You're supercool if you help me solve Q-23 and 24 :D
Answers
Explanation:
23)C10H10=12×10+10×1=130g/mol
23) option (d).
Answer:
23. (D) C₁₀H₁₀
24. (A) C₅H₁₂
Explanation:
23. Let the number of atoms each of Carbon and Hydrogen be x
we know that, the mass of carbon is 12 and hydrogen is 1
So, 12x + 1x = 130 [Rounding off helps us to get answer in a simple way]
=> 13x = 130
=> x = 10
So, number of atoms of C and H are 10
So, the molecular formula is C₁₀H₁₀ [Option D] is the right answer
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24. Burning 10g of sample, we got 30.5g of CO₂ and 14.98g of H₂O
We know that, Combustion of hydrocarbon is of the form,
So, Now, with the provided masses of the carbon dioxide and the water, we can calculate the molar amounts of carbon and hydrogen in the sample.
First we calculate the molar masses:
C = 12.011 x 1 = 12.011 g/mol
O = 15.99 x 2 = 31.99 g/mol
CO₂ = 12.011 + 31.99 = 44.001 g/mol
H = 1.008 x 2 = 2.016 g/mol
O = 15.99 x 1 = 15.99 g/mol
H₂O = 2.01 + 15.99 = 18.006 g/mol
Now we obtain the molar amounts of C and H using the obtaines masses of carbon dioxide and water:
mol C = 30.50g CO₂ x (1mol CO₂)/(44.001 g/mol) x (1mol C)/(1mol CO₂) = 0.6931 mol C
mol H = 14.98g H₂O x (1mol H₂O)/(18.006 g/mol) x (2mol H)/(1mol H₂O) = 1.6638 mol H
Finally, we can obtain the H/C molar ratio by identifying the smaller whole-number ratio for these molar amounts. For this we can first divide each molar amount by the smaller amount:
mol C = 0.6931/0.6931 = 1
mol H = 1.6638/0.6931 = 2.4
As we are still getting a decimal amount for the hydrogen, what we can do is multiply both molar amounts by the smaller whole multiple that can give us a whole number for the hydrogen's molar amount, in this case, that multiple would be 5:
mol C = 0.6931/0.6931 = 1 x 5 = 5
mol H = 1.6638/0.6931 = 2.4 x 5 = 12
Now we can write the empirical formula for the hydrocarbon, which is:
C₅H₁₂[Option A is the right answer]
Hope it helps!!