Question

A sample of propanoic acid CH3,-CH2-COOH,2.80×10 23

Answer 1

It seems like you're providing a quantity of propanoic acid (CH₃-CH₂-COOH) and mentioning a sample size of \(2.80 \times 10^{23}\) molecules. If you need to calculate the mass or some other property of this sample, I can help with that!

### Step 1: Calculate the number of moles

To find the number of moles of propanoic acid in your sample, use Avogadro's number, which is approximately \(6.022 \times 10^{23}\) molecules/mol.

\[
\text{Number of moles} = \frac{\text{Number of molecules}}{\text{Avogadro's number}} = \frac{2.80 \times 10^{23}}{6.022 \times 10^{23}} \approx 0.465 \text{ moles}
\]

### Step 2: Calculate the molar mass of propanoic acid

The molecular formula for propanoic acid (CH₃CH₂COOH) can be broken down as follows:

- Carbon (C): 3 atoms × 12.01 g/mol = 36.03 g/mol
- Hydrogen (H): 6 atoms × 1.008 g/mol = 6.048 g/mol
- Oxygen (O): 2 atoms × 16.00 g/mol = 32.00 g/mol

Adding these together gives the molar mass of propanoic acid:

\[
\text{Molar Mass} = 36.03 \, \text{g/mol} + 6.048 \, \text{g/mol} + 32.00 \, \text{g/mol} = 74.078 \, \text{g/mol}
\]

### Step 3: Calculate the mass of the sample

Now you can find the mass of your sample using the number of moles and the molar mass:

\[
\text{Mass} = \text{Number of moles} \times \text{Molar mass} = 0.465 \, \text{moles} \times 74.078 \, \text{g/mol} \approx 34.41 \, \text{grams}
\]

### Summary

- **Number of Moles**: Approximately 0.465 moles
- **Mass of Sample**: Approximately 34.41 grams

If you need additional calculations or information, feel free to ask!