QuestionA 7.021 g sample of manganese reacts with excess chlorine to form a 16.08 g metal chloride. Find the empirical formula: $\mathrm{Mn}, \mathrm{Cl}$ .

Studdy Solution

STEP 1

Assumptions1. The mass of manganese used is7.021 grams. The mass of the metal chloride formed is16.08 grams3. The reaction is complete and all the manganese reacts with chlorine to form the metal chloride4. The excess chlorine does not contribute to the mass of the metal chloride

STEP 2

First, we need to find the mass of chlorine in the metal chloride. We can do this by subtracting the mass of manganese from the mass of the metal chloride.
$Mass\, of\, Chlorine = Mass\, of\, metal\, chloride - Mass\, of\, Manganese$

STEP 3

Now, plug in the given values for the mass of the metal chloride and the mass of manganese to calculate the mass of chlorine.
$Mass\, of\, Chlorine =16.08\, grams -7.021\, grams$

STEP 4

Calculate the mass of chlorine.
$Mass\, of\, Chlorine =16.08\, grams -7.021\, grams =9.059\, grams$

STEP 5

Next, we need to convert the masses of manganese and chlorine to moles. We can do this by dividing the mass by the atomic mass of each element. The atomic mass of manganese (Mn) is approximately54.94 grams per mole and the atomic mass of chlorine (Cl) is approximately35.45 grams per mole.
$Moles\, of\, Mn = Mass\, of\, Mn / Atomic\, mass\, of\, Mn$ $Moles\, of\, Cl = Mass\, of\, Cl / Atomic\, mass\, of\, Cl$

STEP 6

Now, plug in the given values for the mass of manganese and chlorine and their atomic masses to calculate the number of moles.
$Moles\, of\, Mn =.021\, grams /54.94\, g/mol$ $Moles\, of\, Cl =9.059\, grams /35.45\, g/mol$

STEP 7

Calculate the number of moles of manganese and chlorine.
$Moles\, of\, Mn =7.021\, grams /54.94\, g/mol =0.128\, mol$ $Moles\, of\, Cl =9.059\, grams /35.45\, g/mol =0.256\, mol$

STEP 8

The empirical formula is the simplest whole number ratio of the atoms in a compound. To find this, we divide the number of moles of each element by the smallest number of moles calculated in the previous step.
$Ratio\, of\, Mn = Moles\, of\, Mn / Smallest\, number\, of\, moles$ $Ratio\, of\, Cl = Moles\, of\, Cl / Smallest\, number\, of\, moles$

STEP 9

Now, plug in the values for the number of moles of manganese and chlorine and the smallest number of moles to calculate the ratio.
$Ratio\, of\, Mn =.128\, mol /.128\, mol$ $Ratio\, of\, Cl =.256\, mol /.128\, mol$

STEP 10

Calculate the ratio of manganese and chlorine.
$Ratio\, of\, Mn =0.128\, mol /0.128\, mol =$ $Ratio\, of\, Cl =0.256\, mol /0.128\, mol =2$ The empirical formula of the metal chloride is MnCl2.

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