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masa cząsteczki:
1υ × 3 + 31υ + 16υ × 4 = 3υ + 31υ + 64υ = 98υ
zawartość procentowa fosforu:
31υ : 98υ × 100% ≈ 31,63%
masa H3PO4 = 3 at H * 1 [u] + 1 at P * 31 [u] + 4 at O * 16 [u] = 3 + 31 + 64 = 98 [u]
obliczanie zawartości procentowej fosforu
% P
98 [u} - 100 %
31 [u] - x
x = 31 [ u ] * 100 : 98 {u}
x = 31,63 %