To calculate the empirical formula of the compound, we need to determine the number of moles of each element in the compound.
1. Let's start by finding the number of moles of carbon dioxide (CO2) produced:
- Mass of carbon dioxide = 0.224g
- We know that 1 mole of carbon dioxide has a molar mass of 44g/mol.
- Therefore, the number of moles of carbon dioxide = mass / molar mass = 0.224g / 44g/mol = 0.00509 mol.
2. Next, let's find the number of moles of water (H2O) produced:
- Mass of water = 0.0372g
- We know that 1 mole of water has a molar mass of 18g/mol.
- Therefore, the number of moles of water = mass / molar mass = 0.0372g / 18g/mol = 0.00207 mol.
3. Now, let's calculate the number of moles of nitrogen in the compound:
- Since the compound contains 5.2% nitrogen by mass, we assume we have 100g of the compound.
- Therefore, the mass of nitrogen in the compound = 5.2g (5.2% of 100g)
- We know that 1 mole of nitrogen (N2) has a molar mass of 28 g/mol.
- Therefore, the number of moles of nitrogen = mass / molar mass = 5.2g / 28g/mol = 0.186 mol.
4. We have the mole ratios of carbon dioxide, water, and nitrogen. Now, we need to simplify these ratios to determine the empirical formula.
- Carbon dioxide (CO2) has a ratio of 0.00509 moles.
- Water (H2O) has a ratio of 0.00207 moles.
- Nitrogen (N2) has a ratio of 0.186 moles.
Since the smallest value is 0.00207 moles, we will divide all the ratios by 0.00207 to obtain the simplest whole-number ratios.
- Carbon dioxide (CO2) becomes approximately 2.46 moles.
- Water (H2O) becomes approximately 1 moles.
- Nitrogen (N2) becomes approximately 89.62 moles.
5. Finally, we convert these ratios into whole numbers by multiplying each value to obtain the simplest, whole-number ratio. Here, we assume the nitrogen ratio is 1:
- Carbon dioxide (CO2) becomes 2 (approximately).
- Water (H2O) becomes 3 (approximately).
- Nitrogen (N2) becomes 1.
Therefore, the empirical formula of the compound is:
N2C2H6O3.