Author Topic: One simple, impossible, chemistry problem. (Read 614 times) Tweet Share

run-bandit · « **on:** February 21, 2010, 09:48:01 pm »

   A 0.360 sample of an organic compound containing carbon, hydrogen and oxygen was burnt in excess oxygen. When the gases evolved were passed through anhydrous CaCl2 its mass increased by 0.326g. The remaining gases, when bubbled through a NaOH solution, increased its mass by 0.528g. The mass of oxygen in the sample is:

   A. 0.192
   B. 0.144
   C. 0.024
   D. Unable to be determined from the information.

My class teacher could not answer this!

It is far harder than I thought, could you explain it to me?

kenhung123 · « **Reply #1 on:** February 21, 2010, 09:52:14 pm »

m(H)=2/18*0.326
m(C)=12/44*0.528
m(O)=0.36-m(H)-m(C)
=0.36-0.036-0.144
=0.18g

Mao · « **Reply #2 on:** February 22, 2010, 12:39:56 am »

The organic compound was burnt in excess oxygen, hence the products will be CO2 and H2O

CaCl2 - http://en.wikipedia.org/wiki/Calcium_chloride

$CaCl_2(s) + 2H_2 O \to CaCl_2\cdot 2H_2O (s)$
Hence, the mass gained by CaCl2 is from the water absorbed. n(H₂O) = 0.326/18

$CO_2 + OH^- \to HCO_3^-$
Hence, the mass gained by NaOH solution is from the carbon dioxide absorbed, n(CO₂) = 0.528/44

For any combustion reaction:
$C_xH_yO_z + k O_2 \to x CO_2 + \frac{y}{2} H_2 O$

By this relationship, we can see that $n(C) = n(CO_2)$ , and $n(H) = 2\times n(H_2O)$
Hence, $m(C) = 12.0 \times n(CO_2)$ , $m(H) = 1.01 \times 2 \times n(H_2 O)$

Since $m(Sample) = m(C) + m(H) + m(O)$ , we then have

$m(O) = 0.360 - 12\times \frac{0.528}{44} - 2.02 \times \frac{0.326}{18}$

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Author Topic: One simple, impossible, chemistry problem. (Read 614 times) Tweet Share

run-bandit

One simple, impossible, chemistry problem.

kenhung123

Re: One simple, impossible, chemistry problem.

Mao

Re: One simple, impossible, chemistry problem.

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