To calculate the mass of gas, we must recall the following two equation equations—

The first equation is the ideal gas equation:

** PV = nRT —— (1)**

where,

P is pressure

V is volume

n is moles of gas

T is temperature in Kelvin

R is the universal gas constant with a value of 0.0821 L atm/mol K

And the second equation is:

**Mass (m) = moles (n) x molar mass (Mm), in short m = n x Mm ———–(2)**

We must rearrange equation (2) to isolate moles (n) by dividing both sides of the equation by molar mass (Mm)

Once we do, we will get:

Moles, n = m/Mm, which means you get moles by dividing mass of substance by its molar mass.

So, let’s substitute (m/Mm) for moles in the ideal gas equation. Once we do, we will get

**PV = m/Mm x RT—————————-(3)**

Now, let’s get rid of the molar mass (Mm) on the right side of equation (3) by multiplying through the whole equation by Mm, once we do, we will get

**PV x Mm = (m/ Mm) x Mm x RT **

Once Mm cancels out on the right side of the equation, we will get:

**PV x Mm = m x RT ——————-(4)**

Now, let’s divide both sides of the equation by RT

Once we do, we will get

**(PV x Mm)/RT = M ——————-(5)**

As you can see, we have isolated mass by combining the ideal gas equation (1) and the moles equation (2).

So, to calculate the mass of gas, we will use equation (5), and if we can recall equation (6) from the top of our head, it saves us time with all the derivation.

Now let’s use equation (5) to solve the following problem

Question

Calculate the mass of gas in grams present in a sample that has a molar mass of 70.0 g/mol and occupies a 2.00 L container at 117 kPa and 35.1 °C.

From the question, here is what is given

T = 35.1 °C, but we need to covert to temperature in Kelvin, so we add 273 to the value

So, T becomes — 35.1 °C + 273 = **308.1 K**

P = 117 kPa atm, but since our universal gas constant R, has unit of atm for pressure, we must covert pressure in **kPa** to **atm, **so that units of pressure will agree with each other.** **So, we must recall that— **1 atm = 101.3 kPa**, and once we do, we ‘ll mulitply 117 kPa by the conversion factor (1 atm/101.3 kPa). That is,

117 ~~kPa~~ x 1 atm/101.3 ~~kPa~~ = **1.155 atm**

R = **0.0821 atm L/mol K**

Mm = **70.0 g/mol**

V = 2.00 L

Once we substitute these values into equation (5),

We ‘ll get

**(1.155 atm x 2.00 L x 70.0g/mol)/0.0821 atm L/mol K x 308.1 K = M ——————-(5)**

M = 161.7g/25.295 = 6.393 g, however, since the least number of significant digits in our calculation is 3, it follows that our answer in 3 significant figures is **6.39 g**

Do recall that the forward slash(/) means division.

To learn how to calculate the molar mass of gas click here.

To learn how to calculate the density of gas click here.