Question

4. A purified glucose sample is analyzed by combustion. 0.1486g of CO2 and 0.0609g of water are produced. The sample contains C, H and O. If the molecular weight is 180 amu; Determine the empirical and molecular formula of the compound.

Answer 1

Answer :

combustion reaction of sugar is given in the image attached

Answer & General Formula of sugar (x Hy ₂ General formula of Glucose - ((Hon y=6 combustion reaction of Sety z 2 1 x Hy ozt q (excess xcoat yH₂O

Calculations:

Mass of CO2 obtained =0.1486 g

Molar mass of CO2 = 44 g/mol

Moles of CO2 obtained = mass obtained / molar mass

= 0.1486 / 44 = 0.0034 mol

x = moles of CO2 obtained = 0.0034

Mass of H2O produced = 0.0609 g

Molar mass of H2O = 18 g/mol

Moles of H2O produced = masa produced / molar mass

= 0.0609 / 18 = 0.0034 mol

y/2 = moles of H2O obtained = 0.0034

y = 2 * 0.0034 = 0.0068

x = atomicity of carbon in the Glucose

y = atomicity of Hydrogen in the glucose

x = 0.0034

y = 0.0068

x / y (simplest ratio of C and H) = 0.0034 / 0.0068 = 1/2

So, atomicity of carbon in empirical formula = 1

Atomicity of Hydrogen in empirical formula = 2

( Keep in mind atomicity of carbon and oxygen is same in Glucose)

Atomicity of oxygen in the empirical formula = 1

Empirical formula of Glucose = C1 H2 O1

Empirical formula mass of Glucose = 1 * atomic mass of C + 2 * atomic mass of H + 1 * atomic mass of oxygen

= 1 * 12 + 2 * 1 + 1 * 16 = 30 amu

Molecular formula mass = 180 amu

Molecular formula mass = n * empirical formula mass

n = 180 amu / 30 amu = 6

So, Molecular formula is C6 H12 O6