Calculating Mass and Molecules of Nitrogen at NTP
Atomic mass and Molecular mass • April 2026

Calculating Mass and Molecules of Nitrogen at NTP

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Written By Archive Editorial
Reading Time 5 Min Read

Understanding the Concept

In chemistry, dealing with gases involves the concept of the molar volume. At Normal Temperature and Pressure (NTP), 1 mole of any ideal gas occupies a standard volume of 22.4 liters (or 22,400 cubic centimeters/cc). By using this relationship and Avogadro's number, we can easily convert volume into mass and molecular count.

Given Data

  • Volume of $N_2$ ($V$): 120 cc = 120 mL = 0.12 liters
  • Gas: Nitrogen ($N_2$)
  • Molar mass of $N_2$: $2 \times 14 = 28 \text{ g/mol}$
  • Molar volume at NTP: 22,400 cc/mol
  • Avogadro's constant ($N_A$): $6.022 \times 10^{23} \text{ molecules/mol}$

Step 1: Calculate the Mass of Nitrogen

To find the mass, we first calculate the number of moles ($n$):

$$n = \frac{\text{Volume at NTP}}{\text{Molar Volume}} = \frac{120 \text{ cc}}{22400 \text{ cc/mol}} \approx 0.005357 \text{ moles}$$

Now, use the formula $\text{Mass} = n \times \text{Molar Mass}$:

$$\text{Mass} = 0.005357 \text{ mol} \times 28 \text{ g/mol} \approx 0.150 \text{ g}$$

So, the mass of 120 cc of nitrogen is approximately 0.150 grams.

Step 2: Calculate the Number of Molecules

The number of molecules ($N$) is calculated by multiplying the number of moles ($n$) by Avogadro's constant ($N_A$):

$$N = n \times N_A$$ $$N = 0.005357 \times 6.022 \times 10^{23}$$ $$N \approx 3.226 \times 10^{21} \text{ molecules}$$

Summary

  • Mass of nitrogen: ~0.150 g
  • Number of molecules: ~3.226 × 10²¹ molecules

This simple approach demonstrates the fundamental relationship between gas volume, stoichiometry, and Avogadro's hypothesis, which states that equal volumes of all gases at the same temperature and pressure contain the same number of molecules.

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