Analysis of skunk spray yields a molecule with 44.77% c, 7.46% h and 47.76% s by mass. what is the empirical formula for this mo
1 answer:
Suppose we have 100 gr of the substance. Then by weight, it would contain 44.77 gr of C, 7.46 gr of H and 47.76 gr of S. We need to look up the atomic weights of these atoms; M_H=1, M_C=12, M_S=32. The following formula holds (where n are the moles of the substance, M its molecular mass and m its mass): n=m/M. Substituting the known quantities for each element, we get that the substance has 3.73 moles of C, 7.46 moles of H and 1.49 moles of S. In the empirical formula for the molecule, all atoms appear an integer amout of times. Hence, for every mole of Sulfur, we have 2.5 moles of C and 5 moles of H (by taking the moles ratios). Thus, for every 2 moles of sulfur, we have 5 moles of C and 10 moles of H. Now that all the coefficients are integer, we have arrived at an empirical formula for the skunk spray agent: 
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Answer:
1. Butyraldehyde; 2. but-3-en-1-ol
Explanation:
1. Peak ≥ 160 ppm
The formula C₄H₈O shows that the Index of Hydrogen Deficiency = 1.
It could be caused by either a ring or a double bond.
A peak at ≥ 160 ppm strongly indicates a C=O group, so the rest of the molecule can contain no rings or double bonds.
There are no other heteroatoms, so the compound most be an aldehyde or a ketone.
One compound that meets these criteria is butyraldehyde, CH₃CH₂CH₂CH=O (see Fig. 1.)
2. No rings; all peaks < 160 ppm
If all peaks are < 160 ppm, there can no C=O group.
There is no ring, so there must be a C=C double bond.
There is no other unsaturation, so the O atom must be present as an alcohol or an ether.
One compound that meets these criteria is but-3-en-1-ol, CH₂=CHCH₂CH₂OH (see Fig. 2).
