A dead alkaline battery is found to contain a compound of Mn and O. It's analysis gives 69.6% Mn and 30.4% O
To find the empirical formula of the compound found in the dead alkaline battery, we need to determine the ratio of Mn to O.
Here's how you can calculate the empirical formula step by step:
Step 1: Convert the percentages to grams.
Assume we have a 100g sample of the compound. This means we have 69.6g of Mn and 30.4g of O.
Step 2: Convert grams to moles.
To convert grams to moles, we need to divide the mass of each element by its molar mass.
The molar mass of Mn is approximately 54.938 g/mol, and the molar mass of O is approximately 15.999 g/mol.
Number of moles of Mn = 69.6 g / 54.938 g/mol ≈ 1.267 mol
Number of moles of O = 30.4 g / 15.999 g/mol ≈ 1.900 mol
Step 3: Find the simplest whole-number ratio.
Divide the moles of each element by the smaller number of moles to obtain the simplest ratio.
In this case, the number of moles of Mn is approximately 1.267, and the number of moles of O is approximately 1.900.
Dividing by 1.267 gives us 1 Mn and dividing by 1.267 gives us approximately 1.500 O.
Step 4: Adjust the ratio to whole numbers if necessary.
In this case, the ratio of Mn to O is approximately 1:1.5. To obtain whole numbers, we can multiply the ratio by 2 to get 2:3.
Therefore, the empirical formula of the compound is Mn2O3.