A 1.24-g sample of a hydrocarbon, when completely burned in an excess of O2(g), yields 4.04 g CO2 and 1.24 g H2O Draw a plausible structural formula for the hydrocarbon molecule.
I think I understand how to get it to an empirical formula but I don't know how to get it into a structural formula.
5 answers
based on the empirical formula and its contents, you can form the structural formula. what do you think the empirical formula is?
Sorry, I thought empirical gave only the simplest structure. I thought I would need to know the molecular to get the correct number of carbons and hydrogens.
What empirical formula did you get?
The empirical formula and structural formula are not the same thing... The empirical formula just shows the lowest whole number ratio of elements in a compound. The structural formula is a 2-dimensional or 3-dimensional graphic formula showing a plausible arrangement of elements.
For this problem ...
1. Determine the % Elemental Composition of CO2 and H2O using formula weights.
2. Determine grams Carbon and Oxygen in 4.04 gram yield of CO2
... grams Carbon = Wt%C x 4.04g CO2 = 27.27% of 4.04g = 1.102g C
... grams Oxygen = Wt%O x 4.04g CO2 = 72.73% of 4.04g = 2.938g O
3. Determine grams Hydrogen and Oxygen in 1.24 gram yield of H2O
... grams Hydrogen = Wt%H x 1.24g H2O = 11% of 1.24g = 0.136g H
... grams Oxygen = Wt%O x 1.24g H2O = 89% of 1.24g = 1.104g O
4. Determine mass of Hydrocarbon consumed...
... grams of HC = grams C + grams H =
4. Determine grams of Hydrocarbon consumed = grams C + grams H
= 1.102g + 0.136g = 1.238g HC consumed
5. Determine Empirical Formula from gram mass values ...
a. Convert to % per 100 wt
... %C per 100wt = (1.102/1.238)100% = 89.014%
... %H per 100wt = (0.136/1.238)100% = 10.986%
b. Calculate the empirical ratio ...
% => grams => moles => mole ratio => reduce (divide by smaller mole value) => empirical ratio => empirical formula
%C = 89.014% => 89.014g => (89.014/12) = 7.418 mole C
%H = 10.986% => 10.986g => (10.986/1) = 10.986 mole H
C:H mole ratio = 7.418 mole C : 10.986 mole H
Reduce moles by dividing by smaller mole value => empirical ratio ...
=> 7.418/7.418 : 10.986/7.418 => 1:1.5 => 2:3 empirical ratio...
Empirical Formula => C2H3
Most likely Molecular Formula => 2(C2H3) => C4H6
2C4H6 + 11O2 => 8CO2 + 6H2O which corresponds to moles of product yield (4.04/44 = 0.092 mole CO2) and (1.24/18 = 0.069 mole H2O) given in problem. Such requires 0.1263 mole O2 and 0.023 mole of C4H6.
Structural Formula (using 'Lewis Formula' graphics) ...
H2C = CH - CH = CH2
Hope this helps, Doc :-)
For this problem ...
1. Determine the % Elemental Composition of CO2 and H2O using formula weights.
2. Determine grams Carbon and Oxygen in 4.04 gram yield of CO2
... grams Carbon = Wt%C x 4.04g CO2 = 27.27% of 4.04g = 1.102g C
... grams Oxygen = Wt%O x 4.04g CO2 = 72.73% of 4.04g = 2.938g O
3. Determine grams Hydrogen and Oxygen in 1.24 gram yield of H2O
... grams Hydrogen = Wt%H x 1.24g H2O = 11% of 1.24g = 0.136g H
... grams Oxygen = Wt%O x 1.24g H2O = 89% of 1.24g = 1.104g O
4. Determine mass of Hydrocarbon consumed...
... grams of HC = grams C + grams H =
4. Determine grams of Hydrocarbon consumed = grams C + grams H
= 1.102g + 0.136g = 1.238g HC consumed
5. Determine Empirical Formula from gram mass values ...
a. Convert to % per 100 wt
... %C per 100wt = (1.102/1.238)100% = 89.014%
... %H per 100wt = (0.136/1.238)100% = 10.986%
b. Calculate the empirical ratio ...
% => grams => moles => mole ratio => reduce (divide by smaller mole value) => empirical ratio => empirical formula
%C = 89.014% => 89.014g => (89.014/12) = 7.418 mole C
%H = 10.986% => 10.986g => (10.986/1) = 10.986 mole H
C:H mole ratio = 7.418 mole C : 10.986 mole H
Reduce moles by dividing by smaller mole value => empirical ratio ...
=> 7.418/7.418 : 10.986/7.418 => 1:1.5 => 2:3 empirical ratio...
Empirical Formula => C2H3
Most likely Molecular Formula => 2(C2H3) => C4H6
2C4H6 + 11O2 => 8CO2 + 6H2O which corresponds to moles of product yield (4.04/44 = 0.092 mole CO2) and (1.24/18 = 0.069 mole H2O) given in problem. Such requires 0.1263 mole O2 and 0.023 mole of C4H6.
Structural Formula (using 'Lewis Formula' graphics) ...
H2C = CH - CH = CH2
Hope this helps, Doc :-)
Here's the Lewis Structure Analysis for C4H6 ...
Available valence electrons = 4C + 6H = 4(4) + 6(1) = 22
Electrons in All Bonds = 4C + 6H = 4(8) + 6(2) = 44
No of Cov Bonds = Bonded Electrons - Valence Electrons / 2 = (44-22/2) = 11 Covalent Bonds in molecule...
The most likely structure conforming to the octet rule is ...
H2C = CH - CH = CH2 (containing 11 covalent bonds)
Available valence electrons = 4C + 6H = 4(4) + 6(1) = 22
Electrons in All Bonds = 4C + 6H = 4(8) + 6(2) = 44
No of Cov Bonds = Bonded Electrons - Valence Electrons / 2 = (44-22/2) = 11 Covalent Bonds in molecule...
The most likely structure conforming to the octet rule is ...
H2C = CH - CH = CH2 (containing 11 covalent bonds)