Question

How many molecules are in 10.0 g of `O_2` gas at STP?

Answer 1

How many molecules at STP? The same number of molecules in 10.0 g of LIQUID OXYGEN at LOW TEMPERATURES.

Explanation:

You specified a mass of 10.0 g. Clearly, this is a finite number of dioxygen molecules. This same finite number (which is `N_Axx(10.0*g)/(32.0*g*mol^-1)`, `N_A = "Avogadro's number"`) is the same mass whatever the state, solid, liquid, or gas.

Of course, at STP, dioxygen is a gas, but 10.0 g is still 10.0 g. We could calculate its volume at STP, which is `22.4` `L` `xx` its molar quantity, approx. `8*L`.

Answer 2

There are `1.51xx10^23"molecules O"_2"` in `"10.0 g O"_2"`.

Explanation:

STP is irrelevant - it's a detractor. Mass is not dependent on temperature or pressure. The number of molecules of `"O"_2"` are dependent only on the moles of `"O"_2"` in `"10.0 g O"_2"`.

Determine the molar mass of `"O"_2"`, which is `(2xx15.998"g/mol")="39.998 g/mol"`.

`1"mol of molecules"=6.022xx10^23"molecules"`

You will make the following conversions:

`color(red)("mass O"_2)` `rarr` `color(green)("mol O"_2)` `rarr` `color(blue)("molecules O"_2)`**
by 1) dividing the given mass `"O"_2"` by its molar mass, and 2) multiplying mol `"O"_2"` by `6.022xx10^23"molecules".`

`color(red)10.0cancel(color(red)("g O"_2))xx(color(green)(1)cancelcolor(green)("mol O"_2))/(color(green)(39.998)cancelcolor(green)("g O"_2))xx(color(blue)(6.022xx10^23"molecules O"_2))/(cancelcolor(blue)("mol O"_2))=color(blue)(1.51xx10^23"molecules O"_2")` rounded to three significant figures