Module 8: Applying Chemical Ideas

(a) Calculate lead concentration:

Given information:

• Calibration equation: A = 0.0523c
• Sample absorbance: A = 0.419

Calculation:

Rearrange the calibration equation to solve for concentration:

$c = \frac{A}{0.0523}$

$c = \frac{0.419}{0.0523} = 8.01 \text{ }\mu\text{g/L}$

Answer: The lead concentration is 8.0 μg/L (to 2 significant figures)

(b) Does sample meet WHO guidelines?

Comparison:

• Sample concentration: 8.0 μg/L
• WHO guideline: 10 μg/L maximum

Conclusion: YES, the sample meets WHO guidelines because 8.0 μg/L is below the maximum allowable level of 10 μg/L.

The water is safe to drink with respect to lead content, though it contains 80% of the maximum allowable concentration, so it's approaching the limit.

(c) Why calibration curve is necessary:

A calibration curve is essential for AAS analysis for several reasons:

1. Converting absorbance to concentration: AAS instruments measure absorbance, not concentration directly. The calibration curve provides the mathematical relationship between these two quantities. Without it, we would only have an absorbance value, which is meaningless on its own.

2. Accounts for instrument-specific factors: Different AAS instruments have different sensitivities, lamp intensities, flame temperatures, and optical path lengths. The calibration curve accounts for all these variables specific to the instrument being used. The same sample might give different absorbance values on different instruments, but both would give the correct concentration when using their respective calibration curves.

3. Verifies linearity: The calibration curve confirms that Beer-Lambert law (A ∝ c) is being obeyed in the concentration range being used. If the calibration points don't form a straight line, it indicates problems such as:

• Concentration too high (deviation from linearity)
• Instrument malfunction
• Matrix interference

4. Ensures accuracy: By using multiple standard solutions of known concentration, the calibration curve minimizes random errors and improves the reliability of concentration measurements. A single-point calibration would be less reliable.

5. Element-specific: Different metals absorb light at different wavelengths with different sensitivities. Each element requires its own calibration curve because the slope (molar absorptivity × path length) varies between elements.

Summary:

Without a calibration curve, AAS would be useless for quantitative analysis. The curve transforms raw absorbance data into meaningful concentration values while accounting for all instrument-specific and element-specific factors.

(a) Molar mass:

The parent peak represents the molecular ion (M⁺), which has the highest m/z value.

Parent peak = 58

Therefore, molar mass = 58 g/mol

(b) Mass lost to give fragment at 43:

Mass lost = Parent peak - Fragment peak

Mass lost = 58 - 43 = 15

A loss of 15 corresponds to loss of a CH₃ group (methyl group)

This indicates the molecule contains at least one CH₃ group

(c) Two possible structures with C₃H₆O:

Verification: C₃H₆O has M_r = 3(12) + 6(1) + 16 = 58 ✓

Structure 1: Propanone (acetone)

CH₃-CO-CH₃

Fragmentation:

CH₃COCH₃⁺• (58) → CH₃• + CH₃CO⁺ (43)

Loss of CH₃ gives CH₃CO⁺ (acetyl cation, m/z = 43)

The acetyl cation is very stable, explaining why this is the base peak

Structure 2: Propanal

CH₃-CH₂-CHO

Fragmentation:

CH₃CH₂CHO⁺• (58) → CH₃• + CH₂CHO⁺ (43)

or

CH₃CH₂CHO⁺• (58) → CH₃CH₂• + CHO⁺ (29)

Loss of CH₃ gives C₂H₃O⁺ (m/z = 43)

Why fragment at 43 is the base peak:

In both cases, the fragment at m/z = 43 involves a carbonyl group (C=O). Carbocations adjacent to oxygen are stabilized by resonance, making them more stable and abundant than other fragments. This is why the peak at 43 is the base peak (100% relative abundance). For propanone: CH₃CO⁺ ↔ CH₃C⁺=O (resonance stabilization)

Distinguishing between the two:

To distinguish propanone from propanal, you would need additional information:

• IR spectroscopy: Propanal has a C-H aldehyde stretch (~2720 cm⁻¹), propanone doesn't
• NMR: Different splitting patterns and chemical shifts
• Chemical tests: Propanal would give a positive Tollens' test (aldehyde), propanone wouldn't

Both structures are valid answers based solely on the mass spectrum data provided.

