Question

The number of ethoxy group in an organic compound can be determined by the reactions:
$R\left( OC{{H}_{2}}C{{H}_{3}} \right)+xHI\to R{{\left( OH \right)}_{x}}+xC{{H}_{3}}C{{H}_{2}}I$
$C{{H}_{3}}C{{H}_{2}}I+A{{g}^{+}}+{{H}_{2}}O\to C{{H}_{3}}C{{H}_{2}}OH+AgI\left( s \right)$
When 37g of organic compound (molar mass = 176 $\dfrac{g}{mol}$) was treated as above, 148 AgI was precipitated out. How many ethoxy groups are present in each molecule of the organic compound?
(Ag = 108, I = 127)

Answer 1

The ethoxy group is $-OC{{H}_{2}}C{{H}_{3}}$ , with the mass of AgI precipitated, the number of ethoxy group in each molecule of the organic compound can be found out easily by using simple mathematical calculations.

Complete answer: Given: Molar mass = 176 $\dfrac{g}{mol}$ , mass of AgI precipitated = 148g, atomic weight of silver (Ag) = 108, atomic weight of iodine (I) = 127.
Now, let us first find the moles of AgI,
Moles of AgI = $\dfrac{148}{235}$ = 0.63 moles
Thus, there are 0.63 moles of AgI precipitated in the compound.
Now, to find the mass of the given compound, we have to get the product of the molar mass with the moles of precipitated AgI.
Thus, mass of the given compound = 176 $\times$ 0.63
= 110.88 g
Now, to find the number of ethoxy groups present in each molecule, we have to divide the mass of the precipitated compound with the mass of the organic compound.
Thus, number of ethoxy group in each molecule = $\dfrac{110.88}{37}$
= 2.9 = 3.
Here, 2.9 has been rounded off to 3. Thus there are 3 ethoxy groups present in each molecule of the organic compound.

Note:
The number of ethoxy groups obtained may be in decimals, the decimal number can be rounded off to the closest whole number possible. For example in this problem, we got 2.9 as the answer which is rounded off to 3.