Calculate empirical formula when given mass data

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: