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Chapter 3: Problem 52

(a) The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of \(2.78 \mathrm{mg}\) of ethyl butyrate produces \(6.32 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(2.58 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O}\). What is the empirical formula of the compound? (b) Nicotine, a component of tobacco, is composed of \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{N}\). \(\mathrm{A}\) 5.250-mg sample of nicotine was combusted, producing \(14.242 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(4.083 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O}\). What is the empirical formula for nicotine? If nicotine has a molar mass of \(160 \pm 5 \mathrm{~g} / \mathrm{mol}\), what is its molecular formula?

Short Answer

Expert verified
The empirical formula of ethyl butyrate is CH鈧侽. The empirical formula of nicotine is C鈧匟鈧嘚, and its molecular formula is C鈧佲個H鈧佲倓N鈧�.

Step by step solution

01

Determine moles of each element

We can calculate the moles of carbon and hydrogen using the amounts of CO鈧� and H鈧侽 formed. Firstly, we need to convert the mass of CO鈧� and H鈧侽 into moles: Moles of CO鈧� = \(\frac{6.32 \text{ mg}}{44.01 \text{ mg/mol}} = 0.0001436 \text{ mol}\) Moles of H鈧侽 = \(\frac{2.58 \text{ mg}}{18.02 \text{ mg/mol}} = 0.0001431 \text{ mol}\) Now for each product, we can find the moles of C and H: Moles of C = moles of CO鈧� = \( 0.0001436 \text{ mol}\) Moles of H = 2 脳 moles of H鈧侽 = \( 2 \times 0.0001431 = 0.0002862 \text{ mol}\)
02

Determine the mole ratio of each element

Divide each mole value by the smallest mole value: Mole ratio of C = \(\frac{0.0001436}{0.0001431} \approx 1 \) Mole ratio of H = \(\frac{0.0002862}{0.0001431} \approx 2 \)
03

Write the empirical formula

Based on the mole ratios, the empirical formula of ethyl butyrate is CH鈧侽. ### Part B - Empirical formula and Molecular formula of Nicotine ###
04

Determine moles of each element

We can calculate the moles of carbon and hydrogen using the amounts of CO鈧� and H鈧侽 formed. Firstly, we need to convert the mass of CO鈧� and H鈧侽 into moles: Moles of CO鈧� = \(\frac{14.242 \text{ mg}}{44.01 \text{ mg/mol}} = 0.0003236 \text{ mol}\) Moles of H鈧侽 = \(\frac{4.083 \text{ mg}}{18.02 \text{ mg/mol}} = 0.0002266 \text{ mol}\) Now for each product, we can find the moles of C and H: Moles of C = moles of CO鈧� = \( 0.0003236 \text{ mol}\) Moles of H = 2 脳 moles of H鈧侽 = \( 2 \times 0.0002266 = 0.0004532 \text{ mol}\) To determine moles of N, subtract the mass of C and H from the total mass of nicotine, then divide by the molar mass of N: Mass of C = 0.0003236 mol 脳 12.01 mg/mol = 3.881 mg Mass of H = 0.0004532 mol 脳 1.008 mg/mol = 0.4566 mg Mass of N = 5.250 mg - (3.881 mg + 0.4566 mg) = 0.9124 mg Moles of N = \(\frac{0.9124 \text{ mg}}{14.01 \text{ mg/mol}} = 0.0000651 \text{ mol}\)
05

Determine the mole ratio of each element

Divide each mole value by the smallest mole value: Mole ratio of C = \(\frac{0.0003236}{0.0000651} \approx 5 \) Mole ratio of H = \(\frac{0.0004532}{0.0000651} \approx 7 \) Mole ratio of N = \(\frac{0.0000651}{0.0000651} \approx 1 \)
06

Write the empirical formula

Based on the mole ratios, the empirical formula of nicotine is C鈧匟鈧嘚.
07

Calculate the empirical formula mass

Empirical formula mass = 5(12.01) + 7(1.008) + 14.01 = 80.09 g/mol
08

Divide the molar mass by the empirical formula mass

Molar mass is given as 160 卤 5 g/mol. We will use the average value of 160 g/mol \(\frac{\text{Molar Mass}}{\text{Empirical Formula Mass}} = \frac{160}{80.09} \approx 2\)
09

Write the molecular formula

Multiply the empirical formula by the integer found in step 5: Molecular formula of nicotine = (C鈧匟鈧嘚) 脳 2 = C鈧佲個H鈧佲倓N鈧�

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Key Concepts

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

Understanding Stoichiometry
Stoichiometry is the mathematical relationship between the relative quantities of reactants and products in a chemical reaction. It is based on the conservation of mass and the concept of moles, making it possible to predict the amounts of substances consumed and produced in a reaction.

When performing stoichiometric calculations, the first step is often to balance the chemical equation to ensure that the same number of atoms of each element appears on both sides. In the context of empirical formula determination, stoichiometry helps us to determine the mole ratio of each element in a compound by using the mass of products formed during a reaction, such as CO2 and H2O in a combustion reaction.

