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

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`

Step-by-step explanation:

First you should find the molar mass of the
`N_(2)O`:

`N_(2)O=2(14.01(g)/(mol))+16.00(g)/(mol)`

`N_(2)O=44.02(g)/(mol)`

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

`12.6gN_(2)O*(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*(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`