Chemistry - Mole Concept Question with Solution | TestHub

(A) $\mathrm{C}:\mathrm{H}:\mathrm{O}=\frac{52.17}{12}:\frac{13.04}{1}:\frac{34.78}{16}=4:12:2$ or $2:6:1$

Empirical formula $={\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}$ & molar mass $=46\mathrm{g}/\mathrm{mol}$

Molecular formula $={\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}$

${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}+3{\mathrm{O}}_2\to 2{\mathrm{CO}}_2+3{\mathrm{H}}_2\mathrm{O}$

1 mole 44.8 L at NTP

0.25 mole (11.2 L at NTP)

(B) Mass of C in organic compound = mass of C in CO₂ = $0.44/44\times 12=0.12g$

Mass of H in organic compound = Mass of H in H₂O = $0.18/18\times 2=0.02g$

∴ Mass of O in organic compound = $0.3–(0.12+0.02)=0.16g$

∴ C : H : O = $0.12/12:0.02/1:0.16/16=0.01:0.02:0.01=1:2:1$

∴ Empirical formula = CH₂O, but it contains 2 O atoms per molecule

∴ Molecular formula = C₂H₄O₂

1 mole of C₂H₄O₂ contains 4 Nᴀ hydrogen atoms.

C₂H₄O₂ + 2O₂ → 2CO₂ + 2H₂O

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 = C₃H₄

1 mole of C₃H₄ contains 4Nᴀ hydrogen atoms.

Empirical formula is same as molecular formula.

C₃H₄ + 4O₂ → 3CO₂ + 2H₂O

$n_{C{O}_2}>n_{H_{2}O}$

(D) C: H = $10.5/12:1/1=7/8:1=7:8$. Empirical formula = C₇H₈

Mol wt. = $2\times VD=2\times 46=92$

Mol formula = Empirical formula = C₇H₈

C₇H₈ + 9O₂ → 7CO₂ + 4H₂O

$n_{C{O}_2}>n_{H_{2}O}$