Ionic and Covalent Compounds Lab Worksheet

Activity on Ionic and Covalent
Compounds (Nomenclature)
Goal
•
to understand properties, names, and formulas for ionic and covalent
compounds.
Introduction
Chemical compounds can be divided into two basic types: ionic and covalent. Ionic
compounds contain a metal and a nonmetal. Covalent compounds contain only
nonmetals. On the periodic table of elements, the metals are located on the left
side, and the nonmetals are located on the right side.
Ions are atoms that have either lost electrons or gained electrons. If the ion has lost
electrons, it is positively charged and is called a cation. If the ion has gained
electrons, it is considered negatively charged and is called an anion.
Monatomic ions are ions composed of only one atom. “Mono” comes from the Greek
word “monos” which means “alone”. A polyatomic ion is an ion which consists of
more than one atom.
Naming Ions and Ionic Compounds
Cations and anions are combined together to form a neutral molecule called an ionic
compound. In naming these compounds, monoatomic cations (positive) are named
the same as the metal element, and they come first when naming a binary ionic
compound. A binary ionic compound is an ionic compound consisting of only two
ions. (“Bi” means “two” in Latin.) Monoatomic anions (negative) take the name of
the nonmetal along with the suffix “-ide”, and they come at the end of the ionic
compound’s name. For example, consider the compound Al2O3. The subscripts
indicate that the compound contains two atoms of aluminum and three atoms of
oxygen, but these numbers are not used in the name of the compound. The name of
the compound is aluminum oxide.
Metals with variable charges: When an ionic compound contains a metal of variable
valence (most of the transition metals and a few main group metals like Sn and Pb),
the metal can carry more than one charge, and thus, form more than one compound.
In naming these, the charge is denoted as a roman numeral in parenthesis. Keep in
mind that the roman numerals refer to the charge of the cation, not how many
anions are attached. For example, in FeCl2, iron has a charge of +2. The name of
the compound is iron (II) chloride. In FeCl3, the charge of the iron ion is +3; so the
name of this compound should read iron (III) chloride.
Polyatomic ions: These ions contain two or more nonmetal atoms covalently bonded
together, and the entire group has a positive or negative charge. Polyatomic ions
have special names. Many of them contain oxygen and are called oxyanions. When
different oxyanions are made of the same element, but have a different number of
oxygen atoms, then prefixes and suffixes are used to tell them apart. The “-ate”
suffix is used on the most common oxyanion. For this lab activity, eight common
polyatomic ions will be highlighted. These will be written in bold and numbered 1-8
in the boxes below.
Modifications to the names of polyatomic ions:
Adding an H+ ion: This reduces the negative
1. Carbonate CO32-
charge on the polyatomic ion by one, and the
Hydrogencarbonate HCO3word “hydrogen” is added to the beginning of
the name. For example, the carbonate ion has the formula CO32-. A hydrogen can
be added to the ion to form the hydrogencarbonate, or hydrogen carbonate ion,
which has the formula HCO3-.
Removing an oxygen: This does not affect the charge, but the ending of the name is
changed from “-ate” to “-ite”. The nitrate ion has the formula NO3-. The nitrite ion,
NO2-, is formed by removing one oxygen atom from the
2. Nitrate NO3
nitrate.
Nitrite NO2Additionally, there
Perchlorate ClO4Perbromate BrO4exists the “hypo-“prefix,
3. Chlorate ClO3Bromate
BrO3which signifies one
Chlorite ClO2Bromite
BrO2oxygen atom fewer than
Hypochlorite ClOHypobromite BrOfor “-ite”. There is also
the “per-“prefix, meaning having one more oxygen atom than an “-ate” molecule.
Here are a few non-standard, but common polyatomic ions:
4. Hydroxide OH-
5. Acetate C2H3O2-
6. Ammonium NH4+
Exchanging the central atom with another element in the same Group on the
periodic table. The sulfate ion has the formula SO42-. The sulfur can be replaced by
selenium (also from Group 16), forming a selenite ion, SeO42-.
7. Sulfate SO42Hydrogensulfate HSO4Sulfite SO32Hydrogensulfite HSO3Selenate SeO42Selenite SeO32Tellurate TeO42-
8. Phosphate PO43Hydrogenphosphate HPO42Phosphite PO33Hypophosphite PO23Hydrogenphosphite HPO32Arsenate AsO43Arsenite AsO33-
Writing the formula for an ionic compound:
Again, cations and anions are combined together to form a neutral molecule. A
magnesium ion (Mg2+) should combine with two chlorine ions (Cl-) to form
magnesium chloride.
