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dmitriy555 [2]
1 year ago
14

Calculate your experimentally determined percent mass of water in Manganese(II) sulfate monohydrate. Report your result to 2 or

3 significant figures, e. g. 9.8% or 10.2%.
Chemistry
1 answer:
antoniya [11.8K]1 year ago
4 0

Answer:

10.6%

Explanation:

The determined percent mass of water can be calculated from the formula of the hydrate by  

dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and  

multiplying this fraction by 100.

 

Manganese(ii) sulphate monohydrate is MnSO4 . H2O

1. Calculate the formula mass. When determining the formula mass for a hydrate, the waters of  

hydration must be included.

1 Manganes  52.94 g = 63.55 g  

1 Sulphur  32.07 g =  

32.07 g 2 Hydrogen is  = 2.02 g

4 Oygen       =  

64.00 g 1 Oxygen 16.00 = 16.00 g

151.01 g/mol  18.02 g/mol

   

Formula Mass = 151.01 + (18.02) = 169.03 g/mol

2. Divide the mass of water in one mole of the hydrate by the molar mass of the hydrate and  

multiply this fraction by 100.

Percent hydration = (18.02 g /169.03 g) x (100) = 10.6%

The final result is 10.6% after the two steps calculations

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Since the Carbon C is 17.39% by mass hence the Fluorine F is 82.61% by mass. Divide each mass % by the respective molar masses, that is:

 

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Divide the two by the smaller number, so divide by 1.45

 

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Hope this answers the question. Have a nice day.
7 0
1 year ago
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