/*! 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 96 When hydrocarbons are burned in ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 96

When hydrocarbons are burned in a limited amount of air, both \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) form. When \(0.450 \mathrm{~g}\) of a particular hydrocarbon was burned in air, \(0.467 \mathrm{~g}\) of \(\mathrm{CO}, 0.733 \mathrm{~g}\) of \(\mathrm{CO}_{2}\), and \(0.450 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) were formed. (a) What is the empirical formula of the compound? (b) How many grams of \(\mathrm{O}_{2}\) were used in the reaction? (c) How many grams would have been required for complete combustion?

Short Answer

Expert verified
The empirical formula of the hydrocarbon is C鈧侶鈧�. The mass of O鈧� consumed in the reaction is 1.20 g, and the mass of O鈧� required for complete combustion is 1.34 g.

Step by step solution

01

1. Convert masses to moles

To determine the empirical formula, we'll first need to convert grams of CO, CO鈧�, H鈧侽, and the hydrocarbon to moles by using their respective molar masses: \(1\ mole\ CO = 28.01\ g\), \(1\ mole\ CO_2 = 44.01\ g\), and \(1\ mole\ H_2O = 18.02\ g\). Moles of CO = \(0.467\ g / 28.01\ g/mol = 0.0167\ mol\) Moles of CO鈧� = \(0.733\ g / 44.01\ g/mol = 0.0167\ mol\) Moles of H鈧侽 = \(0.450\ g / 18.02\ g/mol = 0.0250\ mol\)
02

2. Calculate the moles of C, H, and O in the compound

Now we will calculate the moles of carbon, hydrogen, and oxygen separately in the unknown hydrocarbon. Moles of carbon from CO = \(0.0167\ mol\) Moles of carbon from CO鈧� = \(0.0167\ mol\) Total moles of carbon in the hydrocarbon = \(0.0167\ mol + 0.0167\ mol = 0.0334\ mol\) Moles of hydrogen in the hydrocarbon = \(0.0250\ mol * 2 = 0.0500\ mol\), as the combustion forms one water molecule per hydrogen atom. We already have the moles of carbon and hydrogen, so the moles of oxygen can be found by subtracting the moles of carbon and hydrogen from the total moles of the hydrocarbon.
03

3. Determine the hydrocarbon's empirical formula

Now, we should divide each element's moles by the smallest molar value, which is \(0.0334\ mol\), to determine the empirical formula of the hydrocarbon: Carbon: \(0.0334\ mol\;/\;0.0334\ mol = 1\) Hydrogen: \(0.0500\ mol\;/\;0.0334\ mol = 1.5\) To get whole numbers for each element, we can multiply both by 2: Carbon: \(1 * 2 = 2\) Hydrogen: \(1.5 * 2 = 3\) Therefore, the empirical formula of the hydrocarbon is C鈧侶鈧�.
04

4. Calculate the mass of O鈧� consumed in the reaction

For this, we'll be using the stoichiometry and the balanced chemical equation: \[C_x H_y + O_2 \rightarrow aCO + bCO_2 + cH_2 O\] In our case, \(x=2\) and \(y=3\). Then, using the moles of each of the products, we can figure out the moles of O鈧� consumed: Moles of O鈧� (from CO) = \(\frac{1}{2}(0.0167\ mol) = 0.00833\ mol\) Moles of O鈧� (from CO鈧�) = \(0.0167\ mol\) Moles of O鈧� (from H鈧侽) = \(\frac{1}{2}(0.0250\ mol) = 0.0125\ mol\) Total moles of O鈧� consumed = \(0.00833\ mol + 0.0167\ mol + 0.0125\ mol = 0.0375\ mol\) We can now convert the moles of O鈧� consumed to its corresponding mass: Mass of O鈧� consumed = \(0.0375\ mol * 32.00\ g/mol = 1.20\ g\)
05

5. Calculate the mass of O鈧� required for complete combustion

For complete combustion, the stoichiometry should not have the formation of CO. Thus, the balanced combustion reaction becomes: \[C_{2}H_{3} + \frac{5}{2}O_{2} \rightarrow 2CO_{2} + \frac{3}{2}H_{2}O\] Moles of O鈧� required for complete combustion = \(\frac{5}{2}(0.0167\ mol) = 0.0418\ mol\) Now, convert the moles of O鈧� required to its corresponding mass: Mass of O鈧� required = \(0.0418\ mol * 32.00\ g/mol = 1.34\ g\) To summarize: a) The empirical formula of the hydrocarbon is C鈧侶鈧�. b) The mass of O鈧� consumed in the reaction is 1.20 g. c) The mass of O鈧� required for complete combustion is 1.34 g.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Combustion of Hydrocarbons
Combustion of hydrocarbons is a chemical process where hydrocarbons, compounds consisting only of hydrogen and carbon, react with oxygen to form carbon dioxide, carbon monoxide, and water. This reaction releases energy in the form of heat and is a significant process in many applications such as energy generation and engines.

