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Consider separate 100.0 gram samples of each of the following: H2O, N2O, C3H6O2, CO2 Determine which 100.0 gram sample will have the greatest number of oxygen atoms and list them in order from greatest to least.

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Answer : The sample
H_2O has greatest number of oxygen atoms.

The order of number of oxygen atoms from greatest to least will be:


H_2O>CO_2>C_3H_6O_2>N_2O

Explanation :

First we have to calculate the moles of
H_2O,N_2O,C_3H_6O_2\text{ and }CO_2


\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}

Molar mass of
H_2O = 18 g/mol


\text{Moles of }H_2O=(100.0g)/(18g/mol)=5.55mol

and,


\text{Moles of }N_2O=\frac{\text{Mass of }N_2O}{\text{Molar mass of }N_2O}

Molar mass of
N_2O = 44 g/mol


\text{Moles of }N_2O=(100.0g)/(44g/mol)=2.27mol

and,


\text{Moles of }C_3H_6O_2=\frac{\text{Mass of }C_3H_6O_2}{\text{Molar mass of }C_3H_6O_2}

Molar mass of
C_3H_6O_2 = 74 g/mol


\text{Moles of }C_3H_6O_2=(100.0g)/(74g/mol)=1.35mol

and,


\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}

Molar mass of
CO_2 = 44 g/mol


\text{Moles of }CO_2=(100.0g)/(44g/mol)=2.27mol

Now we have to calculate the number of oxygen atoms in
H_2O,N_2O,C_3H_6O_2\text{ and }CO_2

In
H_2O :

As, 1 mole of
H_2O contains
1* 6.023* 10^(23) number of oxygen atoms

So, 5.55 mole of
H_2O contains
5.55* 6.022* 10^(23)=3.34* 10^(24) number of oxygen atoms

In
N_2O :

As, 1 mole of
N_2O contains
1* 6.023* 10^(23) number of oxygen atoms

So, 2.27 mole of
N_2O contains
2.27* 6.022* 10^(23)=1.37* 10^(24) number of oxygen atoms

In
C_3H_6O_2 :

As, 1 mole of
C_3H_6O_2 contains
2* 6.023* 10^(23) number of oxygen atoms

So, 1.35 mole of
C_3H_6O_2 contains
2* 1.35* 6.022* 10^(23)=1.62* 10^(24) number of oxygen atoms

In
CO_2 :

As, 1 mole of
CO_2 contains
2* 6.023* 10^(23) number of oxygen atoms

So, 2.27 mole of
CO_2 contains
2* 2.27* 6.022* 10^(23)=2.73* 10^(24) number of oxygen atoms

The order of number of oxygen atoms from greatest to least will be:


H_2O>CO_2>C_3H_6O_2>N_2O

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