Here is one way to do this calculation. For a Hubble Constant of 65 kilometers/sec/mpc, the age of the universe is 2/3 x 1/H = 9.9 billion years. The horizon size of our universe is then 9.9 billion light years. If we neglect curvature effects, the volume of space this represents is 4/3 pi R^3 = 3.2 x 10^84 cubic centimeters.
The critical density of the universe for this value of the Hubble constant is 3 H^2/8 pi G, which works out to be 8.4 x 10^-30 grams/cubic centimeter or about 0.0000053 atoms of hydrogen/cc. It is believed that only 15 percent of the critical density is in the form of normal atoms, so this leaves 0.0000053 x 0.15 = 0.0000008 hydrogen atoms/cc. Multiplying this by the volume of the visible universe, you get about 3 x 10^78 hydrogen atoms. Since four of every five of these are locked up inside helium nuclei, you get about 6 x 10^77 atoms of helium and hydrogen. In the TOTAL universe, including the part outside our horizon, the answer is either 'infinity' or a number for a closed universe that is about 5 times larger than the above number because the volume of our visible universe would be only about 1/5 the total spacial volume that emerged from the Big Bang...very approximately! could have