How many atoms are in 931.6 g $KMnO_4$ ?

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Answer 1 · 2016-03-03T21:34:50

Quite a few: each formula unit of potassium permanganate contributes 6 atoms.

Molar quantity of $KMnO_4$ $=$ $(931.6*g)/(158.034*g*mol^-1)$ $=$ $5.89*mol$ .

There are thus $5.89$ $mol$ $xx$ $N_A$ $xx$ $"6 atoms"$ $=$ $35.37*mol$ $xx$ $N_A" atoms"$ $=$ $??$

$N_A = "Avogadro's number", 6.022xx10^23*mol^-1$

How many atoms are in 931.6 g KMnO_4KMnO_4?