A sample containing 0.4775 of `(NH_(4))_(2)C_(2)O_(4)` and inert material was dissolved in water and made strongly alkaline with KOH which converted `NH_(4)^(o+)` to `NH_(3)` The liberated `NH_(3)` was distilled of `H_(2)SO_(4)` was back titrated with 11.3 " mL of " 0.1214 M NaOH. Calculate (a) `% of (NH_(4))_(2)C_(2)O_(4)=124.10`
And atomic weight of `N=14.0078`.

Explanation

`(NH_(4))_(2)C_(2)O+(4)+2KOHtoK_(2)C_(2)O_(4)+2NH_(3)+2H_(2)O`
Total `H_(2)SO_(4)` used `=50xx0.05035xx2NH_(2)SO_(4)`
`=5.035m" Eq of "H_(2)SO_(4)`
Excess of `H_(2)SO_(4)=11.3xx0.124M NaOH`
`=1.372 m" Eq of " NaOH`
`=1.372 m" Eq of "H_(2)SO_(4)`
`H_(2)SO_(4)` used `=5.035-1.372`
`=3.663 m" Eq of "H_(2)SO_(4)`
`=3.663 mEq NH_(3)`
`=3.663 m" Eq of "(NH_(4))_(2)C_(2)O_(4)`
Ew of `(NH_(4))_(2)C_(2)O_(4)=(124.1)/(2)=62.06g`
Weight of `(NH_(4))_(2)C_(2)O_(4)=3.663xx10^(-3)xx62.05=0.2273g`
`124.1 g of (NH_(4))_(2)C_(2)O_(4)` contains `14.0078 g of N`.
`0.2273 g of (NH_(4))_(2)C_(2)O_(4)=(14.0078xx0.2273)/(124.1)=0.02565` g
`% of N=(0.02565)/(0.4775)xx100=5.373%`
`% of (NH_(4))_(2)C_(2)O_(4)=(0.2273)/(0.4775)xx100=5.373%`