Question

Butane gas is compressed and used as a liquid fuel in disposable cigarette lighters and lightweight camping stoves. Suppose a lighter contains \(5.50 \mathrm{~mL}\) of butane \((d=0.579 \mathrm{~g} / \mathrm{mL})\) (a) How many grams of oxygen are needed to burn the butane completely? (b) How many moles of \(\mathrm{H}_{2} \mathrm{O}\) form when all the butane burns? (c) How many total molecules of gas form when the butane burns completely?

Short answer

(a) 11.38 g of \(\mathrm{O}_2\)(b) 0.2735 mol of \(\mathrm{H}_2\mathrm{O}\)(c) \(2.964 \times 10^{23}\) molecules of gas

Step-by-step solution

Identify and Write Down Known Values

Volume of butane, \(V = 5.50 \mathrm{~mL}\), density of butane, \(d = 0.579 \mathrm{~g/mL}\).

Calculate the Mass of Butane

Use the formula \( \text{mass} = \text{density} \times \text{volume} \).\[ m = 0.579 \mathrm{~g/mL} \times 5.50 \mathrm{~mL} \ m = 3.1845 \mathrm{~g} \ \approx 3.18 \mathrm{~g} \]

Write and Balance the Combustion Reaction

The combustion of butane \((\mathrm{C}_4\mathrm{H}_{10})\) is given by the equation:\[ 2\mathrm{C}_4\mathrm{H}_{10} + 13\mathrm{O}_2 \rightarrow 8\mathrm{CO}_2 + 10\mathrm{H}_2\mathrm{O} \]

Calculate Moles of Butane

Use the molar mass of butane \((\mathrm{C}_4\mathrm{H}_{10}) = 4(12.01) + 10(1.008) = 58.12 \mathrm{~g/mol}\) to find the moles of butane.\[ n = \frac{3.18 \mathrm{~g}}{58.12 \mathrm{~g/mol}} = 0.0547 \mathrm{~mol} \]

Calculate Mass of Oxygen Needed

From the balanced equation, 2 moles of butane require 13 moles of \(\mathrm{O}_2\). Thus, 1 mole of butane needs 6.5 moles of \(\mathrm{O}_2\). Therefore,\[ \text{Moles of } \mathrm{O}_2 = 0.0547 \mathrm{~mol} \times 6.5 = 0.3555 \mathrm{~mol} \]Using the molar mass of oxygen \((\mathrm{O}_2 = 2 \times 16.00 \mathrm{~g/mol} = 32.00 \mathrm{~g/mol})\):\[ \text{Mass of } \mathrm{O}_2 = 0.3555 \mathrm{~mol} \times 32.00 \mathrm{~g/mol} = 11.376 \mathrm{~g} \approx 11.38 \mathrm{~g} \]

Calculate Moles of Water Formed

From the balanced equation, 2 moles of butane form 10 moles of \(\mathrm{H}_2\mathrm{O}\). Thus, 1 mole of butane forms 5 moles of \(\mathrm{H}_2\mathrm{O}\).\[ \text{Moles of } \mathrm{H}_2\mathrm{O} = 0.0547 \mathrm{~mol} \times 5 = 0.2735 \mathrm{~mol} \]

Calculate Total Molecules of Gas Formed

From the balanced equation, 2 moles of butane produce 18 moles of gas (8 moles of \(\mathrm{CO}_2\) and 10 moles of \(\mathrm{H}_2\mathrm{O}\)). Thus, 1 mole of butane produces 9 moles of gas. Therefore,\[ \text{Moles of gas} = 0.0547 \mathrm{~mol} \times 9 = 0.4923 \mathrm{~mol} \]Using Avogadro's number \((6.022 \times 10^{23} \text{ molecules/mol})\):\[ \text{Molecules of gas} = 0.4923 \mathrm{~mol} \times 6.022 \times 10^{23} \text{ molecules/mol} = 2.964 \times 10^{23} \text{ molecules} \]

Key concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions. It relies on the concept of a balanced chemical equation, where the number of atoms of each element is conserved. In the combustion of butane, the balanced equation is:\[ 2\text{C}_4\text{H}_{10} + 13\text{O}_2\rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O} \]This equation tells us that 2 moles of butane react with 13 moles of oxygen to produce 8 moles of carbon dioxide and 10 moles of water. This relationship allows us to calculate different quantities such as mass, moles, and molecules. For example, from the moles of butane burned, we can compute the moles of oxygen required, the mass of oxygen consumed, and the moles of water produced using stoichiometric coefficients.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. In a combustion reaction like that of butane, the reactants are butane (\(\text{C}_4\text{H}_{10}\)) and oxygen (\(\text{O}_2\)). Burning, or combustion, typically involves a hydrocarbon reacting with oxygen to produce carbon dioxide and water. The general form of this type of reaction is:\[ \text{Hydrocarbon} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]The combustion of butane, as illustrated in the problem, releases energy, emphasizing the exothermic nature of such reactions.

Molecular Calculations

Molecular calculations in this context refer to determining quantities like mass, moles, and numbers of molecules. For example, for butane: Let’s calculate its mass using density (\(d\)) and volume (\(V\)): \[ m = d \times V \]Using its molar mass (\(\text{C}_4\text{H}_{10}\) which is 58.12 g/mol), we can convert mass to moles:\[ n = \frac{m}{\text{molar mass}} \]From the moles of butane, we can find the moles of other substances involved in the reaction using stoichiometric ratios. For instance, 1 mole of butane produces 9 moles of gas (CO2 and H2O), and we can convert this to molecules using Avogadro’s number (6.022 × 10^23 molecules/mol):\[ \text{molecules} = \text{moles} \times \text{Avogadro's number} \]

Gas Laws

Gas laws describe the behavior of gases in terms of volume, pressure, temperature, and quantity. Although not directly covered in the butane combustion problem, understanding gas laws like the Ideal Gas Law is crucial for comprehending how gases behave under different conditions. The Ideal Gas Law is given by:\[ PV = nRT \]where \(P\) is pressure, \(V\) is volume, \(n\) is moles, \(R\) is the gas constant, and \(T\) is temperature. For calculations of gases produced in combustion (CO2 and H2O in gaseous state), the Ideal Gas Law can help relate the quantities of gas to conditions such as standard temperature and pressure (STP).

Thermochemistry

Thermochemistry deals with the energy changes occurring during chemical reactions, particularly those involving heat. Combustion reactions are typically exothermic, releasing energy. The heat released during the complete combustion of butane can be determined using the enthalpy change (\(\text{ΔH}\)) which is often tabulated for standard reactions. For butane:\[ \text{C}_4\text{H}_{10}(g) + 13\text{O}_2(g) \rightarrow 8\text{CO}_2(g) + 10\text{H}_2\text{O}(g) \ \text{ΔH} = -2877 \text{kJ/mol} \]This means that burning one mole of butane releases 2877 kJ of energy. Understanding thermochemistry helps in appreciating energy conservation, reaction spontaneity, and the practical applications of combustion in energy generation and everyday devices.