Molar Mass Of Argon

1. Molar Mass Of Argon In Grams
2. Molar Mass Of Argon Gas
3. Molar Mass Of Argon
4. Molar Mass Of Argon In Kg/kmol
5. Molar Mass Of Argon

››Convert grams Argon to mole

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Molar mass calculator computes molar mass, molecular weight and elemental composition of any given compound. Massa molar of Ar(argon) Massa Molar, Peso Molecular e. Definitions of molecular mass, molecular weight, molar mass and molar weight. Molecular mass (molecular weight) is the mass of one molecule of a substance and is expressed in the unified atomic mass units (u). (1 u is equal to 1/12 the mass of one atom of carbon-12) Molar mass (molar weight) is the mass of one mole of a substance and is. Solution for Use Graham's law of effusion to calculate molar mass. A sample of argon, Ar, effuses through a small hole at a rate of 9.00x10.° mol/h.

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Molar Mass Of Argon In Grams

How many grams Argon in 1 mol?The answer is 39.948.
We assume you are converting between grams Argon and mole.
You can view more details on each measurement unit:
molecular weight of Argon ormol
The molecular formula for Argon is Ar.
The SI base unit for amount of substance is the mole.
1 grams Argon is equal to 0.025032542304996 mole.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between grams Argon and mole.
Type in your own numbers in the form to convert the units!

››Quick conversion chart of grams Argon to mol

1 grams Argon to mol = 0.02503 mol

10 grams Argon to mol = 0.25033 mol

20 grams Argon to mol = 0.50065 mol

30 grams Argon to mol = 0.75098 mol

40 grams Argon to mol = 1.0013 mol

50 grams Argon to mol = 1.25163 mol

100 grams Argon to mol = 2.50325 mol

200 grams Argon to mol = 5.00651 mol

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››Details on molecular weight calculations

In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together.

The atomic weights used on this site come from NIST, the National Institute of Standards and Technology. We use the most common isotopes. This is how to calculate molar mass (average molecular weight), which is based on isotropically weighted averages. This is not the same as molecular mass, which is the mass of a single molecule of well-defined isotopes. For bulk stoichiometric calculations, we are usually determining molar mass, which may also be called standard atomic weight or average atomic mass.

Finding molar mass starts with units of grams per mole (g/mol). When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. The formula weight is simply the weight in atomic mass units of all the atoms in a given formula.

Using the chemical formula of the compound and the periodic table of elements, we can add up the atomic weights and calculate molecular weight of the substance.

A common request on this site is to convert grams to moles. To complete this calculation, you have to know what substance you are trying to convert. The reason is that the molar mass of the substance affects the conversion. This site explains how to find molar mass.

If the formula used in calculating molar mass is the molecular formula, the formula weight computed is the molecular weight. The percentage by weight of any atom or group of atoms in a compound can be computed by dividing the total weight of the atom (or group of atoms) in the formula by the formula weight and multiplying by 100.

Formula weights are especially useful in determining the relative weights of reagents and products in a chemical reaction. These relative weights computed from the chemical equation are sometimes called equation weights.

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Graham's Law of Effusion
Ten Examples

Boyle's Law	No Name Law	Graham's Law Probs 1-10
Charles' Law	Combined Gas Law	Graham's Law Probs 11-25
Gay-Lussac's Law	Ideal Gas Law	All Graham's Law examples and problems
Avogadro's Law	Dalton's Law	Return to KMT & Gas Laws Menu
Diver's Law

Discovered by Thomas Graham of Scotland sometime in the 1830's (the ChemTeam is not sure exactly when).

To discuss this law, please consider samples of two different gases at the same Kelvin temperature.

Since temperature is proportional to the kinetic energy of the gas molecules, the kinetic energy (KE) of the two gas samples is also the same.

In equation form, we can write this:

KE₁ = KE₂

Since KE = ¹⁄₂mv², (m = mass and v = velocity) we can write the following equation:

m₁v₁² = m₂v₂²

Note that the value of one-half cancels.

The equation above can be rearranged algebraically into the following:

$\sqrt{\mathrm{m₁\; /\; m\; ₂}}$ = v₂ / v₁

You may wish to assure yourself of the correctness of this rearrangement (as well as the one just below).

Graham's Law is often stated as follows:

r₁ / r₂ = $\sqrt{\mathrm{MM₂\; /\; MM₁}}$

where MM means the molar mass of the substance in question. Often, in these types of problems, you will be called upon to determine the molar mass of an unknown gas.