(a) Identify functional groups:

Compound A:

• Sharp, strong peak at 1715 cm⁻¹ → C=O (carbonyl group)
• NO broad O-H peak → Rules out alcohol or carboxylic acid
• Functional group: Carbonyl (C=O)
• Since it has C=O but no O-H, it must be either a ketone or aldehyde

Compound B:

• Broad peak at 3350 cm⁻¹ → O-H (hydroxyl group)
• Strong peak at 1100 cm⁻¹ → C-O bond
• NO peak around 1700 cm⁻¹ → No carbonyl (C=O)
• Functional group: Hydroxyl (-OH)
• The combination of O-H + C-O without C=O indicates an alcohol

(b) Suggest structures:

Both compounds have formula C₃H₆O

This gives several possibilities - we use IR to distinguish them

Compound A (carbonyl group):

With C₃H₆O and a carbonyl, the structure must be:

Propanone (acetone): CH₃-CO-CH₃

This is a ketone with the carbonyl in the middle

Alternative possibility: Propanal (CH₃-CH₂-CHO)

Both are valid based on IR alone, but propanone is more common. Additional tests would distinguish them:

• Propanal would show weak C-H aldehyde peaks at 2720 cm⁻¹
• Propanal gives positive Tollens' test, propanone doesn't

Compound B (alcohol):

With C₃H₆O and a hydroxyl group, possible structures are:

Propan-1-ol: CH₃-CH₂-CH₂-OH (primary alcohol)

or

Propan-2-ol: CH₃-CH(OH)-CH₃ (secondary alcohol)

IR spectroscopy alone cannot distinguish between these two alcohols - they would show nearly identical spectra. To distinguish them, you would need:

• NMR spectroscopy (different splitting patterns)
• Oxidation test (propan-1-ol → propanal → propanoic acid; propan-2-ol → propanone only)

Summary:

Compound A: Propanone (or propanal) - contains C=O

Compound B: Propan-1-ol or propan-2-ol - contains -OH

This question demonstrates how IR spectroscopy distinguishes between structural isomers by identifying functional groups, even when it cannot always determine the exact structure without additional techniques.

(a) Assign signals:

Signal A: δ = 1.2 ppm, triplet, integration 3

• Chemical shift 1.2 ppm: Typical for alkyl CH₃ (0.8-1.2 ppm)
• Integration 3: This is a CH₃ group (3 hydrogens)
• Triplet: Indicates 2 H on the neighboring carbon (n+1 = 3, so n = 2)
• Assignment: This is a CH₃ group next to a CH₂ group

Signal B: δ = 3.7 ppm, quartet, integration 2

• Chemical shift 3.7 ppm: Typical for R-CH₂-O (3.3-4.0 ppm), H next to oxygen
• Integration 2: This is a CH₂ group (2 hydrogens)
• Quartet: Indicates 3 H on the neighboring carbon (n+1 = 4, so n = 3)
• Assignment: This is a CH₂ group next to oxygen AND next to a CH₃ group

Signal C: δ = 2.6 ppm, singlet, integration 1

• Chemical shift 2.6 ppm: Typical for -OH in alcohols (2-5 ppm)
• Integration 1: This is 1 hydrogen
• Singlet: No splitting - typical for -OH (rapid exchange prevents splitting)
• Assignment: This is the -OH hydrogen

(b) Explain splitting patterns using n+1 rule:

Signal A (CH₃) is a triplet:

The CH₃ is next to a CH₂ group

The neighboring CH₂ has 2 hydrogens

n = 2, so n+1 = 3 → triplet ✓

Signal B (CH₂) is a quartet:

The CH₂ is next to a CH₃ group

The neighboring CH₃ has 3 hydrogens

n = 3, so n+1 = 4 → quartet ✓

Signal C (OH) is a singlet:

-OH hydrogens typically don't show splitting due to rapid exchange with other -OH groups and traces of water

The hydrogen exchanges so quickly that the neighboring hydrogens "see" an average environment

Result: singlet ✓

Key point: The triplet-quartet pattern is diagnostic for a -CH₃-CH₂- fragment!

(c) Determine structure:

Combining all information:

• Molecular formula: C₂H₆O
• Has CH₃ group (triplet at δ 1.2)
• Has CH₂ group next to O (quartet at δ 3.7)
• Has -OH group (singlet at δ 2.6)
• CH₃ and CH₂ are neighbors (triplet-quartet pattern)

Structure: CH₃-CH₂-OH (Ethanol)

Complete verification:

Why not methoxymethane (CH₃-O-CH₃)?

