Problems #1 - 10

Go to the ten hydrate examples

Go to hydrate problems #11 - 25

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Calculate empirical formula when given mass data

Calculate empirical formula when given percent composition data

Determine identity of an element from a binary formula and a percent composition

Determine identity of an element from a binary formula and mass data

Look at a list of only the questions.

**Problem #1:** Anhydrous lithium perchlorate (4.78 g) was dissolved in water and re-crystalized. Care was taken to isolate all the lithium perchlorate as its hydrate. The mass of the hydrated salt obtained was 7.21 g. What hydrate is it?

**Solution:**

1) The amount of water in the hydrate is:

7.21 g − 4.78 g = 2.43 g

2) The moles of anhydrous LiClO_{4} and water are:

LiClO_{4}⇒ 4.78 g / 106.39 g/mol = 0.044929 mol

H_{2}O ⇒ 2.43 g / 18.015 g/mol = 0.13489 mol

3) Determine whole number ratio:

LiClO_{4}⇒ 0.044929 mol / 0.044929 mol = 1

H_{2}O ⇒ 0.13489 mol / 0.044929 mol = 3LiClO

_{4}·3H_{2}O

**Problem #2:** A substance was found to have the following percentages by mass: 23% zinc; 11% sulfur; 22% oxygen; 44% water. What is the empirical formula?

**Solution:**

1) Assume 100 g of the compound is present, then find the moles of each:

Zn ⇒ 23/65.4 = 0.3517

S ⇒ 11/32 = 0.34375

O ⇒ 22/16 = 1.375

H_{2}O ⇒ 44/ 18 = 2.444

2) Divide the smallest number into the others. The answers will not be exact but enough to tell the formula:

Zn ⇒ 0.3517 / 0.34375 = 1

S ⇒ 0.344 / 0.34375 = 1

O ⇒ 1.375 / 0.34375 = 4

H_{2}O ⇒ 2.44 / 0.34375 = 7The formula is ZnSO

_{4}·7H_{2}O

**Problem #3:** A 5.00 g sample of hydrated barium chloride, BaCl_{2} **·** nH_{2}O, is heated to drive off the water. After heating, 4.26 g of anhydrous barium chloride, BaCl_{2}, remains. What is the value of n in the hydrate's formula?

**Solution:**

1) Calculate moles of anhydrous barium chloride:

4.26 g / 208.236 g/mol = 0.020458 mol

2) Calculate moles of water:

5.00 − 4.26 = 0.74 g0.74 g / 18.015 g/mol = 0.041077 mol

3) Determine whole number ratio:

0.041077 / 0.020458 = 2

4) Formula is:

BaCl_{2}·2H_{2}O

**Problem #4:** A 1.98 g sample of a cobalt(II) chloride hydrate is heated over a burner. When cooled, the mass of the remaining dehydrated compound is found to be 1.55 g. What is the formula for the original hydrate? How can you make sure that all of the water of hydration has been removed?

**Solution:**

1) Determine mass of water driven off:

1.98 g − 1.55 g = 0.43 g

2) Determine moles of anhydrous CoCl_{2} and H_{2}O:

CoCl_{2}⇒ 1.55 g / 129.839 g/mol = 0.01194 mol

H_{2}O ⇒ 0.43 g / 18.015 g/mol = 0.0239 mol

3) Look for lowest whole-number ratio:

CoCl_{2}⇒ 0.01194 mol / 0.01194 mol = 1

H_{2}O ⇒ 0.0239 mol / 0.01194 mol = 2

4) Formula is:

CoCl_{2}·2H_{2}O

5) How can you make sure that all of the water of hydration has been removed?

After weighing the anhydrous CoCl_{2}, you would continue to heat it. Then, you would weigh it again. If the two weights are in agreement, then you are done heating. If the two weights disagree, you continue heating and weighing until you gets weights that agree. In some cases, where an extra amount of care must be taken, you would want three straight weighings that were in agreement.Also, weighs being in agreement does not mean that they are exactly the same. The standards for being in agreement might vary from one instructor to the next, so make sure to consult with your lab teacher on this point.

