Q151CP

Question

Fluorine is so reactive that it forms compounds with materials inert to other treatments. (a) When 0.327 g of platinum is heated in fluorine, 0.519 g of a dark red, volatile solid forms. What is its empirical formula? (b) When 0.265 g of this red solid reacts with excess xenon gas, 0.378 g of an orange-yellow solid forms. What is the empirical formula of this compound, the first to contain a noble gas? (c) Fluorides of xenon can be formed by direct reaction of the elements at high pressure and temperature. Under conditions that produce only the tetra- and hexafluorides, 1.85×10^-4 mol of xenon reacted with 5.00×10^-4 mol of fluorine, and 9.00×10^-6 mol of xenon was found in excess. What are the mass percents of each xenon fluoride in the product mixture?

Step-by-Step Solution

Verified

Answer

The empirical formula is PtF₆
The empirical formula is XePtF₆
The mass percent of XeF₄ is 13.8% and XeF₆ is 86.2%.

1Step 1: Finding the empirical formula

The moles can be denoted as,

$M o l e s o f P t = 0.327 g P t \times \frac{1 m o l P t}{195, 08 g P} = 1.68 \times 10^{- 3} m o l P t M o l e s o f F = 0.519 g p r o d u c t - 0.327 g P t = 0.192 g F M o l e s o f F = 0.192 g F \times \frac{1 m o l F}{19 g F} = 0.0101 m o l F P t_{\frac{1, 68 \times 10^{- 3}}{1, 68 \times 10^{- 3}}} F_{\frac{0.0101}{1, 68 \times 10^{- 3}}} \to P t F_{6}$

2Step 2: Finding the empirical formula

The moles can be denoted as,

$M o l e s o f P t F_{6} = 0.265 g P t F_{6} \times \frac{1 m o l P t F_{6}}{309.07 g P t F_{6}} = 8.57 \times 10^{- 4} m o l P t F_{6} M o l e s o f X e = 0.378 g p r o d u c t - 0.265 g P t = 0.113 g X e M o l e s o f X e = 0.113 g X e \times \frac{1 m o l X e}{131.29 g X e} = 8.61 \times 10^{- 4} m o l X e$

The empirical formula is XePtF₆.

3Step 3: Finding the mass percentage

The moles of Xe used can be,

$M o l e s o f X e u s e d = 1.85 \times 10^{- 4} m o l - 9 \times 10^{- 6} m o l = 1.76 \times 10^{- 4} m o l X e$

The mass can be expressed as,

$M o s s o f X e F_{4} = 2.8 \times 10^{- 5} m o l X e F_{4} \times \frac{207.28 g X e F_{4}}{1 m o l X e F_{4}} = 5.8 \times 10^{- 3} g X e F_{4} M o s s o f X e F_{4} = 1.48 \times 10^{- 4} m o l X e F 6 \times \frac{245.28 g X e F_{6}}{1 m o l X e F_{6}} = 3.63 \times 10^{- 2} g X e F_{6}$

The total mass of the product mixture can be denoted as,

$M a s s o f p r o d u c t = 5.8 \times 10^{- 3} g X e F_{4} + 3.63 \times 10^{- 2} g X e F_{6} = 0.0421 g p r o d u c t$

The mass percent in the product mixture can be,

$M a s s % o f X e F_{4} = \frac{5.8 \times 10^{- 3} g X e F_{4}}{0.0421 g p r o d u c t} = 13.8 % C e F_{4} M a s s % o f X e F_{6} = \frac{3.63 \times 10^{- 2} g X e F_{6}}{0.0421 g p r o d u c t} \times 100 = 86.2 % C e F_{6}$

Q3.122CP

Q152CP

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