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On combustion, 1.0 L of a gaseous compound of hydrogen, carbon, and nitrogen gives \(2.0 \mathrm{L}\) of \(\mathrm{CO}_{2}, 3.5 \mathrm{L}\) of \(\mathrm{H}_{2} \mathrm{O}\) vapor, and \(0.50 \mathrm{L}\) of \(\mathrm{N}_{2}\) at STP. What is the empirical formula of the compound?

Short Answer

Expert verified
The empirical formula is C\(_2\)H\(_7\)N.

Step by step solution

01

Determine Mole Ratios

At STP, the volume of the gas is directly proportional to the number of moles. Therefore, we can use the given volumes to find the mole ratio of the products. We have 2.0 L of CO\(_2\), 3.5 L of H\(_2\)O, and 0.50 L of N\(_2\). This translates to 2 moles of CO\(_2\), 3.5 moles of H\(_2\)O, and 0.5 moles of N\(_2\).
02

Determine Mole Ratios of Elements in the Compound

The moles of CO\(_2\) correspond to the moles of carbon, so we have 2 moles of carbon. From 3.5 moles of H\(_2\)O, we get 2 \(\times\) 3.5 = 7 moles of hydrogen. The 0.5 moles of N\(_2\) correspond to 2 \(\times\) 0.5 = 1 mole of nitrogen.
03

Find Empirical Formula

The number of moles gives us the chemical compound's formula in terms of the simplest whole number ratio. For carbon, hydrogen, and nitrogen respectively, we have 2 : 7 : 1. Thus, the empirical formula of the compound is C\(_2\)H\(_7\)N.

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Key Concepts

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

Combustion Analysis
Combustion analysis is a laboratory technique used to determine the empirical formula of an unknown compound, specifically those containing elements like carbon, hydrogen, and sometimes nitrogen. When a compound undergoes combustion, it reacts with oxygen to produce common gases such as carbon dioxide, water vapor, and nitrogen gas. By measuring the volumes of these gases produced at standard temperature and pressure (STP), we can derive useful information about the original compound.

For instance, in the given exercise, 1.0 liter of a gaseous compound burns to produce specific volumes of combustion gases: 2.0 liters of carbon dioxide (CO\(_2\)), 3.5 liters of water vapor (H\(_2\)O), and 0.50 liters of nitrogen gas (N\(_2\)). These results act as clues that tell us about the relative amounts of atoms within the initial compound. By calculating the mole ratios from these volumes, we can deduce the empirical formula of the compound, which reflects the simplest whole-number ratio of the elements within it.
Mole Ratios
Mole ratios are fundamental for understanding how many moles of reactants and products are involved in a chemical reaction. These ratios arise directly because of Avogadro's law, which states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

In the context of the exercise, we ascertain the mole ratios by looking at the volumes of the gases produced. The volume of carbon dioxide helps identify how much carbon is present, while the amount of water vapor gives us information about the hydrogen content. Specifically, 2.0 liters of CO\(_2\) suggest 2 moles of carbon, 3.5 liters of H\(_2\)O suggest 7 moles of hydrogen, and 0.5 liters of N\(_2\) indicate 1 mole of nitrogen. With this information, the overall mole ratios of carbon, hydrogen, and nitrogen are determined. These ratios are crucial for calculating the empirical formula, representing the simplest ratio of elements in the compound.
Chemical Compounds
Chemical compounds are substances formed from two or more elements chemically combined in fixed proportions. The empirical formula of a compound provides a simple way to express the proportion of atoms of each involved element, serving as the most basic representation of a compound's composition.