Just above, I formatted Graham's Law using MathML. In a variety of places on the Internet, MathML cannot be used. Look for this:

r₁ / r₂ = SQRT(MM₂ / MM₁)

or this:

$\text{r}{1}_2^{}$ / $\text{r}{2}_2^{}$ = MM₂ / MM₁

as alternate ways to write Graham's Law.

There is also this way:

r₁ / r₂ = √(MM₂ / MM₁)

This method uses the square root symbol that is available through HTML code.

Note that I have substituted r for v. In Graham's Law, we will look at the rate of effusion (movement of gas through a small pinhole into a vacuum) more often than we will look at a speed (like a root mean square speed). That means we are mostly looking at amounts that move per unit time, not how fast the individual particles are moving. For example, a rate unit might be mL/min, which is not a unit of speed. Another example would be moles per second.

Last point: often the rate of effusion of one gas is given relative to the rate of effusion of the other gas. That allows you to set the rate of effusion for one of the gases to a numerical value of 1. This is employed often, look for it.

Example #1: 8.278 x 10¯⁴ mol of an unidentified gaseous substance effuses through a tiny hole in 86.9 s Under identical conditions, 1.740 x 10¯⁴ mol of argon gas takes 81.3 s to effuse.

a) What is the molar mass of the unidentified substance (in g/mol)?
b) What is the molecular formula of the substance?
c) Under identical conditions, how many moles of ethene (C₂H₄) gas would effuse in 91.0 s?

Solution:

1) Calculate the rates of effusion:

unknown ---> 8.278 x 10¯⁴ mol / 86.9 s = 9.525892 x 10¯⁶ mol/s
argon ---> 1.740 x 10¯⁴ / 81.3 s = 2.140221 x 10¯⁶ mol/s

Note that these are not speeds. A speed is an amount of distance covered in a unit amount of time. The above is a rate, a number of moles of gas effuse through a pinhole in a unit amount of time.

2) Use Graham's Law:

r₁ / r₂ = $\sqrt{\mathrm{MM₂\; /\; MM₁}}$

Assign the unknown molar mass to be MM₂. I will cancel the exponent on the rates, since they are both 10¯⁶.

2.140221 / 9.525892 = $\sqrt{\mathrm{MM₂\; /\; 39.948}}$

3) Solve:

0.05047844 = MM₂ / 39.948

MM₂ = 2.0165 g/mol (the answer to part a)

The gas is hydrogen, H₂ (the answer to part b; no other gas weighs 2).

4) The solution to part c:

the molecular weight of ethene is 28.0536 g/mol

let us use the data from argon

let the rate for ethene be r₁

r₁ / r₂ = $\sqrt{\mathrm{MM₂\; /\; MM₁}}$

x / 2.140221 x 10¯⁶ = $\sqrt{\mathrm{39.948\; /\; 28.0536}}$

x / 2.140221 x 10¯⁶ = 1.19331

x = 2.55395 x 10¯⁶ mol/s

2.55395 x 10¯⁶ mol/s times 91.0 s = 2.324 x 10¯⁴ mol

Example #2: It takes 354 seconds for 1.00 mL of Xe to effuse through a small hole. Under the same conditions, how long will it take for 1.00 mL of nitrogen to effuse?

Solution:

1) Write the rates for each gas:

xenon ---> 1.00 mL / 354 s
nitrogen ---> 1.00 mL / x

A common temptation is to use the times directly, as in Xe rate = 354 sec and N₂ rate = x. However, please remember that rates have time in the denominator.

I'm going to leave the rates as fractions when entering them into Graham's Law. Since I knew what I was going to do when I wrote this solution, I'm going to assign nitrogen to r₂.

2) Graham's Law:

r₁ / r₂ = $\sqrt{\mathrm{MM₂\; /\; MM₁}}$

[(1/354) / (1/x)] = $\sqrt{\mathrm{28.014\; /\; 131.293}}$

Watch what happens to the left-hand side.

x / 354 = 0.46192

x = 163.5 s

Note that this is a reasonable answer. Since the nitrogen is lighter than xenon, it takes less time for 1.00 mL of nitrogen to effuse. If you had used the times in the numerator of the rate (as opposed to being in the denominator, where they are supposed to be), you would have calculated an answer much larger than 354 s.

Note that the common mistake described can be made by anybody! The ChemTeam made the above mistake on one of the problems in one of the linked files and did not catch it until a student emailed with a question about the mistaken solution. The ChemTeam was thankful (and embarrassed) and quickly fixef the mistake and uploaded the new file.

Example #3: What is the rate of effusion for a gas that has a molar mass twice that of a gas that effuses at a rate of 4.2 mol/min?

