Question

- Furce HO absorber CO, absorber Sample When 3.119 grams of a hydrocarbon, C Hy, were burned in a combustion analysis apparatus, 10.54 grams of CO2 and 2.159 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 78.11 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

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

= 10.54/44

= 0.2395

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

= 2.159/18

= 0.1199

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.2395

so, x = 0.2395

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.1199 = 0.2399

Divide by smallest to get simplest whole number ratio:

C: 0.2395/0.2395 = 1

H: 0.2399/0.2395 = 1

So empirical formula is:CH

Answer: CH

2)

Molar mass of CH,

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

= 1*12.01 + 1*1.008

= 13.018 g/mol

Now we have:

Molar mass = 78.11 g/mol

Empirical formula mass = 13.018 g/mol

Multiplying factor = molar mass / empirical formula mass

= 78.11/13.018

= 6

So molecular formula is:C6H6

Answer: C6H6