Elements W and Y have atomic numbers 11 and 9 respectively. a) Write the formula of the compound formed. b) State the type of bond present. c) Identify the type of structure formed.

Assume a 100 g sample:
Mass of Carbon (C) = 40 g
Mass of Hydrogen (H) = 6.7 g
Mass of Oxygen (O) = 53.3 g

Moles of C $= \frac{\text{mass}}{\text{molar mass}} = \frac{40\, \text{g}}{12\, \text{g/mol}} \approx 3.33\, \text{mol}$
Moles of H $= \frac{\text{mass}}{\text{molar mass}} = \frac{6.7\, \text{g}}{1\, \text{g/mol}} = 6.7\, \text{mol}$
Moles of O $= \frac{\text{mass}}{\text{molar mass}} = \frac{53.3\, \text{g}}{16\, \text{g/mol}} \approx 3.33\, \text{mol}$

Divide each mole value by the smallest number of moles (3.33 mol):
For C: $\frac{3.33}{3.33} = 1$
For H: $\frac{6.7}{3.33} \approx 2.01 \approx 2$
For O: $\frac{3.33}{3.33} = 1$
The empirical formula is $\text{CH}_2\text{O}$.

Empirical formula mass of $\text{CH}_2\text{O} = (1 \times 12.0) + (2 \times 1.0) + (1 \times 16.0) = 12.0 + 2.0 + 16.0 = 30.0$.

The relative molecular mass is given as 180.
Let $n$ be the number of empirical formula units in the molecular formula.

Multiply the subscripts in the empirical formula by $n$:
Molecular formula $= (\text{CH}_2\text{O})_6 = \boxed{\text{C}_6\text{H}_{12}\text{O}_6}$.

10) Explain the difference in their melting points.

Phosphorus trichloride ($\text{PCl}_3$) is a simple molecular substance. It consists of discrete $\text{PCl}_3$ molecules held together by weak intermolecular forces (van der Waals forces and dipole-dipole interactions). Only a small amount of energy is required to overcome these weak forces, leading to its low melting point of -91°C.

Sodium chloride ($\text{NaCl}$) is an ionic compound with a giant ionic lattice structure. It consists of $\text{Na}^{+}$ and $\text{Cl}^{-}$ ions held together by strong electrostatic forces of attraction throughout the lattice. A large amount of energy is required to overcome these strong forces to break down the lattice, resulting in its high melting point of 801°C.