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 :

Answer : The sample [tex]H_2O[/tex] has greatest number of oxygen atoms.

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

[tex]H_2O>CO_2>C_3H_6O_2>N_2O[/tex]

Explanation :

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

[tex]\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}[/tex]

Molar mass of [tex]H_2O[/tex] = 18 g/mol

[tex]\text{Moles of }H_2O=\frac{100.0g}{18g/mol}=5.55mol[/tex]

and,

[tex]\text{Moles of }N_2O=\frac{\text{Mass of }N_2O}{\text{Molar mass of }N_2O}[/tex]

Molar mass of [tex]N_2O[/tex] = 44 g/mol

[tex]\text{Moles of }N_2O=\frac{100.0g}{44g/mol}=2.27mol[/tex]

and,

[tex]\text{Moles of }C_3H_6O_2=\frac{\text{Mass of }C_3H_6O_2}{\text{Molar mass of }C_3H_6O_2}[/tex]

Molar mass of [tex]C_3H_6O_2[/tex] = 74 g/mol

[tex]\text{Moles of }C_3H_6O_2=\frac{100.0g}{74g/mol}=1.35mol[/tex]

and,

[tex]\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}[/tex]

Molar mass of [tex]CO_2[/tex] = 44 g/mol

[tex]\text{Moles of }CO_2=\frac{100.0g}{44g/mol}=2.27mol[/tex]

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

In [tex]H_2O[/tex] :

As, 1 mole of [tex]H_2O[/tex] contains [tex]1\times 6.023\times 10^{23}[/tex] number of oxygen atoms

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

In [tex]N_2O[/tex] :

As, 1 mole of [tex]N_2O[/tex] contains [tex]1\times 6.023\times 10^{23}[/tex] number of oxygen atoms

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

In [tex]C_3H_6O_2[/tex] :

As, 1 mole of [tex]C_3H_6O_2[/tex] contains [tex]2\times 6.023\times 10^{23}[/tex] number of oxygen atoms

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

In [tex]CO_2[/tex] :

As, 1 mole of [tex]CO_2[/tex] contains [tex]2\times 6.023\times 10^{23}[/tex] number of oxygen atoms

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

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

[tex]H_2O>CO_2>C_3H_6O_2>N_2O[/tex]

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