Question

A butane fireplace lighter contains \(56.0 \mathrm{~g}\) of butane. a. Write the equation for the complete combustion of butane. b. How many grams of oxygen are needed for the complete combustion of the butane in the lighter?

Short answer

200.49 g of oxygen is needed.

Step-by-step solution

Write the Combustion Equation

The complete combustion of butane (C4H10) with oxygen (O2) produces carbon dioxide (CO2) and water (H2O). Write the balanced chemical equation: \[ 2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O} \]

Determine the Moles of Butane

Calculate the moles of butane using its molar mass. The molar mass of butane (C4H10) is calculated as follows: \[ \text{C}_4\text{H}_{10} = 4(12.01 \text{ g/mol}) + 10(1.01 \text{ g/mol}) = 58.14 \text{ g/mol} \] Thus, the moles of butane are: \[ \text{moles of butane} = \frac{56.0 \text{ g}}{58.14 \text{ g/mol}} \ \text{moles of butane} \rightarrow 0.9639 \text{ mol} \]

Determine the Moles of Oxygen Needed

From the balanced equation, 2 moles of butane react with 13 moles of oxygen. Therefore, each mole of butane requires 6.5 moles of oxygen. Calculate the required moles of oxygen for 0.9639 moles of butane: \[ \text{moles of oxygen} = 0.9639 \text{ mol} \times \frac{13}{2} \ \text{moles of oxygen} = 6.2654 \text{ mol} \]

Calculate the Grams of Oxygen Needed

Convert the moles of oxygen to grams using the molar mass of oxygen (O2), which is 32.00 g/mol: \[ \text{grams of oxygen} = 6.2654 \text{ mol} \times 32.00 \text{ g/mol} \ \text{grams of oxygen} = 200.49 \text{ g} \]

Key concepts

Balanced Chemical Equation

Understanding a balanced chemical equation is fundamental in chemistry. It shows the reactants and products in a reaction and ensures that mass is conserved. For the combustion of butane (C4H10), the equation involves butane reacting with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The balanced chemical equation is written as follows:

\[ 2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O} \]
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. Balancing chemical equations ensures that the same number of each type of atom is present on both sides of the equation, reflecting the conservation of mass.

Molar Mass Calculation

Calculating the molar mass is an essential step in converting between grams and moles. The molar mass of a compound is the sum of the atomic masses of all atoms in its formula. For butane (C4H10), we calculate its molar mass as follows:

• The atomic mass of carbon (C) is 12.01 g/mol.
• The atomic mass of hydrogen (H) is 1.01 g/mol.

Therefore, the molar mass of butane is:
\[ \text{C}_4\text{H}_{10} = 4(12.01 \text{ g/mol}) + 10(1.01 \text{ g/mol}) = 58.14 \text{ g/mol} \]
This means one mole of butane weighs 58.14 grams. Understanding molar mass allows for conversions between the mass of a substance and the amount in moles, which is crucial for stoichiometric calculations.

Stoichiometry

Stoichiometry deals with the quantitative relationships between reactants and products in a chemical reaction. Given the balanced equation for butane combustion, we can use it to determine the amounts of reactants needed or products formed. From the equation:

\[ 2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O} \]
We can see that 2 moles of butane react with 13 moles of oxygen. To find out how much oxygen is needed for the combustion of a specific amount of butane, we use the following relationship:

\[ \text{0.9639 mol} \times \frac{13 \text{ mol } \text{O}_2}{2 \text{ mol C}_4\text{H}_{10}} = 6.2654 \text{ mol O}_2 \]
So, 0.9639 moles of butane need 6.2654 moles of oxygen. Stoichiometry helps us carry out such calculations, ensuring correct proportions of reactants and products.

Grams to Moles Conversion

Converting between grams and moles is a fundamental skill in chemistry. It allows you to relate the mass of a substance to the number of molecules or atoms present. To convert grams to moles, you use the formula:
\[ \text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \]
For example, if you have 56.0 grams of butane (C4H10) and the molar mass is 58.14 g/mol, the moles of butane are calculated as:

\[ \text{Moles of butane} = \frac{56.0 \text{ g}}{58.14 \text{ g/mol}} = 0.9639 \text{ mol} \]
Similarly, converting moles of oxygen to grams involves the molar mass of oxygen (32.00 g/mol):
\[ \text{Grams of oxygen} = 6.2654 \text{ mol} \times 32.00 \text{ g/mol} = 200.49 \text{ g} \]
This conversion is essential for any calculations involving mass and amount of substances in chemical reactions.