Download 1 mol C 4 H 10 - GW

Ch. 3 Stoichiometry:
Calculations with Chemical
Formulas
Law of Conservation of Mass
• Atoms are neither created nor destroyed
during any chemical reaction. Atoms are
simply rearranged.
Stoichiometry
• The quantitative nature of chemical
formulas and chemical reactions
Reactants
• The chemical formulas on the left of the
arrow that represent the starting
substances
2H2 + O2  2H2O
Reactants
Products
• The substances that are produced in the
reaction and appear to the right of the
arrow
2H2 + O2  2H2O
Products
• Because atoms are neither created nor
destroyed in any reaction a chemical
equation must have the same number of
atoms of each element on either side of
the arrow
Balancing Chemical Equations
CH4 + O2  CO2 + H2O
C=1
C=1
H=4
H=2
O=2
O=3
Balancing Chemical Equations
CH4 + O2  CO2 + 2H2O
C=1
C=1
H=4
H=2 X 2 =4
O=2
O=3
Balancing Chemical Equations
CH4 + O2  CO2 + 2H2O
C=1
C=1
H=4
H=2 x 2 = 4
O=2
O= 2 + 2x1 = 4
Balancing Chemical Equations
CH4 + 2O2  CO2 + 2H2O
C=1
C=1
H=4
H=2 x 2 = 4
O=2 x 2 = 4 O= 2 + 2x1 = 4
Combustion Reactions
• Rapid reactions that produce a flame.
• Most combustion reactions involve O2 as a
reactant
• Form CO2 and H2O as products
Combustion Reactions
C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l)
C= 3
C=1 X 3 = 3
H=8
H=2 X 4 = 8
O= 2 X 5 = 10
O=(2 X3)+(1X4)=10
Combination Reactions
(synthesis)
• 2 or more substances react to form 1
product.
Combination Reactions
(synthesis)
2Mg(s) + O2(g) 2MgO(s)
Mg=1 x 2=2
Mg= 1 x 2=2
O= 2
O=1 x 2 = 2
Decomposition Reaction
• 1 substance undergoes a reaction to
produce 2 or more substances
Decomposition Reaction
CaCO3 (s)  CaO (s) + CO2(g)
Ca=1
Ca=1
C=1
C=1
O=3
O=1+2=3
3 Methods of Measuring
• Counting
• Mass
• Volume
Example 1
• If 0.20 bushel is 1 dozen apples, and a
dozen apples has a mass of 2.0 kg, what
is the mass of .050 bushel of apples?
Example 1
• Count: 1 dozen apples = 12 apples
• Mass: 1 dozen apples = 2.0 kg apples
• Volume: 1 dozen apples = 0.20 bushels
apples
Conversion Factors:
• 1 dozen
2.0 k.g
1 dozen
12 apples
1 dozen
0.20 bushels
Example 1
• 0.50 bushel x 1 dozen
x 2.0 kg =
0.20 bushel 1 dozen
= 5.0 kg
Avogadro’s Number
• Named after the Italian scientist Amedo
Avogadro di Quaregna
• 6.02 x 10 23
Mole (mol)
• 1 mol = 6.02 x 10 23 representative
particles
• Representative particles: atoms,
molecules ions, or formula units (ionic
compound)
Mole (mol)
• Moles= representative x
particles
1 mol
6.02 x 10 23
Example 2 (atoms  mol)
• How many moles is 2.80 x 10 24 atoms of
silicon?
Example 2
• 2.80 x 10 24 atoms Si x 1 mol Si
6.02 x 10 23 atoms Si
= 4.65 mol Si
Example 3 (mol  molecule)
• How many molecules of water is 0.360
moles?
Example 3
• 0.360 mol H2O x 6.02 x 10 23 molecules H2O
1 mol H2O
=2.17 molecules H2O
The Mass of a Mole of an Element
• The atomic mass of an element expressed
in grams = 1 mol of that element = molar
mass
Molar mass S
Molar mass Hg
Molar mass C
Molar
mass Fe
6.02 x 10 23 atoms S
6.02 x 10 23 atoms Hg
6.02 x 10 23 atoms C
6.02 x 10 23 atoms Fe
Example 4 (mol  gram)
• If you have 4.5 mols of sodium, how much
does it weigh?
Example 4
• .45 mol Na x 23 g Na = 10.35 g Na = 1.0 x 10 2 g Na
1 mol Na
Example 5 (grams  atoms)
• If you have 34.3 g of Iron, how many
atoms are present?
