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Chapter 3: Problem 94

A compound contains only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{N}\) . Combustion of 35.0 \(\mathrm{mg}\) of the compound produces 33.5 \(\mathrm{mg} \mathrm{CO}_{2}\) and 41.1 \(\mathrm{mg}\) \(\mathrm{H}_{2} \mathrm{O}\) . What is the empirical formula of the compound?

Short Answer

Expert verified
The empirical formula of the compound is CH鈧哊鈧�.

Step by step solution

01

Determine moles of C in the compound

First, we use the amount of CO鈧� produced to determine the moles of C in the compound. From the balanced combustion reaction, we know that for each mole of CO鈧� produced, there must be one mole of C in the original compound. The molecular weight of CO鈧� is 12.01 g/mol (C) + 2 * 16.00 g/mol (O) = 44.01 g/mol. Thus, we have: Moles of CO鈧� = Mass of CO鈧� / Molecular weight of CO鈧� Moles of CO鈧� = 33.5 mg / 44.01 g/mol Since 1g = 1000mg: Moles of CO鈧� = 33.5 * 10^(-3) g / 44.01 g/mol = 0.00076 mol Since the moles of CO鈧� are equal to the moles of C, the moles of C in the compound is 0.00076 mol.
02

Determine moles of H in the compound

Next, we use the amount of H鈧侽 produced to determine the moles of H in the compound. From the balanced combustion reaction, we know that for each mole of H鈧侽 produced, there must be two moles of H in the original compound. The molecular weight of H鈧侽 is 2 * 1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol. Thus, we have: Moles of H鈧侽 = Mass of H鈧侽 / Molecular weight of H鈧侽 Moles of H鈧侽 = 41.1 mg / 18.02 g/mol Since 1g = 1000mg: Moles of H鈧侽 = 41.1 * 10^(-3) g / 18.02 g/mol = 0.00228 mol And because 1 mole of H鈧侽 contains 2 moles of H, Moles of H = 2 * Moles of H鈧侽 = 2 * 0.00228 mol = 0.00456 mol
03

Determine mass and moles of N in the compound

To determine the mass of N in the compound, we'll use the initial mass of the compound and subtract the masses of C and H. Mass of N = Mass of compound - (Mass of C + Mass of H) Mass of C = Moles of C * Molecular weight of C = 0.00076 mol * 12.01 g/mol = 9.13 * 10^(-3) g Mass of H = Moles of H * Molecular weight of H = 0.00456 mol * 1.01 g/mol = 4.60 * 10^(-3) g Since the given mass of compound is 35.0mg or 0.035g, Mass of N = 0.035 g - (9.13 * 10^(-3) g + 4.60 * 10^(-3) g) = 0.02127 g Now, we can find the moles of N using its molecular weight (14.01 g/mol): Moles of N = Mass of N / Molecular weight of N = 0.02127 g / 14.01 g/mol = 0.00152 mol
04

Determine the empirical formula

Now that we have determined the moles of C, H, and N in the compound, we can find the simplest whole number ratio of these elements. First, we divide each element's moles by the smallest number of moles to get a preliminary ratio: Preliminary ratio of C : H : N = 0.00076 : 0.00456 : 0.00152 Divide by the smallest moles (C here): = 1 : 6 : 2 Thus, the empirical formula of the compound is CH鈧哊鈧�.

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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 critical method used in chemistry to determine the elemental composition of a compound. In this method, a known mass of a compound is completely burned in excess oxygen, and the products of combustion, usually carbon dioxide (\(\text{CO}_2\)), and water (\(\text{H}_2\text{O}\)), are analyzed. By measuring the masses of \(\text{CO}_2\) and \(\text{H}_2\text{O}\) produced, we can back-calculate to find the amounts of elements originally present in the compound.

During combustion analysis:
  • The carbon in the compound converts to \(\text{CO}_2\). Hence, the amount of \(\text{C}\) in the sample can be determined by measuring the \(\text{CO}_2\) produced.
  • The hydrogen in the compound converts to \(\text{H}_2\text{O}\). The amount of \(\text{H}\) in the sample can be deduced by measuring the \(\text{H}_2\text{O}\) produced.
  • Finding \(\text{N}\), if present, involves calculating the remaining mass, considering the compound's initial mass, minus the determined masses of \(\text{C}\) and \(\text{H}\).
This analytical method directly links to determining the empirical formulas of compounds, providing a basis for deeper chemical understanding.
Molecular Weight Calculation
Calculating molecular weight is an essential part of chemistry, particularly for processes like combustion analysis. Molecular weight, often called molar mass, is the sum of the atomic weights of all atoms in a molecule. Each element in the periodic table has an atomic weight, measured in grams per mole (g/mol), which is used in these calculations.

