absorber is measured before the reaction begins.
  • After the reaction is complete, the masses of the absorbers are measured again.
  • The change in mass represents the specific mass of water and carbon dioxide produced by the sample.

Determining Elemental Composition and Formulas

• Isolation of Elements:
  • Because both products ($CO_2$ and $H_2O$) contain oxygen, oxygen cannot be isolated directly from their masses.
  • Instead, the mass of carbon is isolated from carbon dioxide, and the mass of hydrogen is isolated from water.
• Conversion Steps:
  • Mass of Carbon: Convert the mass of captured $CO_2$ to moles of $CO_2$, then use the $1:1$ mole ratio within the molecule to find moles of Carbon ($C$), and finally convert to grams of $C$ using its atomic mass ($12.011\,g/mol$).
  • Mass of Hydrogen: Convert the mass of captured $H_2O$ to moles of $H_2O$, then use the $2:1$ mole ratio (2 moles of $H$ for every 1 mole of $H_2O$) to find moles of Hydrogen ($H$), and finally convert to grams of $H$ using its atomic mass ($1.008\,g/mol$).
• Identifying Oxygen:
  • Subtract the mass of Carbon and the mass of Hydrogen from the total initial mass of the sample.
  • If mass remains, that leftover amount constitutes the mass of Oxygen ($O$) in the original substance.
  • If no mass remains, the substance was a pure hydrocarbon.
• Formula Calculation:
  • Empirical Formula: Once elemental masses are known, convert them to moles, then divide each mole value by the smallest calculated mole value to find the simplest whole-number ratio.
  • Molecular Formula: Compare the molar mass of the empirical formula to the actual molar mass provided for the substance. Divide the given molar mass by the empirical molar mass to find the multiplier ($n$).

Case Study: Combustion Analysis of Butyl Butyrate

• Sample Data:
  • Initial Sample Mass: $2.5\,g$
  • Substance Contents: Carbon, Hydrogen, and Oxygen.
  • Product Masses: $6.1\,g$ of $CO_2$ and $2.5\,g$ of $H_2O$.
• Law of Conservation of Mass Applied:
  • Total mass of reactants must equal the total mass of products.
  • Since the total product mass is $6.1\,g + 2.5\,g = 8.6\,g$, and the sample reactant was $2.5\,g$, the mass of the added oxygen gas ($X$) must be $6.1\,g$.
• Calculating Elemental Masses:
  • Carbon: $\frac{6.1\,g\,CO_2}{44.01\,g/mol} \times 1\,mol\,C \times 12.011\,g/mol = 1.66\,g\,C$
  • Hydrogen: $\frac{2.5\,g\,H_2O}{18.016\,g/mol} \times 2\,mol\,H \times 1.008\,g/mol = 0.28\,g\,H$
  • Oxygen: $2.5\,g\text{ (initial sample)} - 1.66\,g\text{ (C)} - 0.28\,g\text{ (H)} = 0.56\,g\,O$
• Determining Empirical Formula:
  • Moles of Carbon: $\frac{1.66\,g}{12.011\,g/mol} \approx 0.138\,mol$
  • Moles of Hydrogen: $\frac{0.28\,g}{1.008\,g/mol} \approx 0.278\,mol$
  • Moles of Oxygen: $\frac{0.56\,g}{16.00\,g/mol} \approx 0.035\,mol$
  • Dividing by smallest ($0.035$) yields ratios of approximately $4:8:1$.
  • Result: Empirical Formula is $C_4H_8O$.
• Determining Molecular Formula:
  • Empirical mass ($C_4H_8O$) is approximately $72.1\,g/mol$.
  • Given molecular mass is $144.21\,g/mol$.
  • Multiplier $n = \frac{144.21}{72.1} \approx 2$.
  • Result: Molecular Formula is $C_8H_{16}O_2$.

Class Exercise: Analysis of a Hydrocarbon (Problem #65)

• Sample Data:
  • Carbon Dioxide Produced: $46.39\,g$
  • Water Produced: $18.99\,g$
  • Given Molar Mass: $140.27\,g/mol$
• Calculations for Carbon:
  • Moles of Carbon: $\frac{46.39\,g\,CO_2}{44.01\,g/mol} \times 1\,mol\,C = 1.054\,mol\,C$
  • Mass of Carbon: $1.054\,mol \times 12.011\,g/mol = 12.66\,g\,C$
• Calculations for Hydrogen:
  • Moles of Hydrogen: $\frac{18.99\,g\,H_2O}{18.016\,g/mol} \times 2\,mol\,H = 2.108\,mol\,H$
  • Mass of Hydrogen: $2.108\,mol \times 1.008\,g/mol \approx 2.125\,g\,H$
• Determining Formula:
  • Mole ratio: $\frac{2.108\,mol\,H}{1.054\,mol\,C} \approx 2$.
  • Empirical Formula: $CH_2$ (Empirical mass $\approx 14.03\,g/mol$).
  • Molecular Multiplier: $n = \frac{140.27}{14.03} \approx 10$.
  • Result: Molecular Formula is $C_{10}H_{20}$.

