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University of Maine

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Problem 130

Fluorine is so reactive that it forms compounds with several of the noble gases.

(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 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, $1.85 \times 10^{-4}$ mol of xenon reacted with $5.00 \times 10^{-4}$ mol of fluorine, and $9.00 \times 10^{-6} \mathrm{mol}$ of xenon was found in excess. What are the mass percents of each

xenon fluoride in the product mixture?

Answer

a) $\mathrm{PtF}_{6}$

b) $\mathrm{XeF}_{6}$

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## Discussion

## Video Transcript

When platinum reacts with flooring, a new compound containing both platinum and four Florida is formed. If you know the masses of two of them, you confined the mass of the third by using the law of conservation of Mass. So we know that our final product has a massive 500.519 grams and 0.3 to 7 grams of that is platinum. So the massive flooring is just the difference between these two 4.1 92 grams. Once we know the masses of both reactant sui confined the ratio of them in moles which will give us the ratio in the formula unit. So we change each two moles and we do this by using the molar mass. And once we have the moles, we find the lowest whole number ratio. This gives us the sub scripts in the compound. So if I have 0.3 to 7 grams, the platinum I change it to moles by dividing by the molar mass of platinum, which I find on the periodic table. 195 went 01 grams miss equals 0.1677 I change the grams of Floren two moles by dividing by Florence Molar Mass. And this equals 0.1010 to find the lowest whole number ratio. We divide both of these moles by the smaller number which in this case, is the platinum moles. So this is one in this equal six. So that means that my formula his PT at six if we take this compound that we just formed and react it was seen on We make a new compound that has seen on and flooring in it and releases the platinum. If I start with 0.265 grams of this and form 0.378 grams of the product, we can find out how many grams of Xena informs because it will just be the difference. According to Law of conservation of Mass 3.78 minus 0.265 or 0.113 Gramps, This allows me to find out how many moles of zine on our in the product by changing from grams to moles using the molar mass of Zine on which is 131 0.29 grams. We're point zero 008607 Moles of Seen on To find the rules of Florida in the compound, we use our initial mass and change that two moles using the molar Mass, which I find by adding the Moeller massive platinum plus six times the molar mass of flooring. His 309 went, 089 grams. And then I know that for every one mole of the compound there are six moles of Florence because that's what the ratio because the sub scripts of the formula tell us. And so this equals 0.5144 Moles Florida And so, just like we did before, we divide by the smaller one, which is the scene on, and we find a ratio again of one, 26 And so the formula for this compound is XY at six. Finally, to find the ratio of these compounds, when I combines the non in Florida, we know that will make both XY f for the Tetra fluoride and the hexafluoride or XY F six. To find the moles of zine on. I know that I start off with 1.85 times 10 to the minus four moles and I have nine times 10 to the minus six remaining. So there's 1.76 times 10 to the minus four moles of Xena in both compounds. And so if I say that X stands for the moles of XY F four and why stands for the moles of XY F six, I can say that X plus why equals 1.76 times 10 to the minus four. Because in each compound, for every mole that I started with, there's a 1 to 1 ratio. You can look at the same thing with Florin. We see that Floren there are five times 10 to the minus four moles of Floren that go to produce these two compounds because Florian is too, we say there are two times five times 10 to the minus four equals four go to the first compound so four x plus six Why or one times 10 to the minus Third equals four x plus six ply to know we have two equations with moles of the compounds in them which we can solve simultaneously. X plus y equals 1.76 times 10 to the minus for and my second equation tells me that four X plus six y equals one times 10 to the minus three. Well, multiply both equations. Multiply the top equation by four, which gives us for X plus for why equals seven point 04 times 10 to the minus four. And then we'll subtract that from our second equation. This cancels out the four x is and we're left with two. Why equals 2.96 times 10 to the minus for Or why, which is the moles of XY F six equals 1.48 times 10 to the minus four moles. Now we confined our moles of x cf four because X plus y equals 1.76 times 10 to the minus four. So substituting in X equals 1.76 times 10 to the minus four minus 1.48 times 10 to the minus for or X, which is my moles, uh, seen on tetra fluoride. There's 2.85 times 10 to the minus five moles to find the mass percent. We change our moles of each two grams when we do that, using the molar mass of the substance. So if I have 1.48 times 10 to the minus four mules of X cf six. I multiply that by the molar mass of Xena and Hexafluoride, which is 245 0.28 grams for every mole 0.363 grams. Similarly, 2.85 times 10 to the minus five moles The X e f. Four times the molar mass of XY F for which is 207 0.29 grams for every mole equals 0.0 58 grams. So the percent of each is simply the mass of the compound divided by the total mass of the two compounds. Times 100 and the percent of the second compound, the Tetra fluoride, is found to be 13.8%.

## Recommended Questions

A xenon fluoride can be prepared by heating a mixture of Xe and $\mathrm{F}_{2}$ gases to a high temperature in a pressure-proof container. Assume that xenon gas was added to a $0.25-\mathrm{L}$ container until its pressure reached 0.12 atm at $0.0^{\circ} \mathrm{C}$ Fluorine gas was then added until the total pressure reached 0.72 atm at $0.0^{\circ} \mathrm{C}$. After the reaction was complete, the xenon was consumed completely and the pressure of the $\mathrm{F}_{2}$ remaining in the container was $0.36 \mathrm{atm}$ at $0.0^{\circ} \mathrm{C} .$ What is the empirical formula of the xenon fluoride?

