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The molecular formula of a non - stoichi...

The molecular formula of a non - stoichiometric tin oxide containing Sn (II) and Sn (IV) ions is `Sn_(4.44) O_8`.
Therefore , the molar ratio of Sn (II) to Sn (IV) is approximately

A

`1:8`

B

`1:6`

C

`1:4`

D

`1:1`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the molar ratio of Sn(II) to Sn(IV) in the non-stoichiometric tin oxide represented by the formula Sn₄.44O₈, we can follow these steps: ### Step 1: Define the Variables Let: - \( x \) = moles of Sn(II) (Sn²⁺) - \( y \) = moles of Sn(IV) (Sn⁴⁺) ### Step 2: Set Up the First Equation From the molecular formula Sn₄.44O₈, we know that the total moles of tin (Sn) is given by: \[ x + y = 4.44 \] This is our first equation. ### Step 3: Determine the Total Oxidation State The total oxidation state of the compound must equal zero. The oxidation state of oxygen is -2, and since there are 8 oxygen atoms, the total contribution from oxygen is: \[ 8 \times (-2) = -16 \] Let the oxidation state of tin be represented by \( A \). Then, the total oxidation state can be expressed as: \[ 2x + 4y = 16 \] This is our second equation. ### Step 4: Solve the System of Equations We now have two equations: 1. \( x + y = 4.44 \) (Equation 1) 2. \( 2x + 4y = 16 \) (Equation 2) From Equation 1, we can express \( y \) in terms of \( x \): \[ y = 4.44 - x \] Substituting this expression for \( y \) into Equation 2: \[ 2x + 4(4.44 - x) = 16 \] ### Step 5: Simplify and Solve for \( x \) Expanding the equation: \[ 2x + 17.76 - 4x = 16 \] Combining like terms: \[ -2x + 17.76 = 16 \] Rearranging gives: \[ -2x = 16 - 17.76 \] \[ -2x = -1.76 \] Dividing by -2: \[ x = 0.88 \] ### Step 6: Solve for \( y \) Now substitute \( x \) back into Equation 1 to find \( y \): \[ y = 4.44 - 0.88 \] \[ y = 3.56 \] ### Step 7: Determine the Molar Ratio The molar ratio of Sn(II) to Sn(IV) is given by: \[ \text{Ratio} = \frac{x}{y} = \frac{0.88}{3.56} \] Calculating this gives: \[ \frac{0.88}{3.56} \approx \frac{1}{4} \] ### Final Answer Thus, the molar ratio of Sn(II) to Sn(IV) is approximately: **1:4** ---
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