What weight of manganese is present in 2.58 g of Mn3O4?
Calculate the gravimetric factor for: Substance sought (manganese) from Substance Weighed (Mn3O4).
The Correct Answer and Explanation is :
Step 1: Understanding Mn₃O₄ Composition
Manganese(II,III) oxide ((Mn_3O_4)) consists of both (Mn^{2+}) and (Mn^{3+}) ions. The molecular weight of (Mn_3O_4) is calculated as:
- Atomic mass of Mn = 54.938 g/mol
- Atomic mass of O = 16.00 g/mol
[
\text{Molar mass of } Mn_3O_4 = (3 \times 54.938) + (4 \times 16.00) = 164.81 \text{ g/mol}
]
Step 2: Gravimetric Factor Calculation
The gravimetric factor is the ratio of the molar mass of the substance sought (manganese) to the substance weighed ((Mn_3O_4)).
[
\text{Gravimetric Factor} = \frac{\text{Mass of Mn in Mn}_3O_4}{\text{Molar Mass of Mn}_3O_4}
]
The total mass of manganese in (Mn_3O_4):
[
3 \times 54.938 = 164.814 \text{ g/mol}
]
Thus, the gravimetric factor is:
[
\frac{164.814}{164.81} = 0.9999 \approx 1.00
]
Step 3: Determining the Mass of Manganese in 2.58 g of Mn₃O₄
Now, using the gravimetric factor:
[
\text{Mass of Mn} = 2.58 \times 0.9999 = 2.58 \text{ g}
]
Conclusion
- Weight of manganese in 2.58 g of Mn₃O₄ is 2.58 g
- Gravimetric factor for Mn from Mn₃O₄ is approximately 1.00
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