Answer :
To determine the empirical formula of the lead oxide formed, follow these steps:
1. Calculate the mass of oxygen that reacted with the lead:
- The mass of the product is given as 1.154 grams.
- The mass of lead is given as 1.000 grams.
- Therefore, the mass of oxygen can be determined by subtracting the mass of lead from the mass of the product.
[tex]\[ \text{Mass of oxygen} = \text{Mass of product} - \text{Mass of lead} = 1.154 \, \text{g} - 1.000 \, \text{g} = 0.154 \, \text{g} \][/tex]
2. Calculate the moles of lead (Pb) in the sample:
- The molar mass of lead (Pb) is 207.2 g/mol.
- Using the given mass of lead, convert it to moles:
[tex]\[ \text{Moles of lead} = \frac{\text{Mass of lead}}{\text{Molar mass of lead}} = \frac{1.000 \, \text{g}}{207.2 \, \text{g/mol}} \approx 0.00483 \, \text{moles} \][/tex]
3. Calculate the moles of oxygen (O) in the sample:
- The molar mass of oxygen (O) is 16.0 g/mol.
- Using the calculated mass of oxygen, convert it to moles:
[tex]\[ \text{Moles of oxygen} = \frac{\text{Mass of oxygen}}{\text{Molar mass of oxygen}} = \frac{0.154 \, \text{g}}{16.0 \, \text{g/mol}} \approx 0.00962 \, \text{moles} \][/tex]
4. Determine the simplest whole-number ratio of moles of lead to moles of oxygen:
- Calculate the ratio of moles of lead to moles of oxygen:
[tex]\[ \text{Ratio of lead} = \frac{\text{Moles of lead}}{\text{Moles of lead}} = \frac{0.00483}{0.00483} = 1 \][/tex]
[tex]\[ \text{Ratio of oxygen} = \frac{\text{Moles of oxygen}}{\text{Moles of lead}} = \frac{0.00962}{0.00483} \approx 2 \][/tex]
5. Empirical formula determination:
- The ratio of lead to oxygen is approximately 1:2.
- Thus, the simplest whole-number ratio is 1:2.
- Therefore, the empirical formula for the lead oxide is [tex]\( \text{PbO}_2 \)[/tex].
So, the empirical formula of the lead oxide is [tex]\( \mathbf{PbO_2} \)[/tex].
1. Calculate the mass of oxygen that reacted with the lead:
- The mass of the product is given as 1.154 grams.
- The mass of lead is given as 1.000 grams.
- Therefore, the mass of oxygen can be determined by subtracting the mass of lead from the mass of the product.
[tex]\[ \text{Mass of oxygen} = \text{Mass of product} - \text{Mass of lead} = 1.154 \, \text{g} - 1.000 \, \text{g} = 0.154 \, \text{g} \][/tex]
2. Calculate the moles of lead (Pb) in the sample:
- The molar mass of lead (Pb) is 207.2 g/mol.
- Using the given mass of lead, convert it to moles:
[tex]\[ \text{Moles of lead} = \frac{\text{Mass of lead}}{\text{Molar mass of lead}} = \frac{1.000 \, \text{g}}{207.2 \, \text{g/mol}} \approx 0.00483 \, \text{moles} \][/tex]
3. Calculate the moles of oxygen (O) in the sample:
- The molar mass of oxygen (O) is 16.0 g/mol.
- Using the calculated mass of oxygen, convert it to moles:
[tex]\[ \text{Moles of oxygen} = \frac{\text{Mass of oxygen}}{\text{Molar mass of oxygen}} = \frac{0.154 \, \text{g}}{16.0 \, \text{g/mol}} \approx 0.00962 \, \text{moles} \][/tex]
4. Determine the simplest whole-number ratio of moles of lead to moles of oxygen:
- Calculate the ratio of moles of lead to moles of oxygen:
[tex]\[ \text{Ratio of lead} = \frac{\text{Moles of lead}}{\text{Moles of lead}} = \frac{0.00483}{0.00483} = 1 \][/tex]
[tex]\[ \text{Ratio of oxygen} = \frac{\text{Moles of oxygen}}{\text{Moles of lead}} = \frac{0.00962}{0.00483} \approx 2 \][/tex]
5. Empirical formula determination:
- The ratio of lead to oxygen is approximately 1:2.
- Thus, the simplest whole-number ratio is 1:2.
- Therefore, the empirical formula for the lead oxide is [tex]\( \text{PbO}_2 \)[/tex].
So, the empirical formula of the lead oxide is [tex]\( \mathbf{PbO_2} \)[/tex].
