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Chapter 7: Problem 16

In a balanced chemical equation for the complete combustion of a hydrocarbon, what is the ratio of atoms of C in the hydrocarbon to molecules of \(\mathrm{CO}_{2}\) produced?

Short Answer

Expert verified
Answer: The ratio of carbon atoms in a hydrocarbon to the molecules of carbon dioxide produced during complete combustion is 1:1.

Step by step solution

01

Write the general reaction for complete combustion of a hydrocarbon

A hydrocarbon can be represented as \(C_{x}H_{y}\), where x and y are integers representing the number of carbon and hydrogen atoms, respectively. The complete combustion reaction can be represented as: $$C_{x}H_{y} + O_{2} \rightarrow CO_{2} + H_{2}O$$
02

Balance the equation

To balance the chemical equation, we need to make sure that there are equal numbers of each atom on both sides of the equation. We can do this by adjusting the coefficients: $$C_{x}H_{y} + \frac{y}{2}O_{2} \rightarrow xCO_{2} + \frac{y}{2}H_{2}O$$
03

Identify the ratio of carbon atoms in hydrocarbon to CO2 molecules

From the balanced chemical equation, we can see that there are x carbon atoms in the hydrocarbon, and x molecules of CO2 are produced as a result of the complete combustion. Therefore, the ratio of carbon atoms in the hydrocarbon (C) to molecules of CO2 produced is: $$\frac{x\text{ (carbon atoms in hydrocarbon)}}{x\text{ (molecules of }CO_{2}\text{ produced)}}$$
04

Simplify the ratio

The ratio can be simplified as follows: $$\frac{x}{x} = 1$$ This means that for every carbon atom in the hydrocarbon, one molecule of carbon dioxide (CO2) is produced during complete combustion. Therefore, the ratio of carbon atoms in the hydrocarbon to molecules of CO2 produced is 1:1.

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Key Concepts

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

Combustion
Combustion is an exothermic reaction where a substance combines with oxygen to produce heat and light. In the case of hydrocarbons, this reaction typically involves burning these compounds in the presence of excess oxygen. Let's break it down:
  • A hydrocarbon is a compound made up of carbon (C) and hydrogen (H) atoms.
  • During combustion, the carbon atoms in the hydrocarbon react with oxygen (\(O_{2}\)) to form carbon dioxide (\(CO_{2}\)).
  • Hydrogen atoms react to form water (\(H_{2}O\)).
The general word equation for combustion is: hydrocarbon + oxygen 鈫� carbon dioxide + water.

This process is vital because it releases energy that is used for various applications, such as heating homes and driving engines. In essence, combustion plays a critical role in energy production from chemical reactions.
Balancing Chemical Equations
Balancing chemical equations is crucial in chemistry because it ensures the law of conservation of mass is upheld. This law states that matter cannot be created or destroyed in an isolated system.
  • A balanced chemical equation has equal numbers of each type of atom on both the reactant and product sides.
  • To achieve balance, coefficients are adjusted in front of the chemical formulas involved in the reaction.
For instance, the complete combustion of a hydrocarbon (\(C_{x}H_{y} + O_{2} \rightarrow CO_{2} + H_{2}O\)) begins with identifying the number of carbon and hydrogen atoms in the hydrocarbon.

By adjusting the coefficients, you can balance the equation so that:
  • The number of carbon atoms in the hydrocarbon equals the number of CO2 molecules.
  • The number of hydrogen atoms in the hydrocarbon equals two times the number of H2O molecules.
Balancing ensures precision in chemical reactions and is fundamental in predicting the outcomes of chemical processes.
Chemical Stoichiometry
Chemical stoichiometry involves the calculation of reactants and products in chemical reactions. It is the bridge between the balanced chemical equation and the actual quantities of substances involved. Let's explore its components:
  • Stoichiometry uses molarity, coefficients, and the mole concept to determine quantities needed or produced.
  • By knowing the balanced reaction (\(C_{x}H_{y} + \frac{y}{2}O_{2} \rightarrow xCO_{2} + \frac{y}{2}H_{2}O\)), stoichiometry helps calculate the number of moles of oxygen required or moles of water and CO2 produced.
For example, to find the ratio of carbon atoms to CO2 molecules in combustion, stoichiometry illustrates that for each mole of carbon in the hydrocarbon, one mole of CO2 is produced.

This is crucial in chemical manufacturing and laboratories, where stoichiometric calculations provide the foundation for formulating materials, optimizing yields, and minimizing wastage.

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Most popular questions from this chapter

Pigments for Stoplights Cadmium yellow (cadmium sulfide) is a lemon-yellow pigment used in the lenses of stoplights. Its formula is CdS, and it is very insoluble in water. The recommended recipe for cadmium yellow is to mix cadmium nitrate with sodium sulfide in water. The cadmium yellow forms as a solid, while the other product, sodium nitrate, remains dissolved in the water. a. Write a balanced chemical equation for the reaction. b. Calculate the mass of cadmium nitrate you must start with to make 125 g of CdS.

Corn farmers in the American Midwest typically use \(5.0 \times 10^{3}\) kilograms of ammonium nitrate fertilizer per square kilometer of cornfield per year. Some of the fertilizer washes into the Mississippi River and eventually flows into the Gulf of Mexico, promoting the growth of algae and endangering other aquatic life. a. Ammonium nitrate can be prepared by the following reaction: $$\mathrm{NH}_{3}(g)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{NH}_{4} \mathrm{NO}_{3}(a q)$$ How much nitric acid would be required to make the fertilizer needed for \(1 \mathrm{km}^{2}\) of cornfield per year? b. Ammonium ions dissolved in groundwater may be converted into \(\mathrm{NO}_{3}^{-}\) ions by bacterial action: $$\mathrm{NH}_{4}^{+}(a q)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{NO}_{3}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)+2 \mathrm{H}^{+}(a q)$$ If \(10 \%\) of the ammonium component of \(5.0 \times 10^{3}\) kilograms of fertilizer ends up as nitrate ions, how much oxygen would be consumed?

Artificial Bones for Medical Implants The material often used to make artificial bones is the same material that gives natural bones their strength. Its common name is hydroxyapatite, and its formula is \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\) a. Propose a systematic name for this compound. b. What is the mass percentage of calcium in it? c. When treated with hydrogen fluoride, hydroxyapatite becomes fluorapatite \(\left[\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F}\right]\), an even stronger substance. Does the percent mass of Ca increase or decrease as a result of this substitution?

Chemistry of Volcanic Gases Balance the following reactions that occur during volcanic eruptions: a. \(\mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{3}(g)\) b. \(\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \rightarrow \mathrm{S}_{8}(s)+\mathrm{H}_{2} \mathrm{O}(g)\)

A 3.556 g sample of a pure aluminum oxide decomposes under high heat to produce \(1.674 \mathrm{g}\) of oxygen in addition to pure aluminum metal. What is the empirical formula of the aluminum oxide?
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