Question

When a hydrocarbon is burnt completely, the ratio of masses of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) formed is \(44: 27\). The hydrocarbon is (a) \(\mathrm{CH}_{4}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{2}\)

Short answer

The hydrocarbon is (b) C2H6.

Step-by-step solution

Understand the Combustion Process

When a hydrocarbon burns in oxygen, carbon dioxide (CO2) and water (H2O) are formed. The given ratio of the masses of CO2 to H2O is 44:27. This is vital information that can help us determine the empirical formula of the hydrocarbon.

Calculate the Moles of CO2 and H2O

Using the mass ratio, calculate the moles of CO2 and H2O. The molar mass of CO2 is 44 g/mol and for H2O is 18 g/mol. The ratio of moles would be (44 g / 44 g/mol) : (27 g / 18 g/mol) = 1 mole of CO2 : 1.5 moles of H2O.

Determine the Molecular Formula of the Hydrocarbon

The moles of CO2 give the moles of carbon in the hydrocarbon, and the moles of H2O give twice the moles of hydrogen in the hydrocarbon since each molecule of water contains two hydrogen atoms. Thus, the empirical formula of the hydrocarbon based on the moles is CH3. None of the options (a-d) exactly match this formula, so we must find the simplest whole number ratio that resembles the options listed.

Compare with the Given Options

The empirical formula CH3 does not exactly match any of the options. To find a match, we multiply by an appropriate integer to get a formula present in the options. Multiplying by 1, we get CH3, which is not listed. Multiplying by 2, we get C2H6, which matches option (b). Thus, the hydrocarbon is C2H6.

Key concepts

Stoichiometry

Stoichiometry is the field of chemistry that involves calculating the quantities of reactants and products in chemical reactions. It's based on the law of conservation of mass, which states that in a chemical reaction, matter is neither created nor destroyed. For a combustion reaction, where a hydrocarbon burns in the presence of oxygen to form carbon dioxide (CO2) and water (H2O), stoichiometry allows us to use the masses of these products to deduce the quantitative aspects of the chemical reaction.

In the case of burning hydrocarbons, the stoichiometric calculations help us understand how many moles of CO2 and H2O are produced, which directly relates to the amounts of carbon and hydrogen in the original hydrocarbon. This is what makes stoichiometry a powerful tool for solving problems like the one described in the exercise.

Empirical Formula Determination

Determining the empirical formula of a compound is a process that involves finding the simplest whole number ratio of the elements in the compound. For hydrocarbons, which consist only of hydrogen and carbon, the process starts by converting the mass of each element to moles using their respective molar masses.

Once the moles of carbon (from CO2) and hydrogen (from H2O) are calculated, their ratio can be simplified to the lowest whole numbers to give us the empirical formula. For instance, in the problem being solved, the ratio of moles was found to be 1 mole of CO2 to 1.5 moles of H2O, which simplifies to a 1:3 ratio upon dividing by the smallest number of moles.

This tells us that for every one carbon atom, there are three hydrogen atoms, leading to the empirical formula CH3. However, determining the empirical formula is just one step in identifying the molecular formula of a compound, which may be a multiple of the empirical formula.

Molecular Formula Calculation

The molecular formula is the formula that represents the actual number of atoms of each element in one molecule of a compound. It is a multiple of the empirical formula, which shows the simplest ratio of atoms. To find the molecular formula, one must know the empirical formula, the molar mass of the compound, and possibly additional information about the compound's properties or structure.

Often, the molecular formula is a simple integer multiple of the empirical formula. The integer can be found by dividing the molar mass of the compound by the mass of the empirical formula unit. In the example provided, by comparing the empirical formula CH3 with the options given in the problem, it was established through multiplication that the molecular formula C2H6 corresponds with one of the options.

Therefore, the molecular formula of the hydrocarbon in the exercise is C2H6, as it fits with the ratio of atoms derived from stoichiometric calculations, and it is also a compound that is known to exist and has properties consistent with the data given in the problem.