CYCLE 6
Developing Ideas
ACTIVITY 5: Conservation of Mass and the
Small Particle Model--KEY
Purpose
In Cycle 4 and Cycle 5 you investigated physical properties and changes
and observed that at the macroscopic and particle levels, physical changes
are accompanied by the conservation of mass.
In this cycle you investigated chemical properties and changes. Unlike
physical changes, you observed that new materials are produced in every
chemical change. Like physical changes, you observed that while the
volume may change, the mass of the system does not change. Similar to
physical changes, the Small Particle Model can account for your
observations of mass conservation. However, the Small Particle Model
describes chemical changes somewhat differently than physical changes. As
you learned in Activity 2, chemical changes involve changes in small
particle identity when atoms, smaller components of the small particles,
rearrange to form new small particles with unique properties. In this
activity we will use chemical equations to help us answer the following key
question:
How does the Small Particle Model explain
mass conservation during a chemical change?
Initial Ideas
Two students were discussing how they thought the Small Particle Model
would explain their observations that when they put an Alka-SeltzerTM
tablet in a flask sealed with a balloon, the mass of the system did not
change, even though the system increased in volume.
© 2007 PSET
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Cycle 6
I think that the number of
particles is the same during
chemical changes, just like
for physical changes.
Han
I think the number of atoms is
the same, but the number of
“particles” may change during
the chemical change.
Dave
Do you agree with Han or Dave or with neither? Why?
Possible student response: Han, because in Cycle 4 and 5 we saw that the number of
particles did not change.
Possible student response: Dave, because in this cycle we saw that atoms rearrange and
can change the number of particles. However the number of atoms does not change.
B Participate in a whole class discussion. Be prepared to share your ideas
with the rest of the class.
Collecting and Interpreting Evidence
Model Exploration #1: How does the Small Particle Model explain why
mass is conserved when heating lead iodide?
In Cycle 6 Activity 2 you observed that as you heated lead iodide, the
yellow solid changed into two materials, iodine, a purple solid that became
a purple gas when heated, and lead, a silvery solid. Although you did not
observe it directly, mass is conserved during this chemical change. The
mass of iodine produced plus the mass of lead produced equaled the mass
of the lead iodide that you started with.
STEP 1. Review the three x 10 million magnification Ultrascope views from
Activity 2 given below:
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Activity 5: Conservation of Mass and the SPM
STEP 2. Heating lead iodide produces lead and iodine. The smallest particle
of lead iodide is a formula unit containing one lead ion and two iodide ions.
The smallest particle of lead is an individual lead atom. The smallest
particle of iodine is a diatomic molecule, or two atoms of iodine bonded
together.
Draw a picture (using Small Particle Models like those above) that
shows what happens when one lead iodide formula unit is heated.
Since you are using pictures to represent lead iodide, lead, and iodine,
use an arrow symbol to represent “produces”.
Æ
1-
2+
+
1-
Use iodine’s position on the periodic table to determine the number of
electrons gained to form each iodide ion. What is the charge on each
iodide ion in lead iodide? Label each iodide ion with this charge.
Each iodine gains 1 electron and becomes an ion with 1- charge
Assuming that the electrons gained by both iodide ions were lost by the
lead ion, what must be the charge of the lead ion1 in lead iodide?
Label the lead ion with this charge.
For the compound formula unit to be neutral, the lead must lose 2 electrons, one to
each iodine atom.
1
Most transition metal atoms form 2+ ions when they lose two valence electrons from the
energy level equal to their period #. They may also lose electrons from the energy level
equal to their (period # -1), forming ions with various charges. Some examples are Fe3+, V5+,
and Mn7+. Lead, tin, and the other Group IV metalloids are similar to transition metals in
that they may form ions of different charges; however, they do so by losing one, two, or all
four of their valence electrons in the energy level equal to their period #.
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Cycle 6
STEP 5. You represented the decomposition of lead iodide using a picture
model. An easier way to represent chemical changes is to write a chemical
equation. In this case:
PbI2 (s) Æ Pb (s) + I2 (g)
How does the number of lead ions on the left side of the arrow
compare to the number of lead atoms on the right side of the arrow in
your drawing?
The numbers are equal; 1 lead ion and 1 lead atom.
How does the number of iodide ions on the left side of the arrow
compare to the number of iodine atoms on the right side of the arrow
in your drawing?
The numbers are equal; 2 iodine ions and 2 iodine atoms.
How does the number of “particles” (single atoms or atoms bonded
together in molecules or formula units) on the left side of the arrow
compare to the number of particles on the right side of the arrow?
