Match the following List -I with list -IIcombustion yields 0-44 g - as it-s molecule formula- of - CO - 2 -hydrocarbon containing 10-5 g carbon - S - CO - 2 - text gas produced by the combustion of textper gram of hydrogen having vapour density ofA- I - Q-S-U- II - P- S- T- U- III - P- Q- R- U- IV - Q- R- UB- I - Q-S- II - P- T- U- III - P- Q- R- U- IV - Q- R- UC- I - Q-S-U- II - P- T- U- III - Q- R- U- IV - Q- R- UD- I - Q-S-U- II - P- S- T- U- III - P- Q- R- U- IV - Q- U

The correct option is A (I Q,S,U); (II P, S, T, U); (III P, Q, R, U); (IV Q, R, U)(a) nbsp; C : H : O = 51.1712 : 13.041 : 34.7816 = 4 : 12 : 2 or ...

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Byju's Answer

Standard XII

Chemistry

Classification of Amines

Match the fol...

Question

Match the following List -I with list -II
$\begin{matrix} L i s t - I & L i s t - I I (I) A gaseous organic compound containing & (P) One mole of compound contains 4 N_{A} atoms C = 52.17 %, H = 13.04 % a n d O = 34.78 % & of Hydrogen. (by weight) having molar mass 46 g/mol. (II) 0.3 g of an organic compound containing & (Q) The empirical formula of the compound is same C, H and O on combustion yields 0.44 g & as it's molecule formula. o f C O_{2} and 0.18 g of H_{2} O, with two O atoms per molecule (III) A hydrocarbon containing C = 42.857 % & (R) Combustion products of one mole of compound a n d H = 57.143 % (by mole) containing 3C & contains large number of moles of C O_{2} atoms per molecule. & than that of H_{2} O . (IV) A hydrocarbon containing 10.5 g carbon & (S) C O_{2} gas produced by the combustion of per gram of hydrogen having vapour density of 46 & 0.25 mole of compound occupies a volume of 11.2 L at NTP. (T) Empirical formula and molecular formula are not same. (U) C O_{2} a n d H_{2} O are formed on combustion. \end{matrix}$

(I - Q,S,U); (II - P, S, T, U); (III - P, Q, R, U); (IV - Q, R, U)

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(I - Q,S); (II - P, T, U); (III - P, Q, R, U); (IV - Q, R, U)

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(I - Q,S,U); (II - P, T, U); (III - Q, R, U); (IV - Q, R, U)

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(I - Q,S,U); (II - P, S, T, U); (III - P, Q, R, U); (IV - Q, U)

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Solution

The correct option is A (I - Q,S,U); (II - P, S, T, U); (III - P, Q, R, U); (IV - Q, R, U)
(a) $C : H : O = \frac{51.17}{12} : \frac{13.04}{1} : \frac{34.78}{16} = 4 : 12 : 2 o r 2 : 6 : 1$
$∴$ Empirical formula $= C_{2} H_{6} O$ and molar mass = 46 g/mol , Empirical mass = 46 g/mol so n=1
$∴$ Molecular formula $= C_{2} H_{6} O$
$C_{2} H_{6} O + 3 O_{2} \to 2 C O_{2} + 3 H_{2} O$
1 mole 44.8 L at STP
0.25 mole (11.2 L at STP)

(b) Mass of C in organic compound = mass of C in $C O_{2} = \frac{0.44}{44} \times 12 = 0.12 g$
Mass of H in organic compound = Mass of H in $H_{2} O = \frac{0.18}{18} \times 2 = 0.02 g$
$∴$ Mass of O in organic compound = 0.3 - (0.12 + 0.02) = 0.16 g
$∴$ C : H : O $= \frac{0.12}{12} : \frac{0.02}{1} : \frac{0.16}{16} = 0.01 : 0.02 : 0.01 = 1 : 2 : 1$
$∴$ Empirical formula $= C H_{2} O,$ but it contains 2 O atoms per molecule.
$∴$ Molecular formula $= C_{2} H_{4} O_{2}$
1 mole of $C_{2} H_{4} O_{2}$ contains $4 N_{A}$ hydrogen atoms.
$C_{2} H_{4} O_{2} + 2 O_{2} \to 2 C O_{2} + 2 H_{2} O$
1 mole 44.8 L
0.25 mole 11.2 L

(c) C : H = 42.857 : 57.143
= 3 : x (given)
On solving, x = 4 $∴$ molecular formula $= C_{3} H_{4}$ 1 mole of $C_{3} H_{4}$ contains $4 N_{A}$ hydrogen atoms. Empirical formula is same as molecular formula $C_{3} H_{4} + 4 O_{2} \to 3 C O_{2} + 2 H_{2} O$
$n_{C O_{2}} > n_{H_{2} O}$

(d) $C : H = \frac{10.5}{12} : \frac{1}{1} = \frac{7}{8} : 1 = 7 : 8$
therefore Empirical formula $= C_{7} H_{8}$
Mol wt. $= 2 \times V D = 2 \times 46 = 92$
$∴$ Molecular formula = Empirical formula $= C_{7} H_{8}$
$C_{7} H_{8} + 9 O_{2} \to 7 C O_{2} + 4 H_{2} O$
$n_{C O_{2}} > n_{H_{2} O}$

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