Xenon and fluorine will react to form binary compounds when a mixture of these two gases is heated to $400^{\circ} \mathrm{C}$ in a nickel reaction vessel. A 100.0 -mL nickel container is filled with xenon and fluorine, giving partial pressures of 1.24 atm and 10.10 atm, respectively, at a temperature of $25^{\circ} \mathrm{C}$ . The reaction vessel is heated to $400^{\circ} \mathrm{C}$ to cause a reaction to occur and then cooled to a temperature at which $\mathrm{F}_{2}$ is a gas and the xenon fluoride compound produced is a nonvolatile solid. The remaining $\mathrm{F}_{2}$ gas is transferred to another $100.0-\mathrm{mL}$ nickel container, where the pressure of $\mathrm{F}_{2}$ at $25^{\circ} \mathrm{C}$ is 7.62 $\mathrm{atm}$ . Assuming all of the xenon has reacted, what is the formula of the product?

Elemental fluorine and chlorine gases are very reactive. For example, they react with each other to form chlorine monofluoride.

$$\mathrm{Cl}_{2}(g)+\mathrm{F}_{2}(g) \rightarrow 2 \mathrm{ClF}(g)$$

Calculate the mass of chlorine gas required to produce $5.00 \times 10^{-3} \mathrm{g}$ of chlorine monofluoride given an excess of fluorine gas.

Preparation of Fluorine Gas HF is prepared by reacting $\mathrm{CaF}_{2}$ with $\mathrm{H}_{2} \mathrm{SO}_{4}:$

$$

\mathrm{CaF}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(\ell) \rightarrow 2 \mathrm{HF}(g)+\mathrm{CaSO}_{4}(s)

$$

HF can in turn be electrolyzed when dissolved in molten KF to produce fluorine gas:

$$

2 \mathrm{HF}(\ell) \rightarrow \mathrm{F}_{2}(g)+\mathrm{H}_{2}(g)

$$

Fluorine is extremely reactive, so it is typically sold as a $5 \%$ mixture by volume in an inert gas such as helium. How much $\mathrm{CaF}_{2}$ is required to produce $500.0 \mathrm{L}$ of $5 \% \mathrm{F}_{2}$ in helium? Assume the density of $\mathrm{F}_{2}$ gas is $1.70 \mathrm{g} / \mathrm{L}$

A compound containing xenon and fluorine was prepared by shining sunlight on a mixture of Xe $(0.526 \mathrm{g})$ and excess $\mathrm{F}_{2}$ gas. If you isolate $0.678 \mathrm{g}$ of the new compound, what is its empirical formula?

In halogen displacement reactions a halogen element can be generated by oxidizing its anions with a halogen element that lies above it in the periodic table. This means that there is no way to prepare elemental fluorine, because it is the first member of Group 7 A. Indeed, for years the only way to prepare elemental fluorine was to oxidize $F^{-}$ ions by electrolytic means.Then, in $1986,$ a chemist reported that by reacting potassium hexafluoromanganate(IV) $\left(\mathrm{K}_{2} \mathrm{MnF}_{6}\right)$ with antimony pentafluoride $\left(\mathrm{SbF}_{5}\right)$ at $150^{\circ} \mathrm{C},$ he had generated elemental fluorine. Balance the following equation representing the reaction:

$$\mathrm{K}_{2} \mathrm{MnF}_{6}+\mathrm{SbF}_{5} \longrightarrow \mathrm{KSbF}_{6}+\mathrm{MnF}_{3}+\mathrm{F}_{2}$$

A mixture of xenon and fluorine was heated. A sample of the white solid that formed reacted with hydrogen to yield 81 mL of xenon (at STP) and hydrogen fluoride, which was collected in water, giving a solution of hydrofluoric acid. The hydrofluoric acid solution was titrated, and 68.43 mL of 0.3172 M sodium hydroxide was required to reach the equivalence point. Determine the empirical formula for the white solid and write balanced chemical equations for the reactions involving xenon.

Describe each property of the element fluorine as physical or chemical.

a. is highly reactive

b. is a gas at room temperature

c. has a pale, yellow color

d. will explode in the presence of hydrogen

e. has a melting point of $-220^{\circ} \mathrm{C}$

Chlorine gas reacts with fluorine gas to form chlorine trifluoride.

$$\mathrm{Cl}_{2}(g)+3 \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{ClF}_{3}(g)$$

A 2.00 -L reaction vessel, initially at 298 $\mathrm{K}$ , contains chlorine gas at a partial pressure of 337 $\mathrm{mmg}$ and fluorine gas at a partial pressure of 729 $\mathrm{mm} \mathrm{Hg}$ . Identify the limiting reactant and determine the theoretical yield of $\mathrm{ClF}_{3}$ in grams.

Chlorine gas reacts with fluorine gas to form chlorine trifluoride.

$$\mathrm{Cl}_{2}(g)+3 \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{ClF}_{3}(g)$$

A 2.00 -L reaction vessel, initially at 298 $\mathrm{K}$ , contains chlorine gas at a

partial pressure of 337 $\mathrm{mm} \mathrm{Hg}$ and fluorine gas at a partial pressure of

729 $\mathrm{mm} \mathrm{Hg}$ . Identify the limiting reactant and determine the theoretical

yield of $\mathrm{ClF}_{3}$ in grams.