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Chapter 5: Problem 18

Which of the following pairs of compounds have the same empirical formula? a. acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2},\) and benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\) b. ethane, \(\mathrm{C}_{2} \mathrm{H}_{6},\) and butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\) c. nitrogen dioxide, \(\mathrm{NO}_{2}\), and dinitrogen tetroxide, \(\mathrm{N}_{2} \mathrm{O}_{4}\) d. diphenyl ether, \(\mathrm{C}_{12} \mathrm{H}_{10} \mathrm{O},\) and phenol, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\)

Short Answer

Expert verified
The pair of compounds that have the same empirical formula is (c) nitrogen dioxide, \(\mathrm{NO}_{2}\), and dinitrogen tetroxide, \(\mathrm{N}_{2} \mathrm{O}_{4}\).

Step by step solution

01

Find the Empirical Formula for Each Compound

First, we will find the empirical formula for each of the compounds in the given pairs. a. Acetylene: \(\mathrm{C}_{2} \mathrm{H}_{2}\). There are no common divisors for 2 and 2, so the empirical formula remains \(\mathrm{C}_{2} \mathrm{H}_{2}\). Benzene: \(\mathrm{C}_{6} \mathrm{H}_{6}\). Dividing both subscripts by 6, we have the empirical formula \(\mathrm{CH}\). b. Ethane: \(\mathrm{C}_{2} \mathrm{H}_{6}\). Dividing both subscripts by 2, we have the empirical formula \(\mathrm{CH}_{3}\). Butane: \(\mathrm{C}_{4} \mathrm{H}_{10}\). Dividing both subscripts by 2, we have the empirical formula \(\mathrm{C}_{2} \mathrm{H}_{5}\). c. Nitrogen dioxide: \(\mathrm{NO}_{2}\). There are no common divisors for 1 and 2, so the empirical formula remains \(\mathrm{NO}_{2}\). Dinitrogen tetroxide: \(\mathrm{N}_{2} \mathrm{O}_{4}\). Dividing both subscripts by 2, we have the empirical formula \(\mathrm{NO}_{2}\). d. Diphenyl ether: \(\mathrm{C}_{12} \mathrm{H}_{10} \mathrm{O}\). There are no common divisors for 12, 10, and 1, so the empirical formula remains \(\mathrm{C}_{12} \mathrm{H}_{10} \mathrm{O}\). Phenol: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\) (the molecular formula can also be written as \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}\)). There are no common divisors for 6, 6, and 1, so the empirical formula remains \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}\).
02

Compare the Empirical Formulas from Step 1

Now we will compare the empirical formulas we found in step 1 for each pair of compounds. a. Acetylene: \(\mathrm{C}_{2} \mathrm{H}_{2}\) and benzene: \(\mathrm{CH}\). Different empirical formulas, so they do not have the same empirical formula. b. Ethane: \(\mathrm{CH}_{3}\) and butane: \(\mathrm{C}_{2} \mathrm{H}_{5}\). Different empirical formulas, so they do not have the same empirical formula. c. Nitrogen dioxide: \(\mathrm{NO}_{2}\) and dinitrogen tetroxide: \(\mathrm{NO}_{2}\). The same empirical formulas, so they have the same empirical formula. d. Diphenyl ether: \(\mathrm{C}_{12} \mathrm{H}_{10} \mathrm{O}\) and phenol: \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}\). Different empirical formulas, so they do not have the same empirical formula. Therefore, only pair (c) nitrogen dioxide and dinitrogen tetroxide have the same empirical formula.

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Key Concepts

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

Understanding Chemical Composition
When we talk about chemical composition, we're referring to the types and amounts of elements in a compound.

The simplest ratio of these elements is known as the empirical formula, which gives the lowest whole number ratio of elements in a compound.

For example, acetylene has a molecular formula of \( C_2H_2 \), but since there are no common factors to reduce the subscripts further, its empirical formula is the same as its molecular formula. This contrasts with benzene, which has a molecular formula of \( C_6H_6 \) but an empirical formula of \( CH \), reflecting the 1:1 ratio after dividing each subscript by 6.

Understanding the chemical composition is vital in recognizing how similar or different two compounds are. Even if compounds share the same elements, differing ratios imply distinct properties and behaviors.
Comparing Molecular and Empirical Formulas
Molecules can have identical or different molecular formulas and empirical formulas.