Methoxymethane would show:

• Only ONE signal (both CH₃ groups equivalent)
• Singlet at δ ~3.4 ppm (6H total)
• No OH signal

This doesn't match our data ✗

Answer: The compound is ethanol (CH₃CH₂OH)

(a) Calculate theoretical yield:

Step 1: Balanced equation

$\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}$

Mole ratio: 1 mol C₂H₄ : 1 mol C₂H₅OH

Step 2: Calculate moles of ethene

$n(\text{C}_2\text{H}_4) = \frac{m}{M} = \frac{280{,}000 \text{ g}}{28 \text{ g/mol}} = 10{,}000 \text{ mol}$

(Note: Convert kg to g: 280 kg = 280,000 g)

Step 3: Calculate moles of ethanol produced

From stoichiometry: 1 mol C₂H₄ produces 1 mol C₂H₅OH

$n(\text{C}_2\text{H}_5\text{OH}) = 10{,}000 \text{ mol}$

Step 4: Calculate mass of ethanol

$m(\text{C}_2\text{H}_5\text{OH}) = n \times M = 10{,}000 \times 46 = 460{,}000 \text{ g}$

$= 460 \text{ kg}$

Theoretical yield = 460 kg

(b) Calculate percentage yield:

Given:

• Actual yield = 345 kg
• Theoretical yield = 460 kg

Calculation:

$\text{Percentage yield} = \frac{345}{460} \times 100\% = 75.0\%$

Answer: 75%

(c) Reasons for lower actual yield:

Reason 1: Incomplete reaction (equilibrium)

The hydration of ethene is a reversible reaction that reaches equilibrium:

$\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightleftharpoons \text{C}_2\text{H}_5\text{OH}$

At equilibrium, not all ethene is converted to ethanol. Some ethene remains unreacted, and some ethanol decomposes back to ethene and water. The reaction doesn't go 100% to completion, limiting the maximum possible yield.

Reason 2: Side reactions producing unwanted products

Under the reaction conditions (high temperature, acidic catalyst), ethene can undergo competing reactions:

• Polymerization: Multiple ethene molecules join to form polyethylene (waste product)
• Diethyl ether formation: 2C₂H₅OH → (C₂H₅)₂O + H₂O (consumes product)
• Formation of different alcohols or hydrocarbons

These side reactions consume ethene without producing the desired ethanol product, reducing the overall yield.

Additional valid reasons:

• Product loss during separation: Ethanol must be separated from unreacted ethene, water, and by-products. Some ethanol is lost during distillation, filtration, and purification steps.
• Impure starting materials: The ethene feedstock may contain impurities that don't react, effectively reducing the amount of pure ethene available.
• Catalyst deactivation: The acid catalyst (H₂SO₄ or H₃PO₄) may become less effective over time due to contamination or poisoning, slowing the reaction and reducing conversion.

In industrial practice, a 75% yield is acceptable for this process. Companies optimize conditions (temperature, pressure, catalyst concentration) to maximize yield while balancing production costs.

(a) Calculate atom economy:

$\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$

Identify desired product: CaO (calcium oxide)

Calculate M_r of desired product:

M_r(CaO) = 56 g/mol

From equation: 1 mol produced = 56 g

Calculate M_r of reactants:

M_r(CaCO₃) = 100 g/mol

From equation: 1 mol used = 100 g

Calculate atom economy:

$\text{Atom Economy} = \frac{56}{100} \times 100\% = 56\%$

Answer: 56% atom economy

This means 44% of the reactant mass (the CO₂) is wasted. For every 100 g of CaCO₃ used, only 56 g ends up in the useful product.

(b) Difference between yield and atom economy:

Atom economy (56%):

This is a THEORETICAL value based purely on the balanced chemical equation

It tells us that even with perfect conditions and 100% completion, only 56% of the reactant mass can possibly end up as CaO

The other 44% MUST become CO₂ waste - this is inherent to the reaction chemistry

Atom economy depends on the reaction pathway chosen, not on experimental technique

Question answered: "Is this reaction inherently wasteful?"

Percentage yield (85%):

This is a PRACTICAL value that measures experimental success

It tells us that the student obtained 85% of the maximum possible CaO (based on stoichiometry)

The "missing" 15% could be due to:

• Incomplete decomposition of CaCO₃
• Loss of CaO during transfer or handling
• Impurities in starting material
• CaO reacting with water or CO₂ in air

Question answered: "Did the experiment work well?"

