A 0.0245 mg sample of nitrogen reacts with oxygen to form 0.0945 mg of the oxide. What is the empirical formula of this nitrogen oxide compound?

1 Answer
Aug 19, 2017

Answer:

The empirical formula is #"N"_2"O"_5#.

Explanation:

The equation for the reaction is

#A_"r":color(white)(mmmll)14.01color(white)(mml)16.00#
#color(white)(mmmmmll)x"N"_2color(white)(m) + color(white)(ll) y"O"_2 → 2"N"_x"O"_y"#
#"Mass/µg":color(white)(m)24.5color(white)(mmm)70.0"color(white)(mml)94.5#

First, we must calculate the masses of #"N"# and #"O"# from the mass of the oxide.

#"Mass of N in oxide = 24.5 µg"#

#"Mass of O in oxide" = "mass of oxide - mass of N" = "(94.5 - 24.5) µg" = "70.0 µg"#

Now, we must convert these masses to moles and find their ratios.

From here on, I like to summarize the calculations in a table.

#ul(bb("Element"color(white)(X) "Mass/µg"color(white)(X) "Amt/µmol"color(white)(m) "Ratio"color(white)(ml)×2color(white)(mm)"Integers"))#
#color(white)(mm)"N"color(white)(mmmml)24.5color(white)(mmmmll)1.748color(white)(Xmm)1color(white)(mmml)2color(white)(Xmmml)2#
#color(white)(mm)"O" color(white)(XXXml)70.0 color(white)(mmmmll)4.375 color(white)(mmm)2.502 color(white)(ml)5.004color(white)(Xml)5#

The empirical formula is #"N"_2"O"_5#.