Solution:

1) Let us set the molar masses:

gas A = 1
gas B = 2

I'm going to assign gas B to r₁ (and MM₁, of course) since we want to know the effusion rate for the gas that is twice the molar mass of the other. (Personal note: I like putting the unknown into the numerator. It seems to fit better with my brain.)

Notice how I simply use 1 and 2 for the molar masses. We only know that one is twice the other, we do not know the actual values.

2) Use Graham's Law:

r₁ / r₂ = $\sqrt{\mathrm{MM₂\; /\; MM₁}}$

x / 4.2 = $\sqrt{\mathrm{1\; /\; 2}}$

x / 4.2 = 0.70710678

x = 2.97 mol/min

to two significant figures, the answer is 3.0 mol/min

Example #4: It takes 110. seconds for a sample of carbon dioxide to effuse through a porous plug and 275 seconds for the same volume of an unknown gas to effuse under the same conditions. What is the molar mass of the unknown gas (in g/mol)?

Solution:

1) Set the rates, assuming 1.00 mole of each gas effuses:

r₁ = 1.00 mol / 110. s = 0.00909091 mol/s
r₂ = 1.00 mol / 275 s = 0.003636364 mol/s

Pease note that I did not use the times directly in the problem. I used them to create the rates. In step 5 below, I used the two fractions directly as opposed to what I did just above.

2) Set the molar masses:

MM₁ = 44.009 g/mol
MM₂ = x

3) Set up and substitute into Graham's Law:

r₁ / r₂ = $\sqrt{\mathrm{MM₂\; /\; MM₁}}$

0.00909091 / 0.003636364 = $\sqrt{\mathrm{x\; /\; 44.009}}$

4) Square both sides:

6.25 = x / 44.009

x = 275 g/mol

5) Another way:

[(1/110) / (1/275)] = $\sqrt{\mathrm{x\; /\; 44.009}}$

275 / 110 = $\sqrt{\mathrm{x\; /\; 44.009}}$

2.5 = $\sqrt{\mathrm{x\; /\; 44.009}}$

6.25 = x / 44.009

x = 275 g/mol

Example #5: What is the molar mass of a compound that takes 2.65 times as long to effuse through a porous plug as it did for the same amount of XeF₂ at the same temperature and pressure?

Solution:Comment: you have to be careful reading the problem, in order to keep the gases with the correct subscript. Note how I assign a rate of 1 to the XeF₂. I do this because the unknown gas takes 2.65 times longer than the XeF₂ to effuse.

1) Set the rates:

r₁ = 2.65
r₂ = 1

2) Set the molar masses:

MM₁ = x
MM₂ = 169.286 g/mol

3) Set up and substitute into Graham's Law:

r₁ / r₂ = $\sqrt{\mathrm{MM₂\; /\; MM₁}}$

2.65 / 1 = $\sqrt{\mathrm{x\; /\; 169.286}}$

4) Square both sides:

7.0225 = x / 169.286

x = 1188.8 g/mol

Three significant figures seems best, so 1190 g/mol.

Example #6: If a gas effuses 4.25 times faster than iodine gas (I₂), what is its molar mass?

(a) 59.7 g/mol

(b) 163 g/mol

(c) 123 g/mol

(d) 158 g/mol

Solution:

1) Graham's Law:

r₁ / r₂ = $\sqrt{\mathrm{MM₂\; /\; MM₁}}$

2) We will set the rate for I₂ to be 1.00, which makes the rate for the unknown gas be 4.25. The molar mass of I₂ is 253.809 and the molar mass of the unknown is x

1 / 4.25 = $\sqrt{\mathrm{x\; /\; 253.809}}$

I assigned the r₂ and MM₂ to the unknown gas so as to put x in the numerator.

3) Square both sides, then continue on to the value for x:

0.0553633 = x / 253.809

x = 14.0

4) I think that the most reasonable scenario is that the person who did the problem to get the answer key did not square the left-hand side of the equation. In other words, they only squared the right-hand side of the equation. So, they wound up with this:

1 / 4.25 = x / 253.809 <--- I left the 1/4.25 as a fraction rather than making it into a decimal value

And that leads to answer choice (a).

Example #7: Calculate the density of a gas at STP, if a given volume of the gas effuses through an apparatus in 6.60 min and the same volume of nitrogen, at the same temperature andpressure, effuses through this apparatus in 8.50 minutes.

Solution:

1) Let us assume 1.00 L of each gas effused. This allows us to calculate rates:

unknown ---> 1.00 L / 6.60 min = 0.1515152 L/min
N₂ ---> 1.00 L / 8.50 min = 0.117647 L/min

2) Now, we can use Graham's Law:

r₁ / r₂ = SQRT(MM₂ / MM₁)

I will call the unknown r₂ because that will put its molar mass in the numerator.

