Combustion data allows us to solve for the empirical formula. Although the goal is often the smallest whole number ratio
Posted: Tue May 17, 2022 12:02 pm
Combustion data allows us to solve for the empirical formula.
Although the goal is often the smallest whole number ratio, we
will take a different approach so that we do not work in the
wrong direction.
Make sure to consider oxygen and use
masses from the periodic table to only two decimal places.
Divide by the smallest number of moles and you should get the
relative moles of each element.
Report these as whole
numbers (such as CH,) if applicable but for now, it is okay to
report these as decimals (such as C HO).
Once we learn the
molecular mass of the unknown, we will be able to convert our
data into molecular formulas.
Combustion problems for I/MS assignment
Temporarily skip problems shaded.
Complete answers on back
grid
You will add the answers to the skipped problems when
we cover spectroscopy.
#1: A 105.39 mg sample of unknown was submitted for combustion analysis. The sample yielded 330.60 mg of carbon dioxide and 135.37 mg of water. #2: A 574.98 mg sample of unknown was submitted for combustion analysis. The sample yielded 1694.39 mg of carbon dioxide and 504.56 mg of water. #3: A 132.17 mg sample of unknown was submitted for combustion analysis. The sample yielded 220.05 mg of carbon dioxide and 72.08 mg of water. #4: A 30.05 mg sample of unknown was submitted for combustion analysis. The sample yielded 99.02 mg of carbon dioxide and 27.03 mg of water. #5: A 52.29 mg sample of unknown was submitted for combustion analysis. The sample yielded 127.63 mg of carbon dioxide and 52.26 mg of water #6: A 74.07 mg sample of unknown was submitted for combustion analysis. The sample yielded 159.54 mg of carbon dioxide and 65.32 mg of water.
#7: A 585.68 mg sample of unknown was submitted for combustion Any yielded 1800 mg of carbon dioxide and 842.8 mg of water 18. A 46.30 mg sample of unknown was submitted for combustion analysis. The sample yielded 115.53 mg of carbon dioxide and 56.76 mg of water 19: This unknown contains 51.90% and 9.80 %. Il by mass with the remainder being an unknown clement (you have to figure out what element), MO: This unknown contains 60.96% Cand 1535 % by mass with the remainder being an unknown element (you have to figure out what element) DIT: A 150.26 mg sample of unknown was submitted for combustion analysis. The sample yielded 495.11 mg of carbon dioxide and 135.15 mg of water. #12: A 423 25 mg sample of unknown was submitted for combustion analysis. The sample yielded 1001 23 mg of carbon dioxide and 409.96 mg of water. #13: A 178.50 mg sample of unknown was submitted for combustion analysis. The sample yielded 376,53 mg of carbon dioxide and 188.85 mg of water #14: A 156.32 mg sample of unknown was submitted for combustion analysis. The sample yielded 336,68 mg of carbon dioxide and 137.85 mg of water, #15: This unknown contains 65,69% and 15.16 % H by mass with the remainder being an unknown clement (you have to figure out what element). #16: This unknown contains 28 77%C, 4.22% H and 19.16% Oxygen by mass with the remainder being an unknown clement (you have to figure out what element) #17: A 77.86 mg sample of unknown was submitted for combustion analysis. The sample yielded 184.84 mg of carbon dioxide and 94.61 mg of water. #18: This unknown contains 68.90 %C and 15.03 % H by mass with the remainder being an unknown element (you have to figure out what element). #19: A 239,86 mg sample of unknown was submitted for combustion analysis. The sample yielded 646.95 mg of carbon dioxide and 264.89 mg of water #20: A 286.52 mg sample of unknown was submitted for combustion analysis. The sample yielded 605.14 mg of carbon dioxide and 297.33 mg of water
Although the goal is often the smallest whole number ratio, we
will take a different approach so that we do not work in the
wrong direction.
Make sure to consider oxygen and use
masses from the periodic table to only two decimal places.
Divide by the smallest number of moles and you should get the
relative moles of each element.
Report these as whole
numbers (such as CH,) if applicable but for now, it is okay to
report these as decimals (such as C HO).
Once we learn the
molecular mass of the unknown, we will be able to convert our
data into molecular formulas.
Combustion problems for I/MS assignment
Temporarily skip problems shaded.
Complete answers on back
grid
You will add the answers to the skipped problems when
we cover spectroscopy.
- Combustion Data Allows Us To Solve For The Empirical Formula Although The Goal Is Often The Smallest Whole Number Ratio 1 (41.98 KiB) Viewed 63 times
- Combustion Data Allows Us To Solve For The Empirical Formula Although The Goal Is Often The Smallest Whole Number Ratio 2 (49.31 KiB) Viewed 63 times
#7: A 585.68 mg sample of unknown was submitted for combustion Any yielded 1800 mg of carbon dioxide and 842.8 mg of water 18. A 46.30 mg sample of unknown was submitted for combustion analysis. The sample yielded 115.53 mg of carbon dioxide and 56.76 mg of water 19: This unknown contains 51.90% and 9.80 %. Il by mass with the remainder being an unknown clement (you have to figure out what element), MO: This unknown contains 60.96% Cand 1535 % by mass with the remainder being an unknown element (you have to figure out what element) DIT: A 150.26 mg sample of unknown was submitted for combustion analysis. The sample yielded 495.11 mg of carbon dioxide and 135.15 mg of water. #12: A 423 25 mg sample of unknown was submitted for combustion analysis. The sample yielded 1001 23 mg of carbon dioxide and 409.96 mg of water. #13: A 178.50 mg sample of unknown was submitted for combustion analysis. The sample yielded 376,53 mg of carbon dioxide and 188.85 mg of water #14: A 156.32 mg sample of unknown was submitted for combustion analysis. The sample yielded 336,68 mg of carbon dioxide and 137.85 mg of water, #15: This unknown contains 65,69% and 15.16 % H by mass with the remainder being an unknown clement (you have to figure out what element). #16: This unknown contains 28 77%C, 4.22% H and 19.16% Oxygen by mass with the remainder being an unknown clement (you have to figure out what element) #17: A 77.86 mg sample of unknown was submitted for combustion analysis. The sample yielded 184.84 mg of carbon dioxide and 94.61 mg of water. #18: This unknown contains 68.90 %C and 15.03 % H by mass with the remainder being an unknown element (you have to figure out what element). #19: A 239,86 mg sample of unknown was submitted for combustion analysis. The sample yielded 646.95 mg of carbon dioxide and 264.89 mg of water #20: A 286.52 mg sample of unknown was submitted for combustion analysis. The sample yielded 605.14 mg of carbon dioxide and 297.33 mg of water