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The chemical formula for the initial sample is mnso4·h2o and the chemical formula for the final sample is mnso4. use their molecular masses to calculate the theoretical percent mass of water in the initial sample. report your result to 3 significant figures,
e.g. 11.3%

Respuesta :

tr847007
The initial sample has a molecular formula of MnSO₄·H₂O. This molecule is a hydrate as it has a unit of water within its structure for every molecule of MnSO₄. This sample is being dehydrated to remove the water to give.

MnSO₄·H₂O → MnSO₄ + H₂O

MnSO₄·H₂O has a molecular mass of 169.02 g/mol. While MnSO₄ has a molecular mass of 151 g/mol. Water has a molecular mass of 18.02 g/mol. We now can use the ratio of the mass of water to the mass of the initial sample to determine the percentage of each component by mass.

% water by mass:

18.02/169.02 x 100% = 10.7% Water by mass.

% MnO₄ by mass:

151/169.02 x 100% = 89.3% MnSO₄ by mass.

Water makes up 10.7% of the initial mass of MnSO₄·H₂O.

The percentage of water in in  the initial sample of [tex]\rm \bold {MnSO_4.H_2O}[/tex] is 10.7%.

Molar mass of [tex]\rm \bold {MnSO_4.H_2O}[/tex] is 169.02 g/mol.

Molar mass of [tex]\rm \bold {MnSO_4}[/tex] is 151 g/mol

Molar mass of [tex]\rm \bold {H_2O}[/tex] is 18.02 g/mol

The percentage of water is,

[tex]\rm \bold{\frac{18.02}{169.02} \times 100 =10.7}[/tex]

Hence we can say that the percentage of water in in  the initial sample of [tex]\rm \bold {MnSO_4.H_2O}[/tex] is 10.7%.

To know more about molar mass, refer to the link:

https://brainly.com/question/12127540?referrer=searchResults

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