/*! 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 63 Ethylene glycol is used as an au... [FREE SOLUTION] | 91Ó°ÊÓ

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Ethylene glycol is used as an automobile antifreeze and in the manufacture of polyester fibers. The name glycol stems from the sweet taste of this poisonous compound. Combustion of \(6.38 \mathrm{mg}\) of ethylene glycol gives \(9.06 \mathrm{mg}\) \(\mathrm{CO}_{2}\) and \(5.58 \mathrm{mg} \mathrm{H}_{2} \mathrm{O} .\) The compound contains only \(\mathrm{C}, \mathrm{H}\) and \(\mathrm{O}\). What are the mass percentages of the elements in ethylene glycol?

Short Answer

Expert verified
Carbon: 38.8%, Hydrogen: 9.8%, Oxygen: 51.4%.

Step by step solution

01

Convert Masses to Moles

First, we need the molar masses of COâ‚‚ and Hâ‚‚O. Molar mass of COâ‚‚ is 44.01 g/mol, and for Hâ‚‚O, it's 18.02 g/mol.Calculate moles of COâ‚‚:\[\text{Moles of } \mathrm{CO}_2 = \frac{9.06 \text{ mg}}{44.01 \text{ g/mol}} \times \frac{1 \text{ g}}{1000 \text{ mg}} = 0.000206 \text{ mol}\]Calculate moles of Hâ‚‚O:\[\text{Moles of } \mathrm{H}_2\mathrm{O} = \frac{5.58 \text{ mg}}{18.02 \text{ g/mol}} \times \frac{1 \text{ g}}{1000 \text{ mg}} = 0.00031 \text{ mol}\]
02

Determine Masses of Carbon and Hydrogen

From moles of COâ‚‚, calculate mass of carbon:Since one mole of COâ‚‚ gives one mole of C:\[\text{Mass of C} = 0.000206 \text{ mol} \times 12.01 \text{ g/mol} = 0.002475 \text{ g} \]From moles of Hâ‚‚O, calculate mass of hydrogen:Since one mole of Hâ‚‚O gives two moles of H:\[\text{Mass of H} = 0.00031 \text{ mol} \times 2 \times 1.008 \text{ g/mol} = 0.000625 \text{ g}\]
03

Calculate Mass of Oxygen in Ethylene Glycol

The original sample was 6.38 mg. Subtract the mass of carbon and hydrogen to find the mass of oxygen:\[\text{Mass of } \mathrm{O} = 6.38 \text{ mg} - (2.475 \text{ mg of C} + 0.625 \text{ mg of H}) = 3.28 \text{ mg} \]
04

Calculate Mass Percentages

Convert each element's mass to a percentage of the total mass of ethylene glycol.Percentage of carbon:\[\text{Percentage of C} = \frac{0.002475 \text{ g}}{0.00638 \text{ g}} \times 100\% = 38.8\%\]Percentage of hydrogen:\[\text{Percentage of H} = \frac{0.000625 \text{ g}}{0.00638 \text{ g}} \times 100\% = 9.8\%\]Percentage of oxygen:\[\text{Percentage of O} = \frac{0.00328 \text{ g}}{0.00638 \text{ g}} \times 100\% = 51.4\%\]

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Key Concepts

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

Combustion Analysis
Understanding combustion analysis is crucial for determining the composition of chemical substances. During combustion, a compound reacts with oxygen and converts to water, carbon dioxide, along with any other products depending on the elements in the compound. In our case, the combustion of ethylene glycol gives us specific quantities of COâ‚‚ and Hâ‚‚O.

The first step involves converting these product masses into moles using their molar masses. This conversion helps in determining the moles of individual elements present in the original compound because the number of atoms of each element is conserved during combustion. Through this process, we can ultimately establish the mass of each constituent element in the compound. These calculated masses are then used to find their respective mass percentages in the original substance.

This method helps chemists understand the quantitative composition of compounds and is widely used in the analysis of substances.
Ethylene Glycol
Ethylene glycol is a common compound often recognized for its role as an antifreeze. It is a simple diol, meaning it has two alcohol functional groups, and is essential in various industrial applications, including the manufacture of polyester fibers.

Despite its utility, ethylene glycol is hazardous, primarily because of its sweet taste which can be deceiving. It consists of carbon, hydrogen, and oxygen atoms. By using combustion analysis, we can unravel how much of each element it contains.

For example, in this exercise, when a tiny amount of ethylene glycol combusts, it produces defined amounts of carbon dioxide and water, which gives us vital clues about its elemental composition.
Molar Mass
The concept of molar mass is pivotal in chemistry when converting masses into moles, which is the SI unit for amount of substance. It reflects the mass of one mole of any given substance, typically expressed in grams per mole. Knowing the molar mass helps in various calculations, such as those needed during combustion analysis.

For instance, in our analysis, we first find the molar masses of COâ‚‚ and Hâ‚‚O to convert their respective masses into moles. The conversion uses the relationship shown as:
  • For carbon dioxide: \[\text{Moles of } \text{CO}_2 = \frac{\text{mass of } \text{CO}_2}{\text{molar mass of } \text{CO}_2} \]
  • For water: \[\text{Moles of } \text{H}_2\text{O} = \frac{\text{mass of } \text{H}_2\text{O}}{\text{molar mass of } \text{H}_2\text{O}} \]

The precise use of these conversions allows chemists to understand the elemental distribution in compounds.
Elemental Composition
Elemental composition analysis lets us understand what elements make up a compound and in what quantity. It's a process of determining the proportion of each element within a compound.

