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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:** A 15.67 g sample of a hydrate of magnesium carbonate was heated, without decomposing the carbonate, to drive off the water. The mass was reduced to 7.58 g. What is the formula of the hydrate?

**Solution:**

1) Determine mass of water driven off:

15.67 minus 7.58 = 8.09 g of water

2) Determine moles of MgCO_{3} and water:

MgCO_{3}⇒ 7.58 g / 84.313 g/mol = 0.0899 mol

H_{2}O ⇒ 8.09 g / 18.015 g/mol = 0.449 mol

3) Find a whole number molar ratio:

MgCO_{3}⇒ 0.0899 mol / 0.0899 mol = 1

H_{2}O ⇒ 0.449 mol / 0.0899 mol = 5MgCO

_{3}·5H_{2}O

**Problem #2:** A hydrate of Na_{2}CO_{3} has a mass of 4.31 g before heating. After heating, the mass of the anhydrous compound is found to be 3.22 g. Determine the formula of the hydrate and then write out the name of the hydrate.

**Solution:**

1) Determine mass of water driven off:

4.31 minus 3.22 = 1.09 g of water

2) Determine moles of Na_{2}CO_{3} and water:

Na_{2}CO_{3}⇒ 3.22 g / 105.988 g/mol = 0.0304 mol

H_{2}O ⇒ 1.09 g / 18.015 g/mol = 0.0605 mol

3) Find a whole number molar ratio:

Na_{2}CO_{3}⇒ 0.0304 mol / 0.0304 mol = 1

H_{2}O ⇒ 0.0605 mol / 0.0304 mol = 2Na

_{2}CO_{3}·2H_{2}Osodium carbonate dihydrate

Comment: sodium carbonate forms three hydrates and the above is not one of them. This is a problem probably crafted so that you cannot look up possible answers via the InterTubez®. Just sayin'.

**Problem #3:** When you react 3.9267 grams of Na_{2}CO_{3} **·** nH_{2}O with excess HCl(aq), 0.6039 grams of a gas is given off. What is the number of water molecules bonded to Na_{2}CO_{3} (value of n)?

**Solution:**

1) Some preliminary comments:

Ignore the water of hydration for a moment.Na

_{2}CO_{3}+ 2HCl ---> CO_{2}+ 2NaCl + H_{2}OThe key is that there is a 1:1 molar ratio between Na

_{2}CO_{3}and CO_{2}

2) Determine moles of CO_{2}:

0.6039 g / 44.009 g/mol = 0.013722 mol of CO_{2}

3) Use the 1:1 molar ratio referenced above:

This means that the HCl reacted with 0.013722 mole of sodium carbonate.

4) How many grams of Na_{2}CO_{3} is that?

0.013722 mol times 105.988 g/mol = 1.4544 g

5) Determine grams, then moles of water

3.9267 g minus 1.4544 g = 2.4723 g of water2.4723 g / 18.015 g/mol = 0.13724 mol of water

6) For every one Na_{2}CO_{3}, how many waters are there?

0.13724 mol / 0.013722 mol = 10Na

_{2}CO_{3}·10H_{2}O

Comment: this is one of the three sodium carbonate hydrates that exists.

**Problem #4:** If 1.951 g BaCl_{2} **·** nH_{2}O yields 1.864 g of anhydrous BaSO_{4} after treatment with sulfuric acid, calculate n.

**Solution:**

1) Calculate mass of Ba in BaSO_{4}:

1.864 g times (137.33 g/mol / 233.39 g/mol) = 1.0968 g

2) Calculate mass of anhydrous BaCl_{2} that contains 1.0968 g of Ba:

1.0968 g is to 137.33 g/mol as x is to 208.236 g/molx = 1.663 g

3) Calculate mass of water in original sample:

1.951 g minus 1.663 g = 0.288 g

4) Calculate moles of anhydrous BaCl_{2} and water:

1.663 g / 208.236 g/mol = 0.0080

0.288 g / 18.015 g/mol = 0.0160

5) Express the above ratio in small whole numbers with BaCl_{2} set to a value of one:

1 : 2BaCl

_{2}·2H_{2}O

**Problem #5:** Given that the molar mass of Na_{2}SO_{4} **·** nH_{2}O is 322.1 g/mol, calculate the value of n.

**Solution:**

1) The molar mass of anhydrous Na_{2}SO_{4} is:

142.041 g/mol

2) The mass of water in one mole of the hydrate is:

322.1 g - 142.041 g = 180.059 g

3) Determine moles of water:

180.059 g / 18.0 g/mol = 10 mol

4) Write the formula:

Na_{2}SO_{4}·10H_{2}O

**Problem #6:** 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 minus 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 #7:** 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 = 0.352

S ⇒ 11/32 = 0.344

O ⇒ 22/16 = 1.375

H_{2}O ⇒ 44/ 18 = 2.44

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

Zn ⇒ 0.352 / 0.344 = 3

S ⇒ 0.344 / 0.344 = 1

O ⇒ 1.375 / 0.344 = 3

H_{2}O ⇒ 2.44 / 0.344 = 7The formula is ZnSO

_{4}·7H_{2}O

**Problem #8:** 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 minus 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 #9:** 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 minus 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 #10a:** 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 times 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 #10b:** 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 minus 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 #11:** 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 / 10.185 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 #12:** 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 times 0.47238 = 2.3619 g (this is the water lost)

5.00 g minus 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 #13:** 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 #14:** 1.33g 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 minus 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}Othe answer is this:

H

_{2}C_{2}O_{4}⋅2H_{2}O

**Problem #15:** 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 minus 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

**Problem #16:** 6.9832 g of FeSO_{4} **⋅** xH_{2}O is dissolved in water acidified with sulfuric acid. The solution is made up to 250. cm^{3}. 25.00 cm^{3} of this solution required 25.01 cm^{3} of 0.0200 M KMnO_{4} to titrate completely. Calculate x.

**Solution:**

Fe^{2+}gets oxidized by the MnO_{4}^{-}. The products are Fe^{3+}and Mn^{2+}Fe

^{2+}---> Fe^{3+}+ e^{-}

5e^{-}+ 8H^{+}+ MnO_{4}^{-}---> Mn^{2+}+ 4H_{2}Oleads to:

8H

^{+}+ 5Fe^{2+}+ MnO_{4}^{-}---> 5Fe^{3+}+ Mn^{2+}+ 4H_{2}OThe key is the 5 to 1 molar ratio between ferrous ion and permanganate. What I want to get is the number of moles of ferrous ion in the 25.00 cm

^{3}. That will get me to grams of FeSO_{4}in the solution. A subtraction will give me the water in the hydrate.moles KMnO

_{4}---> (0.0200 mol/L) (0.02501 L) = 0.000500 molfor every one mole of permanganate, five moles of ferrous are oxidized.

0.000500 mol of MnO

_{4}^{-}oxidizes 0.0025 of ferrous ionThat's the moles in 25.00 cm

^{3}. We originally had 250. cm^{3}, so 0.0250 mol total of dissolved FeSO_{4}. xH_{2}O0.0250 moles of anhydrous FeSO

_{4}weighs ---> 0.025 mol times 151.906 g/mol = 3.79765 gwater in the hydrate ---> 6.9832 g minus 3.79765 g = 3.18555 g

moles of water ---> 3.18555 g / 18.015 g/mol = 0.17683 mol

I want this molar ratio:

0.0250 to 0.17683

in smallest whole number terms where the FeSO

_{4}part is 1:1 to 7.0732

Close enough for this:

FeSO

_{4}⋅7H_{2}O

**Problem #17:** A 81.4 gram sample of BaI_{2} **⋅** 2H_{2}O was heated thoroughly in a porcelain crucible, until its weight remained constant. After heating, how many grams of the anhydrous compound remained?