Mg2+ + 2 Cl- → MgCl2
The charge neutralization can be done by crisscrossing the number of charges on
the two ions (without the +/- signs). Thus, aluminum oxide is:
Figure 1. Charge neutralization in writing an ionic compound formula
Polyatomic ions should be placed in parenthesis when there are more than one of
those ions. Such as calcium nitrate:
Ca2+ + NO3- → Ca(NO3)2
The trick is to write the metal cation with the proper positive charge and the
nonmetal anion with the correct negative charge then crisscross the number of
charges to produce a neutral molecule.
Naming binary covalent compounds
Binary covalent compounds are made of two nonmetals. The system for naming
covalent compounds is different from the one for ionic compounds, and it is
important to distinguish the two systems. To name a binary covalent compound,
the first element is written as the nonmetal’s element name; the second element is
written in the element’s “-ide” form. Both elements get a prefix that indicates the
number of each element in the compound. If the prefix of the first element is
“mono”, then the prefix is dropped.
Examples:
CO
carbon monoxide
N2O dinitrogen monoxide
CO2 carbon dioxide
NO2 nitrogen dioxide
P2S5 diphosphorus pentasulfide
SCl2 sulfur dichloride
1
2
mono-
6
hexa-
7
hepta-
8
octa-
3
ditri-
4
tetra-
9
nona-
5
penta-
10
deca-
Prefixes for naming covalent compounds
Investigating properties of ionic and covalent compounds
Ionic compounds differ from covalent compounds in many physical properties. Ionic
solids are usually soluble in water, and the solution will conduct electricity due to
the compound’s dissolution into charged cations and anions. Many covalent
compounds are insoluble in water. If they do dissolve, the solution may be nonconducting. Due to low intermolecular attraction, covalent solids melt easily, while
strong attraction between the ions make ionic solids have higher melting points.
Activity on Ionic and Covalent
Compounds (Nomenclature)
Report sheet
Date: _____________________
Name: _________________________
Section #: _________________
Instructor: _____________________
Pre-Lab Questions
1. Write whether the following compounds are ionic or covalent:
Compound
cesium iodide
SF6
SrI2
P2O5
CaO
CoCl2
2. Write the correct formula for the following:
barium bromide
copper (I) oxide
dinitrogen pentoxide
Ionic or Covalent
Lab Activity
A. List the correct ions and chemical formula for each of the following compounds:
Compound
Ions
Formula
1. potassium chloride
K+, Cl-
KCl
2. barium bromide
_________________
_________________
3. copper (II) sulfide
_________________
_________________
4. lithium oxide
_________________
_________________
5. magnesium phosphide
_________________
_________________
6. barium oxide
_________________
_________________
7. iron (III) sulfide
_________________
_________________
8. sodium carbonate
_________________
_________________
9. ammonium sulfate
_________________
_________________
10. calcium phosphate
_________________
_________________
B. Write the names of the following compounds:
Formula
Compound
1. NaCl
_________________________________________
2. FeCl3
iron (III) chloride
3. MgCO3
_________________________________________
4. Cu2O
_________________________________________
5. Ca(OH)2
_________________________________________
6. AlF3
_________________________________________
7. AgNO3
_________________________________________
8. Ca3(PO4)2
_________________________________________
9. NaHCO3
_________________________________________
10. Na2SO3
_________________________________________
C. Name the compounds listed below.
Formula
Name
1. IF5
_________________________________________
2. N2O5
_________________________________________
3. SeO
_________________________________________
4. CO2
_________________________________________
5. B2H6
_________________________________________
6. P2O5
_________________________________________
7. Cl2S
_________________________________________
8. SiBr4
_________________________________________
9. AsCl3
_________________________________________
10. I2O7
_________________________________________
Moles from Masses, Masses from Moles
Goal
•
•
to apply the mole conversion factors to convert grams to moles and moles to
grams.
to determine experimentally the simplest (empirical) formula of an oxide of
magnesium.
Introduction
A. Finding the Simplest Formula
As previously learned, the chemical formula of a compound shows its composition
using the symbols and relative numbers of its constituent elements. The simplest
form of a chemical formula, known as its empirical formula, is based on the simplest
whole-number ratio of its constituent elements. The molecular formula indicates the
exact numbers of atoms present in a molecule of a compound. In most experiments,
determining the empirical formula is the first step in determining the molecular
formula of a compound.
