Download Empirical and Molecular Formulas and Percentage Composition

General Chemistry
Name:__________________
Unit 6 Note Pocket Stoichiometry
Date: _________ Per: ____
Empirical and Molecular Formulas and Percentage Composition
The formula of a compound indicates:
1.
2.
We can use a formula and atomic masses to determine the percentage (by mass) of each atom in a compound.
Consider carbon dioxide (C02). Assume you have one mole of carbon dioxide. One mole of carbon dioxide has a
molar mass of
g. Because we also know the molar mass of carbon is
g, it easy to calculate
what percentage of the carbon dioxide is carbon. Complete the following calculation:
12.01 g C
44.01 g CO2
x 100 =
What percentage of sulfur in S03?
What is the percentage of potassium in potassium sulfate?
What is the percentage of water in the mineral gypsum (CaS0 4•2H2O)?
If you know the percentage composition of each element in a compound, you can determine the formula of the
compound by using (more or less) the same process as above but in reverse. The formula you determine in this
way is called the empirical formula.
The empirical formula is
Let’s say we have a sample of an unknown hydrocarbon and we want to figure out the empirical formula.
Through experimentation, we determine that the compound is:
17.31% hydrogen
83.67% carbon
Steps to determine the empirical formula
1. Assume 100g of the substance.
2. Break down the 100 g by percentage (in this case l7.31g H and 83.67g C).
3. Use the molar mass conversion factor (factor label) of each element to find the moles of each
atom.
4. Find the smallest whole number ratio of the elements by dividing each mole value by the smallest of
the values found in step three above; remember that you may have to round off to get a whole
number because you are dealing with experimental data.
Unit 6 combined student notes
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Once you have the empirical formula, you might not be done. The empirical formula tells you the lowest ratio
of each type of atom, but it might not tell you the actual number of atoms in a compound. The molecular
formula gives you more information. For example, if the empirical formula of a compound is C 3H8 , its molecular
formula may be C3H8 , C6H16 , etc.
The molecular formula is
Suppose we know that the molar mass of a compound is 180 g/mol. Let's determine the empirical formula for
this compound with the following elemental composition:
40.00% carbon
6.72% hydrogen
53.29% oxygen
Steps to determine the molecular formula from the empirical formula
1. Assume 100g of the substance.
2. Break down the 100 g by percentage (in this case 40.00 g C, 6.72 g H and 53.29 g O).
3. Use the molar mass conversion factor (factor label) of each element to find the moles of each
atom.
4. Find the smallest whole number ratio of the elements by dividing each mole value by the smallest of
the values found in step three above; remember that you may have to round off to get a whole
number because you are dealing with experimental data.
5.
Write the empirical formula of the compound in the example.
6. Find the formula mass of the empirical formula
7. Divide the given molecular mass by the calculated empirical formula mass to find a multiple.
8. Multiply the empirical formula by this ratio to find the molecular formula.
Unit 6 combined student notes
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Measurements in Chemistry - understanding and using the mole
Chemical measurements
In chemistry we make some measurements by counting and other measurements by weighing,
determining volume, etc. Which technique we employ is determined, in large part, by our purpose.
It is also necessary, when determining which technique to use, to consider what type of
measurement is easiest to make or even feasible. Consider the chemical reaction below:
C12(g) + 2KI(s) —> 2KCl(s) + I2 (l)
You may at first think about the fact that one molecule of chlorine reacts with two formula units of
potassium iodide to produce two formula units of potassium chloride and one molecule of iodine.
While this is true, it’s impossible to actually study this reaction at the atomic level in a chemistry
laboratory; we cannot count out just one or two atoms or molecules of anything! We have to use
another way to measure out the necessary quantities of chlorine and potassium iodide.
The mole
A mole is a quantity of something just like a dozen (12), gross (144), or ream (500). We use these
more familiar quantities to count out large quantities because it’s more convenient. We can convert
from the individual item to the quantity units by using simple conversion factors. Try the problems
below using factor labeling:
5 dozen eggs = ______________ eggs
1152 pencils = _______________ gross of pencils
5000 pieces of paper = ______________reams of paper
A mole relates the number of atoms or molecules of a substance to the mass in grams. The
conversion factor used is based on the mass of each element in atomic mass units and how that
relates to the mass of the element in grams.
