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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]

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Answer:

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

Explanation:

First you should find the molar mass of the [tex]N_{2}O[/tex]:

[tex]N_{2}O=2(14.01\frac{g}{mol})+16.00\frac{g}{mol}[/tex]

[tex]N_{2}O=44.02\frac{g}{mol}[/tex]

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

[tex]12.6gN_{2}O*\frac{1molN_{2}O}{44.02gN_{2}O}=0.29molesofN_{2}O[/tex]

So, there are 0.29 moles of [tex]N_{2}O[/tex] in 12.6g of [tex]N_{2}O[/tex].

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

[tex]0.29molesN_{2}O*\frac{6.022*10^{23}}{1molN_{2}O}=N_{2}O[/tex]

There are [tex]1.7*10^{23}[/tex] moles of [tex]N_{2}O[/tex] in 12.6g of [tex]N_{2}O[/tex]

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