Transcript Percentage Composition

Percentage Composition
• The percentage of the total mass of a compound
contributed by an element is the percentage of that
element in the compound.
• Examples:
– Elements:
• Copper: Copper is 100% copper because it is a single element.
– Compounds:
• Sodium chloride: Salt is a composition of two elements, chlorine and
sodium.
Example: Salt (NaCl)
• We know that salt is NaCl and will always combine this
way.
• They are always present in this ratio by mass.
• The ratio that they are calculated is the ratio of their
atomic masses.
• The percentage of sodium in any sample of sodium
chloride would be the atomic mass of the element
divided by the formula mass and multiplied by 100.
Mass Na x 100 & Mass Cl x 100
Mass NaCl
Mass NaCl
(39.3% Na, 60.7% Cl)
Example: Aluminum Sulfate
• Find the percentage composition of aluminum
sulfate.
• Solving Process:
– The formula for aluminum sulfate is Al2(SO4)3
– The molecular mass is
2 Al atoms
2 x 27.0 = 54.0 g mol-1
3 S atoms
3 x 32.1 = 96.3 g mol-1
12 O atoms
12 x 16.0 = 192 g mol-1
342.3 g mol-1
(continued)
• Formula mass = 342.3 g mol-1
– The percentage of Al: mass 2 Al____ x 100
mass of Al2(SO4)3
– The percentage of S: mass 3 S____ x 100
mass of Al2(SO4)3
– The percentage of O: mass 12 O____ x 100
mass of Al2(SO4)3
(15.8% Al, 28.2 % S, 56.1% O)
Mass Conservation in Chemical Reactions
• Mass and atoms are
conserved in every
chemical reaction.
• Molecules, formula units,
moles and volumes are
not always conserved
every time.
• Chemical reactions need
to be balanced to satisfy
the law of conservation of
mass.
Empirical Formulas
• Empirical Formula is the SIMPLEST whole # ratio of
the elements present in the compound.
• Using experimental data, we can find the empirical
formula for a substance.
• We only need to know the mass (or percent) of each
element in the laboratory sample.
• Elements in compounds combine in simple whole
number ratios, such as 1:1, 1:2, 2:3, etc.
Example: Empirical Formulas From Mass
•
•
What is the empirical formula for a compound if a 2.50
g sample contains 0.900 g of calcium and 1.60 g of
chlorine?
Solving Process:
1. We must determine the number of moles of each element in
the compound.
(0.900g Ca/ 40.08 g/mol = 0.0225 mol Ca)
(1.60g Cl/ 35.45 g/mol = 0.0451 mol Cl)
2. To get the simplest ratio, divide both numbers of moles by
the smaller one.
(0.0225 mol Ca/0.0225 mol = 1 Ca)
(0.0451 mol Cl/0.0225 mol = 2 Cl)
3. This calculation shows that for each mole of calcium, there
are 2 moles of chlorine.
4. The empirical formula is CaCl2
Example: Empirical Formulas From % Composition
•
•
A compound has a percent composition of 40%
carbon, 6.71% hydrogen and 53.3% oxygen. What is
the empirical formula?
Solving Process:
1. To calculate the ratio of moles of these elements, we
assume an amount of the compound (100g). This makes the
percentages of the compound the same as the mass in
grams.
So we would have 40.0 g of carbon, 6.71 g of Hydrogen
and 53.3 g of oxygen in a 100 g sample.
2. Next change the quantities to moles.
• 40.0g C/ 12.01 g/mol = 3.33 mol of carbon
• 6.71g H/ 1.01 g/mol = 6.64 mole of hydrogen
• 53.3g O/16.0 g/mol = 3.33 mol of oxygen.
3. Dividing each result by smallest moles (3.33), we get
1:1.99:1.
• This makes the empirical formula CH2O
Example: Empirical Formulas From Other
Experimental Methods
• The empirical formula can be found by direct
determination.
– Converting a massed sample of one element to a compound
(to find the mass of the second that combined with the first).
• Ex. 2Mg + O2  2 MgO
• Organic compounds are found by burning a known mass
of the compound in excess oxygen, and then finding the
mass of both carbon dioxide and water formed.
• Ex. C2H4 + 3O2  2CO2 + 2H2O
Molecular Formulas
• To move from an empirical formula (which is the
simplest form) to a molecular formula, we only need
one more piece of information (the molecular mass).
• The molecular formula shows the actual number of
atoms of each element in the compound, as well as the
ratio of atoms.
Example: Molecular Formula
• Earlier, we found that the empirical formula of a
compound was CH2O.
• If we know the molecular mass of the compound is 180
g/mol, then we can determine the molecular formula.
• We can calculate the mass of the empirical formula
(CH2O). It is 30.0 g/mol
• Next, we divide our molecular mass of the true formula
by the molecular mass of the empirical formula
(180g / 30g) = 6. This is our “multiplier”.
• It will take six of these E.F. units to equal 180 or one
molecular formula. (Mulitplier x E.F. = M. F.)
• The molecular formula is C6H12O6