Question

Glucose, a common carbohydrate, is a molecule that contains the elements carbon, hydrogen, and oxygen. Complete combustion of a 0.360 g sample of glucose produced 0.528 g carbon dioxide and 0.216 g of water. Find the empirical formula for glucose

Answer 1

The formation of carbondioxide is given by equation as

C + O2 ----> CO2(g)

As per balanced equation one mole of carbondioxide is obtained from one mole of carbon

if 44 g of CO2 is formed it means it contains 12 g of carbon

if 1g of CO2 is formed it means it contains 12/44 g of carbon

if 0.528g of CO2 is formed it means it contains = 12 X 0.528 / 44 g of carbon

                                                                           = 0.144g of carbon

Similarly for water

H2 + 0.5O2 ---> H2O

so one mole of water is obtained from one mole of H2

if 18g of water is formed it means 2g H2 is present

if 1g of water is formed it means 2/18 g of H2 is present

if 0.216g of water if formed it means 2 X 0.216 / 18g of H2 is present = 0.024g of Hydrogen

In glucose we have Carbon , Hydrogen and oxygen

As given the total mass of sample = 0.360g

Mass of carbon = 0.144 , moles of carbon = 0.144 / 12 = 0.012

Mass of hydrogen = 0.024 , moles of hydrogen = 0.024 /1 = 0.024

mass of oxygen = 0.360 - (0.144+0.024) = 0.192

Moles of oxygen = 0.192 / 16 = 0.012

So mole ratio of C : H : O = 0.012 : 0.024 : 0.012 = 1:2:1

Empirical formula = CH2O