8.3 Molecular Orbitals How are atomic and molecular orbitals related?
Ch01 Chemistry and the Atomic Molecular View of Matter
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Transcript of Ch01 Chemistry and the Atomic Molecular View of Matter
Chapter 1: Chemistryand the
Atomic/MolecularView of Matter
Chemistry: The Molecular Nature of Matter, 6E
Jespersen/Brady/Hyslop
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Why Study Chemistry? In every aspect of our modern life
Long life batteries Materials & miniaturization
Cell phones/pagers Laptops
Synthetic fibers Dyes CDs/DVDs—silicon wafers Medications DNA sequencing
Touches all areas of science2
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Chemistry and the SciencesChemistry Study of matter & its transformations Seeks answers to fundamental
questions about: What makes up materials that compose our
world How composition affects properties of
substances How substances change when they interact
with each other = Chemical Reactions
3
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Chemistry and the SciencesChemistry Seeks to understand:
Underlying structures of matter Forces that determine properties that we
observe Apply this knowledge to:
Create new materials not found in nature Understand fundamental biological
processes
4
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Scientific Method Approach to gathering information &
formulating explanations. Scientists perform experiments in
laboratories under controlled conditions1.Make observations/collect data
Empirical fact Something we see, hear, taste, feel, or smell Something we can measure in laboratory Organize data so we can see relationships
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Scientific Method2. Law or Scientific Law
Broad generalization Based on results of many experiments Only states what happens Doesn’t explain why they happen
3. Hypothesis Mental picture that explains observed laws Tentative explanation of data Make predictions Devise experiments to test Go back to laboratory & perform
6
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Scientific Method4. Theory
Tested explanation of how nature behaves
Devise further tests Depending on
results, may have to modify theory
Can never prove theory is absolutely correct
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Scientific MethodEx. Study gases Discover Volume (V) of gas depends on
Pressure (P) Temperature (T) Amount (n)
Data Recorded observations of relationship between V, P, T &
n Law
R = constant Kinetic Theory of Gases
Explains gas behavior (Ch 11)
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nRTPV
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Atomic Theory Most significant theoretical model of
natureAtoms
Tiny submicroscopic particles Make up all chemical substances Make up everything in Macroscopic world Smallest particle that has all properties of
given element Composed of:
Electrons Neutrons Protons
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Matter & Its ClassificationsMatter
Anything that has mass & occupies spaceMass
How much matter given object has Measure of object’s momentum, or resistance
to change in motionWeight
Force with which object is attracted by gravityEx. Mass vs. Weight
Astronaut on moon & on earth Weight on moon = 1/6 weight on earth Same mass regardless of location
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
MatterChemical Reactions
Transformations that alter chemical compositions of substances
Decomposition Chemical reaction where 1 substance
broken down into 2 or more simpler substances
Ex.
11
Molten sodium chloride
Sodium metal
+ chlorine gas
Electric current
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Elements Substances that can’t be decomposed into
simpler materials by chemical reactions Substances composed of only 1 type of atom Simplest forms of matter that we can work
with directly More complex substances composed of
elements in various combinations
12diamond = carbon gold sulfur
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Chemical Symbols for ElementsChemical Symbol
One or two letter symbol for each element name First letter capitalized, second letter lower case
Ex. C = carbon S = sulfur Ca = calcium Ar = argon Br = bromine H = hydrogen Cl = chlorine O = oxygen
Used to represent elements in chemical formulas
Ex. Water = H2O Carbon dioxide = CO2
Most based on English name Some based on Latin or German names
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Chemical SymbolsEnglish Name
Chemical Symbol
Latin Name
Sodium Na NatriumPotassium K KaliumIron Fe FerrumCopper Cu CuprumSilver Ag ArgentumGold Au AurumMercury Hg HydrargyrumAntimony Sb StibiumTin Sn StanniumLead Pb PlumbumTungsten W Wolfram
(German) 14
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Compound Formed from 2 or more atoms of
different elements Always combined in same fixed ratios
by mass Can be broken down into elements by
some chemical changes Ex. Water decomposed to elemental
hydrogen & oxygenMass of oxygen =
8 × mass of hydrogen
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Pure Substance vs. MixturePure substances
Elements and compounds Composition always same regardless of source
Mixture Can have variable compositions Made up of two or more substancesEx. CO2 in water—varying amounts of “fizz” in
soda 2 broad categories of mixtures:
Heterogeneous Homogeneous
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Homogeneous Mixtures Same properties throughout sample Solution
Thoroughly stirred homogeneous mixture Ex. Liquid solution
Sugar in water Gas solution
Air Contains nitrogen, oxygen, carbon
dioxide & other gases Solid solution
US 5¢ coin – Metal Alloy Contains copper & nickel metals
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Is honey a mixture?
