Respuesta :
Answer:
The empirical formua is KMnO4
Explanation:
Step 1: Data given
Mass of the compound = 100 grams
24.74% = potassium
Molar mass of K = 39.10 g/mol)
34.76% = manganese
Molar mass of Mn= 54.94 g/mol
40.50% = oxygen
Molar mass = 16.00 g/mol
Step 2: Calculate moles of potassium
Moles K = mass K / molar mass K
Moles K = 24.74 grams / 39.10 g/mol
Moles K = 0.6327 moles
Step 3: Calculate moles of Mn
Moles Mn = 34.76 grams / 54.94 g/mol
Moles Mn = 0.6327 moles
Step 4: Calculate moles of O
Moles O = 40.50 grams / 16.00 g/mol
Moles O = 2.53
Step 5: Calculate the mol ratio
We divide by the smallest amount of mol
K: 0.6327 / 0.6327 = 1
Mn: 0.6327/0.6327 = 1
O: 2.53 / 0.6327 = 4
The empirical formua is KMnO4
The empirical formula of the given compound is [tex]\bold {KMnO_4}[/tex]. The empirical formula is the simplest ratios of the elements in the compound.
What is empirical formula?
The formula that is written as the simplest ratio of the elements in the compounds.
First calculate the moles of elements,
[tex]K =\dfrac { 24.74 }{ 39.10}\\\\K = 0.6327 \rm \ moles[/tex]
Moles of Mn,
[tex]Mn = \dfrac {34.76} {54.94}\\\\Mn = 0.6327 \rm \ moles[/tex]
Moles of Oxygen:
[tex]O = \dfrac {40.50}{ 16.00} \\\\O = 2.53 \rm \ moles[/tex]
Now calculate the molar ratios by dividing from smallest number.
[tex]\bold {Potassium = \dfrac {0.6327}{ 0.6327} = 1}[/tex]
[tex]\bold {Manganese = \dfrac {0.6327}{ 0.6327} = 1}[/tex]
[tex]\bold {Oxygen = \dfrac {2.43}{ 0.6327} = 4}[/tex]
Therefore, the empirical formula of the given compound is [tex]\bold {KMnO_4}[/tex].
Learn more about the empirical formula:
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