For instance, in determining the empirical formula of a compound through combustion analysis, we calculate the following:

  • Carbon Dioxide (\(\text{CO}_2\)): The molecular weight calculation for \(\text{CO}_2\) involves adding the atomic weight of carbon (\(12.01\) g/mol) with two times the atomic weight of oxygen (\(16.00\) g/mol each), giving \(44.01\) g/mol in total.
  • Water (\(\text{H}_2\text{O}\)): For \(\text{H}_2\text{O}\), the molecular weight is calculated by adding two times the atomic weight of hydrogen (\(1.01\) g/mol each) and the atomic weight of oxygen (\(16.00\) g/mol), resulting in \(18.02\) g/mol.
  • Nitrogen: If the compound includes nitrogen, its molecular weight (\(14.01\) g/mol) contributes to deducing the stoichiometry of the compound.
These calculations assist in understanding the ratios in which elements bond together in molecules, crucially supporting empirical formula determination.
Mole Concept
The mole concept is a fundamental pillar in chemistry used to relate the mass of a substance to the amount of substance it contains, which can be counted as a number of atoms or molecules. A mole contains \(6.022 \times 10^{23}\) entities (Avogadro's number), and this relationship forms the basis for converting between mass, moles, and number of molecules or atoms.

In the context of combustion analysis, the mole concept aids in finding how much of each element is present in the compound based on the products of combustion:

  • Moles of \(\text{CO}_2\), \(\text{H}_2\text{O}\): By knowing the mass of \(\text{CO}_2\) and \(\text{H}_2\text{O}\) produced, you can calculate their moles by dividing the mass by their respective molecular weights.
  • Moles of \(\text{C}\), \(\text{H}\), and \(\text{N}\) in the original compound: For each mole of \(\text{CO}_2\) produced, there is one mole of \(\text{C}\) in the original compound. Similarly, \(\text{H}\) is determined by \(2\times\) moles of \(\text{H}_2\text{O}\), and \(\text{N}\) is calculated after subtracting total masses of \(\text{C}\) and \(\text{H}\) from the initial mass.
Accurately applying the mole concept ensures precise determination of empirical formulas, shedding light on the simplest ratios of atoms in compounds.

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Most popular questions from this chapter

Consider the following reaction: $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ If a container were to have 10 molecules of \(\mathrm{O}_{2}\) and 10 molecules of \(\mathrm{NH}_{3}\) initially, how many total molecules (reactants plus products) would be present in the container after this reaction goes to completion?

Hexamethylenediamine \(\left(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{N}_{2}\right)\) is one of the starting materials for the production of nylon. It can be prepared from adipic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}\right)\) by the following overall equation: $$ \mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}(l)+2 \mathrm{NH}_{3}(g)+4 \mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{6} \mathrm{H}_{16} \mathrm{N}_{2}(l)+4 \mathrm{H}_{2} \mathrm{O}(l) $$ What is the percent yield for the reaction if 765 g of hexamethylenediamine is made from \(1.00 \times 10^{3} \mathrm{g}\) of adipic acid?

The compound As \(_{2} \mathrm{L}_{4}\) is synthesized by reaction of arsenic metal with arsenic triodide. If a solid cubic block of arsenic \(\left(d=5.72 \mathrm{g} / \mathrm{cm}^{3}\right)\) that is 3.00 \(\mathrm{cm}\) on edge is allowed to react with \(1.01 \times 10^{24}\) molecules of arsenic triodide, what mass of \(\mathrm{As}_{2} \mathrm{L}_{4}\) can be prepared? If the percent yield of \(\mathrm{As}_{2} \mathrm{L}_{4}\) was 75.6\(\%\) what mass of \(\mathrm{As}_{2} \mathrm{I}_{4}\) was actually isolated?

Aspirin \(\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\right)\) is synthesized by reacting salicylic acid \(\left(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}\right)\) with acetic anhydride \(\left(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}\right) .\) The balanced equation is $$ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}+\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3} \longrightarrow \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} $$ a. What mass of acetic anhydride is needed to completely consume \(1.00 \times 10^{2}\) g salicylic acid? b. What is the maximum mass of aspirin (the theoretical yield) that could be produced in this reaction?

Maleic acid is an organic compound composed of \(41.39 \% \mathrm{C},\) 3.47\(\% \mathrm{H}\) , and the rest oxygen. If 0.129 mole of maleic acid has a mass of 15.0 \(\mathrm{g}\) , what are the empirical and molecular formulas of maleic acid?
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