Key difference:

• Atom economy = maximum possible product as % of reactants (reaction design)

• Yield = actual product as % of maximum possible (experimental execution)

You could have 100% yield but still poor atom economy if the reaction produces lots of waste!

(c) How to improve atom economy:

The problem: Currently CO₂ is a waste product with no value

Solution 1: Find a use for the CO₂ byproduct

If CO₂ produced is captured and used for another purpose, it's no longer "waste":

• Use CO₂ in carbonated drinks production
• Use CO₂ in fire extinguishers
• Use CO₂ in greenhouses to enhance plant growth
• Sell CO₂ to other industries

This doesn't change the atom economy calculation, but it makes the process more efficient overall by utilizing what would otherwise be waste.

Solution 2: Choose a different reaction pathway

Use a reaction that doesn't produce CO₂ as a byproduct

Example: If another synthesis route to CaO exists that uses different reactants with higher atom economy

However, for CaCO₃ decomposition specifically, this is the only practical route

Solution 3: Design reaction to make both products useful

In this case, if BOTH CaO and CO₂ were desired products, the atom economy would effectively be 100%

Example: A facility that needs both quicklime (CaO) for cement AND CO₂ for carbonation

Practical note:

For this specific reaction (limestone decomposition), the 56% atom economy is accepted in industry because:

• CaCO₃ (limestone) is very cheap and abundant
• There's no practical alternative route to CaO
• The CO₂ can often be captured and used
• CaO is essential for cement, steel, and other industries

Analysis using Green Chemistry Principles:

Method X evaluation:

Negative aspects:

• Benzene solvent (Principle #3 - Less Hazardous, #4 - Safer Solvents): Benzene is highly toxic and carcinogenic. It poses serious health risks to workers and creates hazardous waste that must be carefully disposed of at significant cost. Using benzene violates the principle of using safer chemicals.
• Low yield - 30% (Principle #1 - Waste Prevention): Only 30% of starting materials end up as product, meaning 70% becomes waste. This is extremely inefficient and wasteful of resources.
• 5-step synthesis (Principle #1, #2 - Atom Economy): Multiple steps means:
  • More reagents consumed at each step
  • More intermediate purifications needed
  • More waste generated overall
  • Lower overall atom economy
  • More opportunities for loss of product

Method Y evaluation:

Positive aspects:

• Ethanol solvent (Principle #4 - Safer Solvents): Ethanol is much safer than benzene - it's less toxic, renewable (can be produced from fermentation), and biodegradable. It can be recovered and reused. This is a major improvement.
• Higher yield - 65% (Principle #1 - Waste Prevention): More than double the yield of Method X means much more efficient use of starting materials and significantly less waste.
• Fewer steps - 2 instead of 5 (Principle #1, #2): Simpler synthesis means:
  • Less waste overall
  • Better atom economy
  • Simpler purification
  • Lower costs (fewer reagents, less time)
• Uses catalyst (Principle #7 - Catalysis): Platinum catalyst can speed up reactions, potentially allowing milder conditions (lower temperature), and the catalyst can be recovered and reused, reducing long-term costs.

Potential concerns:

• Platinum catalyst is expensive: However, it can be recovered and reused many times, so the cost per batch decreases over time
• Initial investment: May require new equipment, but this is offset by long-term savings from higher yield and safer solvents

Recommendation:

Method Y is STRONGLY preferable from a green chemistry perspective.

Justification:

1. Safety and health: Eliminating benzene dramatically improves worker safety and reduces environmental harm. This alone is a compelling reason to choose Method Y. The long-term health costs and liability associated with benzene use far outweigh any short-term cost savings.

2. Waste reduction: Method Y's 65% yield vs 30% means that for every kg of product made:

• Method X: Wastes ~2.3 kg of starting materials
• Method Y: Wastes ~0.5 kg of starting materials

This is a 4.6× reduction in waste, which translates to huge savings in raw material costs and waste disposal costs.

3. Efficiency: With only 2 steps vs 5, Method Y requires less time, labor, equipment, and purification steps. This improves both economic and environmental efficiency.

4. Catalyst benefits: Although platinum is expensive, catalysts are reusable. A single batch of catalyst can be used for thousands of reactions, making the per-unit cost negligible. Additionally, catalysts often allow reactions to proceed under milder conditions (lower temperature/pressure), saving energy.

5. Long-term sustainability: Method Y aligns with modern pharmaceutical industry trends toward greener processes. Many countries are tightening regulations on hazardous solvents like benzene, so Method Y is more future-proof.

