Question

How many oxygen atoms are there in 6.00 g of sodium dichromate, na2cr2o7?

Answer 1

Answer is: there are 9,7·10²² atoms of oxygen.
m(Na₂Cr₂O₇) = 6,00 g.
n(Na₂Cr₂O₇) = m(Na₂Cr₂O₇) ÷ M(Na₂Cr₂O₇).
n(Na₂Cr₂O₇) = 6 g ÷ 262 g/mol.
n(Na₂Cr₂O₇) = 0,023 mol.
n(Na₂Cr₂O₇) : n(O) = 1 : 7.
n(O) = 7 · 0,023 mol.
n(O) = 0,161 mol.
N(O) = 0,161 mol · 6,022·10²³ 1/mol.
N(O) = 9,7·10²².