/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 (a) Polychlorinated biphenyls (P... [FREE SOLUTION] | 91影视

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(a) Polychlorinated biphenyls (PCBs) were compounds used as coolants in transformers and capacitors, but their production was banned by the U.S. Congress in 1979 because they are highly toxic and persist in the environment. When \(1.0 \mathrm{~g}\) of a \(\mathrm{PCB}\) containing carbon, hydrogen, and chlorine was subjected to combustion analysis, \(1.617 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.138 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) were produced. What is the empirical formula? (b) If the molecular weight is \(326.26\), what is the molecular formula? (c) Can combustion analysis be used to determine the empirical formula of a compound containing carbon, hydrogen, oxygen, and chlorine?

Short Answer

Expert verified
(a) C鈧僅鈧匔l鈧, (b) C鈧咹鈧佲個Cl鈧, (c) Yes, but only if oxygen content is known separately.

Step by step solution

01

Determine moles of CO2 and H2O

From the combustion data, we know that the amount of CO2 produced provides information about the carbon content, and the amount of H2O provides information about the hydrogen content. Calculate the moles of each: \[ \text{Moles of CO}_2 = \frac{1.617 \, \text{g}}{44.01 \, \text{g/mol}} = 0.03675 \, \text{mol} \]\[ \text{Moles of H}_2\text{O} = \frac{0.138 \, \text{g}}{18.01528 \, \text{g/mol}} = 0.00766 \, \text{mol} \]
02

Determine moles of C and H

Using the moles of CO2 and H2O, we calculate the moles of carbon and hydrogen:For Carbon: Each mole of CO2 contains one mole of carbon.\[ \text{Moles of C} = 0.03675 \, \text{mol} \]For Hydrogen: Each mole of H2O contains two moles of hydrogen.\[ \text{Moles of H} = 2 \times 0.00766 \, \text{mol} = 0.01532 \, \text{mol} \]
03

Calculate mass of C and H

Convert moles of C and H to mass:For Carbon:\[ \text{Mass of C} = 0.03675 \, \text{mol} \times 12.01 \, \text{g/mol} = 0.4414 \, \text{g} \]For Hydrogen:\[ \text{Mass of H} = 0.01532 \, \text{mol} \times 1.008 \, \text{g/mol} = 0.01544 \, \text{g} \]
04

Calculate mass of Cl

Since the total mass of the PCB is known (1.0 g), determine the mass of Cl by subtraction:\[ \text{Mass of Cl} = 1.0 \, \text{g} - (0.4414 \, \text{g} + 0.01544 \, \text{g}) = 0.5432 \, \text{g} \]
05

Determine moles of Cl

Calculate the moles of chlorine based on its mass:\[ \text{Moles of Cl} = \frac{0.5432 \, \text{g}}{35.453 \, \text{g/mol}} = 0.01533 \, \text{mol} \]
06

Determine empirical formula

Divide each mole value by the smallest number of moles calculated to find the ratio:\[ \text{C: } \frac{0.03675}{0.01532} \approx 2.4 \]\[ \text{H: } \frac{0.01532}{0.01532} = 1 \]\[ \text{Cl: } \frac{0.01533}{0.01532} \approx 1 \]Since we cannot have a fractional atom in a formula, multiply each by a factor to convert all numbers into whole numbers. In this case, multiply all by 5: \[ \text{Empirical formula} = \text{C}_3\text{H}_5\text{Cl}_2 \]
07

Determine molecular formula

The empirical formula is C鈧僅鈧匔l鈧, with a molar mass of \[ (3 \times 12.01) + (5 \times 1.008) + (2 \times 35.453) = 127.05 \, \text{g/mol} \]Divide the molecular weight by the empirical formula weight: \[ \frac{326.26}{127.05} \approx 2.57 \]Rounding gives us 2, meaning the molecular formula is twice the empirical formula:\[ \text{Molecular formula} = \text{C}_6\text{H}_{10}\text{Cl}_4 \]
08

Evaluate combustion analysis for empirical formula determination

Combustion analysis can determine empirical formulas for compounds containing carbon, hydrogen, and other elements if the number of oxygen atoms are also known. However, with unknown oxygen content, additional analysis methods are required.

