(a) From the elemental analysis the empirical formula is C₂H₄O

---

(b) The M⁺ peak at m/z = 88 leads to the conclusion that the molar mass of Compound H is twice the empirical mass and the molecular formula is C₄H₈O₂. The fragment at m/z = 73 is due to the loss of a methyl group (M – CH₃)⁺ leaving C₃H₅O₂⁺. The fragment at m/z =  59 is due to the loss of an ethyl group (M – C₂H₅)⁺ leaving C₂H₃O₂⁺ and the fragment at m/z = 45 is due to the loss of an propyl group (M – C₃H₇)⁺ leaving CHO₂⁺.

---

(c) The sharp absorption at around 3000 cm^-1 is due to C–H. The absorption at about 1740 cm^-1 shows the presence of a carbonyl group, C=O and the absorption at about 1190 cm^-1 the presence of a carbon to oxygen single bond, C-O. 

---

(d) The ¹H NMR spectrum shows that the hydrogen atoms are in four different chemical environments. These equate to one –CH₃ group, two –CH₂- groups and one hydrogen atom bonded singly to a carbon atom to which no other alkyl groups are bonded. From the CHO₂⁺ fragment in the mass spectrometer it would seem that a propyl group is attached directly to the oxygen atom by a single bond. This is backed up by the fact that the signal due to the two hydrogen atoms of the –CH₂- group bonded to the oxygen atom is shifted upfield (4.2 ppm) away from the next –CH₂-  protons at 1.5 ppm followed by the methyl group at 0.9 ppm. The remaining single hydrogen atom is bonded directly to carbon atom of the carboxyl group.

---

All this information taken together confirms that Compound H is propyl methanoate, HCOOCH₂CH₂CH₃.

---