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

Feb 27, 2016

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 ${N}_{A} \times \frac{10.0 \cdot g}{32.0 \cdot g \cdot m o {l}^{-} 1}$, ${N}_{A} = \text{Avogadro's number}$) 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 $22.4$ $L$ $\times$ its molar quantity, approx. $8 \cdot L$.

Feb 27, 2016

There are $1.51 \times {10}^{23} \text{molecules O"_2}$ in $\text{10.0 g O"_2}$.

#### Explanation:

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

Determine the molar mass of $\text{O"_2}$, which is (2xx15.998"g/mol")="39.998 g/mol".

$1 \text{mol of molecules"=6.022xx10^23"molecules}$

You will make the following conversions:

$\textcolor{red}{{\text{mass O}}_{2}}$$\rightarrow$$\textcolor{g r e e n}{{\text{mol O}}_{2}}$$\rightarrow$$\textcolor{b l u e}{{\text{molecules O}}_{2}}$**
by 1) dividing the given mass $\text{O"_2}$ by its molar mass, and 2) multiplying mol $\text{O"_2}$ by $6.022 \times {10}^{23} \text{molecules} .$

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