5.00 g of nitrogen is completely converted into an oxide of nitrogen - VCE - SSCE Chemistry - Question 10 - 2002 - Paper 1

Now, calculate the number of moles of nitrogen and oxygen. The molar mass of nitrogen (N) is approximately 14.01 g/mol and that of oxygen (O) is about 16.00 g/mol:

Moles of Nitrogen: ext{Moles of N} = rac{5.00 ext{ g}}{14.01 ext{ g/mol}} \\ \approx 0.357 ext{ mol}

Moles of Oxygen: ext{Moles of O} = rac{14.3 ext{ g}}{16.00 ext{ g/mol}} \\ \approx 0.894 ext{ mol}

Next, to determine the empirical formula, find the simplest mole ratio by dividing each value of moles by the smaller number of moles:

Ration for Nitrogen: rac{0.357}{0.357} = 1

Ratio for Oxygen: rac{0.894}{0.357} \\ \approx 2.5

To express this as a whole number, multiply both ratios by 2:

$N: 1 \times 2 = 2$ $O: 2.5 \times 2 = 5$

Thus, the empirical formula of the oxide is:

$\text{N}_2\text{O}_5$