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Chapter 8: Problem 139

An organic compound contains \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and \(\mathrm{O} .\) Combustion of \(0.1023 \mathrm{g}\) of the compound in excess oxygen yielded \(0.2766 \mathrm{g} \mathrm{CO}_{2}\) and \(0.0991 \mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) A sample of \(0.4831 \mathrm{g}\) of the compound was analyzed for nitrogen by the Dumas method (see Exercise 129 ). At \(\mathrm{STP}, 27.6 \mathrm{mL}\) of dry \(\mathrm{N}_{2}\) was obtained. In a third experiment, the density of the compound as a gas was found to be \(4.02 \mathrm{g} / \mathrm{L}\) at \(127^{\circ} \mathrm{C}\) and \(256\) torr. What are the empirical and molecular formulas of the compound?

Short Answer

Expert verified
The empirical formula of the compound is C鈧僅鈧凬O鈧� and the molecular formula is C鈧咹鈧圢鈧侽鈧�.

Step by step solution

01

Determine the masses of carbon, hydrogen, nitrogen and oxygen in the compound

Since we are given the masses of CO鈧� and H鈧侽 that are produced upon combustion, we can use these values to find the masses of carbon and hydrogen in the original compound. Mass of \(\text{carbon in CO}_2 = \frac{\text{mass of CO}_2 \times \text{mass of C in CO}_2}{\text{molar mass of CO}_2} = \frac{0.2766 \times 12}{44} = 0.07560\) g Mass of \(\text{hydrogen in H}_2\text{O} = \frac{\text{mass of H}_2\text{O} \times \text{mass of H in H}_2\text{O}}{\text{molar mass of H}_2\text{O}} = \frac{0.0991 \times 2}{18} = 0.01101\) g Since the total mass of the compound is given to be \(0.1023\) g, we can now find the mass of nitrogen and oxygen in the compound. Mass of \(\text{N}_2 = \frac{27.6 \ \text{mL} \ \text{N}_2\times\text{molar mass of N}_2}{22.4\ \text{L/mol} \times 1000\ \text{mL/L}}=\frac{27.6\times28}{22.4 \times 1000}=0.03409\ \text{g}\) Therefore, mass of noncarbon and non-hydrogen atoms = total mass of the compound - mass of carbon - mass of hydrogen \(= 0.1023 - 0.07560 - 0.01101 = 0.01569\ \text{g}\) Since these are only nitrogen and oxygen atoms, mass of oxygen = mass of noncarbon and non-hydrogen atoms - mass of nitrogen \(= 0.01569 - 0.03409 = 0.04978\ \text{g}\)
02

Convert the masses to moles

Next, we need to convert the masses of carbon, hydrogen, nitrogen and oxygen into moles: Moles of C \(= \frac{0.07560\ \text{g}}{12\ \text{g/mol}} = 0.00630\ \text{mol}\) Moles of H \(= \frac{0.01101\ \text{g}}{1\ \text{g/mol}} = 0.01101\ \text{mol}\) Moles of N \(= \frac{0.03409\ \text{g}}{14\ \text{g/mol}} = 0.002435\ \text{mol}\) Moles of O \(= \frac{0.04978\ \text{g}}{16\ \text{g/mol}} = 0.003112\ \text{mol}\)
03

Determine the empirical formula

To find the empirical formula, we need to determine the simplest whole-number ratio of moles of C, H, N, and O: Molar ratios: C: \(\frac{0.00630}{0.002435} = 2.586\) H: \(\frac{0.01101}{0.002435} = 4.518\) N: \(\frac{0.002435}{0.002435} = 1\) O: \(\frac{0.003112}{0.002435} = 1.278\) Since all ratios are close to whole numbers, we can round them to get the empirical formula: C鈧僅鈧凬O鈧�
04

Calculate the empirical formula mass

To calculate the empirical formula mass, we add the molar masses of C鈧僅鈧凬O鈧�: Empirical formula mass \(= 3 \times 12 + 4 \times 1 + 14 + 2 \times 16 = 88\ \text{g/mol}\)
05

Determine the molecular formula

Next, we need to use the given density, temperature, and pressure to find the molar mass of the compound: Molar mass \(= \frac{density \times R \times T}{P} = \frac{4.02\ \text{g/L} \times 0.0821\ \text{L atm/K mol} \times (127 + 273)\ \text{K}}{256\ \text{torr} \times \frac{1\ \text{atm}}{760\ \text{torr}}} = 176\ \text{g/mol}\) Now let's find the ratio of molecular formula mass to empirical formula mass: Ratio \(= \frac{176}{88} = 2\) This indicates that the molecular formula is two times the empirical formula, which gives us: Molecular formula: C鈧咹鈧圢鈧侽鈧�

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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 powerful technique used to determine the elemental composition of organic compounds. It involves burning the substance in excess oxygen and analyzing the resulting combustion products. Typically, carbon dioxide (CO鈧�) and water (H鈧侽) are measured to quantify carbon and hydrogen, while other techniques may be applied for additional elements like nitrogen and sulfur.

