Question

1.50 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 138. g/mol, is burned completely in excess oxygen, and the mass of the products carefully measured: product carbon dioxide water mass 3.35 g 0.59 g Use this information to find the molecular formula of X. x 6 ?

Answer 1

let in compound number of moles of C, H and O be x, y and z respectively

Number of moles of CO2 = mass of CO2 / molar mass CO2

= 3.35/44

= 7.614*10^-2

Number of moles of H2O = mass of H2O / molar mass H2O

= 0.59/18

= 3.278*10^-2

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 7.614*10^-2

so, x = 7.614*10^-2

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*3.278*10^-2 = 6.556*10^-2

Molar mass of O = 16 g/mol

mass O = total mass - mass of C and H

= 1.5 - 7.614*10^-2*12 - 6.556*10^-2*1

= 0.5208

number of mol of O = mass of O / molar mass of O

= 0.5208/16.0

= 3.255*10^-2

so, z = 3.255*10^-2

Divide by smallest:

C: 7.614*10^-2/3.255*10^-2 = 2.33 = 7/3

H: 6.556*10^-2/3.255*10^-2 = 2

O: 3.255*10^-2/3.255*10^-2 = 1

Multiply by 3 to get simplest whole number ratio:

C: 7/3 * 3 = 7

H: 2*3 = 6

O: 1*3 = 3

So empirical formula is:C7H6O3

Molar mass of C7H6O3,

MM = 7*MM(C) + 6*MM(H) + 3*MM(O)

= 7*12.01 + 6*1.008 + 3*16.0

= 138.118 g/mol

Now we have:

Molar mass = 138.0 g/mol

Empirical formula mass = 138.118 g/mol

Multiplying factor = molar mass / empirical formula mass

= 138.0/138.118

= 1

So molecular formula is:C7H6O3

Answer: C7H6O3