Question

(A)

When 5.492 grams of a hydrocarbon, C_xH_y, were burned in a combustion analysis apparatus, 16.64 grams of CO₂ and 8.514 grams of H₂O were produced.
In a separate experiment, the molar mass of the compound was found to be 58.12 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 =

(B) When 5.925 grams of a hydrocarbon, C_xH_y, were burned in a combustion analysis apparatus, 20.03 grams of CO₂ and 4.100 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 =

(C)
A 7.325 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 7.160 grams of CO₂ and 1.466 grams of H₂O are produced.
In a separate experiment, the molar mass is found to be 90.04 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

• Enter the elements in the order C, H, O
  

  empirical formula =

  molecular formula =

Answer 1

A)

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

= 16.64/44

= 0.3782

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

= 8.514/18

= 0.473

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.3782

so, x = 0.3782

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.473 = 0.946

Divide by smallest:

C: 0.3782/0.3782 = 1

H: 0.946/0.3782 = 2.50

Multiply by 2 to get simplest whole number ratio:

C: 1*2 = 2

H: 2.50*2 = 5

So empirical formula is:C2H5

Molar mass of CH3,

MM = 2*MM(C) + 5*MM(H)

= 2*12.01 + 5*1.008

= 29.06 g/mol

Now we have:

Molar mass = 58.12 g/mol

Empirical formula mass = 29.06 g/mol

Multiplying factor = molar mass / empirical formula mass

= 58.12/29.06

= 2

So molecular formula is:C4H10

Answer:

C2H5

C4H10

B)

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

= 20.03/44

= 0.4552

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

= 4.1/18

= 0.2278

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.4552

so, x = 0.4552

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.2278 = 0.4556

Divide by smallest to get simplest whole number ratio:

C: 0.4552/0.4552 = 1

H: 0.4556/0.4552 = 1

So empirical formula is:CH

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:

CH

C6H6

C)

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

= 7.16/44

= 0.1627

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

= 1.466/18

= 8.144*10^-2

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.1627

so, x = 0.1627

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*8.144*10^-2 = 0.1629

Molar mass of O = 16 g/mol

mass O = total mass - mass of C and H

= 7.325 - 0.1627*12 - 0.1629*1

= 5.209

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

= 5.209/16.0

= 0.3256

so, z = 0.3256

Divide by smallest to get simplest whole number ratio:

C: 0.1627/0.1627 = 1

H: 0.1629/0.1627 = 1

O: 0.3256/0.1627 = 2

So empirical formula is:CHO2

Molar mass of CHO2,

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

= 1*12.01 + 1*1.008 + 2*16.0

= 45.018 g/mol

Now we have:

Molar mass = 90.04 g/mol

Empirical formula mass = 45.018 g/mol

Multiplying factor = molar mass / empirical formula mass

= 90.04/45.018

= 2

So molecular formula is:C2H2O4

Answer:

CHO2

C2H2O4