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Key Concepts

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

Combustion Analysis
When understanding how to determine the empirical formula of a substance, combustion analysis is a crucial technique. This method involves burning a sample of a compound and analyzing the amount of resultant products, typically carbon dioxide and water, to determine the amount of carbon and hydrogen present. Here's how it works:
  • When a compound burns, it reacts with oxygen to produce specific products. For organic compounds, these are mostly carbon dioxide (CO鈧) and water (H鈧侽).
  • The mass of produced CO鈧 tells us how much carbon was in the original compound, as each molecule of CO鈧 contains one atom of carbon.
  • Similarly, the mass of H鈧侽 shows how much hydrogen was present, with each water molecule containing two hydrogen atoms.
This analysis is very effective for compounds like polychlorinated biphenyls (PCBs) as you begin with knowing the total mass of the sample and by the end deduce the mass and moles of each component, which then helps in finding the empirical formula.
Molecular Formula Calculation
Determining the molecular formula of a compound is a logical extension after finding its empirical formula, the simplest whole-number ratio of elements. Here's the process: Once you have the empirical formula, you need a piece of additional information: the molecular weight of the compound, which is often provided or must be determined experimentally.
  • The empirical formula's mass can be calculated by adding the atomic weights corresponding to each element in the formula.
  • Next, divide the compound's given molecular weight by the empirical formula's calculated mass. This division yields a number, which should be nearly an integer.
  • This integer tells you how many times the empirical formula must be multiplied to achieve the molecular formula of the compound.
For the PCB in question, multiplying the empirical formula's ratio to achieve a molecular formula provides a detailed representation of the actual molecule's structure. This process emphasizes the importance of accurate molecular mass determination for constructing molecular formulas.
Polychlorinated Biphenyls (PCBs)
Polychlorinated biphenyls (PCBs) are a class of chemical compounds that once had widespread use due to their excellent insulating properties and chemical stability. These compounds contain carbon, hydrogen, and chlorine. However, their persistence in the environment and significant toxicity led to a production ban by various global authorities, including the U.S. Congress in 1979. Despite the ban, PCBs remain a topic of study, especially concerning environmental contamination and remediation efforts. Here's what you should know about PCBs:
  • PCBs have various configurations, reflecting the different numbers of chlorine atoms that can replace hydrogen atoms in biphenyl.
  • They were primarily used in electrical equipment like transformers and capacitors, as they could withstand high temperatures without degrading.
  • Due to their persistence, they possess significant potential for long-range environmental transport, making them a focus of global pollutant studies.
Knowledge of their molecular structure, derived through analyses such as combustion analysis, helps chemists understand the potential transformations and breakdown products of PCBs, contributing to better management and remediation strategies.

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Most popular questions from this chapter

Aluminum reacts with oxygen to yield aluminum oxide. If \(5.0 \mathrm{~g}\) of \(\mathrm{Al}\) reacts with \(4.45 \mathrm{~g}\) of \(\mathrm{O}_{2}\), what is the empirical formula of aluminum oxide?

Compound X contains only carbon, hydrogen, nitrogen, and chlorine. When \(1.00 \mathrm{~g}\) of \(\mathrm{X}\) is dissolved in water and allowed to react with excess silver nitrate, \(\mathrm{AgNO}_{3}\), all the chlorine in \(\mathrm{X}\) reacts and \(1.95 \mathrm{~g}\) of solid \(\mathrm{AgCl}\) is formed. When \(1.00 \mathrm{~g}\) of \(\mathrm{X}\) undergoes complete combustion, \(0.900 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.735 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are formed. What is the empirical formula of \(\mathrm{X}\) ?

Acetic acid \(\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right)\) reacts with isopentyl alcohol \(\left(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\right)\) to yield isopentyl acetate \(\left(\mathrm{C}_{7} \mathrm{H}_{14} \mathrm{O}_{2}\right)\), a fragrant substance with the odor of bananas. If the yield from the reaction of acetic acid with isopentyl alcohol is \(45 \%\), how many grams of isopentyl acetate are formed from \(3.58 \mathrm{~g}\) of acetic acid and \(4.75 \mathrm{~g}\) of isopentyl alcohol? The reaction is $$ \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}+\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O} \longrightarrow \mathrm{C}_{7} \mathrm{H}_{14} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O} $$

What is the empirical formula of stannous fluoride, the first fluoride compound added to toothpaste to protect teeth against decay? Its mass percent composition is \(24.25 \% \mathrm{~F}\) and \(75.75 \% \mathrm{Sn}\).

Calculate the amount of carbon dioxide (in kilograms) emitted when four alternative fuels are burned to provide the same amount of energy as \(10.0\) gallons of gasoline. Compare the carbon dioxide emissions from alternative fuels to gasoline (Worked Example 3.12). (a) Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) Although \(\mathrm{E} 85\) is a blend of ethanol and gasoline, let's use pure ethanol in our calculation for simplification. A gallon of ethanol contains \(68 \%\) of the energy of a gallon of gas, so \(14.7\) gallons of ethanol provides the same amount of energy as \(10.0\) gallons of gasoline. The density of ethanol is \(0.79\) \(\mathrm{kg} / \mathrm{L}\) and \(1 \mathrm{~L}=0.2642\) gal. (b) Liquefied Petroleum Gas/Propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right) \quad\) A gallon of propane contains \(73 \%\) of the energy of a gallon of gas, so \(13.7\) gallons of propane provides the same amount of energy as \(10.0\) gallons of gasoline. The density of liquefied propane is \(0.49 \mathrm{~kg} / \mathrm{L}\) and \(1 \mathrm{~L}=0.2642\) gal. (c) Compressed Natural Gas \(\left(\mathrm{CH}_{4}\right) \quad\) It takes \(25.7 \mathrm{~kg}\) of natural gas, methane \(\left(\mathrm{CH}_{4}\right)\), to provide the same amount of energy as \(10.0\) gallons of gasoline. (d) Electricity from a Coal-Burning Power Plant An electric power plant using bituminous coal produces \(0.94 \mathrm{~kg} \mathrm{CO}_{2} / \mathrm{kWh}\), and an electric vehicle uses \(35 \mathrm{kWh}\) per 100 miles. (A kilowatthour, symbolized \(\mathrm{kWh}\), is a common unit of electrical energy equivalent to \(3.6\) megajoules.) (e) Electricity from a Natural Gas-Burning Power Plant An electric power plant using natural gas produces \(0.55 \mathrm{~kg} \mathrm{CO}_{2} / \mathrm{kWh}\) and an electric vehicle uses \(35 \mathrm{kWh}\) per 100 miles. (f) Which fuel produces the least amount of \(\mathrm{CO}_{2}\) when burned to provide energy for a car: gasoline, ethanol, propane, CNG, electricity from coal, or electricity from natural gas?

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