Question

In the reaction, \(\mathrm{CH}_{4}+2 \mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\), the amount of oxygen needed to completely burn \(4.0 \mathrm{~g}\) of \(\mathrm{CH}_{4}\) is: (a) 4 (b) \(8 \mathrm{~g}\) (c) \(16 \mathrm{~g}\) (d) \(32 \mathrm{~g}\)

Short answer

The amount of oxygen needed is 16 g (option c).

Step-by-step solution

Determine the Molar Mass of Methane (CH4)

Calculate the molar mass of \( \mathrm{CH}_4 \). The molar mass of carbon (C) is approximately \(12.01 \, \text{g/mol}\) and hydrogen (H) is approximately \(1.01 \, \text{g/mol}\). Therefore, the molar mass of \(\mathrm{CH}_4\) is \(12.01 + 4\times1.01 = 16.05 \, \text{g/mol}\).

Calculate Moles of Methane

Use the given mass of methane to find the moles. The moles of \(\mathrm{CH}_4\) can be calculated using the formula \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\). So, \(\frac{4.0 \, \text{g}}{16.05 \, \text{g/mol}} = 0.249 \text{ moles}\).

Use Reaction Stoichiometry

According to the balanced equation \(\mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}\), one mole of \(\mathrm{CH}_4\) reacts with two moles of \(\mathrm{O}_2\). Thus, \(0.249\) moles of \(\mathrm{CH}_4\) needs \(0.249 \times 2 = 0.498\) moles of \(\mathrm{O}_2\).

Convert Moles of Oxygen to Grams

The molar mass of \(\mathrm{O}_2\) is approximately \(32.00 \, \text{g/mol}\). Hence, the mass of \(0.498\) moles of \(\mathrm{O}_2\) is \(0.498 \times 32.00 = 15.936 \, \text{g}\).

Select the Closest Answer Option

Comparing the calculated mass of oxygen needed (\(15.936 \, \text{g}\)) with the given options, the closest answer is \(16 \, \text{g}\).

Key concepts

Molar Mass Calculation

Molar mass is a fundamental concept in chemistry that represents the mass of one mole of a particular substance. It's essentially the sum of the atomic masses of all atoms present in a molecule. For any compound, you can find the molar mass by adding up the weights of its constituent atoms.
To illustrate, let's consider methane (\( \mathrm{CH}_4 \)). Methane consists of one carbon atom and four hydrogen atoms. The atomic mass of carbon is approximately \( 12.01 \, \text{g/mol} \) and hydrogen is approximately \( 1.01 \, \text{g/mol} \). Calculating the molar mass of \( \mathrm{CH}_4 \) involves combining these weights:

• Carbon: \( 1 \times 12.01 = 12.01 \, \text{g/mol} \)
• Hydrogen: \( 4 \times 1.01 = 4.04 \, \text{g/mol} \)

Thus, the molar mass of methane is \( 12.01 + 4.04 = 16.05 \, \text{g/mol} \). This means one mole of methane weighs 16.05 grams. Understanding molar mass is crucial for converting between grams and moles, a common necessity in stoichiometry.

Chemical Reaction Balancing

Balancing chemical equations is essential in stoichiometry because it ensures that the law of conservation of mass is upheld. This means the mass and the number of each type of atom must be the same before and after the reaction. Every chemical reaction equation must thus be balanced.
The provided chemical reaction \( \mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O} \) is a complete combustion reaction. In this equation:

• One molecule of methane (\( \mathrm{CH}_4 \)) reacts with two molecules of oxygen (\( \mathrm{O}_2 \)).
• The products are one molecule of carbon dioxide (\( \mathrm{CO}_2 \)) and two molecules of water (\( \mathrm{H}_2\mathrm{O} \)).

To ensure the equation is balanced:

• Count the atoms on both sides: one carbon atom, four hydrogen atoms, and four oxygen atoms are present both in reactants and products.

Balancing equations is a foundational step in stoichiometry, allowing accurate calculations of reactants and products.

Moles and Mass Relationship

Understanding the relationship between moles and mass is fundamental in stoichiometry. It's about using the molar mass to convert between the amount of substance (in moles) and its mass (in grams).
To perform these conversions:

• The formula used is \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \)
• For instance, if you have 4.0 grams of methane (\( \mathrm{CH}_4 \)), you calculate the moles by dividing the mass of the sample by its molar mass.

For methane, given its molar mass is \( 16.05 \, \text{g/mol} \):

• \( \text{moles of methane} = \frac{4.0 \, \text{g}}{16.05 \, \text{g/mol}} = 0.249 \text{ moles} \)

Once you have the moles, this quantity can be used in stoichiometric calculations to determine how much of the other reactants or products are involved in a reaction. In the original exercise, this approach allows for finding that 0.249 moles of methane reacts with 0.498 moles of oxygen, due to the 1:2 ratio in the balanced equation, translating into 15.936 grams of oxygen.