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
The mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.
The percentage mass is the ratio of the mass of the element or molecule in the given compound.
The percentage can be given as:
[tex]\text{Percent Mass} = \frac{\text{Mass of molecule}}{\text{total mass of compound}} \times 100 \%[/tex]
The mass of the water is 18.02 g/mol and the molar mass of hydrated magnesium sulfate (MnSO4 . H2O) is 169.03 g/mol.
Thus,
[tex]\text{Percent Mass} = \frac{\text{18.02}}{\text{169.03 }} \times 100 \%\\\\\text{Percent Mass} = 10.6 \%}[/tex]
Therefore, the mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.
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