DAT General Chemistry: Reactant Purity and Sequential Reactions

DAT High-Yield Concept

Reactant Purity: Only the active (pure) mass of a reactant participates in stoichiometry. Always multiply total sample mass by percent purity (expressed as a decimal) before executing mole calculations.
Sequential Reactions: When reactions occur in consecutive steps, intermediate species connect the steps. Eliminate intermediates by adjusting stoichiometric coefficients to write an overall balanced equation.
Multi-Step Overall Yield: Overall percent yield across sequential steps is the product of individual fractional yields: $\text{Yield}_{\text{overall}} = y_1 \times y_2 \times \dots \times y_n$ .

1. Reactant Purity in Stoichiometric Calculations

In real-world laboratory and industrial settings, chemical reagents often contain non-reactive impurities. Stoichiometric mole ratios depend strictly on the active chemical species. Calculating product yield from an impure reactant requires converting total sample mass to active mass first.

Reactant Purity Formula:

\text{Mass of Pure Reactant} = \text{Mass of Impure Sample} \times \left( \frac{\text{Percent Purity}}{100\%} \right)

Worked Example: Reactant Purity Calculation

Problem: A $\pu{5.00 g}$ sample of impure limestone ( $\ce{CaCO3}$ ) reacts completely with excess hydrochloric acid to produce $\pu{0.0400 mol}$ of $\ce{CO2}$ gas. Calculate the percent purity of $\ce{CaCO3}$ in the sample. (Molar mass of $\ce{CaCO3} = \pu{100.09 g/mol}$ )

Reaction:

\ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)}

Dimensional Analysis:

1. Determine pure $\ce{CaCO3}$ mass required to form $\pu{0.0400 mol}$ of $\ce{CO2}$ :

\pu{0.0400 mol CO2} \times \frac{\pu{1 mol CaCO3}}{\pu{1 mol CO2}} \times \frac{\pu{100.09 g CaCO3}}{\pu{1 mol CaCO3}} = \pu{4.00 g CaCO3}

2. Calculate percent purity:

\text{Percent Purity} = \frac{\pu{4.00 g pure CaCO3}}{\pu{5.00 g sample}} \times 100\% = \pu{80.0 \%}

Common Exam Trap: Using the total sample mass ( $\pu{5.00 g}$ ) directly in stoichiometric mole ratios. Impurities do not form $\ce{CO2}$ ; only pure $\ce{CaCO3}$ reacts.

2. Sequential Reactions Carried Through Multiple Steps

Many chemical syntheses proceed through a series of consecutive steps where the product of one reaction becomes the reactant for the next step. These intermediate species are completely consumed in a perfect sequence.

Sequential Reaction Rule: To find the overall stoichiometric relationship between initial reactants and final products, multiply step equations by integers so that intermediate species cancel completely upon summation.

Worked Example: Two-Step Sequential Synthesis

Problem: Titanium metal is extracted from titanium(IV) oxide in a two-step process:

Step 1: $\ce{TiO2(s) + 2Cl2(g) + 2C(s) -> TiCl4(l) + 2CO(g)}$

Step 2: $\ce{TiCl4(l) + 2Mg(s) -> Ti(s) + 2MgCl2(s)}$

How many moles of $\ce{Ti(s)}$ are produced starting from $\pu{2.00 mol}$ of $\ce{TiO2(s)}$ with excess reagents?

Step-by-Step Solution:

1. Combine the sequential reactions by identifying the intermediate $\ce{TiCl4(l)}$ . Since $\pu{1 mol}$ of $\ce{TiCl4}$ is produced in Step 1 and consumed in Step 2, sum the equations directly:

\ce{TiO2(s) + 2Cl2(g) + 2C(s) + 2Mg(s) -> Ti(s) + 2CO(g) + 2MgCl2(s)}

2. Apply mole ratio from initial reactant $\ce{TiO2}$ to final product $\ce{Ti}$ :

\pu{2.00 mol TiO2} \times \frac{\pu{1 mol Ti}}{\pu{1 mol TiO2}} = \pu{2.00 mol Ti}

Common Exam Trap: Failing to equalize the coefficient of intermediate species when summing equations. If Step 2 consumed $\pu{2 mol}$ of intermediate for every $\pu{1 mol}$ formed in Step 1, Step 1 would need to be multiplied by 2.

3. Carrying Yield Through Consecutive Steps

Percent yield (the ratio of actual to theoretical yield) applies at each step of a multi-step sequence. Losses accumulate multiplicatively across steps, significantly reducing overall efficiency.

Overall Percent Yield Formula:

\text{Overall Fractional Yield} = y_1 \times y_2 \times \dots \times y_n

\text{Overall Percent Yield} = (y_1 \times y_2 \times \dots \times y_n) \times 100\%

Worked Example: Multi-Step Yield Calculation

Problem: A three-step organic synthesis has individual step yields of $\pu{80.0 \%}$ , $\pu{90.0 \%}$ , and $\pu{75.0 \%}$ . If the theoretical yield calculated from $\pu{10.0 mol}$ of starting material is $\pu{10.0 mol}$ of final product, what is the actual yield obtained?

Calculation:

1. Convert percentage yields to decimal fractions: $y_1 = 0.800$ , $y_2 = 0.900$ , $y_3 = 0.750$ .

2. Calculate overall fractional yield:

\text{Overall Yield} = 0.800 \times 0.900 \times 0.750 = 0.540 \quad (\pu{54.0 \%})

3. Determine actual final yield in moles:

\text{Actual Yield} = \pu{10.0 mol theoretical} \times 0.540 = \pu{5.40 mol}

Common Exam Trap: Arithmetic averaging of individual yields ( $\frac{80+90+75}{3} = 81.7\%$ ). Multi-step yields are multiplicative, never additive or averaged.

Comparison Table: Reaction Calculations

Calculation Type	Key Input Variable	Mathematical Operation	Common DAT Pitfall
Reactant Purity	Percent purity of sample	Multiply total mass by purity fraction	Using impure total mass as pure reactant mass
Sequential Reaction	Intermediate mole ratios	Cancel intermediates across balanced steps	Unbalanced intermediate coefficients between steps
Multi-Step Yield	Individual step yields ( $y_i$ )	Multiply fractional yields ( $y_1 \times y_2 \times \dots$ )	Averaging step percentages instead of multiplying

DAT Mnemonic for Multi-Step Yield: Remember "P-M-C":
Purity = Multiply sample mass first.
Consecutive yields = Compound (multiply) fractions!

Critical Thinking Questions

If an impure reagent contains $\pu{10 \%}$ non-reactive sand and $\pu{90 \%}$ active $\ce{Na2CO3}$ , how does this purity factor affect the stoichiometric volume of $\pu{1.0 M}$ $\ce{HCl}$ required to completely neutralize a $\pu{10.0 g}$ sample?
A four-step synthesis has an individual efficiency of $\pu{50.0 \%}$ per step. What is the overall percentage yield of the synthesis? Why are shorter synthetic routes preferred in pharmaceutical drug discovery?
In a two-step sequential process, if the intermediate produced in Step 1 partially decomposes before reacting in Step 2, how does this affect the stoichiometric calculation based on the Step 1 starting material?

References

Sources returned by Google Search grounding during lesson verification.

4.5: Other Practical Matters in Reaction Stoichiometry - Chemistry LibreTexts — chem.libretexts.org
Stoichiometry of Chemical Reactions — openstax.org

Reactant Purity and Sequential Reactions

Summary

Practice Questions

AI Tutor