Mr(NH₄Cl) = 14 + 4×1 + 35 = 14+4+35= 53 а.е.м.
ω(N) = 14/53×100% = 26,41%
ω(H) = 4/53×100% = 7,54%
ω(Cl) = 100% - 26,41% - 7,54% = 66,05%