**Solution:**

Dehydration of the hydrated barium iodide salt is shown by this reaction:BaI

_{2}⋅2H_{2}O(s) ---> BaI_{2}(s) + 2H_{2}O(g)The following "dimensional analysis" set-up uses the concepts of mass-mass stoichiometry:

81.4 g BaI

_{2}⋅2H_{2}O x (1 mol BaI_{2}⋅2H_{2}O / 427.1 g BaI_{2}⋅2H_{2}O) x (1 mol BaI_{2}/ 1 mol BaI_{2}⋅2H_{2}O) x (391.1 g BaI_{2}/ 1 mol BaI_{2}) = 74.5 g BaI_{2}Brief Explanation:

1) 81.4 g BaI

_{2}⋅2H_{2}O ---> the starting mass2) (1 mol BaI

_{2}⋅2H_{2}O / 427.1 g BaI_{2}⋅2H_{2}O) ---> divide by the molar mass of BaI_{2}⋅2H_{2}O3) (1 mol BaI

_{2}/ 1 mol BaI_{2}⋅2H_{2}O) ---> the 1:1 molar ratio4) (391.1 g BaI

_{2}/ 1 mol BaI_{2}) ---> multiply by the molar mass of BaI_{2}

**Problem #18:** If the hydrated compound UO_{2}(NO_{3})_{2} **⋅** 9H_{2}O is heated gently, the water of hydration is lost. If you heat 4.05 mg of the hydrated compound to dryness, what mass of UO_{2}(NO_{3})_{2} will remain?

**Solution:**

Calculate how many moles of UO_{2}(NO_{3})_{2}⋅9H_{2}O you have in 4.05 gWhen you heat it, you have only UO

_{2}(NO_{3})_{2}remaining. 1 mole of UO_{2}(NO_{3})_{2}⋅9H_{2}O leaves 1 mole of UO_{2}(NO_{3})_{2}.Multiply moles of UO

_{2}(NO_{3})_{2}by the molar mass of UO_{2}(NO_{3})_{2}to get the mass.

**Problem #19:** Here is a Yahoo Answers solution to a waters of hydration problem which I answered. Another person answered it as well and their approach differed from mine. You may want to examine it to see the differences.

**Problem #20:** How many grams of water and anhydrous salt would you get when heating 9.42 g of Fe(NO_{3})_{3} **⋅** 9H_{2}O?

**Solution:**

1) Determine mass percentages of Fe(NO_{3})_{3} and H_{2}O:

mass of one mole of Fe(NO_{3})_{3}⋅9H_{2}O ---> 403.9902 gdecimal percent by mass of Fe(NO

_{3})_{3}---> 241.857 g / 403.9902 g = 0.59867decimal percent by mass of water ---> 1 - 0.59867 = 0.40133

2) Determine masses of water and anhydrous salt:

anhydrous salt ---> 9.42 g times 0.59867 = 5.64 g

water ---> 9.42 g minus 5.64 g = 3.78 g

**Problem #21:** Heating 0.695 g CuSO_{4} **⋅** nH_{2}O gives a residue of 0.445 g. Determine the value of n.

**Solution:**

I copied this problem from Yahoo Answers because the person gave a correct solution in the question. That is, a correct solution right up until the end of the problem.

0.695 - 0.445 = 0.25 g of H_{2}O0.25 g H

_{2}O / 18 g/mol = 0.014 mol0.445 g CuSO

_{4}/ 159.5 g/mol = 0.0028 mol0.0028 mol / 0.014 mol = 0.2

Here's the error:

So since there's no such thing has 0.2 of a molecule I round it to a whole number which in this case is 0 to get:0CuSO

_{4}⋅1H_{2}OThe correct technique is to multiply by 5 so as to get a 1 in front of the CuSO

_{4}:0.2CuSO_{4}⋅1H_{2}O times 5 equals this:CuSO

_{4}⋅5H_{2}O

Comment: the student who made this mistake failed to see the coefficients of 0.2 and 1 as representing moles, focusing only on the ratio as molecules. It is perfectly fine to have a molar ratio of 0.2 to 1, which then becomes 1 to 5, which is the lowest whole-number ratio of moles.

**Problem #22:** Epsom salt is MgSO_{4} **⋅** nH_{2}O. The hydrate was found to contain 71.4% oxygen. Calculate the number of water molecules associated with each formula unit of magnesium sulfate hydrate.