Conclusion:

Method Y is superior in nearly every green chemistry metric: safer solvents, higher yield, fewer steps, lower waste, and uses catalysis. While the platinum catalyst represents an initial investment, the overall environmental and economic benefits strongly favor Method Y. A company choosing Method X would face higher waste disposal costs, regulatory compliance costs, health risks, and inefficient resource use.

(a) Why high pressure increases rate and yield:

Effect on YIELD (equilibrium position):

The balanced equation is:

$\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$

Counting moles of gas:

• Left side (reactants): 1 + 3 = 4 moles of gas
• Right side (products): 2 moles of gas
• Change: 4 moles → 2 moles (decrease in gas moles)

Applying Le Chatelier's Principle:

When pressure is increased, the system shifts to reduce the pressure.

The system can reduce pressure by shifting toward the side with fewer gas moles.

Equilibrium shifts RIGHT (toward NH₃, which has only 2 moles of gas).

Result: More ammonia is produced at equilibrium. The equilibrium yield increases with pressure.

Effect on RATE (kinetics):

Higher pressure means gas molecules are compressed into a smaller volume.

This increases the concentration of N₂ and H₂ molecules.

More molecules per unit volume → more frequent collisions between reactant molecules.

More frequent collisions → faster reaction rate.

Result: The rate at which ammonia is produced increases. The system reaches equilibrium faster.

Summary for (a):

High pressure shifts equilibrium toward NH₃ (fewer moles of gas), increasing yield. High pressure also increases particle concentration, increasing collision frequency and rate. Therefore, high pressure improves BOTH yield and rate - a rare win-win in chemical equilibria!

(b) Why extremely high pressures aren't used:

Although higher pressure would theoretically give even better yields and rates, pressures above 250 atm are not used industrially for several practical and economic reasons:

1. Equipment costs:

Very high pressure requires extremely thick-walled steel vessels to prevent rupture.

The cost of equipment increases dramatically with pressure rating.

At 1000 atm, vessel walls would need to be so thick that construction costs become prohibitive.

2. Energy costs:

Compressing gases to extremely high pressures requires large amounts of energy.

Energy costs increase with pressure.

At 1000 atm, the energy cost would outweigh the benefit from slightly higher yield.

3. Safety concerns:

Higher pressure = higher risk of catastrophic equipment failure (explosion).

A breach at 1000 atm could cause a major industrial accident.

Safety systems and protocols become more complex and expensive.

4. Diminishing returns:

The yield improvement becomes smaller as pressure increases further.

Going from 50 atm to 200 atm gives significant yield improvement.

Going from 200 atm to 1000 atm gives only marginal additional improvement.

The extra yield doesn't justify the extra cost.

Economic optimum: 200-250 atm

This pressure range balances yield benefits against equipment and energy costs.

Further increases don't provide enough additional benefit to justify the costs and risks.

(c) Role of iron catalyst:

What the catalyst DOES:

The iron catalyst provides an alternative reaction pathway with LOWER activation energy.

This means more collisions have sufficient energy to react.

The catalyst speeds up BOTH the forward reaction (N₂ + H₂ → NH₃) AND the reverse reaction (NH₃ → N₂ + H₂) EQUALLY.

Result:

• The system reaches equilibrium much faster

• The rate of ammonia production increases dramatically

• Production becomes economically viable at the compromise temperature

What the catalyst DOES NOT do:

The catalyst does NOT change the equilibrium position (yield).

Why the catalyst doesn't affect equilibrium yield:

A catalyst lowers activation energy for both forward and reverse reactions by the same amount.

This means the rate of the forward reaction increases, BUT the rate of the reverse reaction increases by exactly the same factor.

At equilibrium, rate_forward = rate_reverse. The catalyst speeds up both rates equally, so they still equal each other - just at a faster pace.

The equilibrium position depends on thermodynamics (ΔG, temperature, pressure), not kinetics.

The catalyst only affects kinetics (how fast equilibrium is reached), not thermodynamics (where equilibrium lies).

Analogy:

Imagine two people walking toward each other until they meet (equilibrium). A catalyst is like a faster walking speed - they meet at the same spot (same equilibrium position), just sooner (faster rate).

Why the catalyst is essential:

Without the iron catalyst, the reaction at 450°C would be too slow to be economically viable.

The catalyst allows us to use the compromise temperature and still achieve commercially acceptable production rates.

Even though it doesn't change the yield, it makes the process practical by allowing equilibrium to be reached quickly.