QQuestionChemistry
QuestionChemistry
What is the empirical formula of an oxide of manganese that contains 69.6% of manganese?
(Mn = 55, O = 16)
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Answer
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Step 1:Let's solve this step by step:
Step 2:: Calculate the mass of oxygen in the compound
- If we assume 100 g of the compound, the mass of manganese would be 69.6 g - Mass of oxygen would be 100 g - 69.6 g = 30.4 g
Step 3:: Calculate the number of moles of manganese
n_{Mn} = \frac{69.6}{55} = 1.27 \text{ mol}
Step 4:: Calculate the number of moles of oxygen
n_{O} = \frac{30.4}{16} = 1.90 \text{ mol}
Step 5:: Find the mole ratio by dividing by the smallest number of moles
\frac{n_{O}}{1.27} = 1.50
- Divide both mole quantities by 1.27
Step 6:: Simplify the ratio to whole numbers
Mn : O = 2 : 3
- Multiply by 2 to get whole numbers
Final Answer
The empirical formula is Mn_{2}O_{3}.
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