Problems #21 - 35

Fifteen Examples | Problems #1 - 10 | Problems #11 - 20 | 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

The example and problem questions only, no solutions

**Problem #21:** A 0.256 g sample of CoCl_{2} **⋅** yH_{2}O was dissolved in water, and excess silver salt was added. The silver chloride was filtered, dried, and weighed, and it had a mass of 0.308 g. What is the value of y?

**Solution:**

1) The chemical reaction of interest:

CoCl_{2}+ 2Ag^{+}---> 2AgCl + Co^{2+}The CoCl

_{2}to AgCl molar ratio is 1:2

2) Determine moles of CoCl_{2} that were in solution:

(0.308 g AgCl) / (143.3212 g AgCl/mol) x (1 mol CoCl_{2}/ 2 mol AgCl) = 0.0010745 mol CoCl_{2}

3) Convert moles of CoCl_{2} to grams:

(0.0010745 mol CoCl_{2}) x (129.8392 g CoCl_{2}/mol) = 0.1395 g CoCl_{2}

4) Determine mass, then moles, of water in original CoCl_{2} sample:

(0.256 g total) − (0.1395 g CoCl_{2}) = 0.1165 g H_{2}O(0.1165 g H

_{2}O) / (18.01532 g H_{2}O/mol) = 0.0064667 mol H_{2}O

5) Determine value of y:

y = (0.0064667 mol H_{2}O) / (0.0010745 mol CoCl_{2}) = 6

**Problem #22:** A sample of 0.416 g of CoCl_{2} **⋅** yH_{2}O was dissolved in water, and an excess of sodium hydroxide (NaOH) was added. The cobalt hydroxide salt was filtered and heated in a flame, forming cobalt(III) oxide (Co_{2}O_{3}). The mass of cobalt(III) oxide formed was 0.145 g. What is the value of y?

**Solution:**

1) The following reactions take place:

CoCl_{2}+ 2NaOH ---> Co(OH)_{2}+ 2NaCl

4Co(OH)_{2}+ O_{2}---> 2Co_{2}O_{3}+ 4H_{2}O

2) The combined reaction is:

4CoCl_{2}+ 8NaOH + O_{2}---> 8NaCl + 2Co_{2}O_{3}+ 4H_{2}ONote that the first equation was multiplied through by 4 before adding. This was done so as to allow for the 4Co(OH)

_{2}to be cancelled.

3) Determine moles of CoCl_{2} that dissolved:

(0.145 g Co_{2}O_{3}) / (165.8646 g Co_{2}O_{3}/ mol) x (4 mol CoCl_{2}/ 2 mol Co_{2}O_{3}) = 0.0017484 mol CoCl_{2}

4) Determine mass of CoCl_{2}:

(0.0017484 mol CoCl_{2}) x (129.8392 g CoCl_{2}/ mol) = 0.22701 g CoCl_{2}

5) Determine mass, then moles, of water in the solid CoCl_{2} before dissolving:

(0.416 g total) − (0.22701 g CoCl_{2}) = 0.18899 g H_{2}O(0.18899 g H

_{2}O) / (18.01532 g H_{2}O/mol) = 0.0104905 mol H_{2}O

6) Determine value of y:

y = (0.0104905 mol H_{2}O) / (0.0017484 mol CoCl_{2}) = 6

**Problem #23:** When 5 g of iron(III) chloride hydrate are heated, 2 g of water are driven off . Find the chemical formula of the hydrate.

**Solution:**

1) 3 grams of FeCl_{3} are left after the 2 grams of H_{2}O are driven off.

2) Determine moles of each:

FeCl_{3}---> 3 g / 162.204 g/mol = 0.018495 mol

H_{2}O ---> 2 g / 18.015 g/mol = 0.11102 mol

3) Divide both values by the FeCl_{3} value.

FeCl_{3}---> 0.018495 mol / 0.018495 mol = 1

H_{2}O ---> 0.11102 mol / 0.018495 mol = 6This gives us a molar ratio between FeCl

_{3}and H_{2}O of 1 to 6

4) The formula of the hydrate is this:

FeCl_{3}⋅6H_{2}O

**Problem #24:** Cupric chloride, CuCl_{2}, dehydrates when heated. If 0.235 g of CuCl_{2} **⋅** xH_{2}O gives 0.185 g of CuCl_{2} on heating, what is the value of x?

