Question

Furnace HO absorber CO, absorber Sample When 1.690 grams of a hydrocarbon, CyHy, were burned in a combustion analysis apparatus, 5.303 grams of Co, and 2.171 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 70.13 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon. Enter the elements in the order presented in the question. empirical formula molecular formula <Previous

Answer 1

1)

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

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

= 5.303/44

= 0.1205

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

= 2.171/18

= 0.1206

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.1205

so, x = 0.1205

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.1206 = 0.2412

Divide by smallest to get simplest whole number ratio:

C: 0.1205/0.1205 = 1

H: 0.2412/0.1205 = 2

So empirical formula is:CH2

Answer: CH2

2)

Molar mass of CH2,

MM = 1*MM(C) + 2*MM(H)

= 1*12.01 + 2*1.008

= 14.026 g/mol

Now we have:

Molar mass = 70.13 g/mol

Empirical formula mass = 14.026 g/mol

Multiplying factor = molar mass / empirical formula mass

= 70.13/14.026

= 5

So molecular formula is:C5H10

Answer: C5H10