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.

1 Answers

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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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