Given 30.45 mass % N, the remaining percentage is oxygen.

Given 30.45 mass % N, the remaining percentage is oxygen.
Mass of N = 30.45 g
Mass of O = $100.00 \text{ g} - 30.45 \text{ g} = 69.55 \text{ g}$

Atomic mass of N = 14.01 g/mol
Atomic mass of O = 16.00 g/mol

Divide each mole value by the smallest number of moles (2.1734 mol).

The mole ratio of N:O is 1:2.
The empirical formula is $\boxed{\text{NO}_2}$.

b) Molecular Formula

Empirical formula mass of $\text{NO}_2 = (1 \times 14.01 \text{ g/mol}) + (2 \times 16.00 \text{ g/mol})$

Given molar mass = $90 \pm 5$ g/mol. We will use 90 g/mol for the calculation.

Multiply the subscripts in the empirical formula by $n=2$.
Molecular formula = $(\text{NO}_2)_2 = \text{N}_2\text{O}_4$.
The molar mass of $\text{N}_2\text{O}_4$ is $2(14.01) + 4(16.00) = 28.02 + 64.00 = 92.02$ g/mol, which is within the $90 \pm 5$ g/mol range.
The molecular formula is $\boxed{\text{N}_2\text{O}_4}$.

.45

a) Empirical Formula

Given 79.1 mass % Cl, the remaining percentage is silicon.
Mass of Cl = 79.1 g
Mass of Si = $100.0 \text{ g} - 79.1 \text{ g} = 20.9 \text{ g}$

Atomic mass of Si = 28.09 g/mol
Atomic mass of Cl = 35.45 g/mol

Divide each mole value by the smallest number of moles (0.7440 mol).

The mole ratio of Si:Cl is 1:3.
The empirical formula is $\boxed{\text{SiCl}_3}$.

b) Molecular Formula

Empirical formula mass of $\text{SiCl}_3 = (1 \times 28.09 \text{ g/mol}) + (3 \times 35.45 \text{ g/mol})$

Given molar mass = 269 g/mol.

Multiply the subscripts in the empirical formula by $n=2$.
Molecular formula = $(\text{SiCl}_3)_2 = \text{Si}_2\text{Cl}_6$.
The molar mass of $\text{Si}_2\text{Cl}_6$ is $2(28.09) + 6(35.45) = 56.18 + 212.70 = 268.88$ g/mol, which is very close to 269 g/mol.
The molecular formula is $\boxed{\text{Si}_2\text{Cl}_6}$.