Question

Air contains nitrogen and oxygen in the volume ratio 4:1

Answer 1

Molecular mass of O2 - 32 amu

Molecular mass of N2. - 28 amu

(This means that if you take 32 grams of oxygen or 28 grams of nitrogen it will contain approximately 6.022 X 10 to the power 23 lmolecules of it)

amu ----- atomic mass unit

Let the mass of oxygen taken be Y grams then according to the question the mass of nitrogen is 4Y grams. Now to calculate the no of particles we divide the mass taken by amu.

For oxygen we get , Y/32. ------- result 1

For nitrogen we get, 4Y/28. --------result 2

Now the ratio of no. of particles for oxygen by nitrogen is

Result 1 / Result 2. which is equals to

Y/32 into 4Y/28.

By solving we get the ratio

7/32. -------- required ratio

Answer 2

Explanation:

Average molecular weight of N2 = 28.014 (I will use 28)

Average molecular weight of O2 = 31.998 (I'll use 32)

Weighted average:

There are 4 parts N2 and 1 part O2, for a total of 5 parts:

(4*28 + 1*32)/5 = (112+32)/5 = 144/5 = 28.8

You expect it to be between 28 and 32, since it is a mix of both; and

You expect it to be closer to 28, because there is more nitrogen than oxygen.

The real molar mass of air is slightly different because air does contain a small quantity of other "stuff" - some are atoms, some are molecules -- such as H2O, Argon, CO2... (the list goes on).

But this exercise (if this is for homework) is to get you to practice "weighted averages":

Count the total number of parts (here, it is 4 + 1)

multiply each value by the number of parts it represents,

then divide by the total number of parts.

It is a lot more fun when proportions are given in fractions (or percentages, which are simply fractions over 100).

Your question, using percentages:

80% of nitrogen (80% is the same as 0.8)

20% of oxygen (same as 0.2)

Total number of parts = 0.8 + 0.2 = 1

(we do not need to divide by 1)

Weighted average:

0.8*28 + 0.2*32 = 22.4 + 6.4 = 28.8