A mixture weighing 7.290 mg contained only cyclohexane, C6H12 (FM 84.159), and oxirane, C2H4O (FM 44.053). When the mixture was analyzed by combustion analysis, 21.999 mg of CO2 (FM 44.010) were produced. Find the weight percent of oxirane in the mixture.
The combustion equations are as follows:
C6H12 + 9O2 ==> 6CO2 + 6H2O
2C2H4O + 5O2 ==> 4CO2 + 4H2O
This is an equation with two unknowns. Here are the two equations which you should solve simultaneously.
Let Y = milligrams cyclohexane
and Z = milligrams oxirane
===================
Y + Z = 7.290 is eqn 1.
(6Y*44.01/84.159) + (2Z*44.01/44.053) = 21.999 is eqn 2.
After you find Z, then
%Z = (Z(in mg)/7.290)*100 = ?
10.5 wt%
10.8
Thank you DrBob222!
To find the weight percent of oxirane in the mixture, we need to determine the amount of oxirane and cyclohexane present in the mixture.
First, let's determine the moles of CO2 produced in the combustion analysis. We can use the molar mass of CO2 to do this:
Molar mass of CO2 = 44.010 g/mol
Moles of CO2 = mass of CO2 / molar mass of CO2
= 21.999 mg / 44.010 g/mol
= 0.49981 mmol
Since the combustion of 1 mole of oxirane produces 2 moles of CO2, we can calculate the moles of oxirane present in the mixture:
Moles of oxirane = (moles of CO2 produced) / 2
= 0.49981 mmol / 2
= 0.24991 mmol
Now, we can convert the moles of oxirane to grams using the molar mass of oxirane:
Molar mass of oxirane = 44.053 g/mol
Mass of oxirane = moles of oxirane * molar mass of oxirane
= 0.24991 mmol * 44.053 g/mol
= 11.000 g
Finally, we can calculate the weight percent of oxirane in the mixture:
Weight percent of oxirane = (mass of oxirane / mass of mixture) * 100
= (11.000 g / 7.290 mg) * 100
= 150.82%
Therefore, the weight percent of oxirane in the mixture is approximately 150.82%.
Well, it seems like you're asking for some chemistry help. I'm more of a clown than a chemist, but let's see if I can help you out with a little humor!
To find out the weight percent of oxirane in the mixture, we need to do a little math. Now, I'm not a fan of math, but I'll give it a shot!
First, let's find the moles of CO2 produced. We can do this by dividing the weight of CO2 by its formula weight. So, 21.999 mg CO2 / 44.010 g/mol = approximately 0.4999 moles.
Since we know that 1 mole of oxirane produces 1 mole of CO2, and the formula weight of oxirane (C2H4O) is 44.053 g/mol, we can say that the number of moles of oxirane in the mixture is also approximately 0.4999 moles.
Now, let's calculate the weight of oxirane in the mixture. We can do this by multiplying the moles of oxirane by its formula weight. So, 0.4999 moles x 44.053 g/mol = approximately 22.021 g.
Finally, to find the weight percent of oxirane in the mixture, we divide the weight of oxirane by the weight of the mixture and multiply by 100%. So, (22.021 g / 7.290 mg) x 100% = approximately 302.3%.
Now, I don't know about you, but that seems like a lot of oxirane to me! I hope my clownish attempt at chemistry was helpful. If not, don't worry - I'm here for the laughter, not the calculations!