To determine the empirical formula of a compound given the percentages by mass of nitrogen (N) and oxygen (O), follow these steps:
1. Convert the mass percentages to grams:
Assume we have 100 grams of the compound. This assumption simplifies the calculation because the percentages can be directly treated as grams:
- Mass of Nitrogen (N) = 36.86 grams
- Mass of Oxygen (O) = 63.14 grams
2. Convert grams to moles:
Use the atomic masses of each element to convert grams to moles:
- Atomic mass of Nitrogen (N) = 14.01 g/mol
- Atomic mass of Oxygen (O) = 16.00 g/mol
Calculating the moles of each element:
[tex]\[
\text{Moles of N} = \frac{\text{Mass of N}}{\text{Atomic mass of N}} = \frac{36.86 \text{ grams}}{14.01 \text{ g/mol}} \approx 2.63 \text{ moles}
\][/tex]
[tex]\[
\text{Moles of O} = \frac{\text{Mass of O}}{\text{Atomic mass of O}} = \frac{63.14 \text{ grams}}{16.00 \text{ g/mol}} \approx 3.95 \text{ moles}
\][/tex]
3. Determine the simplest mole ratio:
Divide the number of moles of each element by the smallest number of moles obtained:
[tex]\[
\text{Ratio of N} = \frac{2.63 \text{ moles}}{2.63 \text{ moles}} = 1
\][/tex]
[tex]\[
\text{Ratio of O} = \frac{3.95 \text{ moles}}{2.63 \text{ moles}} \approx 1.5
\][/tex]
Since these numbers are not perfectly whole numbers, we look for simple whole number ratios. The ratio 1 : 1.5 can be multiplied by 2 to approximate whole numbers:
[tex]\[
1 \times 2 = 2
\][/tex]
[tex]\[
1.5 \times 2 = 3
\][/tex]
The simplest whole number ratio of N to O is 2:3.
4. Write the empirical formula:
Using the whole number ratio obtained, the empirical formula is derived:
[tex]\[
\text{Empirical formula} = \text{N}_2\text{O}_3
\][/tex]
Therefore, the empirical formula for the compound with mass percentages 36.86% Nitrogen (N) and 63.14% Oxygen (O) is [tex]\( \mathrm{N_2O_3} \)[/tex].