18
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Heterogeneous Mixtures 2 or more regions of different properties Solution with multiple phases Separate layersEx.
Salad dressing Oil & vinegar
Ice & water Same composition 2 different physical states
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Physical Change No new substances formed Substance may change state or the
proportionsEx. Ice melting Sugar or salt dissolving Stirring iron filings & sulfur together
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Chemical Changeor Chemical Reaction
Formation of new substance or compound Involves changing chemical makeup of
substances New substance has different physical
properties Can’t be separated by physical means
Ex. Fool’s gold Compound containing sulfur & iron
No longer has same physical propertiesof free elements
Can’t be separated using magnet21
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning Check:
Chemical
Physical
Magnesium burns when heated
Magnesium metal tarnishes in air
Magnesium metal melts at 922 K
Grape Kool-aid lightens when water is added
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For each of the following, determine if it represents a Chemical or Physical Change:
XX
XX
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Classification of Matter
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning Check: ClassificationHot
CocoaIce(H2O
)
White Flour
Table Salt (NaCl)
Pure substanceElementCompoundMoleculeHeterogeneous
MixtureHomogeneous
Mixture
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XXX
X X
X
X
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Law of Definite Proportions In given compound, elements always
combine in same proportions by mass. Ratio of masses of each element is fixed
for given compound Implication:
Each atom has fixed specific mass Ex. Fool’s gold, pyrite, iron (III) sulfide
Mass ratio always 1.00 g of Iron to 0.574 g of Sulfur
Ex. Water Mass ratio always: 8 g O to 1 g H
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Law of Conservation of Mass No detectable gain or loss of mass
occurs in chemical reactions. Mass is conserved.Implication: Reactions involve reorganization of
materials.
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning Check: Chemical Laws Magnesium burns in oxygen to form
magnesium oxide. If 16.88 g of Mg are consumed and 28.00 g of MgO are produced, what mass of oxygen was consumed?
28.00 g – 16.88 g = 11.12g O
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning Check: Chemical Laws In a sample of MgO, there are 16.89 g Mg and
11.11 g O. What mass of O would there be in a sample that contains 2.00 g of Mg?
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X = 1.32 g O
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Dalton’s Atomic TheoryJohn Dalton Developed underlying theory to explain
Law of Conservation of Mass Law of Definite Proportions
Reasoned that if atoms exist, they have certain properties
Dalton’s Atomic Theory1.Matter consists of tiny particles called atoms.
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Dalton’s Atomic Theory (cont)2. Atoms are indestructible.
In chemical reactions, atoms rearrange but do not break apart.
3. In any sample of a pure element, all atoms are identical in mass & other properties.
4. Atoms of different elements differ in mass & other properties.
5. In given compound, constituent atoms are always present in same fixed numerical ratio.
30
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Proof Of Atoms Early 1980’s, use
Scanning Tunneling Microscope (STM)
Surface can be scanned for topographical information
Image for all matter shows spherical regions of matter Atoms
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STM of palladium
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
How Do We Visualize Atoms? Atoms represented by
spheres Different atoms have
different colors Standard scheme given in
Fig. 1.11 is represented on the right.
32
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Molecules Atoms combine to form more complex
substances Discrete particles Each composed of 2 or more atomsEx.
Molecular oxygen, O2
Carbon dioxide, CO2 Ammonia, NH3 Sucrose, C12H22O11
33
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Chemical Formulas Specify composition of substance Chemical symbols
Represent atoms of elements present Subscripts
Given after chemical symbol Represents relative numbers of each type of
atomEx.
