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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 10-22 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 = 2C6H14
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 = 1Empirical 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
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