**Solution #1:**

1) For the water:

0.235 minus 0.185 = 0.050 g of water driven off.0.050 g / 18.0 g/mol = 0.0028 mol of water

2) For the CuCl_{2}:

0.185 g / 134.452 g/mol = 0.001376 mol of anhydrous CuCl_{2}

3) We now want to see if there is a small whole number ratio between the two amounts of moles. Let's divide each number of the number of moles of CuCl_{2}

0.001376 mol / 0.001376 mol = 10.0028 mol / 0.001376 mol = 2.03

4) To a reasonable level of accuracy, this is a 2 to 1 ratio. The answer is this:

CuCl_{2}⋅2H_{2}O

**Solution #2:**

CuCl_{2}has a molecular weight of 134.3 and water weighs 18.0.The ratio 235/185 expresses the ratio of the weight of hydrate to anhydrate

Let X equal the number of water molecules. Therefore:

235/185 = [134.3 + X(18.0)] / 134.3

Solve for X to the nearest integer.

**Problem #25:** An hydrate of copper (II) chloride has the formula CuCl_{2} **⋅** xH_{2}O. The water in a 3.41 g sample of the hydrate was driven off by heating. The remaining sample had a mass of 2.69 g. Find the number of waters of hydration (x) in the hydrate.

**Solution #1:**

1) Mass:

CuCl_{2}---> 2.69 g

H_{2}O ---> 3.41 − 2.69 = 0.72 g

2) Moles:

CuCl_{2}---> 2.69 g / 134.452 g/mol = 0.02

H_{2}O ---> 0.72 g / 18.0 g/mol = 0.04

3) Get lowest whole-number mole ratio:

CuCl_{2}---> 0.02 / 0.02 = 1

H_{2}O ---> 0.04 / 0.02 = 2CuCl

_{2}⋅2H_{2}O

**Solution #2:**

1) Calculate percent composition of the hydrate

CuCl_{2}---> 2.69 g / 3.41 g = 78.9%

H_{2}O ---> 0.72 g / 3.41 g = 21.1 %

2) Assume 100 g of substance is present. Calculate mass of the anhydrate and water, then moles of each:

CuCl_{2}---> (100 g) (0.789) = 78.9 g

H_{2}O ---> (100 g) (0.211) = 21.1 gCuCl

_{2}---> 78.9 g / 134.5 g/mol = 0.5866 mol

H_{2}O ---> 21.1 g / 18.0 g/mol = 1.17 mol

3) Divide by the smaller to get whole number mole ratio:

CuCl_{2}---> 0.5866 / 0.5866 = 1

H_{2}O ---> 1.17 / 0.5866 = 2CuCl

_{2}⋅2H_{2}O

**Problem #26:** A sample of hydrated copper(II) sulfate weighs 124.8 g. The sample has been determined to contain 31.8 g of copper(Il) ions and 48.0 g of sulfate ions. Determine how many molecules of water of crystallization are present in the sample and thus deduce the chemical formula of hydrated copper(ll) sulfate.

**Solution:**

31.8 g + 48.0 g = 79.8 g (the mass of the anhydrous CuSO_{4})124.8 g − 79.8 g = 45.0 g (the mass of water in the hydrate)

45.0 g / 18.0 g/mol = 2.5 mol (of water)

79.8 g / 160 g/mol = 0.50 mol (of anhydrous CuSO

_{4})2.5 mol / 0.50 mol = 5 (molecules of water per one formula unit of CuSO

_{4})CuSO

_{4}⋅5H_{2}O

**Problem #27:** A hydrated salt is found to have the empirical formula CaN_{2}H_{8}O_{10}. What is its dot formula?

**Solution:**

This is the general formula for a substance that is hydrated:X⋅nH_{2}Owhere X is the salt and it has n molecules of water attached. The term "dot formula" comes from the use of the "

⋅" to show the water attachment.For this problem then, CaN

_{2}H_{8}O_{10}, all of the hydrogen will be found in the hydrating water. Since water is H_{2}O, there will be n = 8/2 = 4 molecules of water. This requires 4 oxygen atoms; the remaining 6 are part of the salt.Now we have CaN

_{2}O_{6}⋅4H_{2}O. If you look up salts containing calcium you'll find calcium nitrate, Ca(NO_{3})_{2}. You can also look at the salt ionically. Balancing the charges will put you on the path to Ca(NO_{3})_{2}.This is the dot formula for hydrated calcium nitrate:

Ca(NO_{3})_{2}⋅4H_{2}O

**Problem #28:** You have 1.023 g of hydrated CoSO_{4}. After heating and driving off all the water, the mass of the anhydride is 0.603 g. Determine the hydrate's chemical formula (number of waters attached).

