If An Unknown Compound Contains Only The Three Elements C, H, And O, And A Pure Sample Of The Compound Is Analyzed And Found To Be 65.60 Percent C And 9.44 Percent H By Mass, What Is Compound?

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Oddman Profile
Oddman answered
You read the molar masses of the constituent elements from a periodic table or other reference.
Molar masses of elements:
 C - 12.011
 O - 15.999
 H - 1.008
You know the relative proportions of the masses of these in the unknown:
 C - 65.60%
 O - 100-(65.60+9.44) = 100-75.04 = 24.96%
 H - 9.44%

Let X be the molar mass of the unknown compound. Then we know
65.60%(X)/12.011 =  moles of carbon
24.96%(X)/15.999 = moles of oxygen
9.44%(X)/1.008 = moles of hydrogen.

We are interested in the ratios of these, so we compute
 (moles of carbon)/(moles of hydrogen) = (65.60%X/12.011)/(9.44%X/1.008)
  = 65.60(1.008)/(12.011*9.44) = 66.1248/113.384 = .5832
 (moles of oxygen)/(moles of hydrogen) = (24.96%X/15.999)/(9.44%X/1.008)
  = 24.96(1.008)/(15.999*9.44) = 25.1597/151.031 = .1666
Clearly there are more moles of hydrogen than of either of the other two elements.

These fractions don't look very helpful, so perhaps the inverse will be more interesting:
 1/.1666 = 6.00, suggesting 6H for every O.
 1/.5832 = 1.715, not suggesting anything immediately.
How about .5832/.1666 = 3.50, suggesting 7C for every 2O.

So, our empirical formula for X is now C7O2H12.

Check:
The molar mass of X is 7(12.011) + 2(15.999) + 12(1.008)
 = 84.077 + 31.998 + 12.096 = 128.171
%C = 84.077/128.171 = 65.597%
%O = 31.998/128.171 = 24.965%
%H = 12.096/128.171 = 9.437%
These compare favorably with the given numbers and what we computed.
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Note: We also learned that for this problem it would have been more useful to find the ratios of the other elements to moles of oxygen, rather than hydrogen. By using hydrogen as the base, we were looking at ratios of 2/12 and 7/12, not fractions whose decimal values we are usually familiar with. Had we used oxygen as the base, we would have seen ratios of 7/2=3.5 and 12/2=6 -- the numbers we actually ended up looking at.

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