NaCl, saltNaCl, saltNaCl, saltNaCl, salt
Buckyball, CBuckyball, C6060
Ethanol, CEthanol, C22HH66OO
Just to Review…Please convert 12.4 fluid ounces to cm3
1 fl. oz. = 0.0295735 L
1 cm3 = 1 mL
Convert 4.5 centuries to seconds.
The Carbon-12 ScaleThe Carbon-12 Scale
The atomic mass of an element indicates how heavy, on average, an atom of an element is when compared to an atom of another element
– Unit is the atomic mass unit (amu)
The standard scale is based on the
carbon-12 isotope
•Mass of one 12C atom = 12 amu (exactly)•Note that 12C and C-12 mean the same thing
Isotopic Abundance
• Most elements exist in nature as a mixture of two or more isotopes.
• To determine the mass of an element, we must know the mass of each isotope and the atom percent of the isotopes (isotopic abundance)
• The mass spectrometer can determine the isotopic abundance & atomic mass.
Figure 3.1 – Mass Spectrometer
• A mass spectrometer is used to determine atomic masses
Figure 3.1 – Mass Spectrometer
Avogadro’s Number• A sample of any element with a mass equal to
its atomic mass contains the same number of atoms, NA, regardless of the identity of the element.
– NA = 6.022 X 1023 Avogadro’s #
• It represents the number of atoms of an element in a sample whose mass in grams is numerically equal to the atomic mass of the element.
Amadeo Avogadro & Avogadro’s Number
6.02214199 x 1023
(6.022 x 1023 to 4 s.f.)
Avogadro’s NumberAvogadro’s Number
There is Avogadro’s number of particles in a mole of ANY substance.There is Avogadro’s number of particles in a mole of ANY substance.
Amadeo AvogadroAmadeo Avogadro1776-18561776-1856
Examples:
• 6.022 x 1023 H atoms in 1.008 g atomic mass H = 1.008 amu
• 6.022 x 1023 S atoms in 32.07 g atomic mass S = 32.07 amu
Mole: A Perspective on SizeThe Green Pea Analogy:
A Dramatic Reading
Examples of Mole Quantities
EOS
1 mole of stars in the universe = 6.022 x 1023 stars1 mole of pennies = 6.022 x 1023 pennies
(beats the lottery!)
1 mole of glucose molecules = 6.022 x 1023 molecules1 mole of helium atoms = 6.022 x 1023 atoms1 mole of potassium ions (K+) = 6.022 x 1023 ions
Example:water - H2O
Molecular MassMolecular mass: the sum of the atomic masses of all atoms in a molecular formula.
- (units are amu or u)
2 Hydrogen atoms2(1.01 amu)
1 Oxygen atom
+ 16.00 amu
= 18.02 amu
Sample Problem: Glucose
Glucose - C6H12O6
= 6(12.01 u) + 12(1.01 u) + 6(16.00 u)= 180.18 u
EOS
Formula MassFormula mass is the sum of the masses of the atoms or ions present in a formula unit – the unit for an ionic formula
Na+ Cl-
Na+
Na+Cl-
Cl-
Cl-Na+
Crystal ofsodium chloride
One Na+ and one Cl– make a formula unit for sodium chloride
The mass of one formula unit is:= 22.99 amu + 35.45 amu= 58.44 amu
Sample Problem
Example 3.1 Determine the formula or molecular mass for each of the following:
CaI2 (NH4)2S
Al(NO3)3 C6H12O6
Mole--DefinitionChemistry is a quantitative
science—we need a “counting unit” aka the
1 mole = the amount of substance that contains as
many particles (atoms, molecules, formula units) as
there are in 12.0 g of 12C.
1 mole = the amount of substance that contains as
many particles (atoms, molecules, formula units) as
there are in 12.0 g of 12C.
MOLE!MOLE!
518 g of Pb, 2.50 mol
One-mole Amounts
AnalogiesWe can group items and count by grouping:
12 eggs = 1 dozen eggs like6.02 x 1023 items = 1 mole
We can also group items and
count by weighing:
Grass seed and nails —Purchased by the POUND, not by the item.
The molar mass, MM, in grams/mole, is numerically equal to the sum of the masses (in amu) of the atoms in the formula
***Molar mass is the mass of one mole of a particular substance.
Molar Masses of Some Substances
MOLECULAR WEIGHT VS. MOLECULAR WEIGHT VS. MOLAR MASSMOLAR MASS
MOLECULAR WEIGHT VS. MOLECULAR WEIGHT VS. MOLAR MASSMOLAR MASS
Molecular weight = sum of the atomic
weights of all atoms in the molecule (in amu)
Molar mass = molecular weight in grams/mole
We will use molar mass in all problems in this
chapter!!!!
