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As we know, the empirical formula of Ethyl Butyrate is = CxHyOz
Now the atomic weight of these molecules are,
C = 12
H = 1
O = 16
Therefore, the weight of 1 mol of CO2 is = (12+ 2*16) gm.
= 44 gm.
Now, this shows that, 44 gm. of CO2 is = 1 mol CO2
Therefore, 44 mg CO2 is = 1 m mol CO2
Therefore, 9.48 mg CO2 refers to = 9.48/ 44 m mol of CO2
= 0.215 m mol of CO2
Now, in 1 mol of CO2, there is 12 mg of carbon.
Therefore, in 0.215 mol of CO2, the amount of carbon is = 12* 0.215 mg.
= 2.58 mg of carbon.
Similarly, the weight of 1 mol of H2O, is = (2*1 + 16) gm.
= 18 gm.
Therefore, similarly it can be considered that 18 gm. of H2O is = 1 mol of H2O.
Therefore, 18 mg of H2O is = 1 m mol of H2O.
Therefore, 3.87 mg. of H2O is = 3.87/ 18 m mol of H2O.
= 0.215 m mol of H2O.
Now, it can be said that, in 1 m mol of H2O the amount of Hydrogen is = 2 mg.
Therefore, in 0.215 m mol of H2O, the amount of Hydrogen is = 2* 0.215 mg.
0.43 mg.
So, in 4.17 ethyl butyrate the amount of Oxygen is = {4.17 – (2.58 + 0. 43)} mg.
= 1.16 mg.
Now, since the valence of Carbon, Hydrogen and Oxygen had been assumed x, y, and z respectively,
Therefore, x: y: z = 2.58/ 12 : 0.43/ 2 : 1.16/ 16
= 2.96: 5.93: 2 …. (upon calculation)
Therefore, x: y: z = 3: 6: 2
Therefore, the empirical formula of Ethyl Butyrate is, C3H6O. Read more question answer like this one at Assignmenthelp.us.
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