We can solve this problem in two ways: Using mass percents or ratios.
Mass Percent Method
The mass percent of oxygen in #C_2H_6O#, or the percentage of mass taken up by oxygen in #C_2H_6O#, is:
#"1 mole of O (mass)"/"2 moles of C + 6 moles of H + 1 mole of O (mass)"#
#16.00/ ((2*12.01) + (6*1.008) + 16.00)" = 0.3473 = 34.73%#
Oxygen takes up #34.73%# of the mass of #C_2H_6O#.
We can apply this information to #"5.0 g"#: the mass percent of oxygen in #C_2H_6O# must also be #34.73%#, therefore the actual mass of oxygen has to be:
#34.73% * 5.0 = "1.7 g"#
Mole Ratio Method
In #C_2H_6O#, for every #1# mole of #C_2H_6O#, there is #1# mole of oxygen.
The ratio is #1:1#.
So, if we can figure out the number of moles in #"5.0 g"# of #C_2H_6O#, we can find the number of moles of oxygen and then its mass.
#"5.0 g"# of #C_2H_6O# is #0.109# moles, because:
#"No. of moles in sample" = "Sample mass"/"Molar mass"#
#0.109 = 5.0 / ((2*12.01) + (6*1.008) + 16.00)#
Therefore, the number of moles of oxygen in #"5.0 g"# of #C_2H_6O# must also be #0.109#.
The mass of #1# mole of oxygen is #"16.00 g"#, so #0.109# moles is #0.109 * 16.00 = "1.7 g"#.