Equivalencies1 mole of any substance contains Avogadro’s number of particles (and the mass on the periodic table expressed in grams).
1 mol of C = 12.01 g of C= 6.022 x 1023 atoms of C
1 mol of O2 = 32.00 g of O2= 6.022 x 1023 molecules
1 mol of NaCl = 58.44 g of NaCl= 6.022 x 1023 formula units of NaCl
The Mole and Reactions
Example: consider the formation of carbon dioxide
EOS
Answer: one doesn’t!
At the molecular level ...
Problem: how does one mass out a single carbon atom? Note that the mass in grams is ~2.00 x 10–23 g!
The Solution ...
EOS
For carbon, mass out:
2.0 × 10–23 g atom–1 × 6.0 × 1023 atoms mol–1
= 12 g C
Use a measurable amount – molar quantities
RememberRemember HON17 HON17 !!!!!!
Must memorize elements that exist as diatomic elements!!
Mole Conversions
Moles!
MassRepresentative
particles(atoms, molecules, formula
units, ions)
Mole Conversions
Use molar mass
Use Avogadro’s # (6.02 x 1023)
Practice problems1. Calcium carbonate, CaCO3, is the principal
mineral found in marble and limestone. How many moles are in 188.0 g of CaCO3?
Practice problems2. What is the mass, in grams, of 0.329 mol of
spearmint oil, C10H14O?
Practice problems3. Find the mass of a single lead atom.
Practice problems4. How many individual lead atoms are in a
1.000 g sample of this metal?
Practice problems5. (a) Calculate the number of moles of
aluminum in a solid cube that measures 3.40 cm on a side. (d=2.70 g/cm3).
(b) How many atoms of aluminum are in the same sample?
Practice problems6. (a) How many molecules of oxygen, O2, are
in 0.00100 grams of this gas?
(b) How many atoms?
Definition: Describes the proportion of elements in a compound using a percent
Equal to the mass of each element present in a 100 g sample of compound!
Percent Composition by Mass
Mass of elementPercent composition of an element = ×100
Mass of compound
Example: Sodium carbonate is a compound used in the manufacture of soap and glass. Determine the percent composition by mass of each element in this compound.
Example: Determine the percent by mass of water in Al2(SO4)3∙18H2O.
Example: Magnetite, Fe2O3, is one of the principal iron containing ores. How much elemental iron can be obtained from a metric ton (103 kg) of this ore, assuming 100 % recovery? (hint: first find % iron in Fe2O3)
In chemical analysis we determine the % by weight of each element in a given amount of pure compound and derive the EMPIRICAL or SIMPLEST formula.
PROBLEM: A compound of B and H is 81.10% B. What is its empirical formula?
Determining Formulas
Empirical Formula Calculations
Percent to massMass to molesDivide by Small
Multiply ‘til whole
% g moles empirical formulaDivide by smallest
mole value
Percent to mass: (always assume 100 g sample!)
A compound of B and H is 81.10% B. What is its empirical formula?
Mass to moles:
81.10 g B • 1 mol
10.81 g = 7.502 mol B81.10 g B •
1 mol10.81 g
= 7.502 mol B
18.90 g H • 1 mol
1.008 g = 18.75 mol H18.90 g H •
1 mol1.008 g
= 18.75 mol H
A compound of B and H is 81.10% B. What is its empirical formula?
Now, recognize that atoms combine in the ratio of small whole numbers.
Find the ratio of moles of elements in the compound by “dividing by small”
A compound of B and H is 81.10% B. What is its empirical formula?
But we need a whole number ratio!
Must multiply each ratio by the smallest integer available to obtain whole numbers
(multiply ‘til whole)
A compound of B and H is 81.10% B. What is its empirical formula?
Multiply ‘til Whole Hints
Common Possible Endings:
.33 x 3
.25 x 4
.67 x 3
.50 x 2
Sample Problem
Example: Bicarbonate of soda is used in
products like Alka-Seltzer and generally
relieves an upset stomach. Determine the
empirical formula of this compound based on
the following percent composition: 27.36% Na,
1.200% H, 14.30% C, 57.14% O.
Sample ProblemExample: A 25.00 gram sample of an orange compound contains 6.64 g of potassium, 8.84 g of chromium, and 9.52 g of oxygen. Find its empirical formula.
A compound of B and H is 81.10% B. Its empirical formula is B2H5. What is its molecular formula?
Is the molecular formula B2H5, B4H10,
B6H15, B8H20, etc.?
BB22HH66 is one example of this class of compounds. is one example of this class of compounds.
B2H6
We need to do an EXPERIMENT to find the MOLAR MASS.