In our example, the compound contains carbon, hydrogen, nitrogen, and oxygen. When subjected to combustion analysis, it produced specific amounts of CO鈧� and H鈧侽. From the given mass of CO鈧�, we can deduce the mass of carbon by using the relation that each mole of CO鈧� contains a mole of carbon.

Similarly, the mass of hydrogen can be deduced from the mass of H鈧侽, as each mole of water contains two moles of hydrogen. These calculated masses of carbon and hydrogen can then be used to determine their respective moles in the compound.
Dumas Method
The Dumas method is a classical way to determine the nitrogen content in organic compounds. This method involves heating the compound to produce gaseous nitrogen (N鈧�), which can then be measured to determine its amount in the original sample.

In the given exercise, the Dumas method was applied to a sample of the compound, indicating that 27.6 mL of nitrogen gas was collected at standard temperature and pressure (STP). Using the known molar volume of a gas at STP (22.4 L/mol), we can calculate the mass and thus the moles of nitrogen present.
  • This is essential for establishing the complete elemental composition of the compound.
  • It provides a more detailed understanding alongside the combustion analysis data.
This amount of nitrogen tells us how much of it is present in the original compound sample, which is necessary for determining both empirical and molecular formulas.
Gas Density Calculation
Gas density is an important factor when determining the molar mass of a compound in its gaseous state. Through the ideal gas law, we can use the given density, along with temperature and pressure data, to calculate molar mass.

The exercise provides a density value of 4.02 g/L, a temperature of 127掳C, and a pressure of 256 torr. The ideal gas law \( PV = nRT \) helps convert these conditions into the molar mass of the compound.
  • P is the pressure, which must be in atm,
  • V is the volume (considered as 1 L for density purposes),
  • T is the absolute temperature in Kelvin (calculated by adding 273 to 掳C),
  • R is the universal gas constant (0.0821 L atm/mol K).
By using these values, we find how the physical behavior of the gas relates to its chemical properties, further assisting in determining the molecular formula.
Mole Concept
The mole concept is a foundational principle in chemistry that facilitates the quantification of substances. Using Avogadro's number, it allows us to count atoms, molecules, and ions and is crucial for relating mass to number of particles.

In the exercise, we take the masses of carbon, hydrogen, and nitrogen converted from their respective compounds (CO鈧�, H鈧侽, and N鈧�) and switch these into moles using their atomic masses. For example, the atomic mass of carbon is 12 g/mol. Thus, by dividing the mass of carbon derived from CO鈧� by its atomic mass, the moles of carbon in the original compound are determined.

  • The mole ratios of the elements in the compound help us find the empirical formula, which represents the simplest whole-number ratio of the elements.
  • Comparing this empirical formula to the calculated molar mass helps determine the molecular formula, the actual number of atoms in each element in one molecule of the compound.
Through these calculated mole ratios, we establish both the empirical and molecular formulas, deriving deeper insights into the compound's chemical nature.

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Most popular questions from this chapter

The partial pressure of \(\mathrm{CH}_{4}(g)\) is 0.175 atm and that of \(\mathrm{O}_{2}(g)\) is 0.250 atm in a mixture of the two gases. a. What is the mole fraction of each gas in the mixture? b. If the mixture occupies a volume of \(10.5 \mathrm{L}\) at \(65^{\circ} \mathrm{C}\), calculate the total number of moles of gas in the mixture. c. Calculate the number of grams of each gas in the mixture.

A particular balloon is designed by its manufacturer to be inflated to a volume of no more than 2.5 L. If the balloon is filled with 2.0 L helium at sea level, is released, and rises to an altitude at which the atmospheric pressure is only \(500 . \mathrm{mm}\) Hg, will the balloon burst? (Assume temperature is constant.)

The oxides of Group \(2 \mathrm{A}\) metals (symbolized by M here) react with carbon dioxide according to the following reaction: $$\mathrm{MO}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{MCO}_{3}(s)$$ A \(2.85\) \(-\mathrm{g}\) sample containing only \(\mathrm{MgO}\) and \(\mathrm{CuO}\) is placed in a 3.00 -L container. The container is filled with \(\mathrm{CO}_{2}\) to a pressure of \(740 .\) torr at \(20 .^{\circ} \mathrm{C}\). After the reaction has gone to completion, the pressure inside the flask is \(390 .\) torr at \(20 .^{\circ} \mathrm{C}\). What is the mass percent of \(MgO\) in the mixture? Assume that only the \(MgO\) reacts with \(\mathrm{CO}_{2}\)

An ideal gas is contained in a cylinder with a volume of \(5.0 \times 10^{2} \mathrm{mL}\) at a temperature of \(30 .^{\circ} \mathrm{C}\) and a pressure of \(710.\) torr. The gas is then compressed to a volume of \(25 \mathrm{mL}\) and the temperature is raised to \(820 .^{\circ} \mathrm{C}\). What is the new pressure of the gas?

Given that a sample of air is made up of nitrogen, oxygen, and argon in the mole fractions \(0.78 \mathrm{N}_{2}, 0.21 \mathrm{O}_{2},\) and \(0.010 \mathrm{Ar},\) what is the density of air at standard temperature and pressure?
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