The NaOH trap weighed 34.086 g. The P2O5 trap weighed 38.295. The Cathecol reage
ID: 891019 • Letter: T
Question
The NaOH trap weighed 34.086 g. The P2O5 trap weighed 38.295. The Cathecol reagent bottle weighed 04.855.
Now, after I got the weight of the Cathecol (04.855) I had to drag it to the chamber and release it, so all 04.855 was released into the chamber, heat it, and after about a minute, it stopped.
When that finished, I had to take the NaOH trap & weigh that, and the mass of it is 45.732 g.
Then, I had to take the P2O5 trap and weigh that, and the mass of it was 40.674 g.
Next, there is an unknown regeant bottle, and I had to disperse 5g from that and weigh the 5g, and that was 05.189 g. After that, I had to take that and load it into the combustion chamber, heat it and after a minute or so, it stopped.
After that completed, I had to take the NaOH trap and weigh it, and it was 57.949 g.
Then, I had to take the P2O5 trap and weigh it, and the mass of it was 44.013 g
Next, repeat the calculations for the unknown sample above. From the change in the P2O5 and NaOH trap masses, determine the masses of H2O and CO2 produced. Please make sure to use the change in mass of each trap just before, and just after the unknown sample is combusted. Note- do not use the intial masses of the two traps at the start. All of those numbers are included above.
3a - Mass of H20 - ? g
Mass of CO2 - ? g
3b - Convert the above masses into the number of moles of H2O and CO2 produced from the unknown sample.
Moles of H2O - ? moles
Moles of CO2 - ? moles
3c - From the number of moles of H2O and CO2 produced, determine the number of moles of H atoms and C atoms in the sample.
Moles of H atoms - ? moles
Moles of C atoms - ? moles
3d - From the above and the atomic masses of the elements, determine the masses of H and C in the sample.
Mass of H - ? g
Mass of C - ? g
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Question 2 -
The mass of O in the unknown sample can obtained by the difference of the sample mass and the total of calculated masses of H and C. Calculate the mass and number of moles of O atoms in the unknown sample.
Mass of O - ? g
Moles of O - ? moles
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Question 3 - Determine the simpliest whole number mole ratio of C, H and O in the unknown, using ONLY 5 characters max such as C2H3O. If a coefficient for a given element is 1, omit it from the empirical formula.
Type answer as CxHyOz - ?
Hint - The correct empirical formula will use small numbers such as 1, 2, 3, 4
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These are apart of a long question, with many parts. I've included more info from previous questions below. I don't know if this will be helpful or not, but I wanted to include it, regardless.
Catechol is C6H6O2
Molar mass of catechol = 110 g/mole
Thus, moles of catechol in 4.855 g of it = mass/molar mass = 0.044
C6H6O2 + (13/2)O2 -----------> 6CO2 + 3H2O
Now, as per the balanced reaction,
moles of CO2 produced = 6*moles of catechol reacted = 0.265
moles of H2O produced = 3*moles of catechol reacted = 0.132
Molar mass of CO2 = 44 g/mole n and H2O = 18 g/mole
Thus, mass of CO2 produced = moles*molar mass = 0.265*44 = 11.652 g
mass of H2O produced = moles*molar mass = 0.132*18 = 2.383 g
Moles of H atoms = 2*moles of H2O = 0.265
moles of C atoms = moles of CO2 = 0.265
Molar mass of C-atom = 12 g/mole & H-atom = 1 g/mole
mass of H-atom produced = moles*molar mass = 0.265 g
mass of C-atoms = moles*molar mass = 0.265*12 = 3.178 g
Explanation / Answer
Solution :-
Mass of unknown sample = 5.189 g
Mass of NaOH trap before = 45.732 g
Mass of P2O5 trap before = 40.674 g
Mass of NaOH trap after the combustion = 57.949 g
Mass of P2O5 trap after combustion = 44.013 g
Mass of CO2 produced = 57.949 g – 45.732 g = 12.217 g
Mass of water produced = 44.013g –40.674 g =3.339 g
Now lets calculate the moles of CO2 and H2O
Moles of CO2 = 12.217 g / 44.01 g per mol = 0.2776 mol CO2
Moles of H2O = 3.339 g/ 18.0148 g per mol = 0.1853 mol H2O
Now lets calculate moles of the C and H
Mole of C= 0.2776 mol CO2 * 1 mol C / 1 mol CO2 = 0.2776 mol C
Moles of H= 0.1853 mol H2O * 2 mol H / 1 mol H2O = 0.3706 mol H
Now lets calculate the mass of the C and H
Mass of C= 0.2776 mol * 12.01 g per mol = 3.334 g C
Mass of H = 0.3706 mol * 1..079 g per mol = 0.3735 g H
Now lets find mass of Oxygen
Mass of oxygen = 5.189 g –(mass of C+ mass of H)
= 5.189 g – (3.334 g+0.3735g)
=1.4815 g O
Now lets calculate moles of oxygen
Moles of O = 1.4815 g / 16 g per mol = 0.0926 mol O
Now lets find the simplest mole ratio of the elements by dividing the moles of each element by smallest mole value
C=0.2776/0.0926 =3
H=0.3706/0.0926 = 4
O= 0.0926/0.0926 =1
Therefore the empirical formula of the unknown = C3H4O