/*! 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 149 Because of its detrimental effec... [FREE SOLUTION] | 91Ó°ÊÓ

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Because of its detrimental effect on the environment, the lead compound described in Problem 3.148 has been replaced by methyl tert-butyl ether (a compound of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) ) to enhance the performance of gasoline. (This compound is also being phased out because of its contamination of drinking water.) When \(12.1 \mathrm{~g}\) of the compound are burned in an apparatus like the one shown in Figure \(3.6,30.2 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(14.8 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are formed. What is the empirical formula of the compound?

Short Answer

Expert verified
The empirical formula of the compound is \(C_{3}H_{7}O\).

Step by step solution

01

Determine Mass of Each Element

Using the given masses of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\), determine the mass of carbon and hydrogen. From \(\mathrm{CO}_{2}\), each carbon atom contributes a mass of 12.01 g/mol. Therefore, given that the mass of \(\mathrm{CO}_{2}\) is 30.2 g, the mass of Carbon is \(\frac{12.01}{44.01} \times 30.2 = 8.21 \mathrm{~g}\), and the mass of Oxygen is \(12.1 - 8.21 = 3.89 \mathrm{~g}\). From \(\mathrm{H}_{2} \mathrm{O}\), each hydrogen atom contributes a mass of 1.01 g/mol. Therefore, given that the mass of \(\mathrm{H}_{2} \mathrm{O}\) is 14.8 g, the mass of Hydrogen is \(\frac{(2 \times 1.01)}{18.02} \times 14.8 = 1.657 \mathrm{~g}\).
02

Calculate Number of Moles

We calculate the number of moles for each of Carbon, Hydrogen and Oxygen. Using the atomic masses, we get: \(\text{Moles of C} = \frac{8.21}{12.01} = 0.683 \mathrm{~mol}\), \(\text{Moles of H} = \frac{1.657}{1.01} = 1.64 \mathrm{~mol}\), \(\text{Moles of O} = \frac{3.89}{16.00} = 0.243 \mathrm{~mol}\).
03

Simplify Ratio of Atoms to Smallest Integer Values

The empirical formula is given by the ratio of moles \(C: H: O = 0.683: 1.64: 0.243\). The smaller ratio is 0.243 for \(O\), so divide each mole value by this: \(C: H: O = \frac{0.683}{0.243} : \frac{1.64}{0.243} : \frac{0.243}{0.243} = 2.81 : 6.75 : 1\). It seems like the ratio is not an integer. This might happen due to experimental errors. An approximate ratio is \(C: H: O = 3 : 7 : 1\), so the empirical formula is \(C_{3}H_{7}O\).

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Key Concepts

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

Methyl tert-butyl ether
Methyl tert-butyl ether (MTBE) is a chemical compound that contains carbon, hydrogen, and oxygen, and serves as a gasoline additive. It is used to boost octane levels and improve combustion efficiency in engines. However, MTBE has a downside: it poses environmental hazards, especially when it contaminates water supplies.
MTBE was introduced as a replacement for lead-based additives in gasoline, which were harmful due to their toxic effects. While it does help reduce engine knock and improve performance, MTBE contamination in the environment can be persistent and challenging to address.
Remember that when studying chemical compounds like MTBE, scientists often focus on its elemental composition—what proportion of carbon, hydrogen, and oxygen is present. This understanding is crucial for further studies on its reactions, effectiveness, and environmental impact.
Combustion Analysis
Combustion analysis is a laboratory process used to determine the composition of an unknown organic compound. It involves burning a sample and analyzing the combustion products. This method is especially useful for compounds made of carbon, hydrogen, and oxygen, like MTBE.
In a typical combustion analysis, the unknown sample is burned in an excess of oxygen. This burning produces water (H\(_2\)O) and carbon dioxide (CO\(_2\)), substances that can be quantitatively measured. By analyzing these products, chemists can deduce the amounts of each element in the original compound.
Understanding combustion analysis is key to calculating the empirical formula of a compound. This process transforms complex laboratory data into a simple formula, representing the most basic whole number ratio of atoms in a compound.
Mass of Elements Determination
Determining the mass of elements in a compound is a crucial step in chemical analysis. This process involves calculating the weight of each element using the mass of combustion products.
In the case of MTBE, after the combustion, you measure the masses of CO\(_2\) and H\(_2\)O formed. From these, the masses of carbon and hydrogen can be directly derived. For carbon, use the conversion from grams of CO\(_2\) to grams of carbon, while hydrogen is obtained from the mass of H\(_2\)O.
This step often requires careful calculations, ensuring accuracy as each element's mass contributes to the total calculated. Any error here would affect the elemental proportions and thus the final empirical formula, highlighting its significance in elemental analysis.
Moles Calculation
Calculating moles is an essential step in chemistry, notably when determining the empirical formula of a compound. The mole concept is a fundamental measure that bridges macroscopic mass to the number of atoms or molecules in a sample.
For MTBE, once the masses of carbon, hydrogen, and oxygen are determined, the next step is to convert these masses into moles. This conversion is done using the molar mass of each element. For instance, using the mass of carbon divided by its molar mass (12.01 g/mol) gives the moles of carbon in the sample.
Understanding how to calculate moles accurately allows chemists to transform weight data into a chemical formula, translating massive abstract data into practical, analyzable science, providing insight into the compound's fundamental makeup.
Environmental Chemistry
Environmental chemistry is the study of chemical phenomena occurring in the environment. It focuses on understanding how human-generated compounds like MTBE interact with natural systems.
MTBE, once believed to be a safe additive, was found to contaminate groundwater. Its solubility and persistence make it difficult to remediate. This issue highlights the importance of studying chemical interactions in environmental contexts.
By learning about environmental chemistry, we gain insight into preventing and mitigating pollutants' impacts. It also informs policy and regulatory measures, ensuring compounds do not pose unacceptable risks to health and ecosystems. This field underscores the importance of responsible chemical usage and management in our technological advancements.

