Question

This might be kinda tough to answer, but it's a several part question and I'm totally lost on it. I'll try to make it as easy to read/understand as I can.

First, here's the data. - 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

Here's the first question.-- From the change in the P2O5 trap mass before and after the reaction, determine the mass of H2O produced by combustion of the cathechol sample.

Mass of H2O - ? g

Mass of CO2 - ? g

Here's how to solve it....The initial mass of the P2O5 trap needs to be subtracted from the final mass of the P2O5 trap.

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Question 2 - Convert those masses into the number of moles of H2O and CO2 produced from the Cathecol sample.

Moles of H2O - ? moles

Moles of CO2 - ? moles

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Question 3 - From the number of moles of H2O and CO2 produced, determine the number of moles of H atoms and C atoms in the cathecol sample.

Moles of H atoms - ? moles

Moles of C atoms - ? moles

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Question 4 - From the above and atomic masses of the elements, determine the mass of H and C in the sample.

Mass of H atoms - ? g

Mass of C atoms - ? g

Thanks so much for any help in advance. I GREATLY appreciate it!

Answer 1

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