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A 10.00 g sample of a compound containing C, H, and O is burned completely to produce...
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Let us say the empirical formulae of the acid is `C_xH_yO` .
`C_xH_yO +(x+y/2-1)O_2 rarr xCO_2+(y/2)H_2O`
Molar mass in `g/(mol)`
`C = 12`
`H = 1`
`O = 16`
`CO_2 = 44`
`H_2O = 18`
Amount of `CO_2` produced `= 14.67/44 = 0.333`
Amount of `H_2O` produced `= 6/18 = 0.333`
Let us say we have n moles of the compound.
`C_xH_yO:CO_2 = 1:x`
1:x = n:0.333
`x = 0.333/n = 1/(3n)`
`C_xH_yO:H_2O = 1:y/2`
`1:y/2 = n:0.333`
`y = 2/(3n)`
It is given that the mass of the compound was 10g.
`n(12x+1y+16) = 10`
`n((12/(3n))+2/(3n)+16) = 10`
`4+2/3+16n = 10`
`n = 16/3`
`x = 1/16`
`y = 1/8`
`C:H = 1/16:1/8 = 1:2`
So the empirical formulae would be `CH_2O`
Posted by jeew-m on August 10, 2013 at 4:43 AM (Answer #1)
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