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erma4kov [3.2K]
3 years ago
12

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?
Chemistry
1 answer:
uysha [10]3 years ago
5 0

Answer:

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

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}\times \dfrac{\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}\times \dfrac{\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$_{3}$}\times \dfrac{\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}\times \dfrac{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}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\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: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{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$_{6}$}}

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The sample of oxygen gas was collected through water displacement. So, the gas collected will be a mixture of oxygen and water vapor.

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How much heat energy, in kilojoules, is required to convert 72.0 gg of ice at −−18.0 ∘C∘C to water at 25.0 ∘C∘C ? Express your a
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Answer: The enthalpy change is 34.3 kJ

Explanation:

The conversions involved in this process are :

(1):H_2O(s)(-18^0C)\rightarrow H_2O(s)(0^0C)\\\\(2):H_2O(s)(0^0C)\rightarrow H_2O(l)(0^0C)\\\\(3):H_2O(l)(0^0C)\rightarrow H_2O(l)(25^0C)

Now we have to calculate the enthalpy change.

\Delta H=[m\times c_{s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{l}\times (T_{final}-T_{initial})]

where,

\Delta H = enthalpy change = ?

m = mass of water = 72.0  g

c_{s} = specific heat of ice = 2.09J/g^0C

c_{l} = specific heat of liquid water = 4.184J/g^0C

n = number of moles of water = \frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{72.0g}{18g/mole}=4.00moles

\Delta H_{fusion} = enthalpy change for fusion = 6010 J/mole

Now put all the given values in the above expression, we get

\Delta H=[72.0g\times 2.09J/g^0C\times (0-(-18)^0C]+4.00mole\times 6010J/mole+[72.0g\times 4.184J/g^)C\times (25-0)^0C]\Delta H=34279.8J=34.3kJ        (1 KJ = 1000 J)

Therefore, the enthalpy change is 34.3 kJ

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