Example 5
• 34.3 g Fe x 1 mol Fe x 6.02 x 10 23 atoms
55.8 g Fe
1 mol Fe
=3.70 x 10 23 atoms Fe
The Mass of a Mole of a
Compound
• To find the mass of a mole of a compound
you must know the formula of the
compound
• H2O  H= 1 g x 2
O= 16 g
18 g = 1 mole = 6.02 x 10 23
molecules
Example 6 (gram  mol)
• What is the mass of 1 mole of sodium
hydrogen carbonate?
Example 6
•
•
•
•
•
•
Sodium Hydrogen Carbonate = NaHCO3
Na=23 g
H=1 g
C=12 g
O=16 g x3
84 g NaHCO3 = 1 mol NaHCO3
Mole-Volume Relationship
• Unlike liquids and solids the volumes of
moles of gases at the same temperature
and pressure will be identical
Avogadro’s Hypothesis
• States that equal volumes of gases at the
same temperature and pressure contain
the same number of particles
• Even though the particles of different
gases are not the same size, since the gas
particles are spread out so far the size
difference is negligible
Standard Temperature and
Pressure (STP)
• Volume of a gas changes depending on
temperature and pressure
• STP= 0oC (273 K)
101.3 kPa (1 atm)
Standard Temperature and
Pressure (STP)
• At STP, 1 mol = 6.02 X 1023 particles =
22.4 L of ANY gas= molar volume
Conversion Factors
• AT STP
• 1 mol gas
22.4 L gas
22.4 L gas
1 mol gas
Example 7
• At STP, what volume does 1.25 mol He
occupy?
Example 7
• 1.25 mol He x 22.4 L He = 28.0 L He
1 mol He
Example 8
• If a tank contains 100. L of O2 gas, how
many moles are present?
Example 8
• 100. L O2 X 1 mol O2 = 4.46 mol O2
22.4 L O2
Calculating Molar Mass from
Density
• The density of a gas at STP is measured
in g/L
• This value can be sued to determine the
molar mass of gas present
Example 9
• A gaseous compound of sulfur and oxygen
has a density of 3.58 g/L at STP. Calculate
the molar mass.
Example 9
• 1 mol gas x 22.4 L gas X 3.58 g gas =
1 mol gas 1 L gas
Molar Mass= 80.2 g
Percent Composition
• The relative amounts of the elements in a
compound
• These percentages must equal 100
Percent Composition
• %element = mass of element x 100
mass of compound
Example 10
• Find the mass percentage of each element
present in Al2 (CO3)3
Example 10
•
•
•
•
Al2(CO3)3
Al= 27 g x 2 = 54 g / 234 g x 100=23%
C= 12 g x 3 = 36 g/ 234 g x 100= 15%
O = 16 g x 9 = 144 g / 234 g x 100=62%
234 g Al2(CO3)3
Empirical Formula
• The simplest whole number ratio of atoms
in a compound
• The formula obtained from percentage
composition
Ex CH , CH4, H2O, C3H8
NOT C2H4, or C6H12O6 these could be
simplified
Example 11
• Mercury forms a compound with chlorine
that is 73.9% mercury and 26.1% chlorine
by mass. What is the empirical fromula.
Example 11
Assume that you have 100 grams of the
compound therefore:
Hg = 73.9 %  73.9 g
Cl= 26.1%  26.1 g
Example 11
Step 2: Change grams of your compound to
moles
Hg = 73.9 g x 1 mol =0.368 mol Hg
200.6g
Cl= 26.1 g x 1 mol = 0.735 mol Cl
35.5 g
Example 11
Step 3: Find the lowest number of moles
present
Hg = 73.9 g x 1 mol =0.368 mol Hg
200.6g
Cl= 26.1 g x 1 mol = 0.735 mol Cl
35.5 g
0.368 < 0.735
Example 11
Step 4: Divide by the lowest number of
moles to obtain whole numbers
Hg = 73.9 g x 1 mol = 0.368 mol = 1
200.6g 0.368 mol
Cl= 26.1 g x 1 mol = 0.735 mol= 1.99=2
35.5 g 0.368 mol
Example 11
Step 5: Put the whole numbers into the
empirical formula
Hg = 73.9 g x 1 mol = 0.368 mol = 1
200.6g 0.368 mol
Cl= 26.1 g x 1 mol = 0.735 mol= 1.99=2
35.5 g 0.368 mol
HgCl2
Molecular Formulas
• The subscripts in the molecular formula of
a substance are always a whole-number
multiple of the corresponding subscripts in
its empirical formula
• We can obtain the molecular formula from
the empirical formula IF we know the
molecular weight of the compound.