**Solution:**

1) Mass of H_{2}O removed after heating = mass of hydrate − mass of anhydrate

1.023 g − 0.603 g = 0.42 g2) Find the moles of CoSO

moles of CoSO3) Divide moles of CoSO_{4}= 0.603 g / 154.9958 g/mol = 0.00389 mol

moles of H_{2}O = 0.42 g / 18 g/mol = 0.02333 mol

CoSO4) Empirical Formula:_{4}= 0.00389 mol / 0.00389 mol = 1

H_{2}O = 0.02333 mol / 0.00389 mol = 6

CoSO_{4}⋅6H_{2}O

**Problem #29:** A solution of FeSO_{4} **⋅** xH_{2}O has a concentration of 0.220 mol dm¯^{3} and a mass concentration of 61.16 g dm¯^{3}. Find the value of x.

**Solution:**

The 61.16 grams (mass conc.) represents 0.220 moles (molar conc.) of solute.61.16 is to 0.220 as x is to 1

x = 278 <--- this is the mass of one mole of FeSO

_{4}⋅xH_{2}O278 − 151.906 = 126.094 g <--- that's the mass of water in one mole of FeSO

_{4}⋅xH_{2}OThe 151.906 is the mass of one mole of anhydrous FeSO

_{4}126.904 g / 18.015 g/mol = 7 mol

FeSO

_{4}⋅7H_{2}O

**Problem #30:** A 1.281 g sample of anhydrous Na_{2}SO_{4} is exposed to the atmosphere and found to gain 0.391 g in mass. What is the percent, by mass, of Na_{2}SO_{4} **⋅** 10H_{2}O in the resulting mixture of anhydrous Na_{2}SO_{4} and the decahydrate?

**Solution:**

0.391 g / 18.0 g/mol = 0.02172 molThe moles of water is represented by 10. The anhydrous Na

_{2}SO_{4}would be in a 1:10 molar ratio with the water.0.02172 mol / 10 = 0.002172 mol of Na

_{2}SO_{4}(0.002172 mol) (142.041 g/mol) = 0.3085 g of Na

_{2}SO_{4}"bonded" with the water0.3085 g + 0.391 g = 0.6995 g of the hydrate

1.281 + 0.391 = 1.672 g <--- new total mass of hydrate/anhydrate mixture

(0.6995 / 1.672) (100) = 41.8% (to three sig figs)

**Problem #31:** A hydrate is found to have the following percent composition: 48.8% MgSO_{4} and 51.2% H_{2}O. What is the formula of the hydrate?

**Solution:**

1) Assume 100 g of the compound is present. This allows us to change percents to mass. For example, in 100 g of the compound, 48.8 gams of MgSO_{4} will be present.

2) Now, calculate moles:

MgSO_{4}---> 48.8 g / 120.366 g/mol = 0.40543 mol

H_{2}O ---> 51.2 g / 18.015 g/mol = 2.8421 mol

3) Look for a whole number ratio:

MgSO_{4}---> 0.40543 / 0.40543 = 1

H_{2}O ---> 2.8421 / 0.40543 = 7MgSO

_{4}⋅7H_{2}O

**Problem #32:** A student dissolves 9.86 g Epsom salt (a MgSO_{4} hydrate) into 50.0 mL water. From this solution the student isolated 9.30 g of anhydrous BaSO_{4} (s) which is formed in the following reaction. BaSO_{4} is very insoluble in water, and esentially all sulfate ions will precipitate as BaSO_{4} (s). Assume handling losses are small.

MgSO_{4}(aq) + BaCl_{42}(aq) ---> MgCl_{2}(aq) + BaSO_{4}(s)

(a) Determine the formula of the MgSO_{4} hydrate (Epson Salt)

(b) Calculate the % by mass of H_{2}O in the hydrate

**Solution:**

1) Moles of BaSO_{4}:

9.30 g / 233.391 g/mol = 0.0398473 mol

2) From the chemical reaction, BaSO_{4} and MgSO_{4} are in a 1:1 molar ratio, so this many moles of MgSO_{4} formed:

0.0398473 mol

3) Grams of MgSO_{4}

(0.0398473 mol) (120.366 g/mol) = 4.79626 g

4) Water in the hydrate:

9.86 g − 4.79626 g = 5.06374 g

5) Moles of water:

5.06374 g / 18.015 g/mol = 0.2810846 mol

6) We want to know a whole number ratio between MgSO_{4} and H_{2}O where MgSO_{4} is set to 1.

