In a combustion analysis, \(5.535 \mathrm{~g}\) of a hydrocarbon formed \(6.750 \mathrm{~g}\) of water and \(17.550 \mathrm{~g}\) of carbon dioxide. Deduce the empirical formula of the hydrocarbon and state whether it is likely to be an alkane, an alkene, or an alkyne. Explain your reasoning.

Combustion analysis is a widely used method in chemistry to determine the elemental composition of an unknown hydrocarbon. In this analysis, a known mass of a substance is combusted in an excess of oxygen to produce water and carbon dioxide. By measuring the mass of these combustion products, one can infer the amount of hydrogen and carbon present in the original compound.

For example, in a typical combustion analysis exercise, the mass of water and carbon dioxide produced is used to calculate the moles of hydrogen and carbon in the original hydrocarbon. This is done by first determining the molar masses of the compounds (water and carbon dioxide). From that point, stoichiometry comes into play to provide the number of moles of hydrogen and carbon atoms, which helps in deducing the empirical formula of the hydrocarbon.

The mole concept is foundational in understanding chemical reactions and stoichiometry. It provides a bridge between the atomic or molecular scale and the macroscopic world we observe. A mole is defined as the amount of a substance containing as many elementary entities (atoms, molecules, ions, etc.) as there are atoms in exactly 12 grams of carbon-12. This number is known as Avogadro's number and is approximately equal to \(6.022 \times 10^{23}\).

Using the mole concept, one can convert the mass of a substance to the number of moles by dividing by its molar mass. For instance, in combustion analysis, the mass of water and carbon dioxide produced from burning a hydrocarbon is converted to moles, which directly relates to the moles of hydrogen and carbon atoms in the hydrocarbon.

Stoichiometry is essentially the mathematics behind chemistry. It involves the calculation of the quantities of reactants and products in chemical reactions. In the context of combustion analysis, stoichiometry assists in establishing the exact amount of carbon and hydrogen atoms, as each molecule of water and carbon dioxide corresponds to a fixed number of hydrogen and carbon atoms, respectively.

Let's take a simplified view of stoichiometry at work: For one mole of a hydrocarbon undergoing complete combustion, it will yield a specific number of moles of water and carbon dioxide. By using the balanced combustion reaction equation, we can determine the stoichiometric coefficients that represent the proportionality between the moles of reactants and products. This ratio is then used to derive the empirical formula of the original hydrocarbon.

Hydrocarbons, compounds composed of only hydrogen and carbon, have different classifications based on the types of bonds between the carbon atoms. The main categories are alkanes, alkenes, and alkynes.

Alkanes are the simplest types of hydrocarbons, with only single bonds between carbon atoms, resulting in a general formula of \( C_nH_{2n+2} \). Alkenes contain at least one double bond between carbon atoms, adopting the general formula \( C_nH_{2n} \). Alkynes are more reactive due to at least one triple bond, and their general formula is \( C_nH_{2n-2} \).

By understanding the empirical formula obtained from combustion analysis, we can often predict the hydrocarbon classification. For instance, if the ratio of hydrogen to carbon atoms is as per the alkane formula, we can likely conclude that the hydrocarbon under analysis is an alkane. The determination is based on the assumption that most organic compounds tend to have a non-cyclic structure unless indicated otherwise.