**Solution:**

The molar mass of MgSO_{4}is 120.366 g/molThere are 4 moles of O in MgSO

_{4}The grams of O in one mole of MgSO

_{4}is 63.9976 gLet x = grams of oxygen from H

_{2}Otherefore, grams of H

_{2}O is this:x times (18.015 grams of H

_{2}O/15.9994 grams O) = 1.126xgrams of oxygen divided by total weight of compound equals 0.714 (from information given in the problem)

(63.9976 + x) / (120.366 + 1.126x) = 0.714

63.9976 + x = 0.714[120.366 + 1.126x]

63.9976 + x = 85.94 + 0.804x

0.196x = 21.9424

x = 111.951 grams (of oxygen in entire MgSO

_{4}⋅nH_{2}O)111.951 g times (1.000 mole H

_{2}O/15.9994 g O) = 6.997 moles H_{2}OMgSO

_{4}⋅7H_{2}O

**Problem #23:** A sample of hydrated magnesium sulphate, MgSO_{4} **·** nH_{2}O, is found to contain 51.1% water. What is the value of n?

**Solution #1 (the traditional way):**

1) Assume 100 g of the compound is present. This allows the percentages to be easily converted to masses:

MgSO_{4}---> 48.9 g

H_{2}O ---> 51.1 g

2) Convert the masses to moles:

MgSO_{4}---> 48.9 g / 120.366 g/mol = 0.406261 mol

H_{2}O ---> 51.1 g / 18.015 g/mol = 2.836525 mol

3) Divide through by smallest:

MgSO_{4}---> 0.406261 mol / 0.406261 mol = 1

2.836525 mol / 0.406261 mol = 6.98

4) Empirical formula:

MgSO_{4}·7H_{2}O

**Solution #2:**

51.1% is water so 48.9% must be MgSO_{4}Assume one mole of MgSO

_{4}is present. This represents 48.9% of the hydrate.120.366 is to 48.9 as x is to 51.1

x = 125.78 g <--- mass of water in the hydrate

125.78 g/18.015 g/mol = 6.98

MgSO

_{4}·7H_{2}O

**Problem #24:** If a 9.15 g sample of a hydrated salt produced 6.50 g of anhydrous salt (309.650 g/mol) and 2.65 g of water (18.015 g/mol), what is the molecular mass of the hydrated salt?

**Solution:**

6.50 g / 309.650 g/mol = 0.021 mol

2.65 g / 18.015 g/mol = 0.147 molWe want to know how many moles of water are present when one mole of the anhydrous salt is present.

0.147 is to 0.021 as x is to 1

x = 7

molecular mass = 309.650 g/mol + (7 * 18.015 g/mol) = 435.755 g/mol

**Problem #25:** A water of hydration problem on Yahoo Answers.

**Bonus Problem** A hydrate of magnesium chloride is present and the following data is collected:

mass of crucible = 22.130 grams

mass of crucible + hydrate = 25.290 grams

mass of crucible and contents after heating = 23.491 grams

What is the complete formula of this hydrate?

**Solution:**

1) mass of hydrate:

25.290 g - 22.130 g = 3.160 g

2) mass of anhydrate:

23.491 g - 22.130 g = 1.181 g

3) water lost:

3.160 g - 1.181 g = 1.979 g

4) moles MgCl_{2}:

1.181 g / 95.211 g/mol = 1.24 mol

5) moles water lost:

1.979 g / 18.015 g/mol = 0.109853 mol

6) molar ratio of MgCl_{2} to water is:

1 : 11.3Within fairly reasonable experimental error, the formula of the hydrate is:

MgCl

_{2}·12H_{2}O

The Wiki page for magnesium chloride shows hydrates with 12, 8, 6, 4 and 2 waters of hydration exist. The one that exists at room temperature is the one with 6. The one with 12 loses 4 waters of hydration above -16.4 °C. So, while the problem above does not occur near room temperature, it can occur.

Look at a list of only the questions.

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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