MgSO_{4}---> 0.0398473 / 0.0398473 = 1

H_{2}O ---> 0.2810846 / 0.0398473 = 7MgSO

_{4}⋅7H_{2}O

7) Percent by mass of water:

(5.06374 g / 9.86 g) (100) = 51.356% (round off to 51.4% for three sig figs)

**Problem #33:** 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) How much water was driven off?

5.00 − 4.26 = 0.74 g

2) Convert mass to moles for both BaCl_{2} and H_{2}O:

BaCl_{2}---> 4.26 g / 208.23 g/mol = 0.020458 mol

H_{2}O ---> 0.74 g / 18.015 g/mol = 0.041077 mol

3) Find lowest whole-number ratio:

BaCl_{2}---> 0.020458 mol / 0.020458 mol = 1

H_{2}O ---> 0.041077 mol / 0.020458 mol = 2

4) The answer is 2. Although not asked for, here is the formula:

BaCl_{2}·2H_{2}O

**Problem #34:** When 1.40 g of a hydrated metal sulphate, XSO_{4} **·** 7H_{2}O is heated to constant mass, 0.68 g of the anhydrous salt remains. Find the atomic mass of the metal in the subtance. Identify the metal.

**Solution:**

1) Determine mass of water lost:

1.40 g − 0.68 g = 0.72 g

2) How many moles is this?

0.72 g / 18.015 g/mol = 0.0399667 molI think 0.0400 is close enough.

3) XSO_{4} and H_{2}O are in a 1:7 molar ratio. Determine how many moles of XSO_{4} are present:

0.0400 / 7 = 0.0057143 mol

4) Determine the molecular weight of XSO_{4}:

0.68 g / 0.0057143 mol = 119 g/mol

5) Remove the weight of the sulfate:

119 − 96.06 = 22.94 g/molWith a resonable amount of experimental error, X is magnesium.

**Problem #35:** 8.70 grams of hydrated potassium carbonated was heated to constant mass. The salt lost 1.80 g of water. What is the formula of the hydrated carbonate?

**Solution:**

mass of K_{2}CO_{3}---> 8.70 − 1.80 = 6.90 g6.90 g / 138.2 g/mole = 0.050 mole of K

_{2}CO_{3}1.8 g H2O / 18.0 g/mole = 0.10 mole of H

_{2}OK

_{2}CO_{3}·2H_{2}O

**Bonus Problem:** A 0.572 g sample of a compound containing only thorium , sulfate and water of hydration. After heating to drive off the water of hydration, the remainer weights 0.414 g . A second sample weighting 0.249 g is dissolved in water and excess aqueous barium chloride is added to precipitate 0.198 g of barium sulfate . Calculate the empirical formula of the compound.

**Solution:**

1) Determine mass of water driven off in first sample:

0.572 g − 0.414 g = 0.158 g

2) Determine mass of water that would have been driven off from a second sample weighting 0.249 g:

(0.249 g) (0.158 g / 0.572 g) = 0.06878 g

3) Determine the mass of sulfate in the second sample:

(0.198 g BaSO_{4}) (96.06 g sulfate/mol) / (233.4 g BaSO_{4}/mol) = 0.08149 g sulfate

4) Determine the mass of thorium in the second sample:

0.249 g − (0.08149 g + 0.06878 g) = 0.09873 g

5) Determine moles:

thorium ---> 0.09873 g / 232.038 g/mol = 0.00042549 mol

sulfate ---> 0.08149 g / 96.06 g/mol = 0.000848324 mol

water ---> 0.06878 g / 18.015 g/mol = 0.00381793 mol

6) Divide through by smallest:

thorium ---> 0.00042549 mol / 0.000848324 mol = 0.5

sulfate ---> 0.000848324 mol / 0.000848324 mol = 1

water ---> 0.00381793 mol / 0.000848324 mol = 4.5

7) Double to get whole numbers:

Th(SO_{4})_{2}⋅9H_{2}O

Fifteen Examples | Problems #1 - 10 | Problems #11 - 20 | 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