In the combustion analysis of ethylene glycol, after converting reaction product masses to moles, we deduce the exact mass of carbon and hydrogen from COâ‚‚ and Hâ‚‚O respectively. The remaining mass in the combustion process corresponds to oxygen present in the original compound. This step-by-step determination is critical because it reveals how each constituent contributes to the compound's overall mass.

Once we have the mass of each element, we can compute the mass percentages using the formula:
  • \[\text{Percentage of element } X = \left(\frac{\text{mass of } X}{\text{total mass of compound}}\right) \times 100\%\]

These percentages provide insight into how those elements are distributed in the molecular structure, which is crucial for chemists when synthesizing new compounds or analyzing unknown substances.

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Most popular questions from this chapter

A friend is doing his chemistry homework and is working with the following chemical reaction. $$ 2 \mathrm{C}_{2} \mathrm{H}_{2}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ He tells you that if he reacts 2 moles of \(\mathrm{C}_{2} \mathrm{H}_{2}\) with 4 moles of \(\mathrm{O}_{2}\), then the \(\mathrm{C}_{2} \mathrm{H}_{2}\) is the limiting reactant since there are fewer moles of \(\mathrm{C}_{2} \mathrm{H}_{2}\) than \(\mathrm{O}_{2}\). How would you explain to him where he went wrong with his reasoning (what concept is he missing)? After providing your friend with the explanation from part a, he still doesn't believe you because he had a homework problem where 2 moles of calcium were reacted with 4 moles of sulfur and he needed to determine the limiting reactant. The reaction is $$ \mathrm{Ca}(s)+\mathrm{S}(s) \longrightarrow \operatorname{CaS}(s) $$ He obtained the correct answer, Ca, by reasoning that since there were fewer moles of calcium reacting, calcium had to be the limiting reactant. How would you explain his reasoning flaw and why he got "lucky" in choosing the answer that he did?

Ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}\), burns in oxygen to give carbon dioxide, \(\mathrm{CO}_{2}\), and water. Write the equation for the reaction, giving molecular, molar, and mass interpretations below the equation.

Moles within Moles and Molar Mass Part 1: How many hydrogen and oxygen atoms are present in 1 molecule of \(\mathrm{H}_{2} \mathrm{O}\) ? How many moles of hydrogen and oxygen atoms are present in \(1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) ? What are the masses of hydrogen and oxygen in \(1.0 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O} ?\) What is the mass of \(1.0 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) ? Part 2: Two hypothetical ionic compounds are discovered with the chemical formulas \(\mathrm{XCl}_{2}\) and \(\mathrm{YCl}_{2}\), where \(\mathrm{X}\) and Y represent symbols of the imaginary elements. Chemical analysis of the two compounds reveals that \(0.25 \mathrm{~mol} \mathrm{XCl}_{2}\) has a mass of \(100.0 \mathrm{~g}\) and \(0.50 \mathrm{~mol} \mathrm{YCl}_{2}\) has a mass of \(125.0 \mathrm{~g}\) What are the molar masses of \(\mathrm{XCl}_{2}\) and \(\mathrm{YCl}_{2}\) ? If you had 1.0 -mol samples of \(\mathrm{XCl}_{2}\) and \(\mathrm{YCl}_{2}\), how would the number of chloride ions compare? If you had 1.0 -mol samples of \(\mathrm{XCl}_{2}\) and \(\mathrm{YCl}_{2}\), how would the masses of elements \(\mathrm{X}\) and \(\mathrm{Y}\) compare? What is the mass of chloride ions present in 1.0 \(\mathrm{mol} \mathrm{XCl}_{2}\) and \(1.0 \mathrm{~mol} \mathrm{YCl}_{2} ?\) What are the molar masses of elements \(\mathrm{X}\) and \(\mathrm{Y} ?\) How many moles of \(\mathrm{X}\) ions and chloride ions would be present in a 200.0 -g sample of \(\mathrm{XCl}_{2}\) ? How many grams of \(Y\) ions would be present in a \(250.0-\mathrm{g}\) sample of \(\mathrm{YCl}_{2} ?\) What would be the molar mass of the compound \(\mathrm{YBr}_{3} ?\) \(\operatorname{Part} 3:\) A minute sample of \(\mathrm{AlCl}_{3}\) is analyzed for chlorine. The analysis reveals that there are 12 chloride ions present in the sample. How many aluminum ions must be present in the sample? What is the total mass of \(\mathrm{AlCl}_{3}\) in this sample? How many moles of \(\mathrm{AlCl}_{3}\) are in this sample?

A 1.680 -g sample of coal contains \(1.584 \mathrm{~g}\) C. Calculate the mass percentage of \(\mathrm{C}\) in the coal.

Ethyl mercaptan is an odorous substance added to natural gas to make leaks easily detectable. A sample of ethyl mercaptan weighing \(3.52 \mathrm{mg}\) contains \(1.64 \mathrm{mg}\) of sulfur. What is the mass percentage of sulfur in the substance?

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