Ten 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 the formula of a hydrate

Only the examples and problems, no solutions

**Example #1:** 100.0 g of XF_{3} contains 49.2 g of fluorine. What element is X?

**Solution #1:**

49.2 g / 19 g/mol = 2.589 mol of F3 is to 2.589 as 1 is to x

x = 0.863 mol of X present

50.8 g / 0.863 mol = 58.8 g/mol

Cobalt

**Solution #2:**

49.2% by weight of your compound is fluorine. In the molecule of your unknown compound you have three atoms of fluorine, this means that in one mole of this compound you have 57 g of fluorine (from 19.0 x 3) that represent 49.2% by weight. One mole of unknown compound weighs (57/0.492) = 115.85 g. Knowing that 57 g are made by fluorine, the unknown atom will count for 58.85 g in each mole. This means that its atomic weight is 58.85 g/mol, the atomic weight of cobalt.

**Example #2:** A 1.443 g sample of an unknown metal is reacted with excess oxygen to yield 1.683 grams of an oxide known to have the formula M_{2}O_{3}. Calculate the atomic weight of the element M and identify the metal.

**Solution:**

1) Determine grams, then moles of oxygen:

1.683 − 1.443 = 0.240 g0.240 g / 16.00 g/mol = 0.015 mol

2) Determine moles of M:

x is to 2 as 0.015 mol is to 3x = 0.010 mol of M

3) Determine atomic weight of M:

1.443 / 0.010 mol = 144.3 g/molNeodymium

**Example #3:** When the element A is burned in an excess of oxygen, the oxide A_{2}O_{3}(s) is formed. 0.5386 g of element A is treated with oxygen and 0.711 g of A_{2}O_{3} are formed. Identify element A.

**Solution:**

1) mass oxygen:

0.711 g − 0.5386 g = 0.1724 g

2) mole oxygen:

0.1724 g / 16.00 g/mol = 0.010775 mol

3) moles A:

0.010775 mol is to 3 as x is to 2x = 0.007183 mol

4) atomic weight (and identity) of A:

0.5386g / 0.007183 mol = 75.0 g/molArsenic

**Example #4:** A 2.89 g sample of osmium oxide, Os_{x}O_{y}, contains 2.16 g of osmium. What are the values of x and y?

**Solution:**

1) Determine mass:

Os ---> 2.16 g

O ---> 2.89 − 2.16 = 0.73 g

2) Divide each by atomic mass:

Os ---> 2.16 g / 190.23 g/mol = 0.01135 mol

O ---> 0.73 g / 16.00 g/mol = 0.0456 mol

3) Divide by smallest:

Os ---> 0.01135 / 0.01135 = 1

O ---> 0.0456 / 0.01135 = 4x = 1

y = 4Although not asked for, the formula is OsO

_{4}

**Example #5:** 7.8 g of an element X reacts with oxygen to form 9.4 g of an oxide X_{2}O. What is the relative atomic mass of X? What is the element X?

**Solution:**

1) Mass of O in the compound:

9.4 − 7.8 = 1.6 g

2) Moles of O in the compound:

1.6 g / 16 g/mol = 0.1 mol

3) Determine moles of X in the compound:

From X_{2}O, the molar ratio of X to O is 2 to 1

Our sample contains 0.2 mol of X

4) Determine atomic weight of X:

7.8 g / 0.2 mol = 39 g/molPotassium

**Example #6:** A 64.8 g sample of the compound X_{2}O_{5} contains 48.0 g of oxygen atoms. What is the atomic weight of X? What element is X?

**Solution:**

1) Moles of O in the compound:

48.0 g / 16.0 g/mol = 4.00 mol

2) Moles of X in the compound:

x is to 4 as 2 is to 5x = 1.60 mol

3) Mass of X in the compound:

64.8 g − 48.0 g = 16.8 g

4) Atomic weight of X:

16.8 g / 1.60 mol = 10.5 g/molBoron

**Example #7:** The chloride of an unknown metal is believed to have the formula MCl_{3}. A 1.603 g sample of the compound is found to contain 0.03606 mol of Cl. Determine the atomic weight of M and identify it by name.