0.117647 / 0.1515152 = SQRT( x / 28.014)

0.881175 = x / 28.014

x = 24.685 g/mol

3) Since we know everything was at STP, the density is as follows:

24.685 g/mol / 22.414 L/mol = 1.10 g/L

Example #8: If 0.0949 moles of NH₃ effuses in 881 seconds, how many seconds would it take for the same number of moles of B₂H₆ to effuse?

Solution:

1) Let us determine the rate at which ammonia effuses:

0.0949 mol / 881 s = 0.0001077185 mol/s

2) Write Graham's Law:

r₁ / r₂ = SQRT(MM₂ / MM₁)

or

r₁ / r₂ = $\sqrt{\mathrm{MM₂\; /\; MM₁}}$

3) Substitute values into Graham's Law and solve:

r₁ / 0.0001077185 mol/s = $\sqrt{\mathrm{17.0307\; /\; 27.6694}}$

r₁ / 0.0001077185 mol/s = 0.7845423

r₁ = 0.0000845097 mol/s

4) One last step:

0.0949 mol / 0.0000845097 mol/s = 1123 s

Example #9: The rate of diffusion of an unknown gas was determined to be 2.92 times greater than that of NH₃. What is the approximate molar mass of the unknown gas?

Solution:

1) Write Graham's Law:

r₁ / r₂ = SQRT(MM₂ / MM₁)

2) Enter appropriate values into the above equation:

1 / 2.92 = SQRT(x / 17.0307)

The sub-ones are for the ammonia and the sub-twos are the unknown gas.

3) Square both sides (remember example #6!):

0.1172828 = x / 17.0307

x = 2.00 g/mol

Although not asked for, the gas is hydrogen (H₂).

Example #10: If a molecule of C₂H₆ diffuses a distance of 0.978 m from a point source, calculate the distance (m) that a molecule of CH₄ would diffuse under the same conditions for the same period of time.

Solution:

We can turn the 0.978 m into a rate by assuming some measure of time equal to 1, say 1 memtosecond. So, the C₂H₆ diffuses at the rate of 0.978 m / 1 memtosecond. When we do the calculation, the other rate will be our answer since it will be the distance traveled in 1 memtosecond also.

By the way, the unit of 1 memtosecond is completely fake. However, what we do know is that C₂H₆ does diffuse 0.978 m in some span of time and it is that span of time that we are calling 1 memtosecond. This is being done to turn a distance (0.978 m) into a rate (0.978 m / memtosecond).

r₁ / r₂ = $\sqrt{\mathrm{MM₂\; /\; MM₁}}$

x / 0.978 = $\sqrt{\mathrm{30.07\; /\; 16.043}}$

x = 1.83 m

Example #11: A sample of oxygen gas (O₂) effuses into a vacuum 1 times faster than an unknown gas. O₂ has a molecular weight of about 32.00 g mol¯¹. What is the molecular weight of this unknown gas (in g mol¯¹)?

Discussion:The student that posted this question to an 'answers' website wrote: When I tried calculating this problem using the formula, I got the same answer of 32.00 g/mol, seeing as the [effusion] is only '1 time faster'. Is this correct?

Solution (not done by the ChemTeam):

I would read that as unknown gas effuses at a rate = x, while O₂ effuses at a rate of 'x' faster than unknown gas.

Therefore, . . . rate O₂ = x + x = 2x

Use Graham's Law

(rate unknown / rate O₂) = SQRT[32 / Z)]

(x / 2x) = SQRT[32 / Z)]

(1/2)² = 32 / Z

Z = 4 * 32 = 128 g/mol

Example #12: Argon effuses from a container at a rate of 0.0260 L/s. How long will it take for 3.00 L of I₂ to effuse from the same container under identical conditions?

Solution:

1) State Graham's Law:

r₁/r₂ = SQRT[MM₂ / MM₁]

2) State the molar masses:

MM of Ar = 39.948 g/mol

MM of I₂ = 253.8 g/mol

3) Use r₂ and MM₂ for argon.

r₁ / 0.0260 = SQRT[39.948 / 253.8]

Note that the g/mol unit on each MM cancels.

r₁ / 0.0260 = 0.39674

r₁ = 0.01031524 L/s

4) The time needed for 3.00 L to effuse:

3.00 L / 0.01031524 L/s = 291 s (to three sig figs)

Example #13: Calculate the density of a gas at STP, if a given volume of the gas effuses through an apparatus in 6.60 min and the same volume of oxygen at the same temperature and pressure, effuses through this apparatus in 8.50 minutes.

