This is a collection of my solutions to the research.net levels of SpaceChem. They are not particularly optimised, but they work and they may give you ideas for your own solutions. I'm not all the way through the official research.net levels yet (not even close), but will add more as I proceed.
I hope this helps you if you are stuck on something.

A very simple puzzle, so why not optimize for speed?

Not a hard puzzle, let Red do the grunt-work of chopping up the molecule and Blue can take care of the rest.

The trick with this level is to use both waldos to simultaneously move the Arsenic and the Silicon "shell" from the tunnel location to the bonding location. Once the first bond is made, everything is easy.

A simple puzzle, the red and the blue waldos do exactly the same job (to solve twice as fast).

Another simple puzzle, again both waldos do exactly the same job to double the solution speed.

This was a tricky one for me. What you need to do here is to temporarily "store" some of the larger atoms while you take care of the smaller ones. You need at least two storage locations, one for the Beryllium and one for the Heliums. Then you go back and collect the stored atoms for further processing.

Blue makes the Chlorine atoms and Red assembles the molecule.

Red assembles each half of the molecule. Blue has the easy job of putting them together.

Red assembles the molecule, Blue checks to see which molecule it is and then sends it to the correct output.

Here I let Red take care of the Ethanol -> Ethane conversion and Blue assembles the O2.

Here Red assembles the "body" of the Propane, then blue adds the ends.

The size of the two input molecules requires some careful manoeuvring. Get rid of the Methane as quickly as possible so you have room to assemble the Benzoic Acid.

Blue splits the Bromine and Red adds the components to the Ethylene.

There needs to be some complicated manoeuvring to ensure the molecule components don't bump into each-other. Rotate is your friend.

In this puzzle, assembling as much of the molecule in a straight line is the key. Red and Blue must work together to build the body of the molecule, then they need to work together to put on the finishing touches.

A fairly simple puzzle, you need to rotate the molecule so that you can use the same bonding tool to rebind the hydrogen.

Red does most of the work assembling one half of the final molecule. Blue puts them together and outputs.

Here Blue works to galvanise each side of the molecule, rotates and repeats, when all four sides have been done, red outputs the molecule and inputs another one.

Blue splits the O2 & applies it to the molecule. Red takes care of the rest.

Blue and Red work together to re-arrange the atoms to make the water and the Glyoxylic Acid.

Blue makes each half of the molecule and Red assembles them together.

This is one of the toughest SpaceChem puzzles I have ever completed. At first it seems simple: Fission from one of the Chlorines, and fuse that onto a hydrogen of the Borane molecule, rinse and repeat until the Chlorine has "swapped" places with the hydrogen. Unfortunately you quickly discover that when the fission atom becomes Neon it breaks the bond (Neon being a noble gas). And this puzzle has not provided us with a way to repair the bond!
The solution I came up with was to build two Chloroborane molecules from two Borane molecules by fissioning from both atoms of the Chlorine molecule (leaving two hydrogen atoms behind). Then building a hydrochloric acid molecule from two Borane molecules and the left-over hydrogens.

Red does most of the work assembling the Sulfuric Acid molecule, Blue puts on the finishing touches and outputs.

This was another tough one for me. There's so much that needs to be done! My advice is to get Red and Blue working simultaneously as much as possible to save on reactor space as it runs out quickly.

A simple atom swap and output. Both waldos run concurrently to save cycles.

Some manipulation is required as there are only four bonding tools for quite a few operations, but this one is not too hard.

Red gets the bits we want from the Butadeine and Blue gets the bits we want from the Ethylene. The rest is flushed down the Hydrogen hole.

Blue splits up the Sulphur Dichloride and places the required atoms ready for Red. Red simply grabs the Ethylene and adds it to the growing chemical weaponry.

This is a tricky one as there is no atom detecting tool. The secret is that Flourine can only have one bond while Chlorine can have many more (7). The trick to detecting what molecule you have got is to +bond to maximum, and then -bond once. Note: you need to seperate the Carbon atoms, or their limit (4) will ruin the test. After that, one waldo (in my case Red) continues on as if the molecules did not separate and outputs the Freon molecule. After that Blue continues on as if the molecules did separate and does it's thing.

Luckily the structure of the Phenol and the Toluene are so similar, that the assembly routine is identical for each. The atom detector is only needed to choose which output to use.

Flip-flops are the key to this one. Blue controls the action, running one of two programs (decided by the flip-flops). Red only has to drop the lime in the same place.

There are two stages to this solution. In stage one on the left, Red and Blue work together to flip two columns around so that all the Leads and Tins line up. In stage two, on the right, Red and blue work together to swap two rows around to create the desired arrangement.

I love these massive molecule puzzles. In this one, the real trick is to assemble the basic components in a very small space as the growing molecule takes up more and more real-estate.

Using a flip-flop, I am able to test both Hydrogen/Uranium atoms for their Uranium-ness. If the test fails, Blue switches the Uranium for a fresh new Hydrogen. otherwise the program continues, (it either goes to the next test, or outputs the clean water molecule).

In this solution, Blue's single job is to grab the presented Oxygen atom and make water. Red's more complicated job is to present the Oxygen atom each time, and that means it needs to split the Sulfur atom at the end.

Make sure you can dispose of the water quickly and then you are free to spend the rest of your reactor "space" on processing the Urea.

Special bonus Solution: Compare my solution above with this much more optimised one by Youtube user ToughThought.
So efficient!

This is a fairly simple atom swap.

The trick to this puzzle is to make the operations that Red and Blue make identical whether they are working on the first column, the second column or the third.

As there are only two bonding tools, this puzzle requires careful management of the bonding/de-bonding actions. The fuse is an afterthought.

A fairly straightforward long chain polymer puzzle.

