What is the molecular formula of a hydrocarbon containing \(n\) carbon atoms and only one double bond? Can such a hydrocarbon yield a greater mass of \(\mathrm{H}_{2} \mathrm{O}\) than \(\mathrm{CO}_{2}\) when burned in an excess of oxygen?

Short Answer

Expert verified
The molecular formula of a hydrocarbon with \(n\) carbon atoms and one double bond is \(C_{n}H_{2n}\). It cannot yield a greater mass of \(H_{2}O\) compared to \(CO_{2}\) when burned in excess oxygen.

Step by step solution

01

Determine the Molecular Formula

The structure of a hydrocarbon with only one double bond is known as an alkene. The general formula for alkenes is \(C_{n}H_{2n}\). Therefore, for a hydrocarbon with \(n\) carbon atoms and one double bond, the molecular formula would be \(C_{n}H_{2n}\).
02

Write Down the Balanced Combustion Reaction

The general balanced equation for complete combustion of hydrocarbons with oxygen is \[ C_{n}H_{2n} + \frac{3n}{2} O_{2} \rightarrow nCO_{2} + nH_{2}O \] Every mole of \(C_{n}H_{2n}\) reacts with \(\frac{3n}{2}\) moles of \(O_{2}\) to produce \(n\) moles of \(CO_{2}\) and \(n\) moles of \(H_{2}O\).
03

Mass Comparison of CO2 and H2O

Now, let's apply the law of Conservation of Mass to analyze the reaction. From the balanced equation, it is clear that 1 mole of hydrocarbon yields the same number of moles of \(CO_{2}\) as \(H_{2}O\). However, the molar mass of \(H_{2}O\)(18.015 g/mol) is less than the molar mass of \(CO_{2}\)(44.01 g/mol). Therefore, the mass of \(H_{2}O\) produced is less than the mass of \(CO_{2}\) produced in the combustion reaction.

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 Vaia!

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

An oxoacid with the formula \(\mathrm{H}_{x} \mathrm{E}_{y} \mathrm{O}_{z}\) has a formula mass of 178 u, has 13 atoms in its formula unit, contains \(34.80 \%\) by mass, and \(15.38 \%\) by number of atoms, of the element E. What is the element \(\mathrm{E}\), and what is the formula of this oxoacid?

In your own words, define or explain the following terms or symbols: (a) formula unit; (b) \(\mathrm{P}_{4} ;\) (c) molecular compound; (d) binary compound; (e) hydrate.

Liquid ethyl mercaptan, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{S},\) has a density of \(0.84 \mathrm{g} / \mathrm{mL} .\) Assuming that the combustion of this compound produces only \(\mathrm{CO}_{2}, \mathrm{H}_{2} \mathrm{O},\) and \(\mathrm{SO}_{2},\) what masses of each of these three products would be produced in the combustion of \(3.15 \mathrm{mL}\) of ethyl mercaptan?

Determine the empirical formula of (a) benzo-[a]pyrene, a suspected carcinogen found in cigarette smoke, consisting of \(95.21 \%\) C and \(4.79 \%\) H, by mass; (b) hexachlorophene, used in germicidal soaps, which consists of \(38.37 \%\) C \(, 1.49 \%\) H, \(52.28 \%\) Cl, and \(7.86 \%\) O by mass.

The density of a mixture of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and water is 1.78 g/mL. The percent composition of the mixture is to be determined by converting \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) If \(32.0 \mathrm{mL}\) of the mixture gives \(65.2 \mathrm{g}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) then what is the percent composition of the mixture?

See all solutions

Recommended explanations on Chemistry Textbooks

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.

Sign-up for free