A combination reaction is a chemical reaction between two or more elements
or compounds to produce a more complex compound. For example, metals combine
with oxygen to produce their oxides. The reaction observed in this laboratory
experiment is based on the combination reaction between magnesium and oxygen
and represented by the unbalanced chemical equation:
∆
Mg(s) +O2(g) →
MgxOy(s)
In this experiment, magnesium metal is heated to a high temperature until it reacts
with the oxygen (O2) in the air. In the equation above, “x” and “y” are the simplest
whole numbers that we are looking for. We start with a known amount of
magnesium and heat strongly to convert it into its oxide (residue). As the
magnesium burns in the air, it may also combine with nitrogen (N2) in the air.
∆
3Mg(s) + N2(g)
→ Mg3N2(s)
To remove any nitride product, water is added and the product is reheated. Any
nitride product is converted to magnesium oxide and ammonia.
Mg3N2(s) + 3H2O(l)
∆
→
3MgO(s) + 2NH3(g)
To begin the experiment, a piece of magnesium metal will be obtained and the mass
of the magnesium measured using a digital scale. At the end of the experiment, the
mass of the oxide product will be measured. The mass of oxygen that combined
with the magnesium in the reaction is the difference between the mass of the oxide
product and the original mass of the magnesium.
grams of the oxide product – grams of Mg = grams of O
The simplest formula of the magnesium oxide product is determined by calculating
the moles of magnesium and the moles of oxygen using their respective molar
masses.
moles of Mg = grams of Mg(s) ribbon ×
1 𝑚𝑜𝑙𝑒 𝑀𝑔
𝑔
24.3𝑚𝑜𝑙𝑀𝑔
moles of O = grams of O(g) combined ×
1 𝑚𝑜𝑙𝑒 𝑂
𝑔
16.0𝑚𝑜𝑙𝑂
B. Molar Mass
Molar mass is defined as the mass in grams of one mole of a substance. Another
way to say this is, the molar mass is the mass of a given chemical element
or chemical compound (g) divided by the amount of substance (mol). The units
of molar mass are grams per mole, abbreviated as g/mol.
So, the calculations for this experiment involve first, measuring the mass of
magnesium and the magnesium oxide product, determining the mass of oxygen
involved in the reaction, and converting the masses of reactants into moles. From
these values, the simplest whole-number ratio between the two-mole amounts gives
the empirical, or simplest formula. This “whole-number ratio” is obtained by
dividing the number of moles of each reactant by the number of mole of the element
with the fewest number of moles. This simplest formula is then reported as the
chemical formula of magnesium oxide.
To illustrate, suppose in a similar experiment, it was determined that 0.040
mole of Zn had combined with 0.080 mole of Cl to form a compound. Proceeding as
follows:
1. Ratio of moles as 0.040 mole Zn : 0.080 mole Cl
2. Determine the smaller number of moles (0.040 mole Zn).
3. Divide the moles of each element by the smaller number of moles and
round to the nearest whole number.
0.080 𝑚𝑜𝑙 𝐶𝑙
0.040
= 2 moles Cl
0.040 𝑚𝑜𝑙 𝑍𝑛
0.040
= 1 mole Zn
4. Use the whole numbers as subscripts to write the formula of the
compound.
Zn1Cl2 or ZnCl2
Experimental Procedures
•
•
•
•
Complete the Pre-Lab Study Questions on the report sheet.
Watch the video, “Finding the Empirical Formula for MgO.wmv”, followed by
the video, “Finding the Empirical Formula for Magnesium Oxide part 2.wmv”
both found in the online folder, “Moles from Masses, Masses from Moles”.
Use the data from these videos to make calculations and to complete the
report sheet.
Submit your completed report sheet.
