Question

step #1: find the mass of water.
notes: the water can be driven off by heat to form the anhydrous (dehydrated) ionic compound. the mass of water evaporated is obtained by subtracting the mass of the anhydrous solid from the mass of the original hydrate.
mass of h₂o = initial mass of mgco₃ x h₂o hydrate - mass of anhydrous mgco₃
mass of h₂o = 15.673 - 7.589 = 8.09 h₂o

step #2: convert mass of water to moles.
? moles h₂o = mass in grams of h₂o / molar mass of h₂o = 8.093 h₂o x 1 mol h₂o / 15.673 h₂o = 0.516 mol h₂o

step #3: convert mass of anhydrous mgco₃ to moles.
? moles of anhydrous mgco₃ = mass in grams of anhydrous mgco₃ / molar mass of anhydrous mgco₃
moles of ads mgco₃ = 15.673 mgco₃ / (25.45+)

step #4: calculate for x = moles h₂o / moles anhydrous mgco₃
mg(co₃·5h₂o)=magnesium carbonate pentahydrate
x = 0.516 mol h₂o / 0.

formula of the hydrate: 5
name of the hydrate: magnesium carbonate pentahydrate

Explanation:

Step1: Calculate water mass

Subtract anhydrous mass from hydrate mass.
$mass\ of\ H_2O=15.673 - 7.58=8.093\ g$

Step2: Convert water mass to moles

Use molar - mass formula. Molar mass of $H_2O$ is approximately $18.015\ g/mol$.
$n_{H_2O}=\frac{8.093\ g}{18.015\ g/mol}\approx0.449\ mol$

Step3: Convert anhydrous $MgCO_3$ mass to moles

Molar mass of $MgCO_3$: $M = 24.31+12.01 + 3\times16.00=84.32\ g/mol$. Assume mass of anhydrous $MgCO_3$ is $7.58\ g$.
$n_{MgCO_3}=\frac{7.58\ g}{84.32\ g/mol}\approx0.09\ mol$

Step4: Find the ratio $X$

$X=\frac{n_{H_2O}}{n_{MgCO_3}}=\frac{0.449\ mol}{0.09\ mol}\approx5$

Answer:

Formula of the hydrate: $MgCO_3\cdot5H_2O$
Name of the hydrate: magnesium carbonate pentahydrate