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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 molThis 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 CH2OThe 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 = 2Empirical 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 : 3Ir2Se3
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" = 141.943284 / 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 ⇒ 4The empirical formula is Fe3O4
Problem #7: If one molecule with the empirical formula C3H7 has a mass of 1.428 x 10 Solution:
1) Determine mass of one mole:
2) Determine "empirical formula weight:
3) Determine molecular formula:
C6H14 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
1.428 x 10
C3H7 = 43.1
86.0 / 43.1 = 2