Answer :
To find the empirical formula of the compound formed from tin ([tex]\(Sn\)[/tex]) and iodine ([tex]\(I_2\)[/tex]), we need to determine the simplest whole number ratio of the moles of tin to the moles of iodine that reacted. Here's the step-by-step solution:
1. Determine the mass of tin that reacted:
[tex]\[ \text{Mass of } Sn \text{ that reacted} = \text{Mass of } Sn \text{ in original mixture} - \text{Mass of } Sn \text{ recovered after reaction} \][/tex]
[tex]\[ \text{Mass of } Sn \text{ that reacted} = 1.056 \, \text{g} - 0.601 \, \text{g} = 0.455 \, \text{g} \][/tex]
2. Calculate the moles of tin that reacted:
Using the atomic weight of tin ([tex]\(Sn\)[/tex]), which is [tex]\(118.71 \, \text{g/mol}\)[/tex]:
[tex]\[ \text{Moles of } Sn = \frac{\text{Mass of } Sn \text{ that reacted}}{\text{Atomic weight of } Sn} = \frac{0.455 \, \text{g}}{118.71 \, \text{g/mol}} = 0.00383 \, \text{mol} \][/tex]
3. Calculate the moles of iodine that reacted:
Using the molecular weight of iodine ([tex]\(I_2\)[/tex]), which is [tex]\(2 \times 126.90 \, \text{g/mol} = 253.80 \, \text{g/mol}\)[/tex]:
[tex]\[ \text{Moles of } I_2 = \frac{\text{Mass of iodine in original mixture}}{\text{Molecular weight of } I_2} = \frac{1.947 \, \text{g}}{253.80 \, \text{g/mol}} = 0.00767 \, \text{mol} \][/tex]
4. Determine the mole ratio of tin to iodine:
To simplify, we divide each mole quantity by the smallest number of moles calculated:
[tex]\[ \text{Mole ratio of } Sn = \frac{0.00383 \, \text{mol}}{0.00383 \, \text{mol}} = 1.0 \][/tex]
[tex]\[ \text{Mole ratio of } I_2 = \frac{0.00767 \, \text{mol}}{0.00383 \, \text{mol}} = 2.0 \][/tex]
5. Write the empirical formula:
The simplest whole number ratio of moles of tin to moles of iodine is [tex]\(1:2\)[/tex]. This gives us the empirical formula:
[tex]\[ \text{Empirical formula} = SnI_2 \][/tex]
In conclusion, the empirical formula of the compound formed from the reaction of tin and iodine is [tex]\( SnI_2 \)[/tex].
1. Determine the mass of tin that reacted:
[tex]\[ \text{Mass of } Sn \text{ that reacted} = \text{Mass of } Sn \text{ in original mixture} - \text{Mass of } Sn \text{ recovered after reaction} \][/tex]
[tex]\[ \text{Mass of } Sn \text{ that reacted} = 1.056 \, \text{g} - 0.601 \, \text{g} = 0.455 \, \text{g} \][/tex]
2. Calculate the moles of tin that reacted:
Using the atomic weight of tin ([tex]\(Sn\)[/tex]), which is [tex]\(118.71 \, \text{g/mol}\)[/tex]:
[tex]\[ \text{Moles of } Sn = \frac{\text{Mass of } Sn \text{ that reacted}}{\text{Atomic weight of } Sn} = \frac{0.455 \, \text{g}}{118.71 \, \text{g/mol}} = 0.00383 \, \text{mol} \][/tex]
3. Calculate the moles of iodine that reacted:
Using the molecular weight of iodine ([tex]\(I_2\)[/tex]), which is [tex]\(2 \times 126.90 \, \text{g/mol} = 253.80 \, \text{g/mol}\)[/tex]:
[tex]\[ \text{Moles of } I_2 = \frac{\text{Mass of iodine in original mixture}}{\text{Molecular weight of } I_2} = \frac{1.947 \, \text{g}}{253.80 \, \text{g/mol}} = 0.00767 \, \text{mol} \][/tex]
4. Determine the mole ratio of tin to iodine:
To simplify, we divide each mole quantity by the smallest number of moles calculated:
[tex]\[ \text{Mole ratio of } Sn = \frac{0.00383 \, \text{mol}}{0.00383 \, \text{mol}} = 1.0 \][/tex]
[tex]\[ \text{Mole ratio of } I_2 = \frac{0.00767 \, \text{mol}}{0.00383 \, \text{mol}} = 2.0 \][/tex]
5. Write the empirical formula:
The simplest whole number ratio of moles of tin to moles of iodine is [tex]\(1:2\)[/tex]. This gives us the empirical formula:
[tex]\[ \text{Empirical formula} = SnI_2 \][/tex]
In conclusion, the empirical formula of the compound formed from the reaction of tin and iodine is [tex]\( SnI_2 \)[/tex].
