Step 1:
When an organic compound undergoes complete combustion, it produces CO₂ and H₂O. The carbon in the compound is converted into CO₂. Therefore, the amount of carbon in the CO₂ produced will help in calculating the percentage composition of carbon in the compound.
Step 2:
We are given that 220 mg of CO₂ is produced. To find the mass of carbon in CO₂, we use the molar masses:
- Molar mass of CO₂ = 12 (C) + 32 (O₂) = 44 g/mol.
- Molar mass of carbon (C) = 12 g/mol.
Now, the mass of carbon in 220 mg of CO₂ can be calculated using the ratio of the molar masses of carbon and CO₂:
Mass of carbon = (12 / 44) × 220 = 60 mg
Step 3:
The total mass of the compound is 500 mg. To find the percentage of carbon in the compound, we use the formula:
Percentage of carbon = (Mass of carbon / Mass of compound) × 100
Percentage of carbon = (60 / 500) × 100 = 12%
Final Answer:
The percentage composition of carbon in the compound is 12%.
Given below are two statements:
Statement (I):
are isomeric compounds.
Statement (II):
are functional group isomers.
In the light of the above statements, choose the correct answer from the options given below:
Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with
\(K_4\)[Fe(CN)\(_6\)] is : \[ {Cu}^{2+}, \, {Fe}^{3+}, \, {Ba}^{2+}, \, {Ca}^{2+}, \, {NH}_4^+, \, {Mg}^{2+}, \, {Zn}^{2+} \]