• mole = the number of atoms of an element that would be found in, say, 12.01 grams of carbon-12 or
14.01 grams of nitrogen-14 (the molar mass)
• mole = ______________________ atoms (this number is called “________________ number”)
• mole = __________ L of any gas at STP (standard temperature and pressure: 273 K and 1.0 atm)
This means that we can use the atomic masses on the periodic table to convert instantly to grams.
The mass of one mole of atoms is called the molar mass. Use the periodic table to determine the
following:
one atom of Ar _____ amu
one atom of Pb _____ amu
one mole of Ar atoms ____ g = molar mass of Ar
one mole of Pb atoms ______ g molar mass of Pb
one atom of Na = __________ amu
one atom of B ______ amu
one mole of Na atoms ____ g = molar mass of Na
one mole of B atoms ______ g = molar mass of B
Unit 6 combined student notes
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Simple conversions using mole equivalencies
We use mole equivalencies to convert:
• moles to ________
1 mole of element = __________________________
• moles to the number of ________
1 mole of element = __________________________
• moles of gas to _______________
1 mole of gas = _____________________________
For example:
1.50 mol Na = ______ Na
1.50 mol Na x 23.0 g Na = 34.50 g Na = 34.5 g Na (sig figs!)
1 mol Na
Do the following conversions using the mole equivalencies above. Use factor labeling and show your
work.
2.0 mol Ni = _______ g Ni
_________mol C = 24.02 g C
5 mol He = _______ atoms He
___________mol Ca = 4.75 x 1026 atoms Ca
0.115 mol Xe = __________L of Xe
_____________ mol Ne = 67.2 L of Ne
Atomic masses of elements  formula mass of a compound  molar mass of a compound
We can work easily with the mass of a compound in the same way we do with elements. Determine
the mass of a unit of sodium chloride or calcium fluoride by adding up __________________of the
component elements in their proper _______________. The sum is called the
________________
one atom of Na = ___________ amu
one atom of Ca = ____________ amu
one atom of Cl = ____________ amu
two atoms of F = ____________ amu
formula mass of NaCl = _______amu
formula mass of CaF2 = ________ amu
Unit 6 combined student notes
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The molar mass of a substance is determined the same way, but you are finding the mass of
_________________________ instead of the mass of just one _______________ (for a
covalent compound) or one _____________________ (for an ionic compound).
molar mass of NaCl = ________ g
molar mass of CaF2 = ________ g
molar mass of H2O _________ 9
molar mass of K2SO4 = _______ g
molar mass of SO3 _________ g
molar mass of Cl2 = __________ g
More simple conversions using mole equivalencies
2.0 mol S02 = ___________g SO2
_____________ mol F2 = 1.4 L F2
2.44 x 1023 molecules H2O = ______________ mol H2O
0.98 mol Fe = ______________ atoms Fe
___________g 02 = 4.00 L 02
5.643 g CO2 = ______________ molecules CO2
0.500 L NO = ____________ moles NO
Multi-step conversions, using mole equivalencies
Conversions involving moles, like other conversions, can involve more than one step. When this is the
case, it is convenient to use the following “mole map” to help you plan the conversion.
Unit 6 combined student notes
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Example problem:
2.3 x 1026 molecules H2 = _____ L of H2
2.3 x 1026 molecules H2
x
1 mol H2
x
23
6.02 x 10 molecules H2
22.4 L H2
1 mol H2
=
Practice Problems:
1. 34 g BeCl2 = __________ formula units BeCl2
2. 44 g H2O = __________ atoms H2O
3. 3.7 x 1023 molecules I2 = ____________ moles I2
4. 6.90 L CH4 = _____________ g CH4
5. 0.034 g NaF = ____________ formula units NaF
6. 5.66 x 1028 atoms Ar = ____________ L Ar
7. 9.0 x 1022 molecules Br2 = __________________ g Br2
8. 0.05 L CO = ______________ molecules CO
9. How many atoms of uranium are in 5.00 g of uranium?
10. What is the volume of 2.6 g of nitrogen gas?
11. What is the mass of 1.0 x 1024 molecules of sulfur dioxide?
12. What is the mass of Neon in a sign when the gas tubes have a total volume of 166 mL?
Unit 6 combined student notes
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STOICHIOMETRY
Stoichiometry :
In many ways, stoichiometry is the backbone of the most practical part of chemistry; it helps us relate actual
quantities (measured by mass or volume) of reactants to products in a chemical reaction. It’s a lot like using a
chocolate chip cookie recipe to produce a certain number of cookies of a given size. You have to put the
ingredients together in the proper proportions to make a cookie that tastes good!
Consider the following balanced equation:
2Na + Cl2  2NaCl
We can look at this equation in two different ways:
1.
at the atomic scale:
2. at the real-world, measurable scale:
Dimensional analysis can be used to relate the moles of reactants and products in a chemical reaction to one
another. It is essentially a matter of determining the ratio of reactants and products by looking at the
coefficients in the balanced chemical equation. In the reaction above, the ratio of sodium to chlorine is 2:1.


