Question

Dinitrogen monoxide or laughing gas (N2O) is used as a dental anesthetic and as an aerosol propellant. How many moles of N2O are present in 12.6 g of the compound? How many molecules of N2O are present in 12.6 g of the compound? [Use molar masses: N, 14.01 g/mol, O, 16.00 g/mol]

Answer 1

Answer:

In 12.6g of $N_{2}O$ there are 0.29 moles of $N_{2}O$ and $1.7*10^{23}$ molecules of $N_{2}O$

Explanation:

First you should find the molar mass of the $N_{2}O$:

$N_{2}O=2(14.01\frac{g}{mol})+16.00\frac{g}{mol}$

$N_{2}O=44.02\frac{g}{mol}$

Then you should write the conversion factor using the molar mass:

$12.6gN_{2}O*\frac{1molN_{2}O}{44.02gN_{2}O}=0.29molesofN_{2}O$

So, there are 0.29 moles of $N_{2}O$ in 12.6g of $N_{2}O$.

Finally to find the number of molecules, you should use the Avogadro´s number:

$0.29molesN_{2}O*\frac{6.022*10^{23}}{1molN_{2}O}=N_{2}O$

There are $1.7*10^{23}$ moles of $N_{2}O$ in 12.6g of $N_{2}O$