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A compound of $P$ and $F$ was analyzed as follows:Heating $0.2324 \mathrm{g}$ of the compound in a $378-\mathrm{cm}^{3}$ container turned all of it to gas, which had a pressure of $97.3 \mathrm{mmHg}$ at $77^{\circ} \mathrm{C}$. Then the gas was mixed with calcium chloride solution, which turned all of the $F$ to $0.2631 \mathrm{g}$ of $\mathrm{CaF}_{2} .$ Determine the molecular formula of the compound.

According to the information above, the molecular formula of the compound is $\mathrm{P}_{2} \mathrm{F}_{4}$

Chemistry 101

Chapter 5

Gases

Carleton College

Rice University

Drexel University

Lectures

05:03

In physics, a gas is one of the three major states of matter (the others being liquid and solid). A gas is a fluid that does not support tensile stress, meaning that it is compressible. The word gas is a neologism first used by the early 17th-century Flemish chemist J.B. van Helmont, based on the Greek word ("chaos"), the simplest of all the elemental forms of matter.

04:46

In physics, thermodynamics is the science of energy and its transformations. The three laws of thermodynamics state that energy can be exchanged between physical systems as heat and work; that the total energy of a system can be calculated by adding up all forms of energy in the system; that energy spontaneously flows from being localized to becoming dispersed, spread out, or uniform; and that the entropy of an isolated system not in equilibrium will tend to increase over time, approaching a maximum value at equilibrium.

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Given information for the above question is the mass of compound compound of p and f is equal to 0.2324 gram in volume, volume v is equal to is equal to 378 cubic sum is equal to 378. Milliliters is equal to 0.378 liter and temperature temperature t is equal to 77 degree. Celcius is equal to 273 kelvin plus 77 is equal to 350 kelvin and pressure pressure is equal to 97.3 millimeter. H g is equal to 18 m divide by n 6760 milliliterg into 97.3. Milliliterg is equal to. We have value 0.128 at m r is equal to have general journal gas, constant 0.0821 liter, a t m per mole per kelvin. Next, we have 0.2631 gram of calcium, fluoride, c, f, 2. Let'S, let's determine the molecular determine the molecular formula formula of the compound the compoundfirst lets calculate the molar mass up. The compound the compoundcompound here n is equal to given mass out of molar mass. So we have, we have a number of moles moles of the compound is equal to mass of the compound mass of the compound divide by molar mass molar mass of the compoundof, the compound and we have, according to, according to the ideal, ges low, ideal gas la And given information information we have information, we can, we can find the molar mass of the compound compound we have ideal gas equation, is equal to p y is equal to n r t here. P v is equal to mass of compound denoted by small m and molar. Mass of the compound is denoted by capital m into r t. Here we have here, we can rearrange the formula molar mass of the compound is equal to mass of compound into r t divided by p v. So we have here m molar. Mass of the compound is equal to 0.2324 gram into 0.0821 liter. A t m per kelvin per mole and 350 kelvin divided by 0.1288 m into 0.378 liter, so m is equal to be have value. Molar mass of the compound is 138.02 gram per mole. This is the value of molar mass of the compound. So next we have now, let's calculate, let's calculate the moles of the moles of f and bin c, a f 2. So we have n is equal to mass of the compound by molar. Mass of the compound number of moles of f is equal to mass of f fluorine divide by molar mass molar mass of florine. So we have here so we have to find the mass of florine and phosphorus be forced, so we have massa fluorine mass of florine is equal to mass of c, a f 2 calcium chloride, dot molar mass of fluorine f divide by molar mass of c. A f 2 into 2 moles of polorine, so here we have molar. Mass of florine is equal to molar mass of c, a f 2 dot, molar mass of f into molar mass of c, a f 2 into 2 mole fluorine. So mass of florine is equal to by putting values, is equal to 0.2631 gram into 19.0 gram per mole divide by 40.08 plus 2.19 into gram per mole 2 mole florine, so we have mass of chlorine is equal to 0.128128 g of florine. So this is the mass of florine. So next we have mass of phosphorus. So here we have mass of b is equal to mass of compound compound. Minus mass of florine florine, so mass of phosphorus is equal to mass first plus is equal to. We have values here: 0.2324 gram, minus 0.128 gram, so mass of phosphorus is equal to 0.10 double 4 grams. So this is the mass of now we can find can find moles of fluorine and phosphorus. So we have mole, is equal to n, is equal to mass of compound out of molar mass of compound, and we have a number of of moles of florine is equal to mass of fluorine divide by molar mass molar mass of fluorine it so by putting values We have an moles of chlorine, is equal to 0.128 gram out of molar mass. We have 19.0 gram per mole. Moles of florine is equal to 6.7 3.10 raised to the power minus 3. Moles number of moles of chlorine is equal to 6.73 into 10, raise to the power minus 3 mole. So next we have to find here. A number of moles of the number of mole of b is equal to mass of p divided by molar mass molar. Mass of so we have in moles of p, is equal to by putting values here. 0.1044 gram divide by molar mass 30.97 gram per mole. Moles of b is equal to 3.35 to 10 raised to the power minus 3 molesso. Next, we have to find here now we have to find to find the simple ratio by dividingthe number of the number of moles moles. So we have here f, is equal to 6.73 multiplied 10 raised to the power minus 3 mole divide by 3.35, multiply 10 raised to the power minus 3. Mole is equal to 2 mole ratio to perchlorine, and next mole ratio or phosphorus p is equal to 3.35. Multiply 10 raised to the power minus 3 mole divide by 3.35, multiply 10 raised to the power minus 3. Mole is equal to 1. Therefore, p ratio f is equal to 1 ratio 2. So the empirical empirical formula for this for this compound is compound, is p. F: twop. F. 2. Now we can calculate cancellate. The empirical perodical molar mass of the compound empirical molar mass of the compound mass of the compound compound. Here we have here. We have 30.97 gram per mole plus 1.2 .19.0 gram per mole is equal to 68.97 gram per mole, so here number of moles is equal to molar mass divided by empirical molar mass molar mass molar mass is equal to. We have 138.20 gram per mole divide by 68.97 gram per moleaccording to the information above. According to the information above, the molecular formula of the compound is we have molecular formula of the compound is p 2 f, 4 molecular formula.

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