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Calculate empirical formula when given mass data
Calculate empirical formula when given percent composition data
Determine identity of an element from a binary formula and a percent composition
Determine the formula of a hydrate
Problem #1: 100.0 g of XF3 contains 49.2 g of fluorine. What element is X?
Solution #1:
49.2 g / 19 g/mol = 2.589 mol of F3 is to 2.589 as 1 is to x
x = 0.863 mol of X present
50.8 g / 0.863 mol = 58.8 g/mol
cobalt
Solution #2:
49.2% by weight of your compound is fluorine. In the molecule of your unknown compound you have three atoms of fluorine, this means that in one mole of this compound you have 57 g of fluorine (from 19.0 x 3) that represent 49.2% by weight. One mole of unknown compound weighs (57/0.492) = 115.85 g. Knowing that 57 g are made by fluorine, the unknown atom will count for 58.85 g in each mole. This means that its atomic weight is 58.85 g/mol, the atomic weight of cobalt.
Problem #2: A 1.443 g sample of an unknown metal is reacted with excess oxygen to yield 1.683 grams of an oxide known to have the formula M2O3. Calculate the atomic weight of the element M and identify the metal.
Solution:
1) Determine grams, then moles of oxygen:
1.683 minus 1.443 = 0.240 g0.240 g divided by 16.00 g/mol = 0.015 mol
2) Determine moles of M:
x is to 2 as 0.015 mol is to 3x = 0.010 mol of M
3) Determine atomic weight of M:
1.443 / 0.010 mol = 144.3 g/molNeodymium (element #60)
Problem #3: When the element A is burned in an excess of oxygen, the oxide A2O3(s) is formed. 0.5386 g of element A is treated with oxygen and 0.711 g of A2O3 are formed. Identify element A.
Solution:
1) mass oxygen:
0.711 g minus 0.5386 g = 0.1724 g
2) mole oxygen:
0.1724 g / 16.00 g/mol = 0.010775 mol
3) moles A:
0.010775 mol is to 3 as x is to 2x = 0.007183 mol
4) atomic weight (and identity) of A:
0.5386g / 0.007183 mol = 75.0 g/molArsenic
Problem #4: A 2.89 g sample of osmium oxide, OsxOy, contains 2.16 g of osmium. What are the values of x and y?
Solution:
1) Determine mass:
Os ⇒ 2.16 g
O ⇒ 2.89 minus 2 .16 = 0.73 g
2) Divide each by atomic mass:
Os ⇒ 2.16 g/ 190.23 g/mol = 0.01135 molO ⇒ 0.73 g/16.00 g/mol = 0.0456 mol
3) Divide by smaller:
Os ⇒ 0.01135/0.01135 = 1
O ⇒ 0.0456/0.01135 = 4x = 1
y = 4Although not asked for, the formula is OsO4
Problem #5: 7.8 g of an element X reacts with oxygen to form 9.4 g of an oxide X2O. What is the relative atomic mass of X? What is the element X?
Solution:
1) Mass of O in the compound:
9.4 - 7.8 = 1.6 g
2) Moles of O in the compound:
1.6 g / 16 g/mol = 0.1 mol
3) Determine moles of X in the compound:
From X2O, the molar ratio of X to O is 2 to 1
Our sample contains 0.2 mol of X
4) Determine atomic weight of X:
7.8 g / 0.2 mol = 39 g/mol
From the periodic table: X = potassium, compound is K2O
Return to Mole Table of Contents
Calculate empirical formula when given mass data
Calculate empirical formula when given percent composition data
Determine identity of an element from a binary formula and a percent composition