Question #38860

The idea here is that you need to convert everything to grams.

You already know that one of the samples contains `2` gram atoms of nitrogen. Now, a gram atom is used to denote the mass of a substance that contains `6.02 * 10^(23)` atoms.

Note that `6.02 * 10^(23)`, known as Avogadro's constant, represents the definition of a mole.

Simply put, a gram atom is the mass of a substance that contains `1` mole of atoms. In this case, `2` gram atoms of nitrogen will be equivalent to `2` moles of elemental nitrogen.

This means that the mass of the first sample will be

`2 color(red)(cancel(color(black)("moles N"))) * "14 g"/(1color(red)(cancel(color(black)("mole N")))) = "28 g"`

Keep in mind that the relative atomic mass is actually equivalent to its molar mass. So if nitrogen is said to have a relative atomic mass of `14`, its molar mass will be equal to `"14 g mol"^(-1)`.

The relative atomic mass of silver is said to be equal to `108`. This is equivalent to saying that `1` mole of silver atoms has a mass of `"108 g"`.

Now, a sample of any ideal gas that is being kept at a pressure of `"1 atm"` and a temperature of `0^@"C"` contains exactly `1` mole of gas and occupies `"22.4 L"`.

This implies that you're dealing with `1` mole of oxygen gas, `"O"_2`. Elemental oxygen has a molar mass of `"16 g mol"^(-1)`, which means that diatomic oxygen will have a molar mass of

`M_ ("M O"_2) = 2 xx "16 g mol"^(-1) = "32 g mol"^(-1)`

You can thus say that the sample of oxygen gas has a mass of `"32 g"`.

Finally, `6.02 * 10^(23)` atoms of carbon are needed in order to have `1` mole of carbon.

This means that the last sample will contain `1` mole of carbon and have a mass of `"12 g"` since carbon has a molar mass of `"12 g mol"^(-1)`.

Therefore, you will have

`"108 g Ag" > "32 g O"_2 > "28 g N" > "12 g C"`