Problem 67 - Chemistry Matter and Change

Question

Chloroform \(\left(\mathrm{CHCl}_{3}\right),\) an important solvent, is produced by a reaction between methane and chlorine. $$\mathrm{CH}_{4}(\mathrm{g})+3 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{CHCl}_{3}(\mathrm{g})+3 \mathrm{HCl}(\mathrm{g})$$ How much \(\mathrm{CH}_{4},\) in grams, is needed to produce 50.0 \(\mathrm{grams}\) of \(\mathrm{CHCl}_{3} ?\)

Step-by-Step Solution

Answer

6.72 grams of methane (CH4) are needed to produce 50.0 grams of chloroform (CHCl3).

Step 1: Write down the balanced chemical equation

The balanced chemical equation for the reaction is given as \[\mathrm{CH}_{4}(\mathrm{g})+3 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{CHCl}_{3}(\mathrm{g})+3 \mathrm{HCl}(\mathrm{g})\]

Step 2: Calculate the molar masses of CH4 and CHCl3

First, we need to calculate the molar masses of methane (CH4) and chloroform (CHCl3) using their atomic masses from the periodic table: Molar mass of CH4 = (1 x 12.01) + (4 x 1.01) = 16.05 g/mol Molar mass of CHCl3 = (1 x 12.01) + (1 x 1.01) + (3 x 35.45) = 119.38 g/mol

Step 3: Calculate the moles of CHCl3

We need to determine the moles of chloroform needed, using the given mass of 50.0 grams: Moles of CHCl3 = (50.0 g) / (119.38 g/mol) = 0.4185 mol

Step 4: Determine the mole ratio between CH4 and CHCl3

The mole ratio between methane (CH4) and chloroform (CHCl3) can be obtained from the balanced chemical equation: Moles of CH4 / Moles of CHCl3 = 1 / 1

Step 5: Calculate the amount of moles of CH4 required

Using the determined mole ratio, we can find the moles of methane (CH4) needed to produce 0.4185 moles of chloroform (CHCl3): Moles of CH4 = (1 mol CH4 / 1 mol CHCl3) x 0.4185 mol CHCl3 = 0.4185 mol CH4

Step 6: Convert moles of CH4 to grams

Finally, we need to convert the moles of methane (CH4) into grams, using its molar mass: Mass of CH4 = (0.4185 mol CH4) x (16.05 g/mol CH4) = 6.72 g Therefore, 6.72 grams of methane (CH4) are needed to produce 50.0 grams of chloroform (CHCl3).

Key Concepts

Chemical Equation Balancing

Chemical equation balancing is a critical step in solving stoichiometry problems. It ensures the law of conservation of mass is followed, meaning the same number of atoms of each element must be present on both sides of the equation.

For example, understanding how to balance the equation for the production of chloroform from methane and chlorine is essential. In the balanced chemical equation \[\mathrm{CH}_{4}(\mathrm{g})+3 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{CHCl}_{3}(\mathrm{g})+3 \mathrm{HCl}(\mathrm{g})\]we see that for every one molecule of methane (\(\mathrm{CH}_{4}\)), three molecules of chlorine gas (\(\mathrm{Cl}_{2}\)) react to produce one molecule of chloroform (\(\mathrm{CHCl}_{3}\)) and three molecules of hydrochloric acid (\(\mathrm{HCl}\)). By balancing the molecules involved in the reaction, we maintain the balanced state, which is crucial to accurately compute the reactants or products needed in a reaction.

Molar Mass Calculation

Understanding the concept of molar mass is another key element in stoichiometry. The molar mass represents the mass of one mole of a substance and is expressed in grams per mole (g/mol).

To calculate molar mass, we need to sum up the atomic masses of all atoms in a given molecule. For methane (\(\mathrm{CH}_{4}\)), the procedure involves adding the atomic mass of carbon (\(12.01\) g/mol) to four times the atomic mass of hydrogen (\(1.01\) g/mol), since there are four hydrogen atoms. The precise calculation is as follows:\[\text{Molar mass of CH4} = (1 \times 12.01) + (4 \times 1.01) = 16.05 \text{ g/mol}\]Similarly, the molar mass of chloroform (\(\mathrm{CHCl}_{3}\)) is determined by summing the atomic masses of one carbon, one hydrogen, and three chlorine atoms. These calculations are necessary for subsequent stoichiometric computations, such as converting mass to moles or moles to mass.

Mole-to-Mass Conversion

Mole-to-mass conversion is the process of translating between the amount of substance (in moles) and its corresponding mass. This is especially important for quantifying reactants or products in a chemical reaction.

To perform this conversion, we make use of the substance's molar mass as a conversion factor. In our chloroform example, once we have the moles of chloroform needed, this number can be used to determine the mass of methane required. The calculation involves this simple formula:

\[\text{Mass} = \text{Moles} \times \text{Molar Mass}\]
For example, converting 0.4185 moles of methane, given a molar mass of 16.05 g/mol, into grams:

\[\text{Mass of CH4} = (0.4185 \text{ mol} \, \mathrm{CH4}) \times (16.05 \text{ g/mol} \, \mathrm{CH4}) = 6.72 \text{ g}\]
This conversion is crucial for accurately measuring the amount of reactants used or products formed in an experimental setting.