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Chapter 3: Problem 51

(a) Combustion analysis of toluene, a common organic solvent, gives \(5.86 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(1.37 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O}\). If the compound contains only carbon and hydrogen, what is its empirical formula? (b) Menthol, the substance we can smell in mentholated cough drops, is composed of \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{O}\). A \(0.1005-\mathrm{g}\) sample of menthol is combusted, producing \(0.2829 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.1159 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\). What is the empirical formula for menthol? If menthol has a molar mass of \(156 \mathrm{~g} / \mathrm{mol}\), what is its molecular formula?

Short Answer

Expert verified
The empirical formula of toluene is CH2. The empirical formula of menthol is C鈧佲個H鈧傗個O, and its molecular formula is also C鈧佲個H鈧傗個O.

Step by step solution

01

Find the moles of carbon and hydrogen in toluene

To find the moles of carbon (C) in toluene, we will use the mass of CO2 produced during combustion. The mass of CO2 produced is 5.86 mg. Now, we need to convert that mass into moles by dividing it by the molar mass of CO2 (44.01 g/mol). Keep in mind that 1 gram equals 1000 milligrams. So, we get: moles of C = \(\frac{5.86 \mathrm{mg} \times \frac{1 \mathrm{g}}{1000 \mathrm{mg}}}{44.01 \mathrm{g/mol}}\) = 0.0001332 mol Similarly, we find the moles of hydrogen (H) using the mass of H2O produced (1.37 mg) and the molar mass of H2O (18.02 g/mol): moles of H = \(\frac{1.37 \mathrm{mg} \times \frac{1 \mathrm{g}}{1000 \mathrm{mg}}}{18.02 \mathrm{g/mol}}\) = 0.0000760 mol
02

Determine the empirical formula

To find the empirical formula, we take the ratio of moles of H to moles of C: \(\frac{0.0000760 \ \mathrm{mol} \ \mathrm{H}}{0.0001332 \ \mathrm{mol} \ \mathrm{C}}\) 鈮� 0.57 From this, we can determine that the empirical formula contains approximately two hydrogen atoms for every carbon atom. Therefore, the empirical formula of toluene is CH2. (b)
03

Find the moles of carbon, hydrogen, and oxygen in menthol

First, let's find the moles of carbon (C) and hydrogen (H) using the same process as before: moles of C = \(\frac{0.2829 \mathrm{g}}{44.01 \mathrm{g/mol}}\) = 0.00643 mol moles of H = \(\frac{0.1159 \mathrm{g}}{18.02 \mathrm{g/mol}}\) = 0.00643 mol To find the moles of oxygen (O), we will use the molar mass of menthol and the given sample mass. Based on the results for C and H, we know that the remaining mass is due to O. So, we subtract the masses of C and H from the mass of the sample and then divide by the molar mass of oxygen (16.00 g/mol) to find the moles: Mass of O = 0.1005 g - (0.00643 mol C 脳 12.01 g/mol C) - (0.00643 mol H 脳 1.008 g/mol H) = 0.0148 g moles of O = \(\frac{0.0148 \mathrm{g}}{16.00 \mathrm{g/mol}}\) = 0.000926 mol
04

Determine the empirical formula

Now let's find the ratio of moles of H, C, and O: \(\frac{\mathrm{moles \ of} \ \mathrm{H}}{\mathrm{moles \ of} \ \mathrm{C}} = \frac{0.00643 \ \mathrm{mol}}{0.00643 \ \mathrm{mol}}\) = 1 \(\frac{\mathrm{moles \ of} \ \mathrm{O}}{\mathrm{moles \ of} \ \mathrm{C}} = \frac{0.000926 \ \mathrm{mol}}{0.00643 \ \mathrm{mol}}\) 鈮� 0.144 The empirical formula of menthol is then C鈧佲個H鈧傗個O (since 0.144 is close to 1/7, and 10 is the smallest integer multiple that allows this ratio).
05

Find the molecular formula

To determine the molecular formula, we need to compare the empirical formula molar mass to the given molar mass of menthol. The molar mass of the empirical formula is: C鈧佲個H鈧傗個O = (10 脳 12.01 g/mol C) + (20 脳 1.008 g/mol H) + (1 脳 16.00 g/mol O) = 152.3 g/mol Now, divide the given molar mass of menthol (156 g/mol) by the empirical formula molar mass (152.3 g/mol) to get a scaling factor: Scaling factor = \(\frac{156 \mathrm{\ g/mol}}{152.3 \mathrm{\ g/mol}}\) 鈮� 1.02 Since the scaling factor is approximately 1, the molecular formula for menthol is therefore the same as the empirical formula: C鈧佲個H鈧傗個O.

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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 fundamental technique used in chemistry to determine the empirical formula of an organic compound. This method involves burning the compound in the presence of excess oxygen to convert all of its carbon into carbon dioxide (\(\mathrm{CO}_2\)) and all hydrogen into water (\(\mathrm{H}_2\mathrm{O}\)).

By measuring the amounts of \(\mathrm{CO}_2\) and \(\mathrm{H}_2\mathrm{O}\) produced, we can deduce the amounts of carbon and hydrogen that were originally present in the compound. This process lets us establish the ratio of carbon and hydrogen atoms in the compound, leading to its empirical formula.

