Question

Furnace H20 absorber Samp A 4.989 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 7.438 grams of CO2 and 2.284 grams of H20 are produced. In a separate experiment, the molar mass is found to be 118.1 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 Submit Answer hound Formulas from Chemical Reactions: This is group attempt 1 of 5 MacBook Pro rch or tune URL

Answer 1

Let the formula of unknown organic compound be C_xH_yO_z. Let the combustion of the organic compound be

C_xH_yO_z + O₂ = CO₂ + H₂O

Applying POAC(principle of atomic conservation) on each atom of the above reaction which says that number of atoms of an element in a chemical reaction remains conserved.

_​Given,

Mass of unknown organic compound = 4.989 g

Mass of H₂O produced=2.284 g

Mass of CO₂ produced=7.438 g

Molar mass of unknown organic compound = 118.1 g/mol

Moles of unknown organic compound , n₁ =

mass of unknown compound /molar mass of unknown organic compound

Moles of unknown organic compound , n₁ = 4.989 /118.1

n₁=0.0422 mol

Moles of H₂O , n₂= mass of H₂O/molar mass of H₂O

n₂=2.284/18

n₂= 0.1269 mol

Moles of CO₂ ,n₃= mass of CO₂/molar mass of CO₂

n₃= 7.438/44

n₃= 0.1690 mol

Applying POAC on carbon atoms,

1 mole of C_xH_yO_z contains x mole atoms of C

0.0422 mole of C_xH_yO_z contains 0.0422x mole atoms of C

0.0422x atoms of C will be in reactant side.

1 mole of CO₂ contains 1 mole atoms of C.

0.1690 mole of CO₂ contains 0.1690 mole atoms of C.

Number of atoms of C in reactant is equal to number of C atoms in product side.

0.0422x=0.1690

x=4.0047= 4

Applying POAC on hydrogen atoms ,

1 mole of C_xH_yO_z contains y atoms of H

0.0422 mole of C_xH_yO_z contains 0.0422y mole atoms of H

0.0422y atoms of H will be in reactant side.

1 mole of H₂O contains 2 mole atoms of H

0.1269 mole of H₂O contains 0.2538 mole atoms of H

Number of atoms of H in reactant side will be equal to number of atoms of H in product side.

0.0422y=0.2538

y=6.0142=6

We will not apply POAC on oxygen atoms because oxygen atoms are present in excess.

Given, molar mass of C_xH_yO_z =118.1 g/mol

Also, molar mass of C_xH_yO_z is equal to

x*molar mass of C + y*molar mass of H + z*molar mass of O

x*12+y*1+z*16

4×12+6×1+z*16=118.1

54+z*16=118.1

z*16=64.1

z=4.00625=4

Hence, x=4,y=6,z=4

Molecular formula= C₄H₆O₄

highest common factor of x,y,z is 2. Hence,

​​​​​​Empirical formula= C₂H₃O₂

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