Combustion of 6.38 mg of ethylene glycol gives 9.06 mg CO2 and 5.58 mg H20. The compound contains only C, H, and O. What are the mass percentages of the elements in the ethylene glycol?
grams C = 9.06 mg CO2 x 12/44 = ?
Then % C = (grams C/6.38)*100 = ?
grams H = 5.58 mg H2O x (2*1/18) = ?
Then % H = (grams H/6.38)*100 = ?
%O = 100 - %C - %H = ?
NOTE: I don't know how accurate this is to be determined. I have rounded the numbers. For example I've used atomic mass C as 12 and not 12.011 and H as 1 and not 1.008 and O as 16 and not 15.9994.
Student
Study
Well, it looks like we have a chemical mystery on our hands! Let's break it down with a touch of humor, shall we?
To find the mass percentages of the elements in ethylene glycol, we need to determine the masses of carbon (C), hydrogen (H), and oxygen (O) in the compound. And remember, we're dealing with milligrams here, so it's a tiny circus in the lab!
First, we determine the mass of carbon dioxide (CO2). Since 1 molecule of CO2 contains 1 atom of carbon, we know that the mass of carbon in CO2 is also 9.06 mg. Quite a weight for our little carbon friend!
Next, we calculate the mass of water (H2O). In 1 molecule of H2O, we have 2 atoms of hydrogen. So, the mass of hydrogen in H2O would be (2 x 5.58 mg), which equals 11.16 mg. Hydrogen pulled an extra weight this time!
Now, let's find out the mass of oxygen. In ethylene glycol, the remaining mass is attributed to oxygen since it contains only C, H, and O. We have the overall mass of ethylene glycol (6.38 mg) minus the mass of carbon (9.06 mg) and hydrogen (11.16 mg), which gives us 55.56 mg.
Now, let's put on a little math magic show! We add up the masses of carbon, hydrogen, and oxygen: 9.06 mg (C) + 11.16 mg (H) + 55.56 mg (O) = 75.78 mg.
To find the mass percentages, we divide each individual element's mass by the total mass and multiply by 100.
Percentage of C = (9.06 mg / 75.78 mg) x 100 ≈ 11.97%
Percentage of H = (11.16 mg / 75.78 mg) x 100 ≈ 14.71%
Percentage of O = (55.56 mg / 75.78 mg) x 100 ≈ 73.32%
So, we have our circus performers' mass percentages—carbon (11.97%), hydrogen (14.71%), and oxygen (73.32%)—ready for the ethylene glycol grand finale! Tada!
To find the mass percentages of the elements in ethylene glycol, we need to calculate the mass of carbon (C), hydrogen (H), and oxygen (O) in the compound first.
Given:
Mass of CO2 produced = 9.06 mg
Mass of H2O produced = 5.58 mg
1. Calculate the moles of CO2:
Molar mass of CO2 = 12.01 g/mol (C) + 2 * 16.00 g/mol (O) = 44.01 g/mol
Moles of CO2 = Mass of CO2 (in grams) / Molar mass of CO2
= 9.06 mg / (44.01 g/mol * 1000 mg/g)
≈ 0.0002064 mol
2. Calculate the moles of H2O:
Molar mass of H2O = 2 * 1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol
Moles of H2O = Mass of H2O (in grams) / Molar mass of H2O
= 5.58 mg / (18.02 g/mol * 1000 mg/g)
≈ 0.0003098 mol
3. Calculate the moles of C:
The number of moles of C in ethylene glycol is the same as the number of moles of C in the CO2 produced, since all the carbon in ethylene glycol is converted to CO2 during combustion.
Moles of C = Moles of CO2
≈ 0.0002064 mol
4. Calculate the moles of H:
The number of moles of H in ethylene glycol is the same as the number of moles of H in the H2O produced, since all the hydrogen in ethylene glycol is converted to H2O during combustion.
Moles of H = Moles of H2O
≈ 0.0003098 mol
5. Calculate the moles of O:
Moles of O = Moles of CO2 * 2 + Moles of H2O
≈ (0.0002064 mol * 2) + 0.0003098 mol
≈ 0.0007226 mol
6. Calculate the mass percentages:
Mass percentage of C = (Mass of C / Total Mass of ethylene glycol) * 100%
= (Moles of C * Molar mass of C) / (Moles of C * Molar mass of C + Moles of H * Molar mass of H + Moles of O * Molar mass of O) * 100%
= (0.0002064 mol * 12.01 g/mol) / (0.0002064 mol * 12.01 g/mol + 0.0003098 mol * 1.01 g/mol + 0.0007226 mol * 16.00 g/mol) * 100%
Similarly, calculate the mass percentages of H and O using the above formula.
Therefore, the mass percentages of C, H, and O in ethylene glycol can be determined by following these steps and performing the necessary calculations.