Direct reaction of solid iodine (I2) and gaseous chlorine (Cl2) produces an iodine chloride,(IxCly), a bright yellow solid. 0.678 g of iodine (I2) is mixed with 0.851 g of chlorine (Cl2). After the elements react, some chlorine is left over along with a single compound that contains the two
elements (IxCly). The general unbalanced equation for this process is written below.
I2(s) + Cl2 (g) → IxCly (s) + (excess Cl2)
The excess chlorine is removed and allowed to react with elemental sodium producing 0.450 g of NaCl.
The balanced reaction is shown below.
2Na (s) + Cl2(g) → 2NaCl (s)
Based on this information, determine the empirical formula of the compound that was formed when
iodine reacted with chlorine.
I2 = 253.8 g/mol
Cl2 = 70.90 g/mol
NaCl = 58.44 g/mol
To determine the empirical formula of the compound formed when iodine reacted with chlorine, we need to find the moles of iodine (I2) and chlorine (Cl2) used and calculate the ratio between them.
1. Calculate the moles of iodine (I2):
Moles of I2 = Mass of I2 / Molar mass of I2
Moles of I2 = 0.678 g / 253.8 g/mol
Moles of I2 = 0.002676 mol
2. Calculate the moles of chlorine (Cl2):
Moles of Cl2 = Mass of Cl2 / Molar mass of Cl2
Moles of Cl2 = 0.851 g / 70.90 g/mol
Moles of Cl2 = 0.0120 mol
3. Determine the ratio between the moles of iodine and chlorine:
Moles ratio = Moles of I2 / Moles of Cl2
Moles ratio = 0.002676 mol / 0.0120 mol
Moles ratio ≈ 0.223 mol : 1 mol
4. Simplify the ratio to the nearest whole numbers by dividing both sides by the smallest number:
Moles ratio ≈ 0.223 mol : 1 mol
Moles ratio ≈ 1 mol : 4.474 mol
5. Write the empirical formula using the simplified ratio:
Empirical formula ≈ I1Cl4 or ICl4
Therefore, the empirical formula of the compound formed when iodine reacts with chlorine is ICl4.
To determine the empirical formula of the compound formed when iodine reacts with chlorine, we need to calculate the moles of iodine, chlorine, and the compound, and then find the ratio of their moles.
1. Calculate the moles of iodine (I2) and chlorine (Cl2):
Moles of I2 = Mass of I2 / Molar mass of I2
= 0.678 g / 253.8 g/mol
= 0.002674 mol
Moles of Cl2 = Mass of Cl2 / Molar mass of Cl2
= 0.851 g / 70.90 g/mol
= 0.01198 mol
2. Determine the moles of excess chlorine that reacted with sodium (Na):
Moles of NaCl = Mass of NaCl / Molar mass of NaCl
= 0.450 g / 58.44 g/mol
= 0.007705 mol
From the balanced equation between Na and Cl2: 2Na (s) + Cl2(g) → 2NaCl (s)
1 mole of Cl2 reacts with 2 moles of NaCl.
Therefore, moles of excess Cl2 = 0.007705 mol / 2
= 0.003853 mol
Subtracting moles of excess Cl2 from the moles of Cl2 initially present:
Moles of Cl2 remaining = Moles of Cl2 - Moles of excess Cl2
= 0.01198 mol - 0.003853 mol
= 0.008127 mol
3. Determine the moles of compound (IxCly) formed:
From the balanced equation between I2 and Cl2: I2(s) + Cl2 (g) → IxCly (s) + (excess Cl2)
1 mole of I2 reacts with 1 mole of IxCly.
Therefore, moles of IxCly = Moles of I2 = 0.002674 mol
4. Determine the ratio of moles between iodine and chlorine in the compound:
Divide the number of moles of each element by the smallest number of moles.
Moles of IxCly / Moles of I2 = 0.002674 mol / 0.002674 mol = 1
Moles of IxCly / Moles of Cl2 = 0.002674 mol / 0.008127 mol = 0.328
Round the ratio to the nearest whole number:
The empirical formula of the compound formed when iodine reacts with chlorine is approximately ICl3.
.........I2 + Cl2 ==> IxCly + xs Cl2
.....0.678g..0.851g....0..........x
2Na + Cl2 --> 2NaCl
You know NaCl weighed 0.450. Convert that to grams Cl2 and that is the excess amount. Subtract that from the 0.851 g there initially.
Now you have 0.678 g I2 and I obtained 0.578 g Cl2.
Now fine mols on the way to empirical formula
mols Cl2 = 0.578/35.44 = ?
mols I2 = 0.678/126.9 = ?
Now find the ratio with the smallest number being 1.00. The easy way to do that is to divide the smaller number by itself and divide the other number by the small number too. Then round to whole number.
Post your work if you get stuck. I ran through a quickie calculation and obtained ICl3
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