They are not equal. There is one particle on the left and 2 particles on the right.
What are the connections between your model of the chemical change
and the chemical equation? How do they both represent the chemical
change you observed in Activity 2?
They both represent number of atoms and particles. The number of atoms/ions of each
element is the same before and after which corresponds to the left and right of the
chemical equation. They also show how the atoms have rearranged to form different
materials.
Model and Simulator Exploration #2: How does the Small Particle
Model explain why mass is conserved during rusting?
When steel wool rusts, iron atoms in the steel wool interact with oxygen
molecules in the air. Like lead, iron particles are single atoms of iron. Like
iodine, oxygen exists in nature as a diatomic molecule with two oxygen
atoms bonded together.
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Activity 5: Conservation of Mass and the SPM
STEP 1. Open the Cycle 6 Activity 5 Setup 1.
Earlier in the cycle we used the
Chemical Analyzer to study the
decomposition of compounds into
elements. The Chemical Reactor
works much like the Chemical
Analyzer, except that it can be used to
study
how
elements
combine
chemically to form compounds. In
this exploration, we will use the
Chemical Reactor Simulator to help us determine how iron and oxygen
combine chemically to form iron oxide (rust).
The scroll menu and corresponding legend below the Chemical Reactor
provide a list of possible materials that can be reacted. The materials in the
scroll menu can also be identified by scrolling over the colored dot.
STEP 2. Select iron from the first row of materials.
Move the mouse over the Chemical Reactor Input
Receptacle. When a black rectangle appears, click the
mouse button to deposit the iron particle into the
receptacle.
STEP 3. Click the mouse five more times to place a total of six atoms of iron
in the first input receptacle.
STEP 4. Select oxygen from the first row of materials.
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Cycle 6
Move the mouse over the Chemical Reactor Input
Receptacle. When a black rectangle appears, click the
mouse button to deposit the oxygen particle into the
receptacle.
STEP 5. Click the mouse five more times to place a total
of six atoms of oxygen in the second input receptacle. Six atoms of oxygen
represent three (3) diatomic molecules of oxygen.
STEP 6. Run the Chemical Reactor. Stop the simulator once something
appears in the output receptacle.
Look at the input receptacles. How many atoms of iron were used in
the chemical change? How many diatomic molecules of oxygen?
4 atoms of lead, 3 diatomic molecules of oxygen (or 6 oxygen atoms)
Look at the output receptacle. How many formula units of rust are
formed in the chemical change?
2
STEP 8. Activate the Select tool. To investigate the composition of rust,
double click on a particle in the output receptacle and an information
window will appear.
What is the chemical formula for iron oxide?
Fe2O3
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Activity 5: Conservation of Mass and the SPM
As you saw from the simulator, the formula unit of iron oxide is made up of
five atoms: two iron atoms bonded to three oxygen atoms. The chemical
bonding of iron oxide is very similar to aluminum oxide from the ionic
bonding model in Activity 4 homework.
Use oxygen’s position on the periodic table to determine the number of
electrons gained to form each oxide ion. What is the charge on each
oxide ion in iron oxide?
2 electrons gained, 2-
Assuming that the electrons gained by all three oxide ions were lost by
the two iron atoms, what must be the charge on each iron ion in iron
oxide?
If each oxide ion has a 2- charge, a combined charge of 6-, then the two iron ions have
combined charge of 6+. Presumably, the iron ions are exactly the same, so each iron
ion must have a charge of 3+.
Draw a picture (using Small Particle Models like those in Exploration
#1) that shows what happened in the Chemical Reactor.2 Since you are
using pictures to represent iron, oxygen, and iron oxide, use an arrow
symbol to represent “produces”. Label the ion charges in iron oxide.
Fe
Fe
OO
+
2
2- 3+ 2- 3+ 2-
Æ
It is not necessary to include the 2 iron atoms remaining in the input receptacle in your
picture. In real chemical reactions, there is usually a starting material that runs out before
another starting materials, like oxygen in this case. The starting material that runs out first
is “limiting”- that is, it limits the amount of final material that is produced. The other
material is said to be “in excess”, like iron in this case. When we take the mass of the
system, the starting material “in excess” was there initially and is still there in the end, so
mass is conserved.
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Cycle 6
Oxygen gas exists as a diatomic molecule, or two oxygen atoms bonded to
one another. Unlike iron atoms, single oxygen atoms are not stable;
therefore, two iron atoms, one diatomic molecule of oxygen, and half of
another diatomic molecule of oxygen will not form a single iron oxide
formula unit. Instead, two more iron atoms and two more diatomic
molecules of oxygen are needed to produce a total of two iron oxide
formula units.