Activity on Ionic and Covalent
Compounds (Nomenclature)
Report sheet
Date: _____________________
Name: _________________________
Section #: _________________
Instructor: _____________________
Pre-Lab Questions
1. Write whether the following compounds are ionic or covalent:
Compound
cesium iodide
SF6
SrI2
P2O5
CaO
CoCl2
2. Write the correct formula for the following:
barium bromide
copper (I) oxide
dinitrogen pentoxide
Ionic or Covalent
Lab Activity
A. List the correct ions and chemical formula for each of the following compounds:
Compound
Ions
Formula
1. potassium chloride
K+, Cl-
KCl
2. barium bromide
_________________
_________________
3. copper (II) sulfide
_________________
_________________
4. lithium oxide
_________________
_________________
5. magnesium phosphide
_________________
_________________
6. barium oxide
_________________
_________________
7. iron (III) sulfide
_________________
_________________
8. sodium carbonate
_________________
_________________
9. ammonium sulfate
_________________
_________________
10. calcium phosphate
_________________
_________________
B. Write the names of the following compounds:
Formula
Compound
1. NaCl
_________________________________________
2. FeCl3
iron (III) chloride
3. MgCO3
_________________________________________
4. Cu2O
_________________________________________
5. Ca(OH)2
_________________________________________
6. AlF3
_________________________________________
7. AgNO3
_________________________________________
8. Ca3(PO4)2
_________________________________________
9. NaHCO3
_________________________________________
10. Na2SO3
_________________________________________
C. Name the compounds listed below.
Formula
Name
1. IF5
_________________________________________
2. N2O5
_________________________________________
3. SeO
_________________________________________
4. CO2
_________________________________________
5. B2H6
_________________________________________
6. P2O5
_________________________________________
7. Cl2S
_________________________________________
8. SiBr4
_________________________________________
9. AsCl3
_________________________________________
10. I2O7
_________________________________________
Moles from Masses, Masses from Moles
Report sheet
Date: _____________________
Name: _________________________
Section #: _________________
Instructor: _____________________
Pre-Lab Questions
1. What is meant by the “simplest formula” of a compound?
2. Which weighs more, a mole of copper or a mole of gold? Why?
3. What are the molecular weights of the following compounds? (all masses must be
to nearest hundredth)
a.) NaOH
b.) H3PO4
c.) H2O
d.) Mn2Se7
e.) MgCl2
f.) (NH4)2SO4
4. How many moles are present in 34 grams of Cu(OH)2?
Moles from Masses, Masses from Moles
A. Finding the Simplest Formula (from the videos, “Finding the Empirical Formula
for MgO.wmv” and “Finding the Empirical Formula for Magnesium Oxide part
2.wmv”)
DATA
A.1
mass of the dry crucible and lid
______________________________
A.2
mass of crucible, lid, and magnesium
______________________________
A.3
mass of crucible, lid, and oxide product
______________________________
CALCULATIONS (show work.)
A.5
mass of magnesium used
______________________________ g Mg
A.6
mass of magnesium oxide product
___________________________ g MgxOy
A.7
mass of oxygen that reacted
____________________________ g O
A.8
moles of Mg
___________________________ mol Mg
A.9
moles of O
____________________________ mol O
A.10 ______________________ mol Mg
smaller value:
A.11
A.12
_______________________________
𝑚𝑜𝑙 𝑀𝑔
𝑚𝑜𝑙 𝑂
___________________________ mol O
= ______________________________moles Mg (rounded)
= _______________________________moles O (rounded)
A.13 Simplest formula: ___________________________________
Questions and Problems
Q.1 Using the rules for writing the formulas of ionic compounds, write the ions
and the correct formula for magnesium oxide.
Q.2
Write a balanced equation for the reaction of the magnesium and the oxygen.
Q.3
How many grams are there in 1.5 moles of MgCO3?
Q.4
How many moles are in each of the following?
a.) 80.0 g of NaOH
b.) 12.0 g of Ca(OH)2
Q.5
Write the simplest formula for each of the following compounds:
a.) 0.200 mole Al and 0.600 mole Cl
b.) 0.080 mole Ba, 0.080 mole S, and 0.320 mole O
Q.6 When 2.50 g of copper (Cu) reacts with oxygen, the copper oxide product has
a mass of 2.81 g. What is the simplest formula of the copper oxide?
Q.7
Aluminum and oxygen gas react to produce aluminum oxide.
a.) Write a balanced equation for the reaction.
b.) If 12 moles of Al2O3 are produced, how many moles of aluminum reacted?
c.) If 75 g of oxygen react, how many grams of aluminum are required?
CHEM 1405 – Introductory Chemistry I
Solutions and Solubility
Solutions and Solubility
Required materials provided in the Home Science Tools chemistry kit: safety goggles, 6
small test tubes, metal spatula, 3 transfer pipettes, glass stir rod
Required materials not provided in the Home Science Tools chemistry kit: cell phone (with
camera), salt, sugar, vegetable oil, isopropyl alcohol (IPA), Sharpie or another type of
permanent marker
Objectives:
to identify polar and non-polar solvents and solutes
Introduction:
In chemistry, a mixture contains two or more substances that are not
chemically combined. Mixtures fall
into two different categories:
Homogeneous and Heterogeneous.