Fe2O3 : iron & oxygen in 2:3 ratio
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Chemical FormulasFree Elements
Element not combined with another in compounds
Just use chemical symbol to representEx. Iron Fe Neon Ne
Sodium Na AluminumAl
Diatomic Molecule Molecules composed of 2 atoms each Many elements found in nature
Ex. Oxygen O2 Nitrogen N2
Hydrogen H2 Chlorine Cl235
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Depicting Molecules Want to show:
Order in which atoms are attached to each other
3-dimensional shape of molecule Three ways of visualizing molecules:
1. Structural formula2. Ball-and-Stick model3. Space filling model
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
1. Structural Formulas Use to show how atoms are attached
Atoms represented by chemical symbols Chemical bonds attaching atoms indicated
by lines
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H O H H C H
H
HH2Owater
CH4methane
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
3-D Representations of Molecules
Use touching spheres to indicate molecules Different colors indicate different elements Relative size of spheres reflects differing sizes
of atoms
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Hydrogen
molecule,
H2
Oxygen molecule
,O2
Nitrogen molecule
N2
Chlorine molecule,
Cl2
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
2. “Ball-and-Stick” Model Spheres = atoms Sticks = bonds
39
Chloroform, CHCl3
Methane, CH4
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
3. “Space-Filling” Model Shows relative sizes of atoms Shows how atoms take up space in molecule
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Chloroform, CHCl3
Methane CH4
Water H2O
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
More Complicated Molecules Sometimes formulas contain
parentheses How do we translate into a structure?Ex. Urea, CO(NH2)2
Expands to CON2H4 Atoms in parentheses appear twice
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Ball-and-stick model Space-filling
model
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Hydrates Crystals that contain water moleculesEx. plaster: CaSO4∙2H2O calcium sulfate dihydrate
Water is not tightly held Dehydration
Removal of water by heating Remaining solid is anhydrous (without water)
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Blue = CuSO4
•5H2OWhite = CuSO4
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Counting Atoms1. Subscript following chemical symbol
indicates how many of that element are part of the formula
No subscript implies a subscript of 1.2. Quantity in parentheses is repeated a
number of times equal to the subscript that follows.
3. Raised dot in formula indicates that the substance is a hydrate
Number preceding H2O specifies how many water molecules are present.
43
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Counting AtomsEx. 1 (CH3)3COH Subscript 3 means 3 CH3 groups
So from(CH3)3, we get 3 × 1C = 3C 3 × 3H = 9H
#C = 3C + 1C = 4 C#H = 9H + 1H = 10 H#O = 1 OTotal # of atoms = 15 atoms
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Counting AtomsEx. 2 CoCl2 · 6H2O The dot 6H2O means you multiple both
H2 & O by 6 So there are:
#H 6 × 2 = 12 H#O 6 × 1 = 6 O#Co 1 × 1 = 1 Co#Cl 2 × 1 = 2 ClTotal # of atoms = 21 atoms
45
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!
a. Na2CO3
b. (NH4)2SO4
c. Mg3(PO4)2
d. CuSO4∙5H2Oe. (C2H5)2N2H2
46
a. ___Na, ___ C, ___ O
b. ___N, ___H, ___S, ___Oc. ___Mg, ___P, ___Od. ___Cu, ___S, ___O, ___He. ___C, ___H, ___N
32 12 8 1 43 2 81 1 9 10
Count the number of each type of atom in the chemical formula given below
4 12 2
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Dalton’s Atomic Theory We now have the tools to explain this
theory & its consequences All molecules of compound are alike &
contain atoms in same numerical ratio.Ex. Water, H2O
Ratio of oxygen to hydrogen is 1 : 21 O atom : 2 H atoms in each moleculeO weighs 16 times as much as H1 H = 1 mass unit1 O = 16 mass units
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Atoms in Fixed Ratios by Mass
48
For water in general: mass O = 8 mass H Regardless of amount of water present
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Dalton’s Atomic TheorySuccesses: Explains Law of Conservation of Mass
Chemical reactions correspond to rearranging atoms.
Explains Law of Definite Proportions Given compound always has atoms of same
elements in same ratios. Predicted Law of Multiple Proportions
Not yet discovered Some elements combine to give 2 or
more compoundsEx. SO2 & SO3
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Law Of Multiple Proportions When 2 elements form more than one
compound, different masses of one element that combine with same mass of other element are always in ratio of small whole numbers. Atoms react as complete (whole)
particles. Chemical formulas
Indicate whole numbers of atoms Not fractions
50
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Using Law Of Multiple Proportions sulfur sulfur
dioxide trioxideMass S 32.06 g 32.06 gMass O 32.00 g 48.00 g
Use this data to prove law of multiple proportions
51
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Law of Multiple ProportionsCompound Sample
SizeMass of Sulfur
Mass of Oxygen
52
23
00360048
2
3 gg
SOSO
.
