Problem 94
Question
A mixture of \(\mathrm{PCl}_{5}(12.41 \mathrm{g})\) and excess \(\mathrm{NH}_{4} \mathrm{Cl}\) was heated at \(145^{\circ} \mathrm{C}\) for 6 hours. The two reacted in equimolar amounts and evolved \(5.14 \mathrm{L}\) of \(\mathrm{HCl}\) (at STP). Three substances \((A, B, \text { and } C)\) were isolated from the reaction mixture. The three substances had the same elemental composition but differed in their molar mass. Substance A had a molar mass of \(347.7 \mathrm{g} / \mathrm{mol}\) and \(\mathrm{B}\) had a molar mass of \(463.5 \mathrm{g} / \mathrm{mol}\). Give the empirical and molecular formulas for \(\mathrm{A}\) and \(\mathrm{B}\) and draw a reasonable Lewis structure for A.
Step-by-Step Solution
Verified Answer
A has empirical formula C, and molecular formula C. Same for B. Lewis for A aligns standard bonded order showing Pi logic observed explained relationships.
1Step 1: Determine Moles of Reactants and Products
First, calculate the moles of \( \mathrm{PCl}_{5} \) and \( \mathrm{HCl} \) produced. The molar mass of \( \mathrm{PCl}_{5} \) is \( 208.22 \; \mathrm{g/mol} \).\[ \text{Moles of } \mathrm{PCl}_{5} = \frac{12.41 \; \mathrm{g}}{208.22 \; \mathrm{g/mol}} = 0.0596 \; \mathrm{mol} \]Next, use the volume of \( \mathrm{HCl} \) to find its moles, given at standard temperature and pressure (STP) where 1 mole = 22.4 L:\[ \text{Moles of } \mathrm{HCl} = \frac{5.14 \; \mathrm{L}}{22.4 \; \mathrm{L/mol}} = 0.2292 \; \mathrm{mol} \]
2Step 2: Analyze Reaction and Molar Relationship
Since the problem states \( \mathrm{PCl}_{5} \) and \( \mathrm{NH}_{4} \mathrm{Cl} \) react in equimolar amounts, both are in limiting equal moles, which means the evolved moles of \( \mathrm{HCl} \) should directly correlate to the amount of \( \mathrm{PCl}_{5} \) used, and thus \( \mathrm{Cl}_{2} \) and any other products will be smaller components of the mole equation, verifying a major reaction is:\[ \mathrm{PCl}_{5} + \mathrm{NH}_{4} \mathrm{Cl} \rightarrow \mathrm{A} + \mathrm{B} + \mathrm{HCl} \]
3Step 3: Confirm Empirical Formula of A and B
Given that all substances have the same elemental composition, determine a common elemental base. Considerations: molar mass of A is \( 347.7 \; \mathrm{g/mol} \), B is \( 463.5 \; \mathrm{g/mol} \). Assume, then, they share a root \( x \), solve:Assume empirical formula based on Minimal compositional factors being CHN same, evolve formulas and multiply by carbon equivalent for C, derived empirical mass bases conforming mass confirms:- Find Factor impressing the relationship extended HCl removed Cl / P from assumptions- **Empirical Formula** derived compatibility for these assumptions.
4Step 4: Determine Molecular Formula for A and B
Use molar mass:
- For A: Trial and Error, until chemically plausible variation of potentially Pi based compatible assumptions / match those with rounding approaches.
- For B: Verify through applied logical divisibility of calculated relative to found molecular assumptions but relationship noted in mole step mapped,
5Step 5: Draw Lewis Structure for A
Starting with elements (root calcs): Place footing Pi below minimal constitution based: using know where protons electron exchange happens:
- Label H20 acting influence Delving necessary C inclusive resolved in heuristic structure based.
- Properly Pair Electrons to Centers bonding / stable shelled form.
Key Concepts
Empirical FormulaMolecular FormulaLewis Structure
Empirical Formula
An empirical formula represents the simplest whole-number ratio of elements within a molecule. It helps to determine the fundamental unit of a compound, focusing more on the proportion of each element rather than the exact number present in complex molecules.
For a reaction involving multiple unknown substances like our substances A, B, and C, they have identical elemental compositions, making the search for their empirical formulas critical. When we identify the empirical formula, we simplify this composition to its basic form.
For a reaction involving multiple unknown substances like our substances A, B, and C, they have identical elemental compositions, making the search for their empirical formulas critical. When we identify the empirical formula, we simplify this composition to its basic form.
- Start by determining the percentages of different elements present initially.
- Sleep these components into the smallest whole-number ratio.
Molecular Formula
The molecular formula provides specific insight into the exact number of each type of atom in a molecule. It not only maintains the ratio established by the empirical formula but increases in scale to reflect the true compound contained within a sample. This amplifies the molecular footprint established by the empirical formula.
For substances A and B:
For substances A and B:
- Use the molar masses provided (A = 347.7 g/mol and B = 463.5 g/mol) to deduce formulas.
- Understand that the molecular formula is often a whole number multiple of the empirical formula.
Lewis Structure
A Lewis structure visually represents the bonds formed between atoms and any lone pairs of electrons, based upon octet rule adjustments, shared electrons, and geometry. It’s an essential tool in visualizing how a molecule’s electrons are configured to provide firing insight into chemical reactivity and stability.
When determining a reasonable Lewis structure for substance A, one must:
When determining a reasonable Lewis structure for substance A, one must:
- List atoms considering the possible empirical formula and arrange them to satisfy octet stability.
- Represent shared electrons (bonds) and unshared ones (lone pairs).
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