To calculate the density 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, we will get:

**PV x Mm = M x RT ——————-(4)**

Since density, D = mass/volume (D = M/V), we must rearrange equation (4) such that M/V will be on the same side of the equation. To isolate (M/V), we must divide both sides of equation (4) by both RT and V.

When we do, we will get

**(P x Mm)/RT = M/V ——————-(5)**

However, M/V is density **(D)**. Therefore, we ‘ll substitue D for M/V in equation (5) to get

**D = (P x Mm)/RT ……………………(6)**

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

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

Now, let’s use equation (6) to solve the following problem

#### Question

**Calculate the density of gas at STP that has a molar mass of 44.0 g/mol**

From the problem, we must first understand what **STP** means. **STP** means **standard temperature and pressure.**

Therefore,

Our **standard temperature is 0° C **but we must convert degree Celsius to Kelvin by adding 273 to it. As a result, out temperature in Kelvin will be **0° C + 273 = 273 K**

Our standard pressure is 1 atm.

Therefore,

**T = 273 K**

**P = 1 atm**

**R = 0.0821 atm L/moK**

**Mm = 44.0 g/m**

Once we substitute these values into equation (6),

We get: **D = (1 atm x 44.0 g/mol)/ 0.0821 atm L/mol K x 273 **

D = 44.0 g/22.41L = 1.963g/L , which is put to 3 significant figures as: **1.96 g/L**

However, since 3 is the least number of significant digits in our calculation, we will put our final answer to 3 significant figures, which is **1.96 g/L**

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 mass of gas click here