STEP 11. Another way to represent chemical changes is to write a chemical
equation. In this case:
4 Fe (s) + 3 O2 (g) Æ 2 Fe2O3 (s)
How does the number of iron atoms on the left side of the arrow
compare to the number of iron ions on the right side of the arrow in
your drawing?
The numbers are equal ; 4 iron atoms and 4 iron ions.
How does the number of oxygen atoms on the left side of the arrow
compare to the number of oxide ions on the right side of the arrow in
your drawing?
The numbers are equal; 6 oxygen atoms and 6 oxide ions.
How does the number of “particles” (single atoms or atoms bonded
together in molecules or formula units) on the left side of the arrow
compare to the number of particles on the right side of the arrow?
They are not the same; 7 particles on the left and 2 particles on the right.
What are the connections between your model of the chemical change
and the chemical equation? How do they both represent the chemical
change you observed in Activity 1?
They both represent number of atoms and particles. The number of atoms is the same
before and after which corresponds to the left and right of the chemical equation. They
also show how the atoms have rearranged to form different materials.
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Activity 5: Conservation of Mass and the SPM
Balanced Chemical Equations and the Mole
In this activity we introduced chemical equations. The materials listed to
the left of the arrow are called the reactants, those listed to the right of the
arrow are called the products, and the arrow indicates “yields” or
“produces”. Materials are identified by their chemical formulas which
include element symbols (from the periodic table) and subscripts that
indicate the number of atoms of that element. No subscript after an element
symbol implies one of that atom. The physical state (solid, liquid, gas or
aqueous solution) is also indicated for each material. The balanced chemical
equation tells us the ratio in which materials combine to form new
materials.
During this activity you observed that the number of atoms (or ions) of each
element is the same before and after a chemical change, in other words, they
are “balanced.” A balanced chemical equation has the same number of
atoms on both sides of the reaction arrow. Coefficients are the whole
numbers placed before each chemical formula to show the ratio in which
materials combine in the balanced chemical equation. No coefficient before
a chemical formula implies one of that atom, molecule, or formula unit. The
coefficients of a balanced equation represent the ratios of “particles”, but
also macroscopic ratios in terms of moles. A mole is a counting unit, just
like a dozen. A dozen represents 12 of something, and a mole is
approximated to be 6.02 × 1023 of something, an extremely large number!
What is the difference between 2 Fe2O3 and Fe4O6? Why is it important
that you not change subscripts in chemical formulas to balance chemical
equations?
2 Fe2O3 means that we have 2 particles of rust. Fe4O6 is a different compound than rust.
Changing the subscripts changes the composition, and hence, identity of the particle,
meaning its properties will also be different.
Try balancing these equations using coefficients.
2 Na (s) + _____ Cl2 (g) Æ 2 NaCl (s)
Formation of table salt
_____N2 (g) + 3 H2 (g) Æ 2 NH3 (g)
Formation of ammonia
2 SO2 (g) + ______ O2 (g) Æ 2 SO3 (g)
2 C8H18 (g) + 25 O2 (g) Æ 16 CO2 (g) +18 H2O (l)
_____C6H12O6 (aq) + 6 O2 (g) Æ 6 CO2 (g) + 6 H2O (l)
Formation of acid-rain precursor
Burning octane in gasoline
Burning sugar (respiration)
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Cycle 6
In the space below, list some of your strategies for solving these problems:
Start on the left and use coefficients to balance the non-H and non-O atoms on left and right.
Then use coefficients to balance H atoms on both sides; use least common multiples.
Then use coefficients to balance O atoms on both sides; use least common multiples.
Simulator Exploration #3: What is the relationship between atomic
mass and molar mass?
STEP 1. Open the Cycle 6 Activity 5 Setup 2.
You should see balances in both the macro and micro regions of the
Chemical Reactor. We will use the macro region to measure the mass of one
mole of a material in grams using the substance scooper. We will use the
micro region to measure the mass of one particle of material in atomic mass
units (amu).
STEP 2. Scroll down to the bottom row and click
the mouse button once to select iron oxide from the
menu.
STEP 3. Click the mouse button once in an empty space of
the macro region to deposit a lump of iron oxide.
STEP 4. Click on the substance scoop on the
bottom toolbar to select it. The substance scoop
contains one mole of a material.
STEP 5. Then move the mouse (empty scoop)
over the name of the lump you deposited and
click the mouse button once to fill the scoop.
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Activity 5: Conservation of Mass and the SPM
STEP 6. Move the mouse (full scoop) over the balance
pan in the macro region. When a black rectangle
appears, click the mouse button once to deposit the 1mole scoop to the pan.