Here the experiment gives 53.3 g/molCompare with the mass of B2H5 = 26.66 g/unit
Find the ratio of these masses.53.3 g/mol26.66 g/unit of B2H5
= 2 units of B2H5
1 mol
53.3 g/mol26.66 g/unit of B2H5
= 2 units of B2H5
1 mol
Multiply all of the subscripts by the ratio and obtain the molecular formula:
Molecular formula = B4H10
Sample ProblemExample : A certain compound has the empirical formula C2H4O. Its molar mass is about 90 g/mol. What is the molecular formula?
Sample ProblemExample: A hydrate of magnesium iodide has the formula MgI2 ∙ X H2O. To determine the value of X, a student heats a sample of the hydrate until all the water is gone. A 1.628 g sample of hydrate is heated to constant mass of 1.072 g. What is the value of X?
Writing and Balancing Chemical Equations
• All chemical reactions have two parts:
• ReactantsReactants - the substances you start with (on left side of arrow)
• Products Products - the substances you end up with (on right side of arrow)
• The reactants turn into the products.
Reactants Products
In a chemical reaction…• The way atoms are joined is changed.• Atoms aren’t created or destroyed; they just
combine together in new ways.• Can be described using sentences, symbols or
word equations:
Example:
Copper reacts with chlorine to form copper (II) chloride.
Copper + chlorine copper (II) chloride
Cu + Cl2 CuCl2
Symbols Used in Equations• The arrow separates the reactants from the
products; means “reacts” or “yields”
• The plus sign = “and”
• Subscripts are used to describe the number of atoms in a FORMULA.
• Coefficients are used to describe the number of molecules or formula units in the REACTION. They are the only things changed when balancing a reaction.
States of Matter• Solid – (s) after the formula
– Precipitate -- a solid formed in a reaction
• Gas--(g) after the formula
• Liquid—(l) after the formula
• Aqueous – (aq) after the formula - dissolved in water.
States of Matter
used after a product indicates a gas oSame as writing (g)
used after a product indicates a solid or precipitateoSame as writing (s)
Other Symbols used in Equations
• indicates a reversible reaction
• show that heat is supplied to the reaction
• is used to indicate a catalyst used or supplied, in this case, platinum.
heat ,
Pt
Chemical Equations
EOS
Example: consider the formation of water
H2(g) + O2(g) H2O(g)
Law of Conservation of Mass must be obeyed …
therefore, equations must be balanced.
Balancing Equations
Chemical “bookkeeping” of atoms involved in the reaction:
H2(g) + O2(g) H2O(g)
H – 2 O – 2Reactants
H – 2 O – 1Products
COEFFICIENTS must be added so reactant atoms EQUAL product atoms!
Note the imbalance in oxygen atoms
Hints & Tips for Balancing Equations
• Take one element at a time, working from left to right except for H and O. Save H for next to last and O for last.
• IF EVERYTHING BALANCES EXCEPT FOR O, and there is no way to balance O with a whole number, double all the coefficients and try again. (Because O is a diatomic element)
• (Shortcut) polyatomic ions that appear on both sides of the equation can be balanced as independent units!
Balancing Equations Practice
Balance the following chemical equation using the appropriate coefficients:
____ Al(s) + _____ Br2 (l) _____ Al2Br6 (s)
Balancing Equations Practice
Balance the following chemical equation using the appropriate coefficients:
____ Na3PO4 + ____ Fe2O3 ____ Na2O + ____ FePO4
Types of Reactions
• Synthesis (combination) reaction
• Decomposition reaction
• Single replacement reaction
• Double replacement reaction
• Combustion reaction
Types of ReactionsSynthesis or Combination
Equation in Symbols: A + B ABSample Equation:
2Cu (s) + O2 (g) 2 CuO (s)
Predicting Products:
Elements Compounds OR
Compounds
More Complex Compounds
Types of ReactionsDecomposition
Equation in Symbols: AB A + BSample Equation:
2 CuO (s) 2Cu (s) + O2 (g)
Predicting Products:Compounds Elements OR More Complex Compounds
Compounds
Types of ReactionsSingle Replacement
Equation in Symbols: A + BC AB + C• Metal replacing metal• Nonmetal replacing nonmetal
Sample Equation:
Mg (s) + CuCl2 (aq) Cu (s) + MgCl2 (aq)
Predicting Products:Cations replace cations (can also have anions replacing anions)
Types of ReactionsDouble Replacement – 2 ionic compounds
Equation in Symbols: AX + BY BX + AY
Sample Equation: 2AgNO3(aq) + CuCl2 (aq) Cu (NO3)2 (aq) + 2AgCl (s)
Predicting Products:Cations switch places; solid formed
(must be driving force)
Types of ReactionsCombustion
Equation in Symbols: CxHy + O2 CO2 + H2O
Sample Equation:
CH4(g) + O2 (g) CO2 (g) + H2O (l)
Predicting Products:
Hydrocarbons react to form CO2 and
H2O
Stoichiometric Equivalence and Reaction Stoichiometry
CS2 + 3O2 CO2 + 2 SO2
Interpretation in terms of moles:
1 mole of CS2 + 3 moles of O2 form:
1 mole of CO2 + 2 moles of SO2
Stoichiometric Equivalence and Reaction Stoichiometry
CS2 + 3O2 CO2 + 2 SO2
Conversion factors extracted from balanced equation:
1 mole of CS2 3 moles of O2
etc.