**Problem #5a:** A solution was made by dissolving 52.0 g of hydrated sodium carbonate in water and making it up to 5.00 dm^{3} of solution. The concentration of the solution was determined to be 0.0366 M. Determine the formula of hydrated sodium carbonate.

**Solution:**

1) Moles of hydrated sodium carbonate in 5.00 liters:

(0.0366 mol/L) (5.00 L) = 0.183 mol

2) Molecular weight of hydrated sodium carbonate:

52.0 g / 0.183 mol = 284.153 g/mol

3) Mass of water in one mole of hydrate:

284.153 − 105.988 = 178.165 g(105.988 is molar mass of anhydrous sodium carbonate)

4) Moles of water in one mole of hydrate:

178.165 g / 18.018 g/mol = 9.9 mol

5) Formula of hydrated sodium carbonate:

Na_{2}CO_{3}·10H_{2}O

**Problem #5b:** A solution was made by dissolving 71.50 g of hydrated sodium carbonate in water and making it up to 5.00 L of solution. The concentration of the solution was found to be 0.04805 M. What is the water of hydration for this hydrate of sodium carbonate?

**Solution:**

1) Mass of sodium carbonate in 5.00 L:

(0.04805 mol/L) (5.00 L) = x / 105.988 g/molx = 25.4636 g

2) Mass of water in 71.50 g of hydrated Na_{2}CO_{3}:

71.50 g − 25.4636 g = 46.0364 g

3) Moles of each:

moles of water ---> 46.0364 g / 18.015 g/mol = 2.55545 mol

moles of Na_{2}CO_{3}---> 25.4636 g / 105.988 g/mol = 0.24025 mol

4) Smallest whole-number ratio:

The ratio we want in smallest whole-numbers is this ---> 0.24025 mol to 2.55545 molThat is a 1 to 10 ratio.

Na

_{2}CO_{3}·10H_{2}O

**Problem #6:** Determine the formula and name for the hydrate: 73.42% ammonium phosphate and 26.58% water.

**Solution:**

1) Assume 100 grams of the compound is present. Therefore:

73.42 g of (NH_{4})_{3}PO_{4}

26.58 g of H_{2}O

2) Determine the moles of each compound:

73.42 g / 149.0858 g/mol = 0.492468 mol

26.58 g / 18.015 g/mol = 1.475437 mol

3) We want to know how many moles of H_{2}O are present for every one mole of (NH_{4})_{3}PO_{4}:

1.475437 mol / 0.492468 mol = 2.996 = 3

4) Formula and name:

(NH_{4})_{3}PO_{4}·3H_{2}Oammonium phosphate trihydrate.

**Problem #7:** 5.00 g of borax (Na_{2}B_{4}O_{7} **·** 10H_{2}O) was heated to remove the water. What is the mass of anhydrous sodium tetraborate that remains?

**Solution:**

The hydrate's molecular weight is 381.365 g/molThe total amount of water in the molecular weight is 180.148 g.

180.148 / 381.365 = 0.47238 (decimal amount of the hydrate that is water)

(5.00 g) (0.47238) = 2.3619 g (this is the water lost)

5.00 g − 2.3619 g = 2.6381 g of the anhydrous Na

_{2}B_{4}O_{7}remainingRound off to three sig figs: 2.64 g

**Problem #8:** A sample of hydrate lost 14.75% of its original weight during heating. Determine the number of moles of hydration per mole of anhydrous substance if the molecular weight of the anhydrate is 208 grams/mole.