What is the mass in grams of one mole of iron oxide?
160
STEP 7. Select iron oxide from the menu of materials. Then move the
mouse over the balance pan in the micro region. When a black rectangle
appears, click the mouse button once to deposit a formula unit of iron oxide.
What is the mass in amu of one formula unit of iron oxide?
160
Use the decimal numbers listed on the periodic table for iron (Fe) and
oxygen (O) to calculate the mass of Fe2O3. Show your work below.
(2 x 55.84) + (3 x 16.00) = 159.68 = 160
Atomic Mass and Molar Mass
Scientists found that the mass of one mole (6.02 × 1023 atoms) of any element
is equal to its atomic mass in grams. Therefore, the decimal number listed
on the periodic table can represent the average atomic mass of an atom in
units of atomic mass units (amu) or the mass of a mole of atoms in units of
grams (g)—commonly called molar mass. Similarly, the sum of the decimal
numbers for all atoms in a compound (or diatomic element) can represent
the average atomic mass of a molecule or formula unit in amu, or the mass
of a mole of molecules or formula units in grams.
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Cycle 6
Use the decimal numbers on the period table to calculate the total mass
(either in amu or in g) of both sides of the equation below. Show your
work clearly.
4 Fe (s) + 3 O2 (g) Æ 2 Fe2O3 (s)
(4 x 55.84) + (3 x 32) Æ (2 x 159.68)
223.36
+ 96 Æ 319.36
319.36
Æ 319.36
How does the total mass of the reactants compare with the total mass
of the products?
They are equal.
Calculate the molar masses for each reactant and product. Then
calculate the total mass of reactants and products to confirm a balanced
chemical equation (and hence, conservation of mass).
2 Na (s) + Cl2 (g) Æ 2 NaCl (s)
Formation of table salt
2 x (22.99 + 35.45)
2 x 58.44 = 116.88
2 N2 (g) + 3 H2 (g) Æ 2 NH3 (g)
Formation of ammonia
2 x (14.01 + [3 x 1.01])
2 x 17.04 = 34.08
2 SO2 (g) + O2 (g) Æ 2 SO3 (g)
Formation of acid-rain precursor
2 x (32.07 + [3 x 16.00])
2 x 80.07 = 160.14
2 C8H18 (g) + 25 O2 (g) Æ 16 CO2 (g) + 18 H2O (l)
Burning octane in gasoline
16 x (12.01 + [2 x 16]) + 18 x ([2 x 1.01]+ 16)
16 x 44.01
+ 18 x 18.02
704.16
+ 324.36 = 1028.52
C6H12O6 (aq) + 6 O2 (g) Æ 6 CO2 (g) + 6 H2O (l)
6 x (12.01 + [2 x 16]) + 6 x ([2 x 1.01]+ 16)
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6 x 44.01
+ 6 x 18.02
264.06
+ 108.06 = 372.12
Burning sugar (respiration)
Activity 5: Conservation of Mass and the SPM
Summarizing Questions
Discuss these questions with your group and note your ideas. Leave
space to add any different ideas that may emerge when the whole
class discusses their thinking.
S1. Are atoms created or destroyed during chemical changes? What
evidence do you have?
Atoms are not created nor destroyed. The same number of atoms is present before and after
chemical changes as evidenced by the macroscopic conservation of mass. This is
represented at the particle level by a balanced chemical equation where the same number of
atoms of each element appears on both sides of the equation.
S2. How is the Small Particle Model description of the Conservation of
Mass different for physical and chemical changes?
In physical changes, the numbers of particles (and the number of atoms by extension) are
conserved.
In chemical changes, the number of atoms is conserved.
S3. Two students are discussing conservation of mass when wood burns.
I think mass is conserved. Since the
ash has less mass than the original
piece of wood, other products must
be generated during the chemical
change. Since we can’t see them,
perhaps they are gases.
Amara
I disagree. When wood burns, mass
is not conserved. The mass of the
ash is significantly less than the
mass of the original piece of wood.
Victor
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Cycle 6
(a) Who do you agree with? Explain.
I agree with Amara. Just like we couldn’t trap gases when we boiled water to
account for conservation of mass, we can’t trap gases when we burn wood to
account for conservation of mass.
(b) Two chemical equations that you balanced involved burning.
What was common to both of them?
O2 was a reactant and both CO2 and H2O form.
(c) What have both students neglected to consider when discussing
conservation of mass when wood burns?
Neither student considered that O2 was a reactant and is included in the mass of the
reactants.
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