How many oxygen atoms are present in 8.4×10^{- 3} grams of CuSO₄ · 5 H₂O?
Molar Mass(CuSO₄ · 5 H₂O) = 250.61 g/mol

Question

How many oxygen atoms are present in 8.4×10^{- 3} grams of CuSO₄ · 5 H₂O? Molar Mass(CuSO₄ · 5 H₂O) = 250.61 g/mol

Quizplus · Accepted Answer

First, we need to calculate the number of moles of the given compound:
moles = mass/molar mass
moles = (8.4×10^-3 g)/(250.61 g/mol) = 3.35×10^-5 mol

Next, we need to use the mole ratio between CuSOSB14S0 and oxygen atoms. From the chemical formula of the compound, we can see that there are 5 CuSOSB14S0 units for every 1 CuSOSB14S0 · 5 HS0O unit, and each CuSOSB14S0 unit contains 4 oxygen atoms. Therefore, we can calculate the total number of oxygen atoms as follows:
(total number of oxygen atoms) = (moles of CuSOSB14S0 · 5 HS0O) × (5 CuSOSB14S0 units/1 CuSOSB14S0 · 5 HS0O unit) × (4 oxygen atoms/1 CuSOSB14S0 unit)

Substituting the values, we get:
(total number of oxygen atoms) = (3.35×10^-5 mol) × (5×4) = 6.7×10^-4
(total number of oxygen atoms) = 1.8×10^20 (using scientific notation)

Therefore, the answer is C, 1.8×10^20 oxygen atoms.

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