Molar mass of carbon is #(12.011 "g")/"mol"#.
Molar mass of oxygen gas is #(2 * 15.999 "g")/"mol" = (31.998 "g")/"mol"#.
So when you add them together to get the molar mass of #"CO"_2# you get #(44.009 "g")/"mol"#.
We also know that there are around #6.02 * 10^23# atoms in a mole (Avogadro's number).
From this information, we can use dimensional analysis to solve for the number of atoms present in #4.4# g #"CO"_2# gas:
#4.4 "g CO"_2 xx "1 mol CO"_2/(44.009 "g CO"_2) xx (6.02 * 10^23 " atoms CO"_2)/"1 mol CO"_2#
#4.4 cancel("g CO"_2) xx cancel("1 mol CO"_2)/(44.009 cancel("g CO"_2)) xx (6.02 * 10^23 " atoms CO"_2)/cancel("1 mol CO"_2)#
#~~ 6.0187689... * 10^22 " atoms CO"_2#.
However, we can only have two significant figures because of the #4.4 "g"# and also because it is multiplication, so the answer is actually:
#6.0 * 10^22 " atoms CO"_2#.
Hope this helps!
