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A 1.36-g sample of magnesium nitrate, mg(no3)2, contains ________ mol of this compound. a 1.36-g sample of magnesium nitrate, mg(no3)2, contains ________ mol of this compound.

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Answer is: magnesium nitrate contains 0.0091 mol.

m(Mg(NO₃)₂) = 1.36 g; mass.
M(Mg(NO₃)₂) = Ar(Mg) + 2Ar(N) + 6Ar(O) · g/mol.
M(Mg(NO₃)₂) = 24.3 + 2·14 + 6·16 · g/mol.
M(Mg(NO₃)₂) = 148.3 g/mol; molar mass.
n(Mg(NO₃)₂) = m(Mg(NO₃)₂) ÷ M(Mg(NO₃)₂).
n(Mg(NO₃)₂) = 1.36 g ÷ 148.3 g/mol.
n(Mg(NO₃)₂) = 0.0091 mol; amount of substance.

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