21 Chemistry -- Foundation and Fundamentals of Chemistry

1). Molecular formula of C6H12O6. Calculate    a). molecular mass of the compound   b).mass ratio and % composition of elements present in the compound2). Empirical formula of glucose is CH2O. Find the mass % of elements present in the compound.

1). Molecular formula of C6H12O6. Calculate 

   a). molecular mass of the compound

   b).mass ratio and % composition of elements present in the compound

2). Empirical formula of glucose is CH2O. Find the mass % of elements present in the compound.

Questions no. 1(a)

Solutions:

Given, Molecular formula of glucose is C6H12O6

Here, Atomic mass of one C in C6H12O6= 12 amu

         Atomic mass of one H in C6H12O6 = 1 amu

         Atomic mass of one O in C6H12O6 = 16 amu

   Now, molecular mass of the compound = (6*12)+(12*1)+(6*16)=180 units

Question no.1(b).

Solutions:

Given, Molecular formula of glucose is C6H12O6 .

Here, Mass of C in C6H12O6 = 6*12 =72

          Mass of H in C6H12O6=12*2 = 24

          Mass of O in C6H12O6 = 6*16 =96

Now, mass ratio of C:H:O = 72 : 24 : 96 = 3 :1 : 4.

Also, Percentage composition of C in  C6H12O6=(12*6/180 )*100%= 40%.  

          Percentage composition of O in  C6H12O6 = (16*6/180)*100% = 53.33%.

          Percentage composition of H in C6H12O6 = (1*12/180)*100% = 6.67%.


Question no 2.

Solution:

Given, Empirical formula of glucose is CH2O.

The total mass of CH2O = (6+6) +(1+0)*2 +(8+8) = 30.

Now, the mass % of C in CH2O= 12/30 *100% = 40%.

         The mass % of H2 in CH2O = 2/30 *100% = 6.67%.

         The mass % of O in CH2O = 16/30*100% = 53.33 %.

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