Answer to Question #143314 in Chemistry for noelia

Question

A 90.00 g sample of Na₂CO₃⋅xH2O is heated until all of the water is driven off, forming 0.314 mol of the anhydrous salt, Na₂CO₃. The molar mass of Na₂CO₃ is 105.99 g/mol. What is the percent by mass of H2O in the hydrated compound, Na₂CO₃⋅xH2O?

Answer 1

Solution:

The chemical reaction equation:

Na₂CO₃⋅xH₂O --> Na₂CO₃ + xH₂O

According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.

or

m(Na₂CO₃⋅xH₂O) = m(Na₂CO₃) + m(xH₂O)

m(Na₂CO₃) = Moles of Na₂CO₃ × Molar mass of Na₂CO₃

m(Na₂CO₃) = 0.314 mol × 105.99 g/mol = 33.28 g

m(xH₂O) = m(Na₂CO₃⋅xH₂O) - m(Na₂CO₃) = 90.00 g - 33.28 g = 56.72 g

Mass percent of a component = (mass of component / total mass of sample) × 100%

The percent by mass of H₂O in the hydrated compound (Na₂CO₃⋅xH₂O) is:

w(H₂O) = (m(xH₂O) / m(Na₂CO₃⋅xH₂O)) × 100%

w(H₂O) = (56.72 g / 90.00 g) × 100% = 63.02%

w(H₂O) = 63.02%

Answer: The percent by mass of H₂O in the hydrated compound is 63.02%.