Question

The general formula of barium chloride hydrate is BaCl₂ • n H₂O, where n is the number of water molecules. Calculate the theoretical mass percent water for each value of n – the mass of each water molecule is 18.02 amu.

Report 2 decimal places for all your answers. Do NOT include units in your answer.

	BaCl2	BaCl2 • H2O	BaCl2 • 2 H2O	BaCl2 • 3 H2O
Mass of anhydrous BaCl2 (amu)	208.23	208.23	208.23	208.23
Mass of H2O molecules in Hydrate (amu)	0.00	18.02	Answer	Answer
Mass of Hydrate (amu)	208.23	226.25	Answer	Answer
Percent H2O in Hydrate (%)	0.00 %	7.97 %	Answer %	Answer %

Compare the percent water you calculated for the unknown barium chloride hydrate from the experimental data in questions #3 to the theoretical percent water values you calculated from formulas in #4. Do you think that the formula for your hydrate is BaCl₂ • H₂O, BaCl₂ • 2 H₂O, or BaCl₂ • 3 H₂O? Explain your answer

This isn't an incomplete question. This is the entire question that I was told to answer and that's why I'm so confused

Answer 1

the first thing to do is to calculate the mass of each molecule of water, we know that each molecule of water has a mass of 18 amu so

for the molecule with 2 water molecules

mass of H2O = 2 * 18 = 36.04

the mass of hydrate is the mass of BaCl2 and the water so:

208.23 (BaCl2) + 36 = 244.27

percent of H2O in the hydrate is:

36.04 / 244.27 * 100 = 14.75 %

now calculate with 3 molecules

mass of h2o = 3 * 18.02 = 54.06 amu

mass of the compound = 208.23 + 54.06 = 262.29 amu

percent of water = 54.06 / 262.29 * 100 = 20.61 %

now calcualte with 1 molecule of water:

mass of h2o = 18 amu

mass of the compound = 208.23 + 18.02 = 226.25 amu

percent of water = 18.02 / 226.25 * 100 = 7.96 %

*hope it helps =)