Example #1: A sample of copper metal weighing 2.50 g is heated to form an oxide of copper. The final mass of the oxide is 3.13 g. Determine the empirical formula of the oxide.

Solution:

1) Determine mass:

Cu ---> 2.50 g
O ---> 3.13 g − 2.50 g = 0.63 g

2) Determine moles:

Cu ---> 2.50 g / 63.546 g/mol = 0.03934 mol
O ---> 0.63 g / 16.00 g/mol = 0.039375 mol

This is a 1:1 molar ratio between Cu and O.

3) Write the empirical formula:

CuO

Example #2: On analysis, a compound with molar mass 60 g/mol was found to contain 12.0 g of carbon, 2.0 g of hydrogen and 16.0 g of oxygen. What is the molecular formula of the compound? (no calculator allowed!)

Solution:

12.0 g carbon is about 1 mole of carbon; 2.0 g of H is about 2 moles and 16.0 g O is about one mole. So the empirical formula is CH₂O

The molecular weight of this molecule is 30 g/mole, so you divide 60/30 to find how many times you must multiple your empirical formula. The answer is 2 times the above empirical formula, so the molecular formula is C₂H₄O₂

Example #3: Determine the empirical formula of a compound made of two elements, iridium (Ir) and oxygen (O), which was produced in a lab by heating iridium while exposed to air. The following data was collected:

Mass of crucible: 38.26 g
Mass of crucible and iridium: 39.52 g
Mass of crucible and iridium oxide: 39.73 g

Solution:

1) Get mass of each element:

Ir ---> 39.52 − 38.26 = 1.26 g
O ---> 39.73 − 39.52 = 0.21 g

2) Get moles of each element:

Ir ---> 1.26 g / 192.217 g/mol = 0.0065551 mol
O ---> 0.21 g / 16.00 g/mol = 0.013125 mol

3) Look for smallest whole-number ratio:

Ir ---> 0.0065551 / 0.0065551 = 1
O ---> 0.013125 / 0.0065551 = 2

Empirical formula = IrO₂

By the way, this problem ignores any errors that might be produced by the Ir reacting with the nitrogen in the air.

Example #4: A compound contains 16.7 g of iridium and 10.3 g of selenium. What is its empirical formula?

Solution:

1) Moles:

Ir ---> 16.7 / 192.217 = 0.086881
Se ---> 10.3 / 78.96 = 0.130446

2) Seek lowest whole-number ratio:

Ir ---> 0.086881 / 0.086881 = 1
Se ---> 0.130446 / 0.086881 = 1.5

3) Multiply 1 : 1.5 ratio by two:

2 : 3

Ir₂Se₃

Example #5: A 1.000 g sample of red phosphorus powder was burned in air and reacted with oxygen gas to give 2.291 g of a phosphorus oxide. Calculate the empirical formula and molecular formula of the phosphorus oxide given the molar mass is approximately 284 g/mol.

Solution:

1) Calculate moles of P and O:

P ---> 1.000 g / 30.97 g/mol = 0.032289 mol
O ---> 1.291 g / 16.00 g/mol = 0.0806875 mol

2) Determine lowest whole-number ratio:

P ---> 0.032289 mol / 0.032289 mol = 1
O ---> 0.0806875 mol / 0.032289 mol = 2.50

3) Determine empirical formula:

P₂O₅

3) Determine molecular formula:

The "empirical formula weight" = 141.943

284 / 142 = 2

P₄O₁₀

Example #6: A sample of magnetite contained 50.4 g of iron and 19.2 g of oxygen. Calculate the empirical formula.

Solution:

1) Convert grams to moles:

Fe ---> 50.4 g / 55.845 g/mol = 0.9025 mol
O ---> 19.2 g / 16.0 g/mol = 1.2 mol

2) Seek smallest whole-number ratio:

Fe ---> 0.9025 / 0.9025 = 1
O ---> 1.2 / 0.9025 = 1.33 <--- a common error is to round this off to 1

3) A ratio that involves a third (like the 1.33 just above or something like 2.67) can be thought of as fractions with a denominator of 3. Like this:

Fe ---> 3/3
O ---> 4/3

4) Multiply by three to get to whole numbers:

Fe ---> 3
O ---> 4

The empirical formula is Fe₃O₄

Example #7: If one molecule with the empirical formula C₃H₇ has a mass of 1.428 x 10¯²² g, determine the molecular formula of the compound.

