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...

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.

Asked on by spock16

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sanjeetmanna's profile pic

sanjeetmanna | College Teacher | (Level 3) Assistant Educator

Posted on

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.

Compound X

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

Compound Y

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` `

cgrant2's profile pic

cgrant2 | High School Teacher | (Level 1) Valedictorian

Posted on

1. Assume that there is a total of 100 grams of solution used. This will allow us to convert the percentages into grams. 

Compound X: 

63.3g of Mn 

36.7g of O 

Compound Y: 

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 

Compound X 

63.3g x 1mol/54.9g = 1.2 moles of Mn 

36.7g x 1mol/16.0g = 2.3 moles of O 

Compound Y 

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: 

Compound X 

1.2/1.2 = 1 mol of Mn 

2.3/1.2 = 2 moles of O 

Empirical Formula is MnO2 for Compound X 

Compound Y 

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


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