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Most popular questions from this chapter

Balance the following equations using the method outlined in Section 3.7 . (a) \(\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}+\mathrm{O}_{2}\) (b) \(\mathrm{KNO}_{3} \longrightarrow \mathrm{KNO}_{2}+\mathrm{O}_{2}\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{3} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{NH}_{4} \mathrm{NO}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}\) (e) \(\mathrm{NaHCO}_{3} \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}\) (f) \(\mathrm{P}_{4} \mathrm{O}_{10}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}\) (g) \(\mathrm{HCl}+\mathrm{CaCO}_{3} \longrightarrow \mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}\) (h) \(\mathrm{Al}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{H}_{2}\) (i) \(\mathrm{CO}_{2}+\mathrm{KOH} \longrightarrow \mathrm{K}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}\) (j) \(\mathrm{CH}_{4}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (k) \(\mathrm{Be}_{2} \mathrm{C}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Be}(\mathrm{OH})_{2}+\mathrm{CH}_{4}\) (l) \(\mathrm{Cu}+\mathrm{HNO}_{3} \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{NO}+\mathrm{H}_{2} \mathrm{O}\) \((\mathrm{m}) \mathrm{S}+\mathrm{HNO}_{3} \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (n) \(\mathrm{NH}_{3}+\mathrm{CuO} \longrightarrow \mathrm{Cu}+\mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}\)

Octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is a component of gasoline. Complete combustion of octane yields \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}_{2} .\) Incomplete combustion produces \(\mathrm{H}_{2} \mathrm{O}\) and CO, which not only reduces the efficiency of the engine using the fuel but is also toxic. In a certain test run, 1.000 gal of octane is burned in an engine. The total mass of \(\mathrm{CO}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\) produced is \(11.53 \mathrm{~kg} .\) Calculate the efficiency of the process; that is, calculate the fraction of octane converted to \(\mathrm{CO}_{2}\). The density of octane is \(2.650 \mathrm{~kg} / \mathrm{gal}\)

Write the symbols used to represent gas, liquid, solid, and the aqueous phase in chemical equations.

A research chemist used a mass spectrometer to study the two isotopes of an element. Over time, she recorded a number of mass spectra of these isotopes. On analysis, she noticed that the ratio of the taller peak (the more abundant isotope) to the shorter peak (the less abundant isotope) gradually increased with time. Assuming that the mass spectrometer was functioning normally, what do you think was causing this change?

Propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) is a component of natural gas and is used in domestic cooking and heating. (a) Balance the following equation representing the combustion of propane in air.$$\mathrm{C}_{3} \mathrm{H}_{8}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}$$ (b) How many grams of carbon dioxide can be produced by burning 3.65 moles of propane? Assume that oxygen is the excess reactant in this reaction.

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