1.5173 g of organic iron compound containing FE , C ,H and O is burnt in oxygen gas produced 2.838 g OF carbon dioxide and 0.8122 g of water. In separate experiment to determine the mass % of iron, 0.3355 g of the compound yielded 0.0758 g iron (II) oxide. Determine the empirical formula of the compound.

To determine the empirical formula of the compound, we need to find the ratio of the elements (Fe, C, H, and O) in the compound.

1. Start by finding the moles of each element present in the compound.
- Moles of carbon dioxide (CO2):
- Mass of carbon dioxide = 2.838 g
- Molar mass of CO2 = 12.01 g/mol (carbon) + 2 * 16.00 g/mol (oxygen) = 44.01 g/mol
- Moles of CO2 = mass of CO2 / molar mass of CO2
- Moles of water (H2O):
- Mass of water = 0.8122 g
- Molar mass of H2O = 2 * 1.01 g/mol (hydrogen) + 16.00 g/mol (oxygen) = 18.02 g/mol
- Moles of H2O = mass of H2O / molar mass of H2O
- Moles of iron (II) oxide (FeO):
- Mass of FeO = 0.0758 g
- Molar mass of FeO = 55.85 g/mol (iron) + 16.00 g/mol (oxygen) = 71.85 g/mol
- Moles of FeO = mass of FeO / molar mass of FeO

2. Determine the mole ratios of the elements.
- Divide the moles of each element by the smallest number of moles to get the simplest ratio.

3. Calculate the empirical formula.
- The empirical formula represents the ratio of elements in the compound using their whole number subscripts.
- The empirical formula tells us the simplest ratio of the elements' atoms in the compound.

Let's calculate the moles of each element and proceed with the calculations:

1. Moles of CO2:
Moles of CO2 = 2.838 g / 44.01 g/mol = 0.0645 mol

2. Moles of H2O:
Moles of H2O = 0.8122 g / 18.02 g/mol = 0.0451 mol

3. Moles of FeO:
Moles of FeO = 0.0758 g / 71.85 g/mol = 0.0011 mol

Now, let's calculate the mole ratios:

Carbon (C) to Oxygen (O) ratio:
Moles of C = 0.0645 mol
Moles of O = 0.0645 mol (2 oxygen atoms per CO2 molecule) + 0.0011 mol (1 oxygen atom per FeO molecule) = 0.0656 mol
C:O ratio = Moles of C / Moles of O = 0.0645 mol / 0.0656 mol ≈ 0.9848

Hydrogen (H) to Oxygen (O) ratio:
Moles of H = 0.0915 mol (2 hydrogen atoms per H2O molecule)
Moles of O = 0.0656 mol (from above)
H:O ratio = Moles of H / Moles of O = 0.0915 mol / 0.0656 mol ≈ 1.3942

Iron (Fe) to Oxygen (O) ratio:
Moles of Fe = 0.0011 mol
Moles of O = 0.0656 mol (from above)
Fe:O ratio = Moles of Fe / Moles of O = 0.0011 mol / 0.0656 mol ≈ 0.0168

To obtain whole number subscripts, we need to multiply all the ratios by a factor that makes them whole numbers.

Let's multiply all the ratios by approximately 60 to get the simplest whole number ratio:

Carbon (C): 0.9848 * 60 = 59.088 ≈ 59
Hydrogen (H): 1.3942 * 60 = 83.652 ≈ 84
Iron (Fe): 0.0168 * 60 = 1.008 ≈ 1
Oxygen (O): 60

The empirical formula for the compound is therefore:
FeC59H84O60