Calculate empirical formula when given mass data

Return to Mole Table of Contents

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

Determine the formula of a hydrate


Problem #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

Problem #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


Problem #3: A compound made of two elements, iridium (Ir) and oxygen (O), 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.


Problem #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


Problem #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


Problem #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

3) A ratio that involves a third (like the 1.33 just above or something like 2.67) can be thought of this way:

Fe ⇒ 3/3
O ⇒ 4/3

4) Multiply by three to get to whole numbers:

Fe ⇒ 3
O ⇒ 4

The empirical formula is Fe3O4


Problem #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 times 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


Problem #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


Problem #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.


Problem #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.


Return to Mole Table of Contents

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

Determine the formula of a hydrate