3 moles of O2 1 mole of CO2
These ratios are called MOLE RATIOS!
Stoichiometric EquivalentsCoefficients from a balanced chemical equation show molar equivalents of reactants and products
==> form conversion factors
In the formation of water:2 mol H2 = 1 mol O2
2 mol H2 = 2 mol H2O
1 mol O2 = 1 mol H2O
2H2 + O2 2H2O
Example: Using the equation below, determine:
CS2 + 3O2 CO2 + 2SO2
• the number of moles of oxygen required to react with 1.38 mol of carbon disulfide
• the number of moles of SO2 produced from 1.38 moles of carbon disulfide.
Example 3.13
For 2NH3 + H2SO4 (NH4)2SO4 determine:
a) the mass of product possible when 1.43 mol of NH3 are reacted with an excess of sulfuric acid.
b) the mass of NH3 required to react completely with 35.00 g of sulfuric acid.
c) the mass of sulfuric acid required to form 1000 grams of product.
Stoichiometry Diagram
Mass
Particles
Volume Mole Mole
Mass
Volume
Particles
Known Unknown
Substance A Substance B
1 mole = molar mass (g)Use coefficientsfrom balanced
chemical equation1 mole = 22.4 L @ STP
1 mole =
6.022 x
1023 partic
les
(atoms o
r molecu
les)
1 mole = 22.4 L @ STP
1 mole = 6.022 x 10 23 particles
(atoms or molecules)
1 mole =
molar m
ass (g
)
(gases) (gases)
Volume(liquids)
USE
DENSITY
Volume (liquids)
US
E
DE
NSI
TY
Example : How many milliliters of liquid water can be produced by the combustion of 775 mL of octane with oxygen? Assume that the volumes of the octane and the water are measured at 20oC where the densities are 0.7025 g/mL for octane and 0.9982 g/mL for water.
C8H18(l) + O2(g) CO2(g) + H2O(l)
Reaction Yields
Theoretical yieldTheoretical yield – predicted amount of product formed from the limiting reagent, based only on the stoichiometry of the reaction
If all worked perfectly ...
2H2(g) + O2(g) 2 H2O(g)Example: 1 mol H2 will produce 1 mol of water
Actual yield – amount of product produced
In practice, actual < theoretical: errors, poor technique, etc. ...
Limiting Reactants (Reagents)
Chemical reactant that is completely consumed in a reaction and therefore limits the quantity of
product formed.
**Depends on stoichiometry of reaction
Excess Reactant = Reactant left over when limiting reactant is used up
How many meals can be made from 105 sandwiches, 202 cookies, and 107 oranges?
1105 105
1
1202 101
2
1107 107
1
mealsandwiches x meals
sandwich
mealcookies x meals
cookies
mealoranges x meals
orange
Cookies limit the total number of whole
meals with excess sandwiches and
oranges
Steps for Determining Limiting Reactant
1. Write a balanced equation.2. Take first reactant, calculate theoretical
yield of desired product (in grams).3. Repeat #2 for second reactant.4. Compare results. Whichever reactant
gives the LEAST amount of the product is the limiting reactant and determines the theoretical yield. The other is in excess.
Sample Problem
CS2(l) + 3 O2(g) CO2(g) + 2 SO2(g)
Determine the theoretical yield of product (CO2) in grams if one starts with 1.20 mol of CS2 and 3.83
mol O2
Sample ProblemExample: For the following reaction, determine the theoretical yield of product (CO2) if one starts with 105 g of CS2 and 145 g of O2
CS2(l) + 3O2(g) CO2(g) + 2SO2(g)
Percent Yield
EOS
actual yiel
theoretipercen
cal yit yi
eldeld =
dx100
Reaction yields are expressed as a ratio in the form of a percentage:
ExampleFor the reaction, determine the theoretical yield if one starts with 1.20 g of antimony and 2.40 g of iodine.
2 Sb(s) + 3 I2(s) 2 SbI3(s)
Determine theoretical yield first:
ExampleFor the reaction, determine the theoretical yield if one starts with 1.20 g of antimony and 2.40 g of iodine.
2 Sb(s) + 3 I2(s) 2 SbI3(s)
Theoretical Yield = 3.17 g SbI3
If 3.00 g of product are actually formed, what is the percent yield?
Top Related