Laws and Percent Composition

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Laws and Percent Composition
Law of Definite Proportions, 1797
• Definition: States that in samples of any
chemical compound, the masses of the
elements are always in the same proportions
– Created by Joseph Proust
– Also called the Law of Constant Composition
– Relates to experimental data of percent
composition
– Use ratios of mass data to prove that composition
is constant
Example 1
• Two samples of carbon dioxide, obtained
from different sources, were decomposed
into their constituent elements. One sample
produced 4.8 g of oxygen and 1.8 g of carbon,
and the other sample produced 17.1 g of
oxygen and 6.4 g of carbon. Show that these
results are consistent with the law of definite
proportions.
Are they consistent?
Example 2
• Two samples said to be carbon disulfide (CS2)
are decomposed into their constituent
elements. One sample produced 8.08 g S and
1.51 g C, while the other produced 31.3 g S
and 3.85 g C. Are these results consistent
with the law of constant composition?
Are they consistent?
Law of Multiple Proportions
• Definition: Whenever the same two elements
form more than one compound, the different
masses of one element that combine with the
same mass of the other element are in the
ratio of small whole numbers
– Created by John Dalton
http://honolulu.hawaii.edu/distance/sci122/Programs/p25/p253.gif
Chemical Formulas can be
determined from experimental
mass data
• Percent Composition describes the relative
masses of the elements in a compound
Percent Composition
• Using mass data to determine percent
composition
• % of element E = grams of E * 100
grams of compound
Percent Composition from Mass Data
Example 1
• When a 13.6-g sample of a compound
contains only Mg and O is decomposed, 5.40
g of O is obtained. What is the %
composition of this compound?
• When a 13.6-g sample of a compound
contains only Mg and O is decomposed, 5.40
g of O is obtained.
Con’t
Percent Composition from Mass Data
Example 2
• What is the percent composition of a
compound that is formed from 2.70 g of Al +
5.10 g of O (forms aluminum oxide)
What is the percent composition of a compound
that is formed from 2.70 g of Al + 5.10 g of O (forms
aluminum oxide)
• Law of conservation of Mass ….
Percent composition from chemical
formula
• Use when no mass data is given, you can
compare the ratio of the molar mass of each
element to molar mass of the compound.
% mass =
grams of element in 1 mole**
molar mass of compound
x 100
**Need to take into account the subscripts for each
atom
Percent Composition from Formula:
Example 1
• Propane, C3H8, the fuel commonly used in gas
grills, is one of the compounds obtained from
petroleum. Calculate the % composition of
propane.
% Composition of Propane
• Do all elements unless noted….
Percent Composition from Formula:
Example 2
• Ammonium sulfate, (NH4)2SO4 is a commonly
used fertilizer. Calculate the % composition of
ammonium sulfate.
% Composition of (NH4)2SO4
% Composition of (NH4)2SO4
Hydrate: A solid compound containing water
molecules combined in a definite ratio as an
integral part of the crystal.
Copper Sulfate•5H2O
Hydrates
• When hydrates are heated they break down
into the anhydrous salt (ionic portion) and
water
• We can use this information to help us find
the formula of an unknown hydrate
Section 10-5
Finding the Formula of a hydrate
1. Using the law of conservation of mass, find
the mass of the salt and the mass of the
water.
1. This may be given or may need to be determined
experimentally
2. Convert the mass of salt to moles using
molar mass of salt.
3. Convert mass of water to moles of water
using molar mass of water (18.02 g/mol)
4. Determine the ratio of moles of water to
moles of salt to find the formula for the
hydrate.
Example 1: Hydrate
• 9.00 grams of an unknown FeCl3 hydrate were
heated until all the water was driven off,
leaving 5.4 g of anhydrous salt. What is the
formula of the hydrate?
• Law of conservation of mass…
9.00 g  5.4 g salt + X g H2O
X g H2O = 3.6 grams
Convert grams to moles
Convert grams of salt (FeCl3) to moles
Convert grams of water to moles
Find the ratio of moles of water/moles of
salt
The formula of the hydrate is…
Hydrates, Example 2
• What is the formula of an unknown CuBr2
hydrate that decomposes into 5.67 g of
anhydrous salt and 1.83 g of water?
Answer: CuBr2•4H2O
• Convert grams to moles for each
Practice 10-16
A mass of 2.50 g of blue hydrate is placed in a
crucible and heated. After heating, 1.59 g of
white anhydrous copper(II) sulfate remains. What
is the formula for the hydrate? Name the hydrate.
(Hint – calculate grams of water lost and convert
this to moles of water; also calculate grams of
copper(II) sulfate and convert this to moles; lastly,
make the ratio as on the previous slide)
Practice 10-17
A mass of 5.00 g of a hydrate is placed in a
crucible and heated. After heating, 4.26 g
of anhydrous barium chloride remains.
What is the formula for the hydrate? Name
the hydrate.
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