Example 12
• The empirical formula of ascorbic acid is
C3H4O3. The molecular weight of ascorbic
acid is 176 amu. Determine the molecular
formula.
Example 12
Step 1: First determine the molecular weight
of the empirical formula
C 3 H 4O 3
C= 12 amu x 3
H= 1 amu x 4
O= 16 amu x 3
88 amu
Example 12
Step 2: Divide the molecular weight of the
molecular formula by the molecular weight
of the empirical formula
C3H4O3
176 amu = 2
C= 12 amu x 3
88 amu
H= 1 amu x 4
O= 16 amu x 3
88 amu
Example 12
Step 3: Multiply the empirical formula by the
number calculated in step 2
176 amu = 2
88 amu
(C3H4O3) x 2 = C6H8O6
Quantitative Information from a
Balanced Equation
2 H2 (g) +
O2 (g)

2 H2O (l)
2 molecules 1 molecule
2 molecules
Or since we can’t count out 2 molecules
2 mol
1 mol
2 moles
The coefficients in a chemical reaction can be
interpreted as either the relative number of
molecules (formula units) involved in the
reaction OR the relative number of moles
Example 13 (mol mol)
2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)
How many moles of O2 do you need to react
with 5 moles of C4H10?
Example 13
2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)
How many moles of O2 do you need to react
with 5 moles of C4H10?
5 mol C4H10 x 13 mol O2 = 32.5 mol O2
2 mol C4H10
Example 14 (gg)
2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)
How many grams of O2 do you need to react
with 50.0 g of C4H10?
Example 14 (gg)
2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)
How many grams of O2 do you need to react
with 50.0 g of C4H10?
50.0 g C4H10 x 1 mol C4H10 x 13 mol O2 x 32 g O2=179 g O2
58 g C4H10
2 mol C4H10 1 mol O2
Limiting Reactants (Reagents)
• The reactant that is completely consumed
• It determines or limits the amount of
product that forms
• The other reactant(s) are called excess
reagents
Example 14 (limiting reactants)
2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)
If you have 25.0 g O2 and 25.0 g of C4H10,
what is the limiting reactant?
Example 14 (limiting reactants)
• 2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)
25.0 g C4H10 x 1 mol C4H10 x 8mol CO2 = 1.72 mol CO2
58 g C4H10
2 mol C4H10
25.0 g C4H10 x 1 mol C4H10 x 8 mol CO2 = 0.481 mol CO2
32 g O2
13 mol O2
0.481 mol < 1.72 mol C4H10 is the limiting
reactant
Theoretical Yield
• The quantity of the product that is
calculated to form when all of the limiting
reactant reacts.
Example 15 (Theoretical Yield)
• 2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)
25.0 g C4H10 x 1 mol C4H10 x 8mol CO2 = 1.72 mol CO2
58 g C4H10
2 mol C4H10
25.0 g C4H10 x 1 mol C4H10 x 8 mol CO2 = 0.481 mol CO2
32 g O2
13 mol O2
0.481 mol < 1.72 mol C4H10 is the limiting
reactant Calculate the theoretical yeild
Example 15 (Theoretical Yield)
25.0 g C4H10 x 1 mol C4H10 x 8 mol CO2 = 0.481 mol CO2
32 g O2
13 mol O2
0.481 mol CO2 X 44 g CO2 = 21.2 g CO2
1 mol CO2
If all of the limiting reactant (25.0 g C4H10) reacts than 21.2
g of CO2 will form.
Percent Yield
Percent Yield = Actual Yield X 100
Theoretical yield
Example 15 (% Yield)
• A student calculates that they should
theoretically make 105 g of iron in an
experiment. When they perform the
experiment only 87.9 g of iron were
produced. What is the percent yield?
Example 15 (% Yield)
A student calculates that they should
theoretically make 105 g of iron in an
experiment. When they perform the
experiment only 87.9 g of iron were
produced. What is the percent yield?
87.9 g / 105 g x 100 = 83.7%
(actual)
(theoretical)