Solution:

1) Determine mass of one mole:

(1.428 x 10¯²² g/molecule) (6.022 x 10²³ molecules/mol) = 86.0 g/mol

2) Determine "empirical formula weight:"

C₃H₇ = 43.1

3) Determine molecular formula:

86.0 / 43.1 = 2

C₆H₁₄

Example #8: What is the empirical formula and molecular formula for lactic acid if the percent composition is 40.00% C, 6.71% H, 53.29% O, and the approximate molar mass is 90 g/mol?

Solution:

1) Assume 100 g of the compound is present. This turns percents into mass.

2) Calculate moles:

C ---> 40.00 g / 12.011 g/mol = 3.33
H ---> 6.71 g / 1.008 g/mol = 6.66
O ---> 53.29 g / 16.00 g/mol = 3.33

3) Divide by lowest value:

C ---> 3.33 / 3.33 = 1
H ---> 6.66 / 3.33 = 2
O ---> 3.33 / 3.33 = 1

Empirical formula = CH₂O

4) Empirical formula weight

12 + 2 + 16 = 30

5) What is the molecular formula?

90/30 = 3 (there are three "units" of the empirical formula in the molecular formula)

C₃H₆O₃

Example #9: A 0.338 g sample of antimony was completely reacted with 0.295 g of chlorine gas to form antimony chloride. Determine the empirical formula of the antimony chloride.

Solution:

Sb ---> 0.338 g / 121.760 g/mol = 0.002776 mol
Cl ---> 0.295 g / 35.453 g/mol = 0.008321 mol

0.008321 mol / 0.002776 mol = 3

SbCl₃

Note the use of the atomic weight of chlorine, not the molecular weight of Cl₂. We are interested in determining how many Cl atoms bond per one Sb atom, not how many Cl₂ molecules bond with one Sb. If you use 70.906 g/mol, you get 1.5 chlorine to one Sb (and forgetting that it's 1.5 Cl₂ molecules) you conclude the formula is Sb₂Cl₃.

Example #10: When the element antimony, Sb, is heated with excess sulfur, a reaction occurs to give a compound containing only antimony and sulfur. On further heating, excess sulfur is burnt off forming gaseous sulfur dioxide, SO₂, and the substance left is pure compound of antimony and sulfur. In one experiment, 2.435 g of antimony was used, and the mass of the pure compound of antimony and sulfur was found to be 3.397 g. What is the empirical formula of the antimony-sulfur compound?

Solution:

1) Calculate mass of sulfur reacted:

3.397 − 2.435 = 0.962 g

2) Calculate moles:

Sb ---> 2.435 g / 121.76 g/mol = 0.0200 mol
S ---> 0.962 g / 32.065 g/mol = 0.0300 mol

3) Divide through by smallest:

0.02 / 0.02 = 1
0.03 / 0.02 = 1.5

4) Multiply by 2 to get whole number ratio

Sb₂S₃

Comment: you can see the 0.2 to 0.3 ratio is the same as a 2 to 3 ratio and arrive at the empirical formula that way. The "divide by smallest" is the classic way to solve problems of this type.

Bonus Example: Which mass ratio would be obtained if methanol is analyzed?

	C	H	O
A.	3g	1g	4g
B.	3g	2g	8g
C.	6g	8g	16g
D.	12g	8g	32g

Solution:

1) The formula for methanol is:

CH₃OH

2) This means the following atom ratio:

one carbon : four hydrogen : one oxygen

3) Use the atomic masses of each element to obtain the mass ratio for the above atom ratio:

12 grams C : 4 grams H : 16 grams O

4) We are looking for an answer that has the weight in the ratio of 12 : 4 : 16. Reduced to lowest terms, that ratio is:

3 : 1 : 4

Answer choice A