When comparing molecular formulas, we look at the exact number of each atom in a molecule; these can vary greatly among compounds. The molecular formula is crucial in understanding the molecular structure and properties of a compound.

Empirical vs. Molecular Formula

On the other hand, the empirical formula tells us the simplest whole-number ratio of atoms. It doesn’t necessarily provide information about the molecule’s actual size or structure, only the proportional relation of elements present.

For instance, nitrogen dioxide (\( NO_2 \) is its own empirical formula), and dinitrogen tetroxide (\( N2O4 \) that simplifies to the empirical formula \( NO_2 \) when we divide by 2) illustrate how compounds can differ molecularly but be identical empirically, with significant implications in their physical and chemical properties.
Stoichiometry: The Heart of Chemical Reactions
At its core, stoichiometry is the study of the quantitative relationships or ratios between substances as they participate in chemical reactions.

It is grounded in the law of conservation of mass where the total mass of reactants equals the total mass of products in a chemical reaction. Understanding ratios of substances allows chemists to predict the amounts of products that will form in a reaction or the amount of reactants needed to create a desired amount of product.

Stoichiometric Calculations

For stoichiometric calculations, the empirical formula can serve as a foundation for calculating the relative weights of reactant and product molecules.

Using the empirical formula aids in simplifying complex reactions into more manageable equations that correspond to the actual amounts of substances involved. This knowledge is indispensable in both the laboratory and industrial applications where precise measurements are crucial for successful chemical synthesis.

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Most popular questions from this chapter

Ammonia reacts with \(\mathrm{O}_{2}\) to form either \(\mathrm{NO}(g)\) or \(\mathrm{NO}_{2}(g)\) according to these unbalanced equations: $$\begin{array}{l}\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\\\\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\end{array}$$ In a certain experiment 2.00 moles of \(\mathrm{NH}_{3}(g)\) and 10.00 moles of \(\mathbf{O}_{2}(g)\) are contained in a closed flask. After the reaction is complete, 6.75 moles of \(\mathbf{O}_{2}(g)\) remains. Calculate the number of moles of \(\mathrm{NO}(g)\) in the product mixture: (Hint: You cannot do this problem by adding the balanced equations because you cannot assume that the two reactions will occur with equal probability.)

Gallium arsenide, GaAs, has gained widespread use in semiconductor devices that convert light and electrical signals in fiber-optic communications systems. Gallium consists of \(60 . \%^{69} \mathrm{Ga}\) and \(40 . \%^{71} \mathrm{Ga}\). Arsenic has only one naturally occurring isotope, \(^{75}\)As. Gallium arsenide is a polymeric material, but its mass spectrum shows fragments with the formulas GaAs and \(\mathrm{Ga}_{2} \mathrm{As}_{2}\). What would the distribution of peaks look like for these two fragments?

Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane: $$2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g)+2 \mathrm{CH}_{4}(g) \longrightarrow 2 \mathrm{HCN}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ If \(5.00 \times 10^{3} \mathrm{kg}\) each of \(\mathrm{NH}_{3}, \mathrm{O}_{2},\) and \(\mathrm{CH}_{4}\) are reacted, what mass of HCN and of \(\mathrm{H}_{2} \mathrm{O}\) will be produced, assuming \(100 \%\) yield?

A common demonstration in chemistry courses involves adding a tiny speck of manganese(IV) oxide to a concentrated hydrogen peroxide \(\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)\) solution. Hydrogen peroxide decomposes quite spectacularly under these conditions to produce oxygen gas and steam (water vapor). Manganese(IV) oxide is a catalyst for the decomposition of hydrogen peroxide and is not consumed in the reaction. Write the balanced equation for the decomposition reaction of hydrogen peroxide.

Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticizers. It contains only \(\mathrm{C}, \mathrm{H}\) and O. Combustion of \(19.81 \mathrm{mg}\) terephthalic acid produces \(41.98 \mathrm{mg} \mathrm{CO}_{2}\) and \(6.45 \mathrm{mg} \mathrm{H}_{2} \mathrm{O} .\) If 0.250 mole of terephthalic acid has a mass of \(41.5 \mathrm{g},\) determine the molecular formula for terephthalic acid.
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