Answer to Question #193309 in Inorganic Chemistry for mary

Question

An oxide of nitrogen contains 63.1% oxygen and has a molar mass of 76.0 g/mol. What is the empirical

formula for this compound?

Answer 1

Take a hypothetical 100-gram sample of the gas and convert each fractional weight to moles.

$\dfrac{63.1g}{15.9994(\frac{g}{mole})}=3.9438979 \;moles\;of\;O$

$\dfrac{36.9\;g}{14.0067(\frac{g}{mole})}=2.6344525\;moles\;ofN$

Divide by the smaller number of moles:

$O:\dfrac{3.9438979}{2.63445351}=1.49704593$

$N:\dfrac{2.63445351}{2.63445351}=1$

Multiply by 2 to make integer ratios:

N₂O₃ This is the empirical formula.

Answer:

N₂O₃ has a molar mass of 76.0117 $(\frac{g}{mole})$, so it is molecular formula too.