A homogeneous mixture is a mixture
in which the components that make
up the mixture are uniformly
distributed throughout the mixture.
A heterogeneous mixture is a
mixture in which the components of
the mixture are not uniform.
A solution is a type of homogeneous
Figure 1. Types of Mixtures
mixture. In a solution, one
substance, called a solute, is dissolved in another substance, known as a solvent. The
component of a solution with the greater amount is the solvent, and the component with
the lesser amount is the solute. Because of its polar nature, water acts a dominant,
universal solvent and can dissolve many solutes such as common table salt (sodium
chloride, NaCl) or the sugar one puts in coffee (sucrose, C12H22O11). Solutes, as well as
solvents, can be either solid, liquid, or gas. The resulting solution can also be in the form
of a solid, liquid, or gas.
Page 1 of 5
Dr. Prem Adhikari
08/2020
CHEM 1405 – Introductory Chemistry I
Solutions and Solubility
The polarity of the component molecules (solute and solvent) determines the nature of the
resulting solution or mixture. The molecular polarity of the solute, as well as the solvent,
is determined by the differences in the electronegativity of its bond pairs and its molecular
geometry, or shape.
a) A nonpolar covalent compound is a chemical compound wherein the atoms
comprising the molecule equally share their bond electrons.
b) A polar covalent compound is a molecule in which the distribution of electrons
between the covalently bonded atoms is not even.
c) An ionic compound is a chemical compound composed of ions held together
by electrostatic forces termed ionic bonding. The compound is neutral overall but
consists of positively charged ions called cations and negatively charged ions called
anions.
Figure 2. Types of Molecular Bonding
Polar molecules have both
centers of positive and
negative charge that are not in
the same location in the
molecule. Non-polar molecules
have the center of positive and
negative charge in the same
place.
If a molecule has more than
two atoms, its shape
(molecular geometry) can
affect the polarity in a crucial
way. Water has a bent
structure, making it a polar
molecule. Carbon dioxide’s
linear molecular shape cancels
its bond polarities resulting in
a nonpolar molecule.
Figure 3. Molecular Shape Affecting Polarity
Page 2 of 5
Dr. Prem Adhikari
08/2020
CHEM 1405 – Introductory Chemistry I
Solutions and Solubility
Weak, intermolecular forces can also contribute to solutes dissolving better in a solvent.
Dispersion forces, also known as London dispersion forces or Van der Waals forces, are
attractive forces between the atoms of molecules and are present in all molecules. Their
presence is due to temporary dipoles created by the momentary differences in the
distribution of molecular electron clouds. A dipole-dipole force is caused by a permanent
dipole present in the polar ends of a compound. Ion-dipole forces occur where the cations
or anions of an ionic compound interact with the end of another molecular dipole.
Figure 4. Intermolecular Forces
As seen in the flowchart below, if a solvent and a solute share the same polarity, they are
likely to result in a solution. The ability of the solute to dissolve in a solvent is known as
solubility. The greater the similarity in polarity between the solute and solvent, the
higher the solubility – the phrase, “like dissolves like” is helpful to remember when
predicting the solubility of a mixture.
Page 3 of 5
Dr. Prem Adhikari
08/2020
CHEM 1405 – Introductory Chemistry I
Solutions and Solubility
Flowchart – “Matter: Elements, Compounds, and Mixtures”
Page 4 of 5
Dr. Prem Adhikari
08/2020
CHEM 1405 – Introductory Chemistry I
Solutions and Solubility
Experimental Procedures:
*** A VIDEO RECORDING must be produced of the student
executing the following steps of the lab experiment as the steps are
performed. ***
[This video will either be (1) attached as a file to the report sheet for submission or
(2) provided as a link on the report sheet to a YouTube video.]
1. Using the permanent marker, label six test tubes 1 through 6.
2. Add a small amount (enough to fully cover the bottom of the tube) of salt to test
tube #1 and to test tube #4.
3. Add about the same amount of sugar to test tube #2 and to test tube #5
4. Using a transfer pipette, add a small amount (enough to fully cover the bottom of
the tube) of vegetable oil to test tube #3 and to test tube #6.
5. Using a different transfer pipette, add 2-3 squirts of isopropyl alcohol (CH3OH) to test tubes #1,
#2, and #3.