For instance, consider a combustion analysis where 5.86 mg of \(\mathrm{CO}_2\) and 1.37 mg of \(\mathrm{H}_2\mathrm{O}\) are produced. You first convert these masses into moles using their respective molar masses. For \(\mathrm{CO}_2\), the molar mass is 44.01 g/mol, and for \(\mathrm{H}_2\mathrm{O}\), it is 18.02 g/mol. This conversion reveals the moles of carbon and hydrogen present and ultimately gives us the necessary data to determine the empirical formula of the compound.
Moles Calculation
In chemistry, calculating moles is a crucial step in finding the empirical or molecular formulas of compounds. A mole is a basic unit in chemistry that represents an amount of a substance containing 6.022 x 10虏鲁 constituent particles, usually atoms or molecules.

To calculate the number of moles from a given mass, divide the mass of the substance by its molar mass (grams per mole). This formula is expressed as:
  • \( \text{Moles} = \frac{\text{Mass in grams}}{\text{Molar mass in g/mol}} \)
Applying this to a situation like combustion analysis, you would use the given mass of \(\mathrm{CO}_2\) or \(\mathrm{H}_2\mathrm{O}\), along with their molar masses, to determine the moles of carbon or hydrogen.

Consider menthol, for example. When 0.2829 g of \(\mathrm{CO}_2\) is produced, you convert it to moles to find the number of carbon atoms. Similarly, converting the 0.1159 g of \(\mathrm{H}_2\mathrm{O}\) to moles gives you the number of hydrogen atoms. Such calculations help to establish the mole ratio needed to construct empirical formulas.
Organic Compounds Analysis
Organic compounds are primarily composed of three elements: carbon, hydrogen, and often oxygen. Analyzing these compounds involves determining which elements are present and in what proportions.

In the context of combustion analysis, you burn the compound to convert all carbon to \(\mathrm{CO}_2\) and hydrogen to \(\mathrm{H}_2\mathrm{O}\). If oxygen is also present, its amount is often determined indirectly by subtracting the mass of carbon and hydrogen from the total mass of the compound.

This approach was used in the analysis of menthol's composition. Combining the mole calculations for carbon and hydrogen with the remaining mass converted to moles for oxygen provides the full picture of the compound's composition. The empirical formula reflects these mole ratios in the smallest whole-number terms. In the case of menthol, after determining the molecule's composition, the empirical formula of \(\mathrm{C}_{10}\mathrm{H}_{20}\mathrm{O}\) was deduced, and the molecular formula turned out to be the same due to the similar mass potential.

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Most popular questions from this chapter

(a) You are given a cube of silver metal that measures \(1.000 \mathrm{~cm}\) on each edge. The density of silver is \(10.5\) \(\mathrm{g} / \mathrm{cm}^{3} .\) How many atoms are in this cube? (b) Because atoms are spherical, they cannot occupy all of the space of the cube. The silver atoms pack in the solid in such a way that \(74 \%\) of the volume of the solid is actually filled with the silver atoms. Calculate the volume of a single silver atom. (c) Using the volume of a silver atom and the formula for the volume of a sphere, calculate the radius in angstroms of a silver atom.

Determine the empirical and molecular formulas of each of the following substances: (a) Ibuprofen, a headache remedy, contains \(75.69 \% \mathrm{C}\), \(8.80 \% \mathrm{H}\), and \(15.51 \%\) O by mass, and has a molar mass of \(206 \mathrm{~g} / \mathrm{mol}\). (b) Cadaverine, a foul smelling substance produced by the action of bacteria on meat, contains \(58.55 \% \mathrm{C}\), \(13.81 \% \mathrm{H}\), and \(27.40 \% \mathrm{~N}\) by mass; its molar mass is \(102.2 \mathrm{~g} / \mathrm{mol}\) (c) Epinephrine (adrenaline), a hormone secreted into the bloodstream in times of danger or stress, contains \(59.0 \% \mathrm{C}, 7.1 \% \mathrm{H}, 26.2 \% \mathrm{O}\), and \(7.7 \% \mathrm{~N}\) by mass; its \(\mathrm{MW}\) is about \(180 \mathrm{amu}\).

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of \(\mathrm{NH}_{3}\) to \(\mathrm{NO}\) : $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ In a certain experiment, \(1.50 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) reacts with \(2.75 \mathrm{~g}\) of \(\mathrm{O}_{2}\) (a) Which is the limiting reactant? (b) How many grams of \(\mathrm{NO}\) and of \(\mathrm{H}_{2} \mathrm{O}\) form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed? (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

The allowable concentration level of vinyl chloride, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\), in the atmosphere in a chemical plant is \(2.0 \times 10^{-6} \mathrm{~g} / \mathrm{L}\). How many moles of vinyl chloride in each liter does this represent? How many molecules per liter?

The source of oxygen that drives the internal combustion engine in an automobile is air. Air is a mixture of gases, which are principally \(\mathrm{N}_{2}(\sim 79 \%)\) and \(\mathrm{O}_{2}(\sim 20 \%)\). In the cylinder of an automobile engine, nitrogen can react with oxygen to produce nitric oxide gas, NO. As NO is emitted from the tailpipe of the car, it can react with more oxygen to produce nitrogen dioxide gas. (a) Write balanced chemical equations for both reactions. (b) Both nitric oxide and nitrogen dioxide are pollutants that can lead to acid rain and global warming; collectively, they are called "NO \(_{x}\) " gases. In 2004, the United States emitted an estimated 19 million tons of nitrogen dioxide into the atmosphere. How many grams of nitrogen dioxide is this? (c) The production of \(\mathrm{NO}_{\mathrm{x}}\) gases is an unwanted side reaction of the main engine combustion process that turns octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), into \(\mathrm{CO}_{2}\) and water. If \(85 \%\) of the oxygen in an engine is used to combust octane, and the remainder used to produce nitrogen dioxide, calculate how many grams of nitrogen dioxide would be produced during the combustion of 500 grams of octane.
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