Question

When $ 1.50g $ of copper is heated in air, it reacts with oxygen to achieve a final mass of $ 2.14g $ . How do you calculate the empirical formula of this copper oxide?

Answer 1

Hint: Given is the mass of copper and mass of final mass which is the sum of the both oxygen and copper mass. Thus, oxygen mass can be determined from the above two masses. Calculate the number of moles of both oxygen and copper from atomic mass and mass. Mole ratio will be calculated from moles given the empirical formula.

Complete Step By Step Answer:
Given that $ 1.50g $ of copper is heated in air, means the mass of copper is $ 1.50g $
Mass of final mass after copper reacted with oxygen is $ 2.14g $
Mass of oxygen will be $ 2.14 - 1.50 = 0.64g $
The molar mass of copper is $ 63.55gmo{l^{ - 1}} $
The molar mass of oxygen is $ 16gmo{l^{ - 1}} $
The number of moles of copper is $ \dfrac{{1.50}}{{63.55}} = 0.0236mol $
The number of moles of oxygen is $ \dfrac{{0.64}}{{16}} = 0.04mol $
The number of moles of copper and oxygen are in the ratio of $ 0.0236:0.04 $ this ratio will be equal to nearly $ 3:5 $
As both these ratios will have nearly the same value
Thus, from the mole ratio the empirical formula of copper and oxygen will be $ C{u_3}{O_5} $
When $ 1.50g $ of copper is heated in air, it reacts with oxygen to achieve a final mass of $ 2.14g $ then the empirical formula of copper oxide will be $ C{u_3}{O_5} $ .

Note:
The empirical formula is the simplest mole ratio of the atoms present in the molecular formula. We obtained the ratio in the decimal points but the simplest mole ratio must be in positive integers, Thus, the ratio was taken as $ 3:5 $ the value is same as $ 0.0236:0.04 $