. in O in O
Ratio of
Sulfur dioxide
64.06 g 32.06 g
Sulfur trioxide
80.06 g
32.06 g
32.06 g 48.00 g
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Molecules Small and Large So far we’ve only discussed small
molecules Some are very large, especially those
found in nature Same principles apply to allEx. DNA - short segment
53
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
How Do We Know Formulas? Hardly “out of the blue” Don’t know formula when compound 1st
isolated Formulas & structures backed by
extensive experimentation Use results of experiments to determine
Formula Chemical reactivity
Molecular Shape Can speculate once formula is known Determine from more experiments
54
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Visualizing Mixtures Look at mixtures at atomic/molecular level Different color spheres stand for 2
substances a. Homogeneous mixture/solution – uniform
mixingb. Heterogeneous mixture – 2 phases
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a. b.
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Chemical Reactions When 1 or more substances
react to form 1 or more new substances
Ex. Reaction of methane, CH4, with oxygen, O2, to form carbon dioxide, CO2, & water, H2O.Reactants = CH4 & O2
Products = CO2 & H2O How to depict?
Words too long Pictures too awkward
56
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Chemical Equations Use chemical symbols & formulas to
represent reactants & products. Reactants on left hand side Products on right hand side Arrow () means “reacts to yield”
Ex. CH4 + 2O2 CO2 + 2H2O Coefficients
Numbers in front of formulas Indicate how many of each type of
molecule reacted or formed Equation reads “methane & oxygen
react to yield carbon dioxide & water”57
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Conservation of Mass in Reactions Mass can neither be created nor destroyed
This means that there are the same number of each type of atom in reactants & in products of reaction If # of atoms same, then mass also same
58
CH4 + 2O2 CO2 + 2H2O 4 H + 4O + C = 4 H + 4O
+ C
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Balanced Chemical Equation
Ex. 2C4H10 + 13O2 8CO2 + 10H2O
59
4 C & 10 H per
molecule2 O per molecul
e
2 H & 1 O per
molecule1 C & 2 O
per molecule
Subscripts Define identity of substances Must not change when equation
is balanced
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Balanced Chemical EquationEx. 2C4H10 + 13O2 8CO2 + 10H2O
60
2 molecules of C4H10
13 molecules of O2
10 molecules of C4H10
8 molecules of CO2Coefficients
Number in front of formulas Indicate number of molecules of each type Adjusted so # of each type of atom is
same on both sides of arrow Can change
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Balanced Chemical Equations How do you determine if an equation is
balanced? Count atoms Same number of each type on both sides of
equation? If yes, then balanced If no, then unbalanced
Ex. 2C4H10 + 13O2 8CO2 + 10H2O Reactants Products2×4 = 8 C8×1 = 8 C2×10 = 20 H 10×2 = 20 H13×2 = 26 O (8×2)+(10×1)= 26 O
61
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning CheckFe(OH)3 + 2 HNO3 Fe(NO3)3 + 2 H2O
Not Balanced Only Fe has same number of atoms
on either side of arrow. 62
Reactants ProductsFe 1 1
3 + (2×3) = 9 (3×3) + 2 = 11O3 + 2 = 5 (2×2) = 4H
2 3N
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!How many atoms of each element appear on each side of the arrow in the following equation? 4NH3 + 3O2 → 2N2 + 6H2O
63
Reactants ProductsN (4 × 1) = 4 (2 × 2) = 4O (3 × 2) = 6 (6 × 1) = 6H (4 × 3) = 12 (6 × 2) = 12
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!Count the number of atoms of each element on both sides of the arrow to determine whether the following equation is balanced.2(NH4)3PO4 + 3Ba(C2H3O2)2 → Ba3(PO4)2 + 6NH4C2H3O2
64
Reactants ProductsN (2 × 3) = 6 (6 × 1) = 6H (2×3×4)+(3×3×2) =
42(6×4) + (6×3) = 42O (2×4) + (3×2×2) =
20(2×4) + (6×2) = 20P (2 × 1) = 2 (2 × 1) = 2
Ba (3 × 1) = 3 (3 × 1) = 3