1) Вычислить массовые доли элементов в (CH3COO) 2Ba.

2) Вычислить массовые доли элементов (NHh) 2 CO3.

1)

w (C) = (M (C) / M (CH3COO) 2Ba) * 100%=24*2 / ((24+3+32) * 2+137) = (48/255) * 100%=18.82%

w (H) = (M (H) / M (CH3COO) 2Ba) * 100%=3*2 / ((24+3+32) * 2+137) = (6/255) * 100%=2.35%

w (O) = ((M (O) / M (CH3COO) 2Ba) * 100%=32*2 / ((24+3+32) * 2+137) = (64/255) * 100%=25.1%

w (Ba) = (M (Ba) / M (CH3COO) 2Ba) * 100%=137 / ((24+3+32) * 2+137) = (137/255) * 100%=53.73%

2)

w (C) = (M (C) / M ((NH4) 2CO3) * 100%=12 / ((14+4) * 2+12+48) = (12/96) * 100%=12.5%

w (H) = (M (H) / M ((NH4) 2CO3) * 100%=4 / ((14+4) * 2+12+48) = (8/96) * 100%=8.33%

w (N) = (M (N) / M ((NH4) 2CO3) * 100%=28 / ((14+4) * 2+12+48) = (28/96) * 100%=29.17%

w (O) = (M (O) / M ((NH4) 2CO3) * 100%=48 / ((14+4) * 2+12+48) = (48/96) * 100%=50%