the ratio of sodium to sodium chloride is
the ratio of chlorine to sodium chloride is
If you’re starting out with 0.650 moles of sodium and you want to know how many moles of chlorine you need
to fully react with the sodium, you can use dimensional analysis:
0.650 mol Na
x
1 mol Cl2
=
2 mol Na
If you want to produce 6.4 moles of sodium chloride, how many moles of chlorine do you need to use?
Because we can literally measure out the moles of sodium by weighing it (one mole would have a mass of
22.99g) and we can measure out the moles of chlorine by volume (each mole of chlorine gas would have a
volume of 22.4 L), we can literally put the reactants together in their proper proportions. This makes
stoichiometry an incredibly powerful but very simple tool - one that you must master in order to be successful
in chemistry!
When solving a stoichiometry problem, you’ll always be given a quantity of either a product or reactant that
you have to relate to a quantity of another product or reactant. To do this, you’ll use the balanced chemical
equation as well as the mole equivalencies you’ve been using in mole problems. These equivalencies are:



1 mole =
1 mole =
1 mole =
Unit 6 combined student notes
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The equivalencies you’ll use depend on the specific problem, but the logic is always the same:
quantity of given  moles of given  moles of unknown  quantity of unknown
Use the “new and improved” mole map shown below to help you map out a problem-solving plan for any
stoichiometry problem.
liters of
substance A
grams of
substance A
liters of
substance B
moles of
substance A
moles of
substance B
particles of
substance A
grams of
substance B
particles of
substance B
Given the following equation:
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l)
What mass of oxygen will react with 36.0 g of methane (CH 4)?
36.0 g CH4 x 1 mol CH4
16.04 g CH4
x
2 mol O2
1 mol CH4
x
32.00 g O2
=
1 mol O2
What mass of water would be produced when oxygen reacts with the 36.0 g of methane?
How many liters of carbon dioxide would be produced from the reaction if 44.0 g of oxygen are consumed?
What volume of methane is needed to produce 26.8 g of water?
How many molecules of methane are needed to produce 2.4 moles of carbon dioxide?
Unit 6 combined student notes
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Limiting Reactant/Reagent Problems:




a stoichiometry problem where quantities of reactants are not available in the EXACT molar ratios
one reactant is “in excess” and the other is the “limiting reactant”
given mass/volume/moles of more than one reactant
you find the actual amount of product(s) produced by determining which reactant LIMITS the
reaction
Steps for Identifying Limiting Reactants:
1.
Write the balanced chemical equation.
2.
Solve two (or more) stoichiometry problems – one for each reactant given.
3.
Compare the results from step #2 above. The reactant that would produce the smaller amount
of product is the limiting reactant; the other reactant is in excess.
or
1.
2.
3.
Write the balanced chemical equation.
Find the number of moles of each reactant given.
Compare the ratio of moles from step #2 above to the ratio of moles in the balanced chemical
equation from step #1.
Example Problem:
What is the limiting reactant when 1.5 moles of hydrogen react with 0.5 moles of oxygen to produce water?
Example Problem:
When 2.60 g of hydrogen react with 10.20 g of oxygen, what mass of water will be produced? What is the
limiting reactant?
Unit 6 combined student notes
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Percent Yield Problems




the amount of product produced from a reaction based on a calculation is the expected yield
the amount of product that is really obtained during the reaction (in a lab or industrial setting) is the
actual yield
reasons for differences between actual and expected yield:
1. some reactants didn’t fully react
2. there were side reactions that differ from the reaction on which the expected yield was based
3. some product was lost during recovery or transfer
it is useful to determine what percent of the expected yield was actually obtained
percent yield =
actual yield
expected yield
x 100%
Example Problem:
A reaction occurs between 500.0 g of carbon disulfide and an excess amount of oxygen gas. The mass of
sulfur dioxide produced is 768 g. What is the percent yield for this reaction?
Step 1. Write the balanced chemical equation.
Step 2. Determine the expected yield using stoichiometry
Step 3. Compare the actual yield to the expected yield using the percent yield equation.
Unit 6 combined student notes
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