How is the ideal gas amount calculated?

Question

Calculate the number of moles ($n$) and molecules of oxygen in your container from the ideal gas law ($pV=nRT$), assuming $T=300 \ \mathrm{K}$ ($R=0.0821 \ \mathrm{L \cdot atm/(mol \cdot K)}$). Convert the pressure into $\mathrm{atm}$ ($1 \ \mathrm{atm} = 14.7 \ \mathrm{psi}$) and the volume into $\mathrm{L}$ ($1 \ \mathrm{L} = 1000 \ \mathrm{mL}$). $$ \begin{align} P &= 50.00 \ \mathrm{psi} \rightarrow \mathrm{atm}? \\ V &= 10.29 \ \mathrm{mL} \rightarrow \mathrm{L} ? \\ R &= 0.08206 \ \mathrm{L \cdot atm}/(\mathrm{K \cdot mol}) \\ T &= 300 \ \mathrm{K} \\ n &= \ ? \\ \end{align} $$

Ok, I'm having problems with significant figures here when I convert the $\mathrm{mL}$ to liters, $10.29 \ \mathrm{ml} / 1000 \ \mathrm{ml} = 0.01029$ (do I leave it like this to use as my $V$ in my problem?) $P = 1 \ \mathrm{atm} = 14.7\ \mathrm{psi}$ so $50.00/14.7=$ I rounded to $3.40 \ \mathrm{atm}$

So then I set up my problem. I'm looking for $n$, the amount of moles of oxygen.

$$n=\frac{3.40 \ \text{atm} \times 0.01029 \ \text{L}}{0.08206 \ \mathrm{L\cdot atm}/(\mathrm{K\cdot mol})\times 300 \ \text{K}}$$

The answer I get is $127.90$ moles/molecules of oxygen? is this correct or what am I doing wrong here?

Answer 1

I always try to use SI units for any kind of problem involving equations and also try to balance units.
$$pV=nRT$$
$$50~\mathrm{psi}\times\frac{1~\mathrm{atm}}{14.69~\mathrm{psi}}\times \frac{101325~\mathrm{N/m^2}}{1~\mathrm{atm}}\times10.25~\mathrm{ml}\times \frac{1~\mathrm l}{1000~\mathrm{ml}}=n\times8.314\ \frac{\mathrm J}{\mathrm{mol~K}}\times300~\mathrm K$$
$$3534.99~\mathrm{N/m^2~l}=n\times2494.2\ \mathrm{J/mol}$$ $$3534.99~\mathrm{N/m^2}\times1\ \mathrm l\times\frac{1\ \mathrm{m^3}}{1000\ \mathrm l}=n\times2494.2\ \mathrm{J/mol}$$ $$3.53499~\mathrm{N\cdot m}\times \frac{1\ \mathrm J}{1~\mathrm{N\cdot m}}=n\times2494.2~\mathrm{J/mol}$$ $$n=\frac{3.535~\mathrm J}{2494.2\ \mathrm{J/mol}}$$ $$n=0.001417~\mathrm{mol}$$ $$\text{number of molecules}=0.001417\ \mathrm{mol}\times\frac{6.02\times10^{23}\ \text{molecules}}{1~\mathrm{mol}}$$ $$\text{number of molecules}=8.53\times 10^{20}\ \text{molecules}$$