Importance in Empirical Formula Determination

Stoichiometry is essential when deriving an empirical formula. It involves converting the masses of substances to moles, using molar masses, and then finding the simplest whole-number ratio between moles of each element. This ratio reflects the relative number of atoms of each element in the compound.

A strong grasp of stoichiometry is crucial for students to accurately perform these conversions and to understand the underlying principles guiding these calculations.
Combustion Analysis Explained
Combustion analysis is a laboratory technique often used to determine the empirical formula of an organic compound. In this process, a known mass of a compound undergoes complete combustion in excess oxygen to produce CO2 and H2O.

The masses of these products are measured, allowing us to calculate the moles of carbon and hydrogen in the original compound. If there are other non-metal elements like nitrogen or sulfur, their presence can also be determined using further chemical analysis.

Role in Empirical Formula Determination

Through combustion analysis, students can learn how to isolate and quantify the carbon and hydrogen components of organic molecules. This is a critical step in working out the empirical formula, as shown in the provided exercise with ethyl butyrate and nicotine. Their corresponding CO2 and H2O outputs indicate carbon and hydrogen content, which then can be articulated into a formula reflecting the simplest whole-number ratio of atoms in the compound.
Molecular Formula Calculation
The molecular formula of a compound represents the actual number of atoms of each element present in a molecule, which might be an integer multiple of the empirical formula. Determining the molecular formula involves calculating the empirical formula mass and then comparing it with the experimentally determined molar mass of the compound.

Once the empirical formula is found, the mass of this formula is summed up from the atomic masses of each element. This sum is then divided into the known molar mass of the compound. The result is typically a whole number, which indicates how many empirical formula units are present in the molecular formula.

Calculating Nicotine's Molecular Formula

In the example of nicotine, after determining the empirical formula as C5H7N, you would calculate its mass. Dividing nicotine's molar mass by the empirical formula mass gives a ratio used to multiply the subscripts in the empirical formula, yielding the molecule's actual atomic composition. This final step is crucial for understanding the compound's true structure and for predicting its chemical properties.

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Most popular questions from this chapter

Hydrogen cyanide, \(\mathrm{HCN}\), is a poisonous gas. The lethal dose is approximately \(300 \mathrm{mg} \mathrm{HCN}\) per kilogram of air when inhaled. (a) Calculate the amount of HCN that gives the lethal dose in a small laboratory room measuring \(12 \times 15 \times 80 \mathrm{ft}\). The density of air at \(26{ }^{\circ} \mathrm{C}\) is \(0.00118 \mathrm{~g} / \mathrm{cm}^{3}\), (b) If the \(\mathrm{HCN}\) is formed by reaction of \(\mathrm{NaCN}\) with an acid such as \(\mathrm{H}_{2} \mathrm{SO}_{4}\), what mass of \(\mathrm{NaCN}\) gives the lethal dose in the room? \(2 \mathrm{NaCN}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{HCN}(g)\) (c) HCN forms when synthetic fibers containing Orlon \({ }^{8}\) or Acrilan burn. Acrilan has an empirical formula of \(\mathrm{CH}_{2} \mathrm{CHCN}\), so \(\mathrm{HCN}\) is \(50.9 \%\) of the formula by mass. \(\mathrm{A}\) rug measures \(12 \times 15 \mathrm{ft}\) and contains \(30 \mathrm{oz}\) of Acrilan \(^{8}\) fibers per square yard of carpet. If the rug burns, will a lethal dose of \(\mathrm{HCN}\) be generated in the room? Assume that the yield of \(\mathrm{HCN}\) from the fibers is \(20 \%\) and that the carpet is \(50 \%\) consumed.

(a) What is the mass, in grams, of a mole of \({ }^{12} \mathrm{C}\) ? (b) How many carbon atoms are present in a mole of \({ }^{12} \mathrm{C}\) ?

When a mixture of \(10.0 \mathrm{~g}\) of acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and \(10.0 \mathrm{~g}\) of oxygen \(\left(\mathrm{O}_{2}\right)\) is ignited, the resultant combustion reaction produces \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). (a) Write the balanced chemical equation for this reaction. (b) Which is the limiting reactant? (c) How many grams of \(\mathrm{C}_{2} \mathrm{H}_{2}, \mathrm{O}_{2}, \mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are present after the reaction is complete?

Glycine, an amino acid used by organisms to make proteins, is represented by the molecular model below. (a) Write its molecular formula. (b) Determine its molar mass. (c) Calculate the mass of 3 moles of glycine. (d) Calculate the percent nitrogen by mass in glycine. [Sections \(3.3\) and \(3.5]\)

What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{CH}_{2}\), molar mass \(=84 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{NH}_{2} \mathrm{Cl}\), molar mass \(=51.5 \mathrm{~g} / \mathrm{mol}\)
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