Question

A 17.65g of a hyrdocarbon undergoes complete combustion to yield 25.01g of CO2 and 10.59g of H2O. Determine the mass of each element in the sample and the empirical formula

Answer 1

Grams of CO2 obtained = 25.01g

Moles of CO2 = Mass/ Molar mass

Molar mass of Carbon = 44g

So, moles of CO2 = 25.01g/44g/mol = 0.5684 moles

Now, 1 mole of CO2 is made up of 1mole of C and 2 moles of O . So, 0.5684 moles of CO2 contains 0.5684moles of Carbon.

So, moles of C = 0.5684

Mass of C = moles* molar mass of C = 0.5684*12.011g = 6.827grams

Grams of H2O = 10.59g

Moles of H2O = Mass/Molar mass

Molar mass of H2O = 18g/mol

Moles of H2O = 10.59g/18g/mol = 0.5883moles

Now 1 mole of H2O contains 2 moles of Hydrogen and 1 mole of Oxygen.

0.5883 moles of H2O contains 2*0.5883 moles = 1.177moles of Hydrogen

So, moles of hydrogen = 1.177moles

Mass of Hydrogen = moles*molar mass = 1.177*1g = 1.177grams

Now, total mass of hydrocarbon = 17.65g

So, mass of C + mass of O + mass of H = 17.75g

6.827g + mass of O + 1.177g = 17.75g

Mass of O = 17.75-(6.827+1.177)

= 17.75 - 8.004 = 9.746g

moles of O = mass/molar mass

= 9.746g/15.994g/mol = 0.6093

So,

Moles of C = 0.5684

Moles of O = 0.6093

Moles of H = 1.177

Divide by the smallest molar amount-

C = 0.5684/0.5684 = 1

O = 0.6093/0.5684 = 1.07=1

H = 1.177/0.5684 = 2.07 =2

So,the formula will be,

CH2O