Phosphorous combines in several different forms. The density of a phosphorous vapor at 310 celsius and 101.3kPa is 2.64g/L. What is the molecular formula of phosphorous?

I'd start off using the Ideal Gas Equation

PV = nRT

Recall that P will be pressure, V is volume, n is number of moles, R is the gas constant, and T is temperature.

Be careful in choosing your gas constant, as the numerical value will change based on the units. The one I always used was 0.08206 L *atm *mol^{-1} *K^{-1}, but when working these problems on your own, you can use the one you're most familiar with or whichever one is used most in your class.

Since I see kPa in this problem, we might as well use R = 8.3145 m^{3} *Pa *mol^{-1} *K^{-1} to make things a tad easier. Once you figure out what equation can apply to your problem, I always recommend doing simple conversions immediately. Celsius to Kelvin is a simple conversion where you won't have to worry about repeating decimals. It's just K = C +273.15. I also often see classes not even include the .15 at the end, making it as simple as only adding 273. Make sure to follow whatever your teacher is doing so that you are using the same values as them. I'm only going to add 273 so that I don't mess up your significant figures.

K=310+273=583

Convert kPa to Pa to ensure it matches the gas constant units.

1kPa=1000Pa

101.3kPa*(1000Pa/1kPa)=101300Pa

You can also skip doing any math and just move the decimal over by three spots if you remember the basic unit conversion.

The density is 2.64 g/L, but the liters might be an issue for our gas constant, which is using m^{3}. The conversion will be simple because 1m^{3}=1000L. I will leave the grams unit there because the "n" in the formula can help us out with it later. While n is just the number of moles, you can play around with it using molar mass, which happens to have units of **grams** per mole.

2.64g/L*1000L/m^{3}= 2640g/m^{3}

Now we have our new temperature, pressure, and density.

T=583 K

P=101300Pa

d=2640g/m^{3}

Remember that density just means mass divided by volume. We will use this information.

Ok, now that those conversions are done, back to PV=nRT.

n is the number of moles, but you aren't usually lucky enough to just know that information. Another way to find how many moles you have is to divide the mass in grams of whatever compound you have by its molar mass, which is measured in grams per mole. The units of grams cancel out, leaving you only with moles. Thus, you can rewrite n.

n=mass (g)/molar mass(g/mol)

Because division is crazy, the mol will end up "on top" as a regular unit, not a -1 superscript unit when you're done and the grams cancel. I could explain why this works if you'd like, but you're here for chem, not arithmetic.

So let's rewrite PV=nRT as

PV=(m/molarmass)RT

We want to find the molar mass, so let's rearrange and isolate it.

molarmass*PV=mRT

molarmass=mRT/PV

Now, let's bring in density, d=2640g/m^{3}. Density equals mass divided by volume. You could just this to rearrange and substitute in for volume or mass, or you can notice that in this molar mass equation that m is being divided by V. In that case, you can just replace m/V with density since d=m/V.

molarmass=dRT/P

Put in the density, gas constant, temperature, and pressure to find molar mass.

molarmass=(2640g/m^{3}*8.3145 m^{3} *Pa *mol^{-1} *K^{-1} *583 K)/101300Pa

molarmass=126.327870089 g/mol

Significant figures from the given information is three.

molarmass=126g/mol

This is the molar mass of the vapor. We can figure out the molecular formula of phosphorous by dividing by the molar mass of phosphorous, which is 30.97g/mol. By dividing the molar mass of the vapor by the molar mass of single atoms of phosphorous, we can figure out the ratio of atoms, just like how diatomic gasses have a ratio of two when they are floating about. That's also why this problem starts off saying that phosphorous could combine in several forms. We are trying to determine which form it's taking in this situation.

molarmass of vapor/molarmass of phosporous=126g/mol/30.97g/mol=4

Units cancel out, and we are just left with the ratio. There are 4 phosphorous atoms grouped together in this vapor. The molecular formula in this case is P_{4}.

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A 2.650g sample of a gas occupies a volume of 428ml at 98.9kPa and 24.3 celsius. Analysis of this compound shows it to be 15.5% Carbon, 23.0% Cl, and 61.5% Fluorine. What is the molecular formula of this compound.

With percent mass, we often assume an overall sample mass of 100g to get the ball rolling. Luckily, you have one given. Let's calculate the grams of each element present based on 2.650g then convert that into moles.

