Question

How do I calculate the molecular weight of a compound given 0.0653g of the compound in 8.31g of ethanol with a solution concentration of 0.0368m, and how do I determine if the concentration is molarity or molality?

Answer 1

To calculate the molecular weight (molar mass) of the compound you have in solution, you can use the formula related to molality (m), which is defined as the number of moles of solute per kilogram of solvent. The process is as follows:

### Step 1: Understand the Given Information
- Mass of the compound (solute): 0.0653 g
- Mass of ethanol (solvent): 8.31 g
- Concentration of the solution: 0.0368 m (which indicates molality)

### Step 2: Determine if the Concentration is Molarity or Molality
The notation "m" stands for molality, which is defined as:
\[
\text{molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}
\]
Since the problem states that the concentration is in molality, we know that it is not molarity (which uses volume of the solution).

### Step 3: Convert the Mass of Solvent to Kilograms
To use the molality formula, convert the mass of ethanol from grams to kilograms:
\[
8.31 \, \text{g} = 0.00831 \, \text{kg}
\]

### Step 4: Calculate the Moles of Solute
Using the definition of molality:
\[
0.0368 \, \text{m} = \frac{\text{moles of solute}}{0.00831 \, \text{kg}}
\]
Rearranging to find the moles of solute:
\[
\text{moles of solute} = 0.0368 \, \text{m} \times 0.00831 \, \text{kg} = 0.000305 \, \text{moles}
\]

### Step 5: Calculate the Molecular Weight
Molecular weight (or molar mass) is defined as the mass of one mole of a substance. Using the mass of the compound and the moles calculated:
\[
\text{Molecular Weight (g/mol)} = \frac{\text{mass of compound (g)}}{\text{moles of solute}}
\]
Substituting the values:
\[
\text{Molecular Weight} = \frac{0.0653 \, \text{g}}{0.000305 \, \text{moles}} \approx 214.75 \, \text{g/mol}
\]

### Conclusion
- The molecular weight of the compound is approximately **214.75 g/mol**.
- The concentration given is **molality (m)**.

You now have both the molecular weight and the determination that the concentration is molality, not molarity.