Answer :
To answer the given question, let's walk through the steps required to determine the total number of oxygen atoms deposited in the pond and the IUPAC name of the compound formed.
1. Identify the Given Data:
- Total weight of substance: 0.262 g
- Percentage composition by mass:
- Sodium (Na): 17.56%
- Chromium (Cr): 39.69%
- Oxygen (O): Remaining percentage
- Molar masses:
- Sodium ([tex]\( Na \)[/tex]): 23 g/mol
- Chromium ([tex]\( Cr \)[/tex]): 52 g/mol
- Oxygen ([tex]\( O \)[/tex]): 16 g/mol
- Avogadro's Number: [tex]\( 6.02 \times 10^{23} \, \text{mol}^{-1} \)[/tex]
2. Calculate the Mass of Each Element in the Compound:
- Mass of Sodium ([tex]\( \text{Na} \)[/tex]) = [tex]\( 0.262 \, \text{g} \times \frac{17.56}{100} = 0.0460072 \, \text{g} \)[/tex]
- Mass of Chromium ([tex]\( \text{Cr} \)[/tex]) = [tex]\( 0.262 \, \text{g} \times \frac{39.69}{100} = 0.1039878 \, \text{g} \)[/tex]
- Remaining Mass for Oxygen ([tex]\( \text{O} \)[/tex]) = [tex]\( 0.262 \, \text{g} - 0.0460072 \, \text{g} - 0.1039878 \, \text{g} = 0.112005 \, \text{g} \)[/tex]
3. Calculate the Moles of Each Element in the Compound:
- Moles of Sodium ([tex]\( \text{Na} \)[/tex]) = [tex]\( \frac{0.0460072 \, \text{g}}{23 \, \text{g/mol}} = 0.002000313 \, \text{mol} \)[/tex]
- Moles of Chromium ([tex]\( \text{Cr} \)[/tex]) = [tex]\( \frac{0.1039878 \, \text{g}}{52 \, \text{g/mol}} = 0.001999765 \, \text{mol} \)[/tex]
- Moles of Oxygen ([tex]\( \text{O} \)[/tex]) = [tex]\( \frac{0.112005 \, \text{g}}{16 \, \text{g/mol}} = 0.007000313 \, \text{mol} \)[/tex]
4. Calculate the Total Number of Oxygen Atoms:
- Number of Oxygen Atoms = Moles of Oxygen ([tex]\( \text{O} \)[/tex]) × Avogadro's Number
- Number of Oxygen Atoms = [tex]\( 0.007000313 \, \text{mol} \times 6.02 \times 10^{23} \, \text{mol}^{-1} = 4.214188125 \times 10^{21} \)[/tex]
5. Naming the Compound:
Given the atomic ratios and mass percentages, the compound is most likely Sodium Chromate, with the chemical formula [tex]\( \text{Na}_2\text{CrO}_4 \)[/tex].
Summary:
- Total number of oxygen atoms deposited in the pond: [tex]\( 4.214188125 \times 10^{21} \)[/tex]
- IUPAC Name: Sodium Chromate ([tex]\( \text{Na}_2\text{CrO}_4 \)[/tex])
1. Identify the Given Data:
- Total weight of substance: 0.262 g
- Percentage composition by mass:
- Sodium (Na): 17.56%
- Chromium (Cr): 39.69%
- Oxygen (O): Remaining percentage
- Molar masses:
- Sodium ([tex]\( Na \)[/tex]): 23 g/mol
- Chromium ([tex]\( Cr \)[/tex]): 52 g/mol
- Oxygen ([tex]\( O \)[/tex]): 16 g/mol
- Avogadro's Number: [tex]\( 6.02 \times 10^{23} \, \text{mol}^{-1} \)[/tex]
2. Calculate the Mass of Each Element in the Compound:
- Mass of Sodium ([tex]\( \text{Na} \)[/tex]) = [tex]\( 0.262 \, \text{g} \times \frac{17.56}{100} = 0.0460072 \, \text{g} \)[/tex]
- Mass of Chromium ([tex]\( \text{Cr} \)[/tex]) = [tex]\( 0.262 \, \text{g} \times \frac{39.69}{100} = 0.1039878 \, \text{g} \)[/tex]
- Remaining Mass for Oxygen ([tex]\( \text{O} \)[/tex]) = [tex]\( 0.262 \, \text{g} - 0.0460072 \, \text{g} - 0.1039878 \, \text{g} = 0.112005 \, \text{g} \)[/tex]
3. Calculate the Moles of Each Element in the Compound:
- Moles of Sodium ([tex]\( \text{Na} \)[/tex]) = [tex]\( \frac{0.0460072 \, \text{g}}{23 \, \text{g/mol}} = 0.002000313 \, \text{mol} \)[/tex]
- Moles of Chromium ([tex]\( \text{Cr} \)[/tex]) = [tex]\( \frac{0.1039878 \, \text{g}}{52 \, \text{g/mol}} = 0.001999765 \, \text{mol} \)[/tex]
- Moles of Oxygen ([tex]\( \text{O} \)[/tex]) = [tex]\( \frac{0.112005 \, \text{g}}{16 \, \text{g/mol}} = 0.007000313 \, \text{mol} \)[/tex]
4. Calculate the Total Number of Oxygen Atoms:
- Number of Oxygen Atoms = Moles of Oxygen ([tex]\( \text{O} \)[/tex]) × Avogadro's Number
- Number of Oxygen Atoms = [tex]\( 0.007000313 \, \text{mol} \times 6.02 \times 10^{23} \, \text{mol}^{-1} = 4.214188125 \times 10^{21} \)[/tex]
5. Naming the Compound:
Given the atomic ratios and mass percentages, the compound is most likely Sodium Chromate, with the chemical formula [tex]\( \text{Na}_2\text{CrO}_4 \)[/tex].
Summary:
- Total number of oxygen atoms deposited in the pond: [tex]\( 4.214188125 \times 10^{21} \)[/tex]
- IUPAC Name: Sodium Chromate ([tex]\( \text{Na}_2\text{CrO}_4 \)[/tex])