When these substances combust, especially in limited oxygen conditions, incomplete combustion occurs, often producing both carbon monoxide (CO) and carbon dioxide (CO鈧�). Complete combustion, on the other hand, typically results only in CO鈧� and H鈧侽.
  • **Incomplete Combustion:** Occurs when there's not enough oxygen to allow the fuel (hydrocarbon) to burn completely, forming CO and possibly soot.
  • **Complete Combustion:** Occurs when sufficient oxygen is present, allowing the fuel to burn fully, resulting in CO鈧� and H鈧侽 only.
Understanding this process is crucial as it informs us about the efficiency of fuel usage and environmental impacts like emissions.
Stoichiometry
Stoichiometry is a key principle in chemistry that involves quantitative relationships in chemical reactions. It allows chemists to calculate the precise amounts of reactants and products involved in a chemical reaction. By utilizing the balanced chemical equation, stoichiometry provides the proportions of each reactant and product.

In our combustion scenario, stoichiometry helps determine how much oxygen is necessary to completely burn a given amount of a hydrocarbon. The amount of CO, CO鈧�, and H鈧侽 gathered from an experiment helps in establishing relationships using molar mass. For instance, by converting grams to moles using their respective molar masses, we can further understand how much of each element participates in the reaction.
  • **Reactant-to-Product Ratios:** Shows how reactants change into products.
  • **Mole Conversions:** Used to interchange between mass and moles using molar mass.
Thus, stoichiometry serves as a bridge in translating between the macroscopic quantities we measure in lab experiments to the sub-microscopic scales chemists work with.
Chemical Reactions
Chemical reactions occur when substances undergo chemical changes to form new substances. Reactants start the reaction, and they transform to produce products. In the context of hydrocarbon combustion, the chemical reactions can be understood as the reordering of atoms where hydrocarbons and oxygen react to form other substances like CO鈧�, CO, and H鈧侽.

For combustion, the general chemical equation can be represented as: \[ C_xH_y + O_2 \rightarrow zCO_2 + wH_2O + ext{(possibly)} \ qCO \]This equation summarizes the transformation of hydrocarbons and oxygen into combustion products. Chemical reactions occur based on specific principles:
  • **Conservation of Mass:** Mass is neither created nor destroyed in a chemical reaction.
  • **Reactivity:** The tendency of a substance to undergo chemical reactions.
Understanding chemical reactions is fundamental to mastering chemistry as it involves predicting the products and quantities that result from given reactants.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Balance the following equations: (a) \(\mathrm{Li}(s)+\mathrm{N}_{2}(g) \longrightarrow \mathrm{Li}_{3} \mathrm{~N}(s)\) (b) \(\mathrm{La}_{2} \mathrm{O}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{La}(\mathrm{OH})_{3}(a q)\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{PH}_{3}(g)\) (e) \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow\) \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) (f) \(\mathrm{AgNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{~s})+\mathrm{NaNO}_{3}(a q)\) (g) \(\mathrm{CH}_{3} \mathrm{NH}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow\) \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)\)

Without doing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing number of atoms: \(0.50 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}, 23 \mathrm{~g} \mathrm{Na}, 6.0 \times 10^{23} \mathrm{~N}_{2}\) molecules.

Very small crystals composed of 1000 to 100,000 atoms, called quantum dots, are being investigated for use in electronic devices. (a) A quantum dot was made of solid silicon in the shape of a sphere, with a diameter of \(4 \mathrm{~nm}\). Calculate the mass of the quantum dot, using the density of silicon \(\left(2.3 \mathrm{~g} / \mathrm{cm}^{3}\right)\) (b) How many silicon atoms are in the quantum dot? (c) The density of germanium is \(5.325 \mathrm{~g} / \mathrm{cm}^{3} .\) If you made a \(4 \mathrm{~nm}\) quantum dot of germanium, how many Ge atoms would it contain? Assume the dot is spherical.

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(10.4 \% \mathrm{C}, 27.8 \% \mathrm{~S}\), and \(61.7 \% \mathrm{Cl}\) (b) \(21.7 \%\) C, \(9.6 \% \mathrm{O}\), and \(68.7 \% \mathrm{~F}\) (c) \(32.79 \% \mathrm{Na}, 13.02 \% \mathrm{Al}\), and \(54.19 \% \mathrm{~F}\)

Balance the following equations, and indicate whether they are combination, decomposition, or combustion reactions: (a) \(\mathrm{Al}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{AlCl}_{3}(s)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{Li}(s)+\mathrm{N}_{2}(g) \longrightarrow \mathrm{Li}_{3} \mathrm{~N}(s)\) (d) \(\mathrm{PbCO}_{3}(s) \longrightarrow \mathrm{PbO}(s)+\mathrm{CO}_{2}(g)\) (e) \(\mathrm{C}_{7} \mathrm{H}_{8} \mathrm{O}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.