Question

consider separate 100.0 gram samples of each of the following: H2O , N2O , C3H6O2 , CO2. Rank them from greatest to least number of oxygen atoms.

Answer 1

Convert mass in to moles using the relation

number of moles present =given mass/molar mass

for H2O : Number of moles of H2O present = 100 g/18.015 g = 5.55 mols

1 mole H2O has 1mol O atoms

so mols of O aroms = 5.55 mols***

For N2O : Number of mols of N2O present = 100 g/44.013 g/mol = 2.27 mol
mols of O atoms present = 2.27 mols***

For C3H6O2 : Number of mols of C3H6O2 present =100g/74.08 g/mol
= 1.35 mols
1 mole C3H6O2 has 2 mol O atoms [see the chemical formula]
SO moles of O atoms present = 2*1.35 mol = 2.7 mols***

For CO2 : Number of mols of CO2 present = 100 g/44.01 g/mol
= 2.27 mols
1 mol CO2 has 2 mol O
So mols of O present = 2*2.27 = 4.54 mols***

more the number of moles higher the number of O atoms .Because

1 mol atoms contain 6.022 x10^23 atoms in number.

So Highest number of O atoms will be in H2O . least will be N2O . We can arrange
them in the following order

H2O > CO2 > C3H6O2 > N2O