How many moles of PbCrO4 are in 50.0 g of PbCrO4?
The correct answer and explanation is:
To find how many moles of PbCrO₄ (lead(II) chromate) are in 50.0 g, we use the formula: Moles=Mass (g)Molar Mass (g/mol)\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}
Step 1: Calculate the molar mass of PbCrO₄
Use the atomic masses from the periodic table:
- Pb (lead) = 207.2 g/mol
- Cr (chromium) = 52.0 g/mol
- O (oxygen) = 16.0 g/mol × 4 = 64.0 g/mol
Molar Mass of PbCrO₄=207.2+52.0+64.0=323.2 g/mol\text{Molar Mass of PbCrO₄} = 207.2 + 52.0 + 64.0 = 323.2\ \text{g/mol}
Step 2: Use the formula to find moles
Moles of PbCrO₄=50.0 g323.2 g/mol≈0.155 mol\text{Moles of PbCrO₄} = \frac{50.0\ \text{g}}{323.2\ \text{g/mol}} \approx 0.155\ \text{mol}
✅ Correct Answer:
0.155 mol of PbCrO₄\boxed{0.155\ \text{mol of PbCrO₄}}
🔍 Explanation (300 words):
In chemistry, the number of moles represents the quantity of a substance based on the number of particles (atoms, ions, or molecules) it contains. The mole is a central concept in stoichiometry because it allows chemists to relate mass to the number of particles using the substance’s molar mass — the mass of one mole of a compound.
In this problem, we are given 50.0 grams of lead(II) chromate (PbCrO₄), and we need to find how many moles this mass represents. The first step is to calculate the molar mass of PbCrO₄ by adding the atomic masses of its elements:
- Pb contributes 207.2 g/mol,
- Cr adds 52.0 g/mol,
- and the four oxygen atoms contribute 64.0 g/mol total.
This gives a total molar mass of 323.2 g/mol.
Next, we apply the basic formula for converting grams to moles: Moles=massmolar mass=50.0323.2≈0.155\text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{50.0}{323.2} \approx 0.155
So, 50.0 grams of PbCrO₄ contains approximately 0.155 moles. This calculation is critical in laboratory settings when preparing solutions, conducting reactions, or analyzing sample compositions. Accurate mole calculations ensure proper stoichiometric ratios and successful experimental outcomes.