Solution #1:

1) Write Graham's Law:

$\text{r}{1}_2^{}$ / $\text{r}{2}_2^{}$ = MM₂ / MM₁

2) Identify values to be used:

r₁ (oxygen) ---> 1 / 8.50 min

Molar Mass Of Argon Gas

r₂

Molar Mass Of Argon

(unknown) ---> 1 / 6.60 min

MM₁ = 32.00 g/mol
MM₂ = x

Note how '1' is used as a placeholder for the 'given volume' specified in the problem statement. Sometimes, I will also just assume the given value was 1 L. It doesn't matter what the actual volume is, because the only requirement is that the two volumes be equal.

3) Insert values and solve:

(1 / 8.50)² / (1 / 6.60)² = x / 32.00

6.60² / 8.50² = x / 32.00

43.56 / 72.25 = x / 32.00

x = 19.293 g/mol

4) Given that molar volume is 22.414 L at STP, calculate the density of the gas:

19.293 g/mol / 22.414 L/mol = 0.862 g/L

Solution #2:

This is the solution given by the person (not the ChemTeam) who answered the question on an 'answers' website:

According to Graham's law of effusion:

Effusion rate ∝ 1/$\sqrt{\mathrm{(Molar\; mass)}}$ . . . . . . . [1]

When going through an apparatus:

Effusion rate ∝ 1/$\sqrt{\mathrm{(Time\; taken)}}$ . . . . . . . [2]

Compare [1] and [2]:

$\sqrt{\mathrm{(Molar\; mass)}}$ ∝ (Time taken), i.e.

$\sqrt{\mathrm{(Molar\; mass\; of\; the\; gas)}}$ / $\sqrt{\mathrm{(Molar\; mass\; of\; O₂)}}$ = (Time taken for the gas) / (Time taken for O₂)

$\sqrt{\mathrm{(Molar\; mass\; of\; the\; gas)}}$ / $\sqrt{\mathrm{(32.0\; g/mol)}}$ = (6.60 min) / (8.50 min)

Molar mass of the gas = 32.0 x (6.60/8.50)² g/mol

Molar mass of the gas = 19.3 g/mol

Molar volume of a gas at STP = 22.4 L/mol

Density of the gas at STP = (19.3 g/mol) / (22.4 L/mol) = 0.862 g/L

Bonus Example #1: The rate of effusion of an unknown gas at 480 K is 1.6 times the rate of effusion of SO₂ gas at 300 K. Calculate the molecular weight of the unknown gas.

Solution:

For a discussion about the equation that will be used to solve this problem, please go here.

1) We will use the second equation, but without the V₁ and V₂.

Molar Mass Of Argon In Kg/kmol

r₁ / r₂ = $\sqrt{\mathrm{(T₁M₂)\; /\; (T₂M₁)}}$

2) We set the rate of effusion for SO₂ to be equal to 1. That means the rate of effusion for the unknown gas is 1.6. Let us use r₂ for the SO₂:

1.6 / 1 = $\sqrt{\mathrm{(480\; <b>·</b>\; 64.063)\; /\; (300\; <b>·</b>\; x)}}$

3) Square both sides:

2.56 = (480 · 64.063) / (300 · x)

4) Multiply and cross-multiply:

30750.24 = 768x

x = 40 g/mol

Bonus Example #2: Heavy water, D₂O (molar mass = 20.0276 g mol¯¹), can be separated from ordinary water, H₂O (molar mass = 18.0152 g mol¯¹), as a result of the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rates of diffusion for H₂O when compared to D₂O.

Solution:

1) Write Graham's Law:

$\text{r}{1}_2^{}$ / $\text{r}{2}_2^{}$ = MM₂ / MM₁

Not the usual ChemTeam writing of Graham's Law, but it still works.

2) The D₂O is slower, so I'm going to assign it to r₂ and give it a value of 1. That means that the lighter H₂O rate will be in the numerator and be a value greater than one.

x² / 1

Molar Mass Of Argon

² = 20.0276 / 18.0152

x² = 1.11170567

x = 1.05437454

3) H₂O diffuses at a rate 1.05 times that of D₂O.

Boyle's Law	No Name Law	Graham's Law Probs 1-10
Charles' Law	Combined Gas Law	Graham's Law Probs 11-25
Gay-Lussac's Law	Ideal Gas Law	All Graham's Law examples and problems
Avogadro's Law	Dalton's Law	Return to KMT & Gas Laws Menu
Diver's Law