Answer to Question #141911 in Physical Chemistry for Alvin

Question

2.4g of a compound of coarbon, hydrogen and oxygen gave on combustion 3.52g of co2 and 1.44g of h2o.The Mr of the compound was found to be 60. What are the masses of c H o and find the empirical formula and the molecular of the formula

Answer 1

1) moles C

$3.52 g (CO_2)\times \frac{1 mole (CO_2)}{44.01 g (CO_2)}\times \frac{1 mole (C)}{1 mole (CO_2)}= 0.0800 moles (C)$

mass C

$0.0800 moles (C)\times \frac{12.01 g (C)}{1 mole (C)}=0.961 g(C)$

2) moles H

$1.44 g (H_2O)\times\frac{1 mole (H_2O)}{18.02 g (H_2O)}\times \frac{2 mole (H)}{1 mole (H_2O)}=0.160 moles (H)$

mass H

$0.160 moles (H)\times \frac{1.01 g(H)}{1 mole (H)}= 0.161 g (H)$

3) mass O

$2.4=m(O)+m(C)+m(H)$

$2.4= m(O) +0.961 + 0.161$

m$(O) = 1.278g$

moles O

$1.278 g (O) \times \frac{1 mole (O)}{16.00 g (O)}=0.08 moles (O)$

4) Empirical formula

$n(C):n(H):n(O) = 0.08:0.16:0.08=1:2:1$

$EF=CH_2O$

5) Molecular formula

M$(CH_2O)=30.03\frac{g}{mol}$

$\frac{M(MF)}{M(EF)}=\frac{60}{30}=2$

$MF=EF\times2= C_2H_4O_2$