Another fairly easy long chain polymer. Just use a flip flop to govern which Oxygen atom to use.

With Red's help, Blue constructs each half of the molecule. After that, Red assembles the two halves and adds any additional bonds required.

Blue assembles each half of the molecule and Red puts them together.

Blue assembles the ends of the molecule while red takes care of the middle parts. I assembled this one on its "side" to reduce the travel time for the waldos and also to give myself a bit more room to manoeuvre during the assemble (the work area is wider than it is tall).

Again, Blue assembles the ends of the molecule while red takes care of the middle parts. While I could easily have used the atom detector to judge that the molecule was big enough, I opted to use a bunch of syncs instead.

The difficulty of managing this one, is that most (or all) of the (Triol) Cyanuric Acid molecule (Beta) needs to be in the reactor while you work on the Benzine (Alpha). Many operations require rotating large molecules, so a complicated dance ensues.
(Apologies for the herky-jerk video, the normal SpaceChem video generator tool didn't want to output the entire solution, so I had to use my dodgy screencapture program.)

This one's pretty tough, there's a lot to do and only one bonding/de-bonding location. I had some trouble with running out of reactor space, but eventually got it and here's my solution.

This is another tough one (that took me a while to get around to). There are so many steps to complete to make this molecule. Ensure Blue and Red work together and use plenty of flip flops so you can reuse paths, otherwise you will run out really fast.

Blue does most of the work, fusing the Carbons together and re-attaching the Hydrogens. Red finishes off and starts another.

One solution to this puzzle is to chop off one Hydrogen and then take the molecule down to fuse seven hydrogens into the carbon, re-attach the first Hydrogen and away you go. The problem that crops up is that when the middle atom passes through being Neon, it breaks all the bonds (Neon being a noble gas with no possible bonds). The quick hack I came up with was to fuse the seven Hydrogens first before being fused to the Carbon. Works pretty well (and speeds it up too),

This one is pretty self explanatory. Red splits off the Carbon-Dioxide, then Blue finishes off the Ethanol molecule.

Red assembles each half of the molucule and Blue puts them together.

As luck would have it, the atoms that you get as the result of a Magnesium being split in half, split in half again and split in half again, are exactly the atoms you need to assemble a Methyl Lithium molecule. This puzzle is easier than it should be.

This is a real head scratcher. I had to go away and think about this one a good long while.
The solution I came up with was to use (and re-use) a single Hydrogen from the beta source as a test tool. If the atom from the alpha source bonds, then it must also be a Hydrogen, otherwise, Red disposes of the Helium. Blue then adds the Hydrogen (if there is one) to a continuous snake of atoms. Being Hydrogens, the "snake" can only grow to two long, but that is enough. When the snake is long enough, Red can take the H2 molecule away and the process continues.

Get rid of the iron straight away, and then Red and Blue can work on the Ozone molecule together.

If it's a methane, Red takes it straight to disposal, otherwise, Red and Blue work together to seperate the oxygen. Having Blue take the assembled methane to disposal (instead of Red) can save on return cycles between molecules.

The only hard part about this puzzle is timing it so the two resulting molecules don't run into each-other.

The hard part about this one is juggling the limited bonding tools (only four).

Fusing the hydrogens up to oxygen will break the bond (when the atom becomes helium), but there are some bonding tools available to repair the damage.

This probably isn't the most optimised solution ever, but it works. The method I used is after swapping out the hydrocarbon with the nitrogen to change the benzene into pyridine, you are left with either a nitrogen or a hydrocarbon. If it's the hydrocarbon, I join the two halves together and output the acetylene and start again. If it's a nitrogen, I join them together and spin them around, thus making a hydrogen cyanide ready to go - ready except I need to move it out of the way for a second as the next cycle restarts.

This starts out simple. Red fuses two hydrogens to make the oxygen and then positions the molecule so that Blue can start pumping in entire Ethylene molecules to make the chlorine atoms. The tricky part is that you need to fuse a carbon atom before fusing in two hydrogens, otherwise the atom may become a noble gas (helium or neon) and break the bond. I managed this with a little bit of flip-flop-fu.

To solve this puzzle, you need to split the neon into a Boron atom and five Hydrogen atoms, then fuse the Boron with one Hydrogen to create the necessary Carbon. It requires some careful space management using flip-flops to achieve this and to assemble the Methane ready for output.

I'm not super happy with this solution because it wont run forever without eventually causing an error. The general plan is to split the lead atoms and re-fuse them back into gold, leaving a Lithium atom left over. There's no way to dispose of the Lithium atom so my plan is to fuse all the left-overs together and hope it doesn't build up too fast before causing an error. I would love to see a permanent solution if anyone has ideas...

This one was a welcome relief after the much harder ones I've been doing recently. Red does most of the work assembling the Urea, while Blue adds the Hydrogens to the Water.

This is a pretty straightforward atom arrangement puzzle. I combined the bonder tools together to get multiple jobs done simultaneously.

The hardest part about this puzzle is moving the Borane and Ammonia molecules around while assembling the much larger Borazine molecule. It gets crowded very quickly.
In this solution I have Red stripping away the unneeded Hydrogens and combining the remains. Blue assembles each half of the Borazine molecule together.

Sorry, no video for this one yet. Unfortunately, the built in screen capture program doesn't show very much detail for the multiple reactor missions and I don't yet have a decent screencapture program working.

Sorry, no video for this one yet. Not until I can get a decent screencapture program working.

I have not solved this puzzle yet.

This one looked like so much fun, I had to attempt it early. In my solution, Red changes each Sulphur Dioxide molecule into a U-shaped O4 molecule. Blue assembles the rest of the snake.

I'm sorry, I have not completed any more levels of SpaceChem. Rest assured I am quite addicted, and it's only a matter of time.