Q: If the mass of water lost is 1.89 g and the mass of hydrate is 10.21 g, what is the percent by mass of water in the hydrate?

$$\frac{1.89}{10.21} \times 100 = 18.5\%$$

Question 1

A student heated a hydrate sample in a crucible. The mass of the empty crucible was 17.95 g, the mass of crucible plus hydrate was 27.16 g, and the mass of crucible plus anhydrous salt after heating was 26.07 g. What is the mass of water lost during heating?

Accepted Answer

$$10.21 - 8.32 = 1.89\ \mathrm{g}$$

Question 2

Given the mass of crucible plus hydrate is 27.16 g and the mass of the empty crucible is 17.95 g, calculate the mass of the hydrate sample.

Accepted Answer

$$27.16 - 17.95 = 10.21\ \mathrm{g}$$

Question 3

After heating, the mass of crucible plus anhydrous salt is 26.07 g. If the mass of the empty crucible is 17.95 g, what is the mass of anhydrous salt?

Accepted Answer

$$26.07 - 17.95 = 8.32\ \mathrm{g}$$

Question 4

If the mass of water lost is 1.89 g and the mass of hydrate is 10.21 g, what is the percent by mass of water in the hydrate?

Accepted Answer

$$\frac{1.89}{10.21} 	imes 100 = 18.5\%$$

Question 5

A hydrate sample loses water upon heating. Which of the following best describes the chemical process occurring?

Accepted Answer

Decomposition of the hydrate into anhydrous salt and water vapor

Question 6

If the molar mass of water is $$18.02\ \mathrm{g/mol}$$, how many moles of water were lost if $$1.89\ \mathrm{g} of $$water was driven off?

Accepted Answer

$$\frac{1.89}{18.02} = 0.105\ \mathrm{mol}$$

Question 7

A student determines the mass of anhydrous salt to be $$8.32\ \mathrm{g}$$ after heating. If the molar mass of the anhydrous salt is $$159.61\ \mathrm{g/mol}$$, how many moles of anhydrous salt are present?

Accepted Answer

$$\frac{8.32}{159.61} = 0.0521\ \mathrm{mol}$$

Question 8

If the hydrate is $$\mathrm{$$CuSO_4$$ \cdot xH_2O}$$ and you found $$0.0521\ \mathrm{mol} of $$anhydrous $$\mathrm{$$CuSO_4$$}$$ and $$0.105\ \mathrm{mol} of $$water lost, what is the value of $$x in $$the hydrate?

Accepted Answer

$$x = \frac{0.105}{0.0521} = 2.02$$

Question 9

Which of the following best explains why repeated heating is performed in the determination of water content in a hydrate?

Accepted Answer

To ensure all water is removed and a constant mass is achieved

Question 10

A student accidentally leaves the crucible open to the air after heating. What error might this introduce in the calculation of water content?

Accepted Answer

The anhydrous salt may absorb moisture, causing the calculated water content to be too low

Question 11

If a hydrate sample is overheated and decomposes, which of the following is a likely consequence for the calculated percent water?

Accepted Answer

The percent water will be artificially high due to loss of other volatile components

Question 12

Which equation correctly represents the dehydration of copper(II) sulfate pentahydrate?

Accepted Answer

$$\mathrm{$$CuSO_4$$ \cdot 5H_2O \rightarrow $$CuSO_4$$ + 5H_2O}$$

Question 13

A hydrate sample is found to contain $$18.5\%$$ water by mass. If the sample mass is $$10.21\ \mathrm{g}$$, how much water is present?

Accepted Answer

$$10.21 	imes 0.185 = 1.89\ \mathrm{g}$$

Question 14

Which of the following is the best reason for using a crucible in the dehydration of a hydrate?

Accepted Answer

A crucible can withstand high temperatures and does not react with the sample

Question 15

If the hydrate formula is $$\mathrm{$$CuSO_4$$ \cdot xH_2O}$$ and you experimentally determine $$x \approx 2$$, what is a likely source of error?

Accepted Answer

Incomplete dehydration or absorption of moisture after heating

Question 16

Which calculation would you use to determine the percent water in a hydrate sample?

Accepted Answer

$$\frac{	ext{mass of water lost}}{	ext{mass of hydrate}} 	imes 100$$

Question 17

A student finds the mass of water lost to be $$1.89\ \mathrm{g}$$ and the mass of anhydrous salt to be $$8.32\ \mathrm{g}. $$What is the ratio of moles of water to moles of anhydrous salt?

Accepted Answer

$$\frac{0.105}{0.0521} = 2.02$$

Question 18

Why is it important to cool the crucible in a desiccator before weighing after heating?

Accepted Answer

To prevent absorption of moisture from the air and ensure accurate mass measurement

Hydrate Analysis: Determining Water Content and Formula of a Hydrate

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