Calculate the number of atoms of each type that are present in `3.42 g` of sucrose `(C_(12) H_(22) O_(11))`.

Explanation

Molecular mass of sucrose
`=12 xx 12 + 22 xx 1 + 11 xx 16 = 342 u`
Molar mass `= 342 g "mol"^(-1)`
Now, Number of molecules `(N) =m/M xx N_(0)`
`therefore` Number of molecules of sucrose in 3.42 g
`=(3.42)/342 xx 6.022 xx 10^(23) = 6.022 xx 10^(21)` molecules.
Now, `rArr` 1 molecular of sucrose `(C_(12) H_(22)O_(11))` conrtains= 12 atoms of C
So, `6.022 xx 10^(21)`
`= 72.26 xx 10^(21)` atoms of C
`rArr 1` molecule of sucrose `(C_(12)H_(22)O_(11))` contains = 22 atoms of H
`therefore 6.022 xx 10^(21)` molecules will contain
`=22 xx 6.022 xx 10^(21)`
`= 132.5 xx 10^(21)` atoms of hydrogen
`rArr 1` molecule of sucrose `(C_(12)H_(22)O_(11))` contains = 11 atoms of oxygen.
`therefore 6.022 xx 10^(21)` molecules will contain
`= 11 xx 6.022 xx 10^(21)`
`=66.24 xx 10^(21)` atoms of oxygen