Question

When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were produced. In a separate experiment to determine the mass percent of iron, 0.4131 g of the compound yielded 0.09333 g of Fe2O3. What is the empirical formula of the compound?

Answer 1

`\boxed{\text{C$_(15)$H$_(21)$FeO$_(6)$}}`

Step-by-step explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

`\text{Mass of C} = \text{2.3162 g } \text{CO}_(2)* \frac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_(2)}= \text{0.632 07 g C}`

(b) Mass of H

`\text{Mass of H} = \text{0.66285 g }\text{H$_(2)$O}* \frac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_(2)$O}}} = \text{0.074 157 g H}`

(c)Mass of Fe

(i)In 0.4131g of X

`\text{Mass of Fe} = \text{0.093 33 g Fe$_(2)$O$_(3)$}* \frac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_(2)$O$_(3)$}} = \text{0.065 277 g Fe}`

(ii) In 1.2383 g of X

`\text{Mass of Fe} = \text{0.065277 g Fe}* (1.2383)/(0.4131) = \text{0.195 67 g Fe}`

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

`\text{Moles of C = 0.63207 g C}*\frac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} * \frac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} * \frac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} * \frac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}`

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

`\text{C: } (0.052629)/(0.003 5038)= 15.021\\\\\text{H: } (0.073568)/(0.003 5038) = 20.997\\\\\text{Fe: } (0.003 5038)/(0.003 5038) = 1\\\\\text{O: } (0.021025)/(0.003 5038) = 6.0006`

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

`\text{The empirical formula is } \boxed{\textbf{C$_(15)$H$_(21)$FeO$_(6)$}}`