Laboratory Procedure: Empirical Formula of Magnesium Oxide

• Theoretical Chemistry:
  • Magnesium metal ($Mg$, a solid) combines with oxygen gas ($O_2(g)$) in a combination reaction (synthesis) when heated.
  • The resulting product is Magnesium Oxide ($MgO$).
  • Balanced equation: $2Mg(s) + O_2(g) \rightarrow 2MgO(s)$.
• Safety Protocol:
  • Rapid oxidation of Magnesium creates a very intense, bright white light.
  • Warning: Do not look directly at the burning magnesium as it can cause temporary blindness (similar to seeing stars from bright headlights).
• Procedural Steps:
  • Crucible Preparation: "Fire" the crucible by heating it until it glows red; let it cool to room temperature before weighing to avoid errors from oils or moisture.
  • Tongs Handling: Use crucible tongs ("little crab pinchers"). Do not touch items with hands to prevent contamination from finger oils.
  • Heating Magnesium: Polish magnesium ribbon with wool/sandpaper to remove oxidation. Curl the ribbon around a pen/pencil to ensure it fits at the bottom of the crucible for even heating.
  • Controlled Oxidation: Occasionally lift the crucible lid slightly to allow air to enter. If it burns too brightly, close the lid.
  • Source of Error: Potential error arises from incomplete reaction or random/systematic errors in balances. Students might calculate incorrect ratios like $Mg_3O_5$ in practice, but the theoretical ratio is $1:1$.

Introduction to Chemical Equations and Reaction Types

• Equation Components:
  • Reactants: Substances on the left side of the arrow.
  • Products: Substances on the right side of the arrow.
  • Arrow ($\rightarrow$): Represents "produces," "yields," or "equals."
  • Delta symbol ($\Delta$): Above the arrow, indicates heat is added to the system.
• Phase Labels:
  • $(s)$: Solid
  • $(l)$: Liquid
  • $(g)$: Gaseous
  • $(aq)$: Aqueous (dissolved in water/homogeneous solution)
• Coefficients vs. Subscripts:
  • Subscripts: Indicate the number of atoms chemically combined in a formula. These cannot be changed while balancing because doing so changes the identity of the chemical (e.g., $H_2O$ vs. $H_2O_2$).
  • Coefficients: Numbers placed in front of formulas to balance the total number of atoms on each side of the equation.
• Reaction Categories:
  • Combination/Synthesis: Two or more reactants form one product ($A + B \rightarrow AB$).
  • Decomposition: One reactant breaks down into two or more products ($AB \rightarrow A + B$).
  • Single Displacement/Replacement: One element takes the place of an element in a compound ($A + BC \rightarrow AC + B$). Effectiveness is determined by the Activity Series.
  • Double Displacement/Replacement: Ions in two compounds switch partners ($AB + CD \rightarrow AD + CB$). Effectiveness is determined by Solubility Rules.
  • Neutralization: A type of double displacement where an Acid (containing $H^+$) and a Base (containing $OH^-$) react to produce a Salt (an ionic compound) and Water.
  • Combustion: A hydrocarbon reacts with oxygen to produce carbon dioxide and water.
• Redox Reactions (Reduction-Oxidation):
  • Involve the transfer of electrons between substances.
  • Oxidation: Loss of electrons (metals typically undergo this).
  • Reduction: Gain of electrons (nonmetals typically undergo this).
  • Mnemonic: "LEO says GER" (Loss of Electrons = Oxidation; Gain of Electrons = Reduction).

Predicting Products and Balancing Examples

• Ammonia Synthesis: Nitrogen gas + Hydrogen gas $\rightarrow$ Ammonia gas.
  • Equation: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$.
• Double Replacement (Precipitation): Calcium Chloride + Sodium Phosphate.
  • Equation: $3CaCl_2(aq) + 2Na_3PO_4(aq) \rightarrow Ca_3(PO_4)_2(s) + 6NaCl(aq)$.
• Single Replacement: Aluminum + Hydrochloric Acid.
  • Since Aluminum is higher than Hydrogen on the Activity Series, it kicks out Hydrogen.
  • Equation: $2Al(s) + 6HCl(aq) \rightarrow 2AlCl_3(aq) + 3H_2(g)$.
• Acid-Base Neutralization: Hydrochloric Acid + Barium Hydroxide.
  • Equation: $2HCl(aq) + Ba(OH)_2(aq) \rightarrow BaCl_2(aq) + 2H_2O(l)$.

Questions & Discussion

• The Danger of Mixing Household Chemicals:
  • Mixing Ammonia and Bleach (Sodium Hypochlorite) produces high-toxicity gases like chloramine, not strictly mustard gas, though it causes chest pain and choking flumes.
  • Mixing bleach with rubbing alcohol can produce chloroform.
• Ammonia in Daily Life: Found in hair dye (responsible for the eyes-watering scent) and cat urine (produced after urea breaks down).
• Properties of Water and pH:
  • Pure/Distilled water has a pH around $7.0$, though removal of ions can make it slightly acidic ($\approx 6.5$).
  • "Alkaline Water" (pH $9.5+$) contains minerals like Calcium or Barium hydroxides to raise the pH.
• Hair Dye Exceptions: Some kits are labeled "Ammonia-free" to avoid the strong odor and reactive properties of the gas.