Problem 102 - Chemistry The Central Science

Question

When hydrocarbons are burned in a limited amount of air, both CO and \(\mathrm{CO}_{2}\) form. When 0.450 g of a particular hydrocarbon was burned in air, 0.467 \(\mathrm{g}\) of \(\mathrm{CO}, 0.733 \mathrm{g}\) of \(\mathrm{CO}_{2},\) and 0.450 \(\mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}\) were formed. (a) What is the empirical formula of the compound? (b) How many grams of O \(_{2}\) were used in the reaction? (c) How many grams would have been required for complete combustion?

Step-by-Step Solution

Answer

The empirical formula of the given hydrocarbon is C₂H₃. In the reaction, 1.200 g of O₂ were used and 1.867 g of O₂ would have been required for complete combustion.

Step 1: Determine the moles of C, H, and O

We are given the masses of CO, CO₂, and H₂O formed. First, we need to convert these masses to moles. For CO: \(0.467 \ g \ of \ CO * \frac{1 \ mol }{28.01 \ g} = 0.01667 \ mol\) For CO₂: \(0.733 \ g \ of \ CO₂ * \frac{1 \ mol }{44.01 \ g} = 0.01667 \ mol\) For H₂O: \(0.450 \ g \ of \ H₂O * \frac{1 \ mol }{18.01 \ g} = 0.025 \ mol\)

Step 2: Determine the empirical formula

Now, we will find the amount of C, H, and O in the given hydrocarbon. C: Both CO and CO₂ resulted from the combustion. Total moles of C in the hydrocarbon: \(0.01667 \ mol \ CO + 0.01667 \ mol \ CO₂ = 0.03334 \ mol \ C\) H: The H₂O resulted from the combustion. Total moles of H in the hydrocarbon: \(0.025 \ mol \ H₂O * \frac{2 \ mol \ H}{1 \ mol \ H₂O} = 0.050 \ mol \ H\) The empirical formula is determined by the ratio of moles of C and H. Dividing each by the smallest value (0.03334): C: \(\frac{0.03334}{0.03334} = 1 \ mol\) H: \(\frac{0.050}{0.03334} = 1.5 \ mol \ \approx 1.5\) We need integer values, so we multiply by 2 to obtain the empirical formula: Empirical formula: C₂H₃

Step 3: Determine the mass of O₂ used

Now, we need to find the mass of O₂ used in the reaction. First, we will find the moles of O₂ used in forming CO, CO₂, and H₂O: For CO: \(0.01667 \ mol \ CO * \frac{1 \ mol \ O}{1 \ mol \ CO} = 0.01667 \ mol \ O\) For CO₂: \(0.01667 \ mol \ CO₂ * \frac{2 \ mol \ O}{1 \ mol \ CO₂} = 0.03334 \ mol \ O\) For H₂O: \(0.025 \ mol \ H₂O * \frac{1 \ mol \ O}{1 \ mol \ H₂O} = 0.025 \ mol \ O\) Total moles of O used: \(0.01667+0.03334+0.025 = 0.075 \ mol \ O\) Now, we can find the mass of O₂ used: \(0.075 \ mol \ O * \frac{16.00 \ g}{1 \ mol} = 1.200 \ g \ of \ O_{2}\)

Step 4: Determine the mass of O₂ required for complete combustion

For complete combustion, only CO₂ and H₂O should be formed. The complete combustion empirical formula is as follows: C₂H₃ + xO₂ → 2CO₂ + \(\frac{3}{2}\)H₂O Now, we will balance the equation: C₂H₃ + \(\frac{5}{2}\)O₂ → 2CO₂ + \(\frac{3}{2}\)H₂O To find the mass of O₂ required for complete combustion, we need to find the moles of O₂ needed: 0.03334 mol C in C₂H₃: \(\frac{0.03334 \ mol \ C * 2 \ mol \ O_{2}}{2 \ mol \ C} = 0.03334 \ mol \ O_{2}\) 0.050 mol H in C₂H₃: \(\frac{0.050 \ mol \ H * 0.5 \ mol \ O_{2}}{1 \ mol \ H} = 0.025 \ mol \ O_{2}\) Total moles of O₂ required: \(0.03334 + 0.025 = 0.05834 \ mol \ O_{2}\) Finally, we convert moles of O₂ to mass: \(0.05834 \ mol \ O_{2} * \frac{32.00 \ g}{1 \ mol} = 1.867 \ g \ of \ O_{2}\)

Key Concepts

Empirical Formula

The empirical formula of a compound is the simplest whole number ratio representation of the elemental composition of that compound. It can be determined by combustion analysis, a laboratory method where a known mass of a compound is burnt, and the masses of the products formed are used to calculate the mole ratio of the elements in the compound.

During combustion, elements like carbon and hydrogen in the hydrocarbon react with oxygen to form water (H₂O) and carbon oxides (CO and CO₂). The moles of carbon and hydrogen can be back-calculated from the products. Once we have the moles of each element, dividing them by the smallest number of moles among them provides ratios that can be used to determine the empirical formula. For complex ratios, they may need to be multiplied by a common factor to get whole numbers, as seen in the exercise's step-by-step solution.

Mole Concept

The mole concept is a central principle in chemistry that relates the mass of a substance to the amount of substance present. One mole is defined as the amount of a substance that contains a number of entities (atoms, molecules, ions, etc.) equal to Avogadro's number, approximately \(6.022 \times 10^{23}\).

By using the molar mass of a compound (the mass of one mole of the compound), we can convert between mass and moles. This conversion is crucial in stoichiometry as it allows chemists to count atoms and molecules during a chemical reaction. For example, in the provided exercise, the mass of carbon monoxide, carbon dioxide, and water is converted into moles to find out the amounts of carbon and hydrogen participating in the reaction.

Stoichiometry

Stoichiometry deals with the quantitative relationships or ratios between the reactants and products in a chemical reaction. It is based on the conservation of mass and the concept of moles. Stoichiometric calculations involve using a balanced chemical equation to calculate the mass, moles, or volume of reactants needed to produce a given amount of product, and vice versa.

For instance, in the exercise, we calculate the amount of oxygen used during the incomplete combustion of the hydrocarbon, using the stoichiometric coefficients in the balanced chemical equations for the formation of CO, CO₂, and H₂O.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products, with new chemical bonds being formed and old ones breaking. These reactions are represented by chemical equations that must be balanced, indicating that the matter is conserved. There are various types of chemical reactions, including combustion—a rapid reaction that produces heat and usually light.

The provided problem deals with a combustion reaction, albeit incomplete, leading to the formation of both CO and CO₂. Understanding the types of reactions and the conditions under which they occur, such as limited oxygen supply leading to incomplete combustion, is important in predicting and explaining the outcomes of reactions.

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that is completely consumed first and thus limits the amount of product that can be formed. In the context of combustion, if not enough oxygen is present, it becomes the limiting reactant, and incomplete combustion occurs, producing both carbon monoxide and carbon dioxide.

By determining the amounts of CO and CO₂ produced in a reaction, and knowing the moles of the hydrocarbon present, it's possible to evaluate what reactant was in excess and how much of the limiting reactant was actually consumed. This is essential for calculating theoretical and actual yields, as well as for scaling up reactions for industrial purposes.