Question

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

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

Step-by-step solution

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

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

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

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

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.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions. It helps us understand how quantities of substances interact. In simpler terms, stoichiometry allows us to interpret the balanced chemical equations quantitatively. These calculations are based on the mole ratios derived from the balanced equation and the conservation of mass principle.

When applying stoichiometry to combustion reactions, such as the burning of octane, we use the mole ratios of reactants (like octane and oxygen) to products (such as carbon dioxide and water). For example, in the combustion of octane, the mole ratio of CO_2 to H_2O is 8:9. This arises from the balanced equation and the volumes given, illustrating how the number of molecules directly relates to the volumes of gases involved, thanks to Avogadro's law.

Mole Concept

The mole concept is another key aspect of understanding chemical reactions. It provides a bridge between the atomic scale and what we can measure; one mole is Avogadro's number (approximately 6.022 x 10^23) of particles (atoms, molecules, ions, etc.). This concept allows chemists to count entities in a practical way by weighing substances.

In the example of octane combustion, despite dealing with large numbers of molecules, we can easily determine the amount of substance needed by examining their mole ratios. For instance, since the reaction produces 8 moles of CO_2 and 9 moles of H_2O, this indicates how many moles of octane were consumed in the combustion process. Thus, the mole concept assists us in quantifying reactants and products, which is vital for calculating chemical quantities and predicting reaction outcomes.

Hydrocarbon Combustion

Hydrocarbon combustion is a type of chemical reaction where hydrocarbons such as octane react with oxygen to release energy. The general outcome of complete hydrocarbon combustion is carbon dioxide and water. The energy released in this reaction is what powers engines in cars when using fuels like gasoline, which are composed largely of hydrocarbons.

In a typical combustion reaction, the principle is straightforward: hydrocarbons are oxidized, releasing heat, and producing a fixed ratio of products. For octane, the simplified balanced equation can be depicted as: C_8H_18 + 12.5 O_2 -> 8 CO_2 + 9 H_2O This equation tells us that one molecule of octane produces 8 molecules of CO_2 and 9 molecules of H_2O upon complete combustion. The observation of volumes at constant temperature and pressure means that the number of moles can be directly inferred from these volumes. Therefore, hydrocarbon combustion not only provides essential energy but also demonstrates the principles of stoichiometry and the mole concept in action.