Five Examples

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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.

**Example #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

**Example #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'.

**Example #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}(s) + 2HCl(aq) ---> 2NaCl(aq) + CO_{2}(g) + H_{2}O(ℓ)The key is that there is a 1:1 molar ratio between Na

_{2}CO_{3}and CO_{2}. (Also, note that we assume that the gas is pure CO_{2}and that there is no water vapor whatsoever. All of the water stays as a liquid. We also assume that no CO_{2}dissolves in the water.)

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.

**Example #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 mol

0.288 g / 18.015 g/mol = 0.0160 mol

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

BaCl_{2}---> 0.0080 mol / 0.0080 mol = 1

H_{2}O ---> 0.0160 mol/ 0.0080 mol = 2BaCl

_{2}·2H_{2}O

**Example #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

**Bonus Example:** 3.20 g of hydrated sodium carbonate, Na_{2}CO_{3} **⋅** nH_{2}O was dissolved in water and the resulting solution was titrated against 1.00 mol dm^{-3} hydrochloric acid. 22.4 cm^{3} of the acid was required. What is the value of n?

**Solution:**

1) Sodium carbonate dissolves in water as follows:

Na_{2}CO_{3}⋅nH_{2}O(s) ---> 2Na^{+}+ CO_{3}^{2-}+ nH_{2}O(ℓ)

2) The addition of HCl will drive all of the CO_{3}^{2-} ion to form CO_{2} gas. One mole of carbonate ion will produce n moles of water.

CO_{3}^{2-}+ 2H^{+}---> CO_{2}(g) + H_{2}O(ℓ)

3) Determine moles of HCl and from that moles of carbonate:

MV = moles(1.00 mol/L) (0.0224 L) = 0.0224 mole of HCl

Two moles of HCl react for every one mole of carbonate. Therefore:

0.0224 mole / 2 = 0.0112 mol of carbonate

4) Determine the mass of 0.0112 mol of Na_{2}CO_{3}.

0.0112 mol Na_{2}CO_{3}x (105.988 g Na/ 1 mol) = 1.187 g Na_{2}CO_{3}This is how many moles of anhydrous sodium carbonate dissolved.

5) Mass of hydrated salt minus mass of anhydrous salt = mass of water

3.20 g - 1.187 g = 2.013 g H_{2}O

6) Convert to moles of water

2.013 g H_{2}O x (1 mol/18.015 g) = 0.11174 mol H_{2}O

7) Determine smallest whole-number ratio between sodium carbonate and water:

Na_{2}CO_{3}: 0.0112 mol / 0.0112 = 1

H_{2}O: 0.11174 mol / 0.0112 = 9.97 = 10Na

_{2}CO_{3}⋅10H_{2}O

Look at a list of only the questions.

Go to hydrate problems #1 - 10

Go to hydrate 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