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Question: When \( 1.50g \) of copper is heated in air, it reacts with oxygen to achieve a final mass of \( 2.1...

When 1.50g1.50g of copper is heated in air, it reacts with oxygen to achieve a final mass of 2.14g2.14g . How do you calculate the empirical formula of this copper oxide?

Explanation

Solution

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 1.50g1.50g of copper is heated in air, means the mass of copper is 1.50g1.50g
Mass of final mass after copper reacted with oxygen is 2.14g2.14g
Mass of oxygen will be 2.141.50=0.64g2.14 - 1.50 = 0.64g
The molar mass of copper is 63.55gmol163.55gmo{l^{ - 1}}
The molar mass of oxygen is 16gmol116gmo{l^{ - 1}}
The number of moles of copper is 1.5063.55=0.0236mol\dfrac{{1.50}}{{63.55}} = 0.0236mol
The number of moles of oxygen is 0.6416=0.04mol\dfrac{{0.64}}{{16}} = 0.04mol
The number of moles of copper and oxygen are in the ratio of 0.0236:0.040.0236:0.04 this ratio will be equal to nearly 3:53:5
As both these ratios will have nearly the same value
Thus, from the mole ratio the empirical formula of copper and oxygen will be Cu3O5C{u_3}{O_5}
When 1.50g1.50g of copper is heated in air, it reacts with oxygen to achieve a final mass of 2.14g2.14g then the empirical formula of copper oxide will be Cu3O5C{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 3:53:5 the value is same as 0.0236:0.040.0236:0.04