A chloro compound "A".
(i) forms aldehydes on ozonolysis followed by the hydrolysis.
(ii) when vaporized completely 1.53 g of A, gives 448 mL of vapour at STP.
The number of carbon atoms in a molecule of compound A is __________.
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To quickly find molar mass from gas data at STP, you can use the relationship:
\[ \text{Molar Mass} = \frac{\text{Mass of gas (g)} \times 22400}{\text{Volume of gas (mL)}} \]
This combines the two steps of finding moles and then molar mass into one. For this problem: \((1.53 \times 22400) / 448 = 76.5\) g/mol.
Step 1: Understanding the Question
We are given two pieces of information about an unknown chloro compound 'A' and asked to find the number of carbon atoms in its molecule. Step 2: Key Formula or Approach
We will first use the data from point (ii) to determine the molar mass of compound A. Then, using this molar mass and the information from point (i), we can deduce the molecular formula and structure. Step 3: Detailed Calculation Calculate the molar mass of A from vapor data:
At Standard Temperature and Pressure (STP), 1 mole of any ideal gas occupies 22.4 liters (or 22400 mL).
Volume of vapour = 448 mL = 0.448 L.
\[ \text{Moles of A} = \frac{\text{Volume at STP (L)}}{22.4 \text{ L/mol}} = \frac{0.448 \text{ L}}{22.4 \text{ L/mol}} = 0.02 \text{ mol} \]
We are given that the mass of this amount of A is 1.53 g.
\[ \text{Molar Mass of A} = \frac{\text{Mass}}{\text{Moles}} = \frac{1.53 \text{ g}}{0.02 \text{ mol}} = 76.5 \text{ g/mol} \]
Deduce the molecular formula:
The compound 'A' is a chloro compound, so its formula is C\(_x\)H\(_y\)Cl.
The molar mass is: 12x + y + 35.5 = 76.5 (using Cl=35.5)
\[ 12x + y = 41 \]
We can test integer values for x (number of carbon atoms):
If x = 1, y = 41 - 12 = 29 (not possible).
If x = 2, y = 41 - 24 = 17 (not possible).
If x = 3, y = 41 - 36 = 5. The formula is C\(_3\)H\(_5\)Cl. This is a valid formula. Verify with information from (i):
The compound 'A' (C\(_3\)H\(_5\)Cl) forms aldehydes on ozonolysis. This implies that 'A' is an alkene. The degree of unsaturation for C\(_3\)H\(_5\)Cl is \(\frac{(2 \times 3 + 2) - (5+1)}{2} = \frac{8-6}{2} = 1\), which corresponds to one double bond.
A possible structure is 3-chloro-prop-1-ene (CH\(_2\)=CH-CH\(_2\)Cl). Ozonolysis of this compound would yield formaldehyde (HCHO) and 2-chloroethanal (Cl-CH\(_2\)-CHO), both of which are aldehydes. This is consistent with the given information. Step 4: Final Answer
The molecular formula of compound A is C\(_3\)H\(_5\)Cl. Therefore, the number of carbon atoms in a molecule of compound A is 3.