Carbon

2.650g*.155=0.41075g

0.41075g*1/12.01mol/g=0.0342mol

Chlorine

2.650g*.230=0.6095g

0.6095g*1/35.45mol/g=0.0172mol

Fluorine

2.650g*.615=1.62975g

1.62975g*1/19.00mol/g=0.0858mol

With this information, we can get the empirical formula. Divide all of the mole quantities by the smallest one, which is chlorine.

Carbon

0.0342mol/0.0172mol=1.988=2

Chlorine

0.0172mol/0.0172mol=1

Fluorine

0.0858mol/0.0172mol=4.988=5

The empirical formula is C_{2}ClF_{5}. This tells us the ratios of the elements within the compound, but we still aren't sure what the actual molecular formula is. It could be any factor of this ratio, like C_{4}Cl_{2}F_{10 }or C_{6}Cl_{3}F_{15}.

I'm going to leave that there and go back to the problem. You're given mass of the sample, volume, pressure, and temperature. That signals to me that we are going to be using the Ideal Gas Equation again.

PV=nRT Recall that n can be rewritten as n=m/molarmass

This becomes very useful because when dealing with percent mass, we will want to have the molar mass of the entire compound to go from the empirical formula to the molecular formula.

Let's use R = 8.3145 m^{3} *Pa *mol^{-1} *K^{-1} since you again have kPa in your given. Convert given information to appropriate units.

428ml*1L/1000ml*1m^{3}/1000L= 0.000428m^{3}

98.9kPa*1000Pa/1kPa=98900Pa

24.3C+273=297.3K

PV=nRT

PV=(m/molarmass)RT

molarmass=mRT/PV

molarmass=2.650g*8.3145 m^{3} *Pa *mol^{-1} *K^{-1} *297.3K/ 98900Pa*0.000428m^{3}

molarmass=154.7522101g/mol=155g/mol

Go back to empirical formula:C_{2}ClF_{5}. Find the molar mass of that by adding atomic masses.

(2*12.01)+35.45+(5*19.00)=154.47g/mol

The numbers are so close, that I won't bother even dividing them. If you divided the molar mass by the empirical formula, you'd basically get 1. That means that, in this case, the empirical formula and the molecular formula are the same. This will not always be the case, so you need to always do the math to make sure. The answer will be C_{2}ClF_{5}.

I ran out of space, otherwise I actually typed out the work for all of the problems.

Jaimie C.

Suppose 105L of NH3 (g) and 285L of O2 (g) are allowed to react until this reaction is complete. All volume measurements are made at 200 celsius and 30.4kPa. Which gas, ammonia or oxygen, remains at the end of the reaction? How many moles of it would there be? Balanced equation: 4NH3 (g) + 7O2 (g) --> 6H2O (g) + 4NO2 (g) This seems to be a limiting reactant problem. But you can't really start a limiting reactant problem if you don't have moles. Notice you have all the information needed to use the Ideal Gas Equation, and use that equation to fine the moles you need to proceed to determining the limiting reactant. NH3 PV=nRT P=30.4kPa=30400Pa V=105L=0.105m3 n=? R = 8.3145 m3 *Pa *mol-1 *K-1 T=200C=473K 30400Pa*0.105m3=n*8.3145 m3 *Pa *mol-1 *K-1 *473K n=0.812 mol NH3 O2 P=30400Pa V=285L=0.285m3 n=? R = 8.3145 m3 *Pa *mol-1 *K-1 T=473K 30400Pa*0.285m3=n*8.3145 m3 *Pa *mol-1 *K-1 *473K n=2.20 mol O2 Now that we are in moles, we can use a variety of methods to figure out the limiting reactant. I am going to do the mole of reaction method. One mole of reaction is when your starting materials react, based on the coefficients in a balanced equation. 1 mol-rxn = 4 mol NH3 = 7 mol O2 The moles of reaction are directly related to how many times the reaction can occur, so the reactant that has less will be the limiting, as the reaction can occur less times based on it. 0.812 mol NH3 * 1mol-rxn/4molNH3= 0.203 mol-rxn 2.20 mol O2 *1 mol-rxn/7molO2= 0.314 mol-rxn Thus, NH3 is the limiting reactant since the reaction can occur less times with it. Using this and the coefficients of the equation, we can calculate how much oxygen will be leftover after the ammonia reacts to completion. 0.812 mol NH3 *7molO2/4molNH3 = 1.42molO2 used up 2.20-1.42=0.78 mol O2 remaining01/21/20