Please note that water is composed by two hydrogens and one oxygen. The stable isotope of oxygen has atomic number $8$, but atomic mass $16$, i.e. it is composed out of eight protons and eight neutrons. Thus in normal water the ratio of hydrogen contribution to the over all in mass is $2/18$.
When deuterium takes the place of hydrogen we have heavy water, and that means that two more neutrons are added to the original mass balance, an effect again of $2/18$ extra mass. So the water is more dense but by this small factor.
In this article the advantages of heavy water in reactors are given.
Using water as a moderator will absorb enough neutrons that there will be too few left over to react with the small amount of 235U in the fuel, again precluding criticality in natural uranium.
heavy water seems to solve the problem:
An alternative solution to the problem is to use a moderator that does not absorb neutrons as readily as water. In this case potentially all of the neutrons being released can be moderated and used in reactions with the 235U, in which case there is enough 235U in natural uranium to sustain criticality. One such moderator is heavy water, or deuterium-oxide.
The reason is that deuterium has already bound a neutron to the hydrogen atom, and thus fewer neutrons needed for criticality are absorbed.
If you throw a bunch a uranium ore in one blob, nothing happens.
If you chemically purify the ore so that the only element present is uranium, still nothing happens.
The runaway chain reaction needed for a uranium-powered bomb involves U-235, an isotope having three fewer neutrons than the most common natural isotope U-238. According to Wikipedia,
The required material must be 85% or more of ${}^{235}\mathrm{U}$ and is known as weapons grade uranium...
Now U-235 and U-238 have essentially indistinguishable chemistry.1 How do you purify the material to be mostly U-235? The traditional technique is to make the gaseous compound uranium hexafluoride, $\mathrm{UF}_6$, and use some mass-based physical process to separate out the ${}^{235}\mathrm{UF}_6$ from the ${}^{238}\mathrm{UF}_6$.2 This is slow, tedious work. In fact, the Oak Ridge facility built to do this during WWII was by some measures the largest building in the world at the time. It extracted U-235 one tiny bit at a time, and even after months of operation, there was just enough material for a single bomb, with nothing left to spare to even test the device.
For plutonium weapons, there is a similar complication. Bombs are made from Pu-239. Pu-240 is too unstable, and if there is too much of it in your bomb, it leads to a premature reaction (by a small fraction of a second), scattering most of the fuel rather than detonating it. The problem is even more difficult given that plutonium isn't found naturally in large quantities - it is produced entirely as a byproduct in uranium-based reactors.
In the end, it takes a large industrial infrastructure to manufacture either type of fission bomb since you need large amounts of an isotopically pure substance.
1 Even the chemical difference between normal hydrogen, consisting of 1 proton, and deuterium, which is a proton and a neutron, is small. Changing the mass of the nucleus, even by a factor of 2, does very little. Now imagine changing the mass by a factor of only about 1%.
2 Note that there is only one natural isotope of fluorine, F-19, so the only difference in masses for the molecules comes from the uranium.
Best Answer
This radiation, representing most of the initial energy output by a nuclear weapon, is swiftly absorbed by the surrounding matter. The latter in turn heats almost instantly to extremely high temperature, so you have the almost instantaneous creation of a ball of extremely high kinetic energy plasma. This in turn means a prodigious rise in pressure, and it is this pressure that gives rise the blast wave.
The same argument applies to the neutrons and other fission fragments / fusion products immediately produced by the reaction. But it is the initial burst of radiation that overwhelmingly creates the fireball in an atmospheric detonation, and the fireball that expands to produce most of the blast wave.