**Solution:**

Let's assume we had one mole of the hydrate present. We know that the one mole of anhyrate weighs 208 grams and represents 85.25% of the weight.208 is to 85.25 as x is to 100

x = 244 (the molar mass of the hydrate)

244 − 208 = 36 (the mass of water in one mole of hydrate)

36/18 = 2 (two moles of hydration per mole of anhydrous substance)

**Problem #9:** 1.33 g of hydrated ethanedioic acid (H_{2}C_{2}O_{4} **⋅** nH_{2}O) were dissolved in distilled water and the solution made up to 250.0 mL in a graduated flask. 25.0 mL of this solution were titrated by 21.1 mL of 0.100M NaOH. Calculate the number of moles of water of crystallization in the hydrated ethanedioic acid.

Equation: H_{2}C_{2}O_{4} + 2NaOH ---> Na_{2}C_{2}O_{4} + 2H_{2}O

**Solution:**

(0.100 mol/L) (0.0211 L) = 0.00211 mol of NaOH used in titrationThere is a 1 to 2 molar ratio between H

_{2}C_{2}O_{4}and NaOH0.00211 mol / 2 = 0.001055 mol of H

_{2}C_{2}O_{4}in the 25.0 mL that was titrated.25.0 mL is to 0.001055 mol as 250.0 mL is to x

x = 0.01055 mol <--- this is how many moles of H

_{2}C_{2}O_{4}⋅nH_{2}O dissolved in the 250.0 mL1.33 g / 0.01055 mol = 126.066 g/mol <--- the molar mass of H

_{2}C_{2}O_{4}⋅nH_{2}O126.066 − 90.0338 = 36.0322 g <--- the mass of water in one mole of H

_{2}C_{2}O_{4}⋅nH_{2}O36.0 / 18.0 = 2 <--- moles of water in one mole of H

_{2}C_{2}O_{4}⋅nH_{2}OH

_{2}C_{2}O_{4}⋅2H_{2}O

**Problem #10:** A solution was made by dissolving 71.5 g of hydrated sodium carbonate in water and making up to 5.00 dm^{3} of solution. The concentration of a 25.0 cm^{3} portion was determined to be 0.04805 M. Use the information to find the waters of hydration of the hydrated sodium carbonate.

**Solution:**

1) Mass of dissolved Na_{2}CO_{3} in 25.0 cm^{3}:

(0.04805 mol/L) (0.0250 L) = x / 105.988 g/molx = 0.12732 g

2) Mass of dissolved Na_{2}CO_{3} in 5.00 dm^{3}

0.12732 g is to 0.0250 L as x is to 5.00 Lx = 25.464 g

3) Water in 71.5 g of hydrated Na_{2}CO_{3}:

71.5 g − 25.464 g = 46.036 g

4) Determine molar ratio:

moles of water ---> 46.036 g / 18.015 g/mol = 2.555426 mol

moles of Na2CO3 ---> 25.464 g / 105.988 g/mol = 0.2402536 molThe ratio we want in smallest whole-numbers is this ---> 0.2402536 mol to 2.555426 mol

That is a 1 to 10 ratio

5) The formula of the hydrate is:

Na_{2}CO_{3}⋅10H_{2}O

**Bonus Problem:** A certain quantity of sodium carbonate decahydrate was heated to remove the water. The mass of the anhydrous compound that remained was 2.764 g. What was the original mass of the hydrated sodium carbonate?

**Solution:**

2.764 g / 105.988 g/mol = 0.0260784 molThe Na

_{2}CO_{3}and H_{2}O are in a 1:10 molar ratio in the decahydrate. Therefore:(0.0260784 mol) (10) = 0.260784 mol of H

_{2}O(0.260784 mol) (18.015 g/mol) = 4.698 g <--- mass of the water

Total mass of the hydrated compound:

2.764 + 4.698 = 7.462 g

Look at a list of only the questions.

Go to the ten hydrate examples

Go to hydrate problems #11 - 25

Return to Mole Table of Contents

Calculate empirical formula when given mass data

Calculate empirical formula when given percent composition data

Determine identity of an element from a binary formula and a percent composition

Determine identity of an element from a binary formula and mass data