If a 8.5 g sample of an unknown compound decomposes to produce 7.29 g n 2 o 5 and 1.21 g h 2 o, what is its empirical formula?

Answer is: empirical formula is HNO₃.
m(unknown compound) = 8,5 g.
m(N₂O₅) = 7,29 g.
m(H₂O) = 1,21 g.
n(N₂O₅) = m(N₂O₅) ÷ M(N₂O₅).
n(N₂O₅) = 7,29 g ÷ 108,01 g/mol.
n(N₂O₅) = 0,0674 mol.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 1,21 g ÷ 18 g/mol.
n(H₂O) = 0,0673 mol.
n(H₂O) : n(N₂O₅) = 0,0673 mol : 0,0674 mol.
n(H₂O) : n(N₂O₅) = 1 : 1.
Compound has two hydroge, two nitrogen and six oxygen, can divided with two.

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer