Question

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.

Answer 1

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

$\text{Moles of }H_2O=\frac{100.0g}{18g/mol}=5.55mol$

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

$\text{Moles of }N_2O=\frac{100.0g}{44g/mol}=2.27mol$

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

$\text{Moles of }C_3H_6O_2=\frac{100.0g}{74g/mol}=1.35mol$

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

$\text{Moles of }CO_2=\frac{100.0g}{44g/mol}=2.27mol$

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 $1\times 6.023\times 10^{23}$ number of oxygen atoms

So, 5.55 mole of $H_2O$ contains $5.55\times 6.022\times 10^{23}=3.34\times 10^{24}$ number of oxygen atoms

In $N_2O$ :

As, 1 mole of $N_2O$ contains $1\times 6.023\times 10^{23}$ number of oxygen atoms

So, 2.27 mole of $N_2O$ contains $2.27\times 6.022\times 10^{23}=1.37\times 10^{24}$ number of oxygen atoms

In $C_3H_6O_2$ :

As, 1 mole of $C_3H_6O_2$ contains $2\times 6.023\times 10^{23}$ number of oxygen atoms

So, 1.35 mole of $C_3H_6O_2$ contains $2\times 1.35\times 6.022\times 10^{23}=1.62\times 10^{24}$ number of oxygen atoms

In $CO_2$ :

As, 1 mole of $CO_2$ contains $2\times 6.023\times 10^{23}$ number of oxygen atoms

So, 2.27 mole of $CO_2$ contains $2\times 2.27\times 6.022\times 10^{23}=2.73\times 10^{24}$ number of oxygen atoms

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

$H_2O>CO_2>C_3H_6O_2>N_2O$