An organic compound weighing 500 mg, produced 220 mg of CO$_2$ on complete combustion. The percentage composition of carbon in the compound is ............... % (nearest integer).
(Given molar mass in g mol$^-$ of C: 12, O: 16)
(Given molar mass in g mol$^-$ of C: 12, O: 16)
Show Hint
Correct Answer: 12
Approach Solution - 1
When the compound undergoes complete combustion, carbon from the compound reacts with oxygen to produce CO\(_2\). The molar mass of CO\(_2\) is: \[ \text{Molar mass of CO}_2 = 12 + 2 \times 16 = 44 \text{ g/mol} \] The molar mass of carbon (C) is 12 g/mol. In 44 g of CO\(_2\), 12 g is carbon. Therefore, the mass of carbon in 220 mg of CO\(_2\) can be calculated as: \[ \text{Mass of C} = \left(\frac{12}{44}\right) \times 220 = 60 \text{ mg} \]
Step 2: Calculate the percentage of carbon in the compound:
The mass of carbon in the original organic compound is 60 mg. The total mass of the compound is 500 mg. The percentage of carbon in the compound is given by: \[ \text{Percentage of C} = \left(\frac{60}{500}\right) \times 100 = 12% \]
Approach Solution -2
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
Substituting the values:
Percentage of carbon = (60 / 500) × 100 = 12%
Final Answer:
The percentage composition of carbon in the compound is 12%.
Top Questions on Combustion and Flame
Questions Asked in JEE Main exam
Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance \( R_p = 1 \, \Omega \) as shown in the figure. An external resistance of \( R_e = 2 \, \Omega \) is connected via the sliding contact.
The current \( i \) is :
- JEE Main - 2025
- Semiconductor electronics: materials, devices and simple circuits
Find the equivalent capacitance between A and B, where \( C = 16 \, \mu F \).
- JEE Main - 2025
- Capacitors
A line charge of length \( \frac{a}{2} \) is kept at the center of an edge BC of a cube ABCDEFGH having edge length \( a \). If the density of the line is \( \lambda C \) per unit length, then the total electric flux through all the faces of the cube will be : (Take \( \varepsilon_0 \) as the free space permittivity)
- JEE Main - 2025
- Electric Flux
- Given below are two statements: Statement I : In a vernier callipers, one vernier scale division is always smaller than one main scale division. Statement II : The vernier constant is given by one main scale division multiplied by the number of vernier scale divisions. In light of the above statements, choose the correct answer from the options given below.
- JEE Main - 2025
- Units and measurement
A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of \( 2 \times 10^5 \, \text{m/s} \). When the electric field is switched off, the proton moves along a circular path of radius 2 cm. The magnitude of electric field is \( x \times 10^4 \, \text{N/C} \). The value of \( x \) is \(\_\_\_\_\_\). (Take the mass of the proton as \( 1.6 \times 10^{-27} \, \text{kg} \)).
- JEE Main - 2025
- Electromagnetic Field (EMF)