In real-world applications like the given exercise, an understanding of chemical compounds is essential. When a compound undergoes combustion, its elemental composition dictates the particular gases produced. Deciphering the elemental structure from these products requires translating observed data into an empirical formula. For example, from the combustion products provided in the exercise, we derived that the simplest integer ratio of atoms present within the compound is C\(_2\)H\(_7\)N. This represents the empirical formula, indicating two carbon atoms, seven hydrogen atoms, and one nitrogen atom in the compound.
Stoichiometry
Stoichiometry involves the calculation of the reactants and products in chemical reactions. It is a quantitative relationship derived from the balanced chemical equation, revealing the proportions of elements and compounds involved.

For the exercise, stoichiometry is applied by analyzing the volumes of combustion products and converting them into moles. The given volumes at STP directly correspond to mole quantities because of Avogadro's principle. Once we calculate the number of moles of each product, stoichiometry enables us to backtrack and determine the empirical formula of the initial compound. We balance carbon, hydrogen, and nitrogen from the data provided.

By following this approach, we convert numbers into meaningful chemical insights. The ultimate goal is to simplify and decode the stoichiometric information to present the empirical formula, a practical application of stoichiometry in chemistry.

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Most popular questions from this chapter

A sample of nitrogen gas has a pressure of \(67.5 \mathrm{mm} \mathrm{Hg}\) in a \(500 .\) -mL flask. What is the pressure of this gas sample when it is transferred to a \(125-\mathrm{mL}\) flask at the same temperature?

You are given a solid mixture of \(\mathrm{NaNO}_{2}\) and \(\mathrm{NaCl}\) and are asked to analyze it for the amount of NaNO_ present. To do so, you allow the mixture to react with sulfamic acid, \(\mathrm{HSO}_{3} \mathrm{NH}_{2}\), in water according to the equation \(\begin{aligned} \operatorname{NaNO}_{2}(\mathrm{aq})+\mathrm{HSO}_{3} \mathrm{NH}_{2}(\mathrm{aq}) \rightarrow & \\ \mathrm{NaHSO}_{4}(\mathrm{aq}) &+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{N}_{2}(\mathrm{g}) \end{aligned}\) What is the weight percentage of \(\mathrm{NaNO}_{2}\) in \(1.232 \mathrm{g}\) of the solid mixture if reaction with sulfamic acid produces \(295 \mathrm{mL}\) of dry \(\mathrm{N}_{2}\) gas with a pressure of \(713 \mathrm{mm}\) Hg at \(21.0^{\circ} \mathrm{C} ?\)

A miniature volcano can be made in the laboratory with ammonium dichromate. When ignited, it decomposes in a fiery display. $$ \left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{s}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s}) $$ If 0.95 g of ammonium dichromate is used and the gases from this reaction are trapped in a 15.0 -L flask at \(23^{\circ} \mathrm{C},\) what is the total pressure of the gas in the flask? What are the partial pressures of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} ?\)

\(\mathrm{Ni}(\mathrm{CO})_{4}\) can be made by reacting finely divided nickel with gaseous CO. If you have \(\mathrm{CO}\) in a 1.50-L flask at a pressure of \(418 \mathrm{mm}\) Hg at \(25.0^{\circ} \mathrm{C},\) along with \(0.450 \mathrm{g}\) of \(\mathrm{Ni}\) powder, what is the theoretical yield of \(\mathrm{Ni}(\mathrm{CO})_{4} ?\)

A halothane-oxygen mixture \(\left(\mathrm{C}_{2} \mathrm{HBrClF}_{3}+\mathrm{O}_{2}\right)\) can be used as an anesthetic. A tank containing such a mixture has the following partial pressures: \(P\) (halothane) \(=170 \mathrm{mm} \mathrm{Hg}\) and \(P\left(\mathrm{O}_{2}\right)=570 \mathrm{mm} \mathrm{Hg}\) (a) What is the ratio of the number of moles of halothane to the number of moles of \(\mathrm{O}_{2} ?\) (b) If the tank contains \(160 \mathrm{g}\) of \(\mathrm{O}_{2}\), what mass of \(\mathrm{C}_{2} \mathrm{HBrClF}_{3}\) is present?

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