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Chapter 1: Problem 33

Empirical formula of a compound is \(\mathrm{A}_{2} \mathrm{~B}_{4} .\) If its empirical formula weight is half of its vapour density, determine the molecular formula of the compound. (a) \(\mathrm{A}_{4} \mathrm{~B}_{8}\) (b) \(\mathrm{A}_{8} \mathrm{~B}_{16}\) (c) \(\mathrm{A}_{2} \mathrm{~B}_{4}\) (d) \(\mathrm{A}_{3} \mathrm{~B}_{6}\)

Short Answer

Expert verified
Answer: The molecular formula of the compound is \(\mathrm{A}_{8} \mathrm{~B}_{16}\).

Step by step solution

01

Determine the relationship between the empirical formula and vapor density

The given information states that the empirical formula weight of the compound is half of its vapor density. This can be represented as: Empirical Formula Weight = 1/2 * Vapor Density Remember that molecular weight is twice the vapor density, so we also have: Molecular Weight = 2 * Vapor Density
02

Express the ratio between empirical formula weight and molecular weight

Divide the empirical formula weight by the molecular weight: Empirical Formula Weight / Molecular Weight = (1/2 * Vapor Density) / (2 * Vapor Density) = 1/4
03

Multiply the empirical formula by the ratio to find the molecular formula

Given the ratio from step 2, we multiply the empirical formula \(\mathrm{A}_{2} \mathrm{~B}_{4}\) by the ratio 4 (the inverse of 1/4): Molecular Formula = 4 * \(\mathrm{A}_{2} \mathrm{~B}_{4}\) = \(\mathrm{A}_{8} \mathrm{~B}_{16}\) So, the molecular formula of the compound is \(\mathrm{A}_{8} \mathrm{~B}_{16}\) which corresponds to the given option (b).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Empirical Formula
The empirical formula of a compound provides the simplest whole-number ratio of the elements within the compound. For instance, the empirical formula of water is H2O, which indicates there are two hydrogen atoms for every one oxygen atom.

To determine the empirical formula from the elemental composition, one would divide the moles of each element by the smallest number of moles to achieve the simplest whole-number ratio. This process becomes crucial when trying to decipher the true molecular structure of a substance from experimental data.
The Role of Vapor Density in Molecular Formula Determination
Vapor density is a measure that compares the weight of a certain volume of a substance's vapor to the weight of an equal volume of hydrogen at the same temperature and pressure. It is used as an indirect way to gauge the molecular weight of volatile substances.

As in the given textbook exercise, vapor density can provide valuable information to find the molecular weight or formula of a compound. The empirical formula weight, when given as half of vapor density, acts as a stepping stone to establish this relationship and further determine the molecular formula.
Molecular Weight Explained
Molecular weight, or molecular mass, is the sum of the atomic weights of the atoms in a molecule. It's usually measured in atomic mass units (amu) or Daltons (Da). Understanding molecular weight is critical for stoichiometry, wherein it assists with calculating the grams of reactants and products in a chemical reaction.

In the context of the exercise, the molecular weight is twice the vapor density, allowing us to use such a relation to compute the actual molecular formula. This mathematical connection simplifies what can often be a complex process of deducing chemical formulas.

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Most popular questions from this chapter

Match the entries given in Column A with appropriate ones in Column \(B\). $$ \begin{array}{|l|l|l|} \hline \begin{array}{l} \text { A. Empirical formula of } \\ \text { glucose } \end{array} & \text { () } & \begin{array}{l} \text { a. Less intermolecular } \\ \text { forces } \end{array} \\ \hline \begin{array}{l} \text { B. Percentage of carbon } \\ \text { in methane } \end{array} & \text { () } & \text { b. } 75 \% \\ \hline \text { C. Ideal gas } & \text { () } & \text { c. } 17.6 \% \\ \hline \text { D. High critical } & \text { () } & \begin{array}{l} \text { d. } \text { Large } \\ \text { intermolecular } \\ \text { forces of attraction } \end{array} \\ \hline \begin{array}{l} \text { E. Percentage of } \\ \text { hydrogen in } \\ \text { ammonia } \end{array} & \text { () } & \text { e. } \mathrm{CH}_{2} \mathrm{O} \\ \hline \begin{array}{l} \text { F. } \text { Empirical formula of } \\ \text { oxalic acid } \end{array} & \text { () } & \text { f. } \mathrm{CHO}_{2} \\ \hline \end{array} $$

State whether the following statements are true or false. \(1 \mathrm{~g}\) atom of nitrogen contains \(6.023 \times 10^{23}\) atoms of nitrogen.

Fill in the blanks. Pressure exerted by water vapour in moist gas is called

A certain mass of a gas occupies a volume of 600 \(\mathrm{ml}\) at a certain temperature and pressure. If the temperature is increased by \(80 \%\) what will be the volume occupied by the same mass of gas at the same pressure? (a) \(1080 \mathrm{ml}\) (b) \(108 \mathrm{ml}\) (c) \(120 \mathrm{ml}\) (d) \(102 \mathrm{ml}\)

The weight of an empty china dish is \(39 \mathrm{~g}\) and when a saturated solution of potassium nitrate is poured into it, its weight is \(108 \mathrm{~g}\) at \(50^{\circ} \mathrm{C}\). After evaporating the solution to dryness, if the weight of the dish along with the crystals is \(72 \mathrm{~g}\) then the solubility of potassium nitrate at \(50^{\circ} \mathrm{C}\) is (a) \(83.9\) (b) \(95.6\) (c) \(91.6\) (d) \(87.4\)
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