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Chapter 2: Problem 121

A combustion reaction involves the reaction of a substance with oxygen gas. The complete combustion of any hydrocarbon (binary compound of carbon and hydrogen) produces carbon dioxide and water as the only products. Octane is a hydrocarbon that is found in gasoline. Complete combustion of octane produces 8 L of carbon dioxide for every 9 L of water vapor (both measured at the same temperature and pressure). What is the ratio of carbon atoms to hydrogen atoms in a molecule of octane?

Short Answer

Expert verified
The simplified ratio of carbon atoms to hydrogen atoms in a molecule of octane is \(4:9\).

Step by step solution

01

Write down the volume ratio of CO_2 and H_2O

From the given information, we know that the volume ratio of CO_2 to H_2O is 8:9. 8 L CO_2 : 9 L H_2O
02

Apply the proportionality of volume and moles

Since the volumes of CO_2 and H_2O are measured at the same temperature and pressure, their volumes are proportional to their moles. We can express the ratio of moles as: 8 moles CO_2 : 9 moles H_2O
03

Deduce the stoichiometry coefficients n and m

From the combustion reaction, we have: C_xH_y + O_2 -> n CO_2 + m H_2O From the mole ratio deduced in step 2, we know that n = 8 and m = 9. So, our reaction becomes: C_xH_y + O_2 -> 8 CO_2 + 9 H_2O
04

Derive the ratio of carbon atoms x to hydrogen atoms y in octane

From the updated combustion reaction, we notice that each CO_2 molecule has one carbon atom and each H_2O molecule has two hydrogen atoms. Therefore, there must be 8 carbon atoms and 18 hydrogen atoms in one molecule of octane: x = 8 y = 18 So the ratio of carbon atoms (x) to hydrogen atoms (y) in octane is: 8 : 18
05

Simplify the ratio

The simplest form of the x : y ratio is obtained by dividing both x and y by their greatest common divisor (gcd), which in this case is 2: \(8 \div 2 = 4\) \(18 \div 2 = 9\) So, the simplified ratio of carbon atoms to hydrogen atoms in a molecule of octane is 4:9.

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Most popular questions from this chapter

You have two distinct gaseous compounds made from element X and element Y. The mass percents are as follows: Compound I: \(30.43 \% \mathrm{X}, 69.57 \% \mathrm{Y}\) Compound \(\mathrm{II} : 63.64 \% \mathrm{X}, 36.36 \% \mathrm{Y}\) In their natural standard states, element X and element Y exist as gases. (Monatomic? Diatomic? Triatomic? That is for you to determine.) When you react “gas X” with “gas Y” to make the products, you get the following data (all at the same pressure and temperature): 1 volume "gas \(\mathrm{X}^{\prime \prime}+2\) volumes "gas $\mathrm{Y}^{\prime \prime} \longrightarrow$ 2 volumes compound I 2 volumes \(^{4}\) gas \(\mathrm{X}^{\prime \prime}+1\) volume "gas \(\mathrm{Y}^{\prime \prime} \longrightarrow\) 2 volumes compound II Assume the simplest possible formulas for reactants and products in the chemical equations above. Then, determine the relative atomic masses of element X and element Y.

Consider 100.0 -g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 \(\mathrm{g}\) of carbon, and the other has 42.9 \(\mathrm{g}\) of carbon. How can these data support the law of multiple proportions if 42.9 is not a multiple of 27.2\(?\) Show that these data support the law of multiple proportions.

a. Classify the following elements as metals or nonmetals: $$\begin{array}{lll}\mathrm{Mg} & \mathrm{Si} & \mathrm{Rn} \\ \mathrm{Ti} & \mathrm{Ge} & \mathrm{Eu} \\ \mathrm{Au} & \mathrm{B} & \mathrm{Am} \\\ \mathrm{Bi} & \mathrm{At} & \mathrm{Br}\end{array}$$ b. The distinction between metals and nonmetals is really not a clear one. Some elements, called metalloids, are intermediate in their properties. Which of these elements would you reclassify as metalloids? What other elements in the periodic table would you expect to be metalloids?

Indium oxide contains 4.784 \(\mathrm{g}\) of indium for every 1.000 \(\mathrm{g}\) of oxygen. In \(1869,\) when Mendeleev first presented his version of the periodic table, he proposed the formula $\operatorname{In}_{2} \mathrm{O}_{3}$ for indium oxide. Before that time it was thought that the formula was InO. What values for the atomic mass of indium are obtained using these two formulas? Assume that oxygen has an atomic mass of 16.00 .

Name the compounds in parts a–d and write the formulas for the compounds in parts e–h. a. \(\operatorname{NaBr}\) b. \(\mathrm{Rb}_{2} \mathrm{O}\) c. \(\mathrm{CaS}\) d. d. \(\mathrm{AlI}_{3}\) e. strontium fluoride f. aluminum selenide g. potassium nitride h. magnesium phosphide
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