Tim Forbes answered question 2/18/2011 How many molecules are in 43.14 kg NaHCO3? M = 23 + 1.008 + 12.01 + 3*16 = approx 84. N = m/M = 43.14*1000/84 (gmols) no of molecules = Avogadro's constant x number of moles = 6.02*10^23*n