A compund X contains 63.3% manganese and 36.7% oxygen by mass. When X is heated, oxygen gas is evolved...
...and a new compound Y containing 72% manganese and 26% oxygen is formed.
Determine the empirical formula of X and Y and write the balanced chemical equation for the conversion of X to Y.
First step to convert percentage into mass
Compound X contain
63.3% Mn = 63.3 g of Mn
36.7% O = 36.7 g of O
Compound Y contain
72% Mn = 72 g of Mn
28% O = 28 g of O
[Note here we have to consider 28% of O Since the sum of the percentage should be 100%]
Second step convert mass into moles.[moles = mass/molar mass]
For Compound X
63.3 g of Mn = 63.3/54.94 = 1.15 moles.
36.7 g of O = 36.7/16 = 2.293 moles.
For compound Y.
72 g of Mn = 72/54.94 = 1.31 moles.
28 g of O = 28/16 = 1.75 moles.
Third step is dividing the moles with the lowest moles.
1.15 moles of Mn = 1.15/1.15 = 1
2.293 moles of O = 2.293/1.15 = 1.9 = 2
So the emprical formula for compound X has 1 Mn and 2 O, Therefore the emprical formula is MnO2
1.31 moles of Mn = 1.31/1.31 = 1
1.75 moles of O = 1.75/1.31 = 1.3 = 1
So the emprical formula for compound Y has 1 Mn and 1 O, Therefore the emprical formula is MnO
Balanced chemical reaction
2MnO2 → 2 MnO + O2` `
1. Assume that there is a total of 100 grams of solution used. This will allow us to convert the percentages into grams.
63.3g of Mn
36.7g of O
72g of Mn
26g of O
2. Now find the amount of moles for each element using the molecular weight and the grams given for each element:
Mn's molecular weight: 54.9 g/mol
O's molecular weight: 16.0 g/mol
63.3g x 1mol/54.9g = 1.2 moles of Mn
36.7g x 1mol/16.0g = 2.3 moles of O
72g x 1mol/54.9g = 1.3 moles of Mn
26g x 1mol/16.0g = 1.6 moles of O
3. Divide the moles by the lowest number of moles for each element:
1.2/1.2 = 1 mol of Mn
2.3/1.2 = 2 moles of O
Empirical Formula is MnO2 for Compound X
1.3/1.3 = 1 mol of Mn
1.6/1.3 = 1 mol of O
Empirical Formula is MnO for Compound Y
Balanced equation = 2MnO2 ---> 2MnO + O2