Question

Could someone please help me with this question?
5.00 g of nitrogen is completely converted into an oxide of nitrogen. The mass of the oxide formed is 19.3g. What is the empirical formula of the oxide?

Answer 1

LOL its N2O5

Answer 2

its like trial and error method
just check for the formaion of different oxides oxide
N2 + O2 ------> 2 NO
in the above reaction 28 gram of N2 gives rise too 60 gram of NO
then if u burn 5 gram of N2 = 60*5 / 28 =10.71 gram of NO

then i checked for
N2+O---->N2O
in the above reaction 28 gram of N2 gives rise too 44 gram of N2O
then if u burn 5 gram of N2 = 44*5 / 28 = 7.8 gram of N2O

then i tried for the following and got ur result
N2 +O2------> N2O5

in the above reaction 28 gram of N2 gives rise too 108 gram of N2O5( nitrogen petoxide)
then if u burn 5 gram of N2 = 108*5 / 28 =19.28 gram of N2O5


the answer is N2O5

Answer 3

\[N _{2} +\frac{ x }{ 2 } O _{2} \rightarrow N _{2}O _{x}\]

Answer 4

1 mole of N2 gives 1 mole of oxide
there for
n1=n2
m1/Mr1 = m2/Mr2
5/28 =19.3/(28+16x)

Answer 5

\[x \approx5\]

Answer 6

LOL nice :)