**Solution:**

(0.03606 mol) (35.453 g/mol) = 1.27844 g1.603 g − 1.27844 g = 0.32456 g

The M to Cl molar ratio is 1 to 3. Therefore, moles M in the sample:

0.03606 mol / 3 = 0.01202 mol

atomic mass:

0.32456 g / 0.01202 mol = 27.0 g/mol

Aluminum

**Example #8:** The 64.8 g sample of the compound X_{2}O_{5} contains 48.0 grams of oxygen atoms. What is the molar mass of X? What element is X?

**Solution:**

64.8 − 48 = 16.8 g (the mass of X in the sample)48.0 g / 16.0 g/mol = 3.00 mol of O in sample.

The molar ratio of X to O is 2 to 5.

2 is to 5 as y is to 3.00

y = 1.2 <--- this is the number of moles of X in the sample.

16.8 g / 1.2 mol = 14.0 g/mol

Nitrogen

**Example #9:** A 47.3 g sample of the compound X_{3}(PO_{4})_{2} contains 8.78 g of phosphorus. Identity X

**Solution #1:**

(8.78 / 47.3) * 100 = 18.56% (percentage of P in compound)and this has a mass of 2 x 31 g = 62 g (in one formula unit of the compound)

molar mass of the compound ---> 62 x (100 / 18.56) = 334.1 g / mol

mass of just PO

_{4}in compound ---> (94.971 x 2) g = 189.9 gmolar mass of just the metal ---> (334.1 − 189.9) / 3 = 48.0 g / mol

Titanium

**Solution #2:**

8.78 g / 30.97 g/mol = 0.2835 mol0.2835 mol as to 2 as y is to 3

y = 0.42525 mol (this is how many moles of X are present in the compound)

0.2835 mol is to 2 as z is to 8

z = 1.134 mol (this is how much oxygen is present)

16.0 g/mol times 1.134 mole = 18.144 g

47.3 g − (8.78 + 18.144) = 20.376 g (grams of X present in the sample)

20.376 g / 0.42525 mol = 47.9 g/mol

**Solution #3:**

You have 8.78 g of P with an atomic weight of 30.97 g/mole so you have 0.2835 moles.From the equation, 1 mole of X

_{3}(PO_{4})_{2}contains 2 moles P so you have 0.2835/2 moles of the mystery compound.You also know the mass of the compound is 47.3 g so the MW is 47.3 g / 0.14175 moles = 333.7 g/mole

In one mole of the compound, you have 3 moles X, 2 moles P and 8 moles O. Eight moles of O is 128 g and two moles P is 61.9 g.

The mass of the mystery material assuming you had 1 mole = 333.7 g − 128 g − 61.9 g = 143.8.

Since the formula has three X, the atomic weight of X = 47.9 g/mol.

**Example #10:** A 30.6-g sample of the compound M_{2}O_{3} contains 14.4 g of oxygen atoms. What is the molar mass of element M? Identify the element M most probably is.

**Solution:**

1) Set up the following ratio and proportion:

30.6 14.4 –––––– = –––– 2x + 48 48 14.4 g / 48 g ---> The moles of O in 30.6 g of M

_{2}O_{3}48 g ---> There are 48 g in three moles of oxygen. The three comes from oxygen's subscript in the formula. (I suppose I should write 48 g/3mol.

2x + 48 ---> The molar mas of M

_{2}O_{3}. x is the atomic weight of M, note that is multiplied by two because of the subscripted two in the formula.

2) Cross multiply and divide:

(14.4) (2x + 48) = (48) (30.6)(14.4) (2x + 48) = 1468.8

2x + 48 = 102

2x = 54

x = 27 g/mol

Aluminum

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