Answer to Question #240300 in General Chemistry for dan

Question

The following experimental data were recorded for the dehydration of an unknown hydrate, BaCl2 • n H2O.

The cation is barium and the anion is chloride.

Mass of test tube (dry) 8.8875 g

Mass of test tube + sample 9.1399 g

Mass of test tube + sample after heating 9.1011 g

What is the percentage of water, by mass, in the hydrate?

Calculate and report your answer as a percentage, but do not input the percent sign. Report your answer to three significant figures.

Percentage of water by mass = ? %

Answer 1

$m_{hydrate}=9.1399-8.8875 = 0.2524 \;g \\ m_{salt}=9.1011-8.8875 = 0.2136 \;g \\ m_{water loss}= 0.2524 -0.2136 = 0.0388 \;g$

Percentage of water $= \frac{0.0388}{0.2524} \times 100 = 15.4$

Answer: 15.4