A 15.67 g sample of a hydrate of magnesium carbonate was heated, without decomposing the carbonate, to drive off the water. The mass was reduced to 7.58 g. What is the name of the hydrate

The correct answer is MgCO₃.5H₂O.

Explanation:

Based on the given information,

The mass of hydrate of magnesium carbonate given is 15.67 grams, and the mass of anhydrous magnesium carbonate given is 7.58 grams.

Now the mass of water will be,

= 15.67 grams – 7.58 grams

= 8.09 grams

The molecular mass of magnesium carbonate is 84.32 grams, the moles of MgCO₃ can be determined by using the formula,

Number of moles = Mass/Molecular mass

Moles of MgCO₃ = 7.58/84.32

= 0.0898 moles

Similarly, the number of moles of water will be,

Moles of H₂O = 8.099/18.02

= 0.4489 moles

After getting the values, there is a need to choose the smaller value and use it to divide both the values to get the ratio,

MgCO₃ = 0.0898/0.0898 = 1

H₂O = 0.4489/0.0898 = 4.9 or 5

Thus, the molecular formula for the hydrate will be MgCO₃.5H₂O.

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.