6. Using a third pipette (different from both the one used for vegetable oil and the one used for the
alcohol) add 2-3 squirts of water (H2O) to test tubes #4, #5, and #6.
7. Closely examine the six test tubes and determine the solubility of each mixture.
8. Record the results in the Solutions and Solubility table on the report sheet.
Waste Disposal:
Clean all six test tubes by rinsing them out with water and allowing them to air dry for use
in other labs.
Page 5 of 5
Dr. Prem Adhikari
08/2020
CHEM 1405 – Introductory Chemistry I
Solutions and Solubility
Solutions and Solubility
Report sheet
Name:_________________________________ Date:__________________________________
Lab Partner(s)__________________________________________________________________
Instructor:______________________________ Section:_______________________________
Pre-lab Questions:
1. What is the objective of this lab? __________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
2. What are polar and nonpolar chemicals? ___________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
3. Explain solubility. _______________________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
Page 1 of 3
Dr. Prem Adhikari
08/2020
CHEM 1405 – Introductory Chemistry I
Solutions and Solubility
Remember to attach the video recording of the student performing this
section of the lab experiment to the report sheet either as a file attachment or
as a link to a YouTube video. Failure to provide this video will result in a
grade of ZERO on this lab assignment.
Solutions and Solubility Table (soluble or insoluble?)
Solute
Solvent
salt
sugar
vegetable oil
isopropyl alcohol
test tube #1
test tube #2
test tube #3
test tube #4
test tube #5
test tube #6
water
Questions:
1. Describe the following compounds as either polar or covalent :
a) Salt (NaCl) ______________________________________________
b) Sugar (C12H22O11) ________________________________________
c) Vegetable Oil ____________________________________________
d) Isopropyl Alcohol _________________________________________
e) Water (H2O) _____________________________________________
Page 2 of 3
Dr. Prem Adhikari
08/2020
CHEM 1405 – Introductory Chemistry I
Solutions and Solubility
2. Briefly explain why salt is soluble in water.
_________________________________________________________________________________
_________________________________________________________________________________
3. What type of intermolecular force is responsible for helping salt to dissolve in
water? Why?
_________________________________________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
4. List any drawbacks or difficulties encountered while performing this experiment.
Suggest any improvements.
_________________________________________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
REMEMBER:
Attach the video of the student’s execution of the experimental procedures
with the submitted report sheet.
Page 3 of 3
Dr. Prem Adhikari
08/2020
CHEM 1405 – Introductory Chemistry I
Chemical Reactions: Super Slime
Chemical Reactions: Super Slime
Required materials provided in the Home Science Tools chemistry kit: polyvinyl alcohol
[CH2CH(OH)-]n, 250mL beaker, 100mL graduated cylinder, metal spatula/scoop, hot plate,
glass stir rod, 150mL beaker, sodium tetraborate (borax), 10mL graduated cylinder
Required materials not provided in the Home Science Tools chemistry kit: towel or mitten,
Ziploc bag (sealable plastic bag), food coloring (optional).
Objectives:
to explore the change in physical properties of a polymer (polyvinyl alcohol,
PVA) when cross-linked through a chemical reaction with sodium tetraborate creating
“super slime.”
Introduction:
A polymer is a chemical substance consisting of large molecules whose
structure is composed of multiple repeating units strung together to form long chains.
Examples of some polymers include anything plastic, proteins such as hair, nails, shells,
and DNA, silly putty, Styrofoam cups and plates, and blocks of wood.
Polyvinyl alcohol, commonly abbreviated PVA, is one such polymer with many, vinyl alcohol
units linked together to form a long chain-like structure. PVA chemically reacts with
sodium borate, known as borax, to form a cross-linked polymer network, i.e. “super slime.”
In chemistry, a cross-link is a bond that links one polymer chain to another. When borax
is dissolved in water, it dissociates into sodium ions, Na+ (aq), borate ions, B(OH)4- (aq), and
boric acid, H3BO3 (aq).
When this aqueous borax
solution is added to an
aqueous PVA solution,
the borate ions, B(OH)4-,
chemically react with the
PVA polymer to yield the
PVA/borate cross-linked
polymer (“super slime”)
and water (see Figure 1.).
Figure 1. Illustration of the Chemical Reaction between PVA (aq) and borax (aq)
Page 1 of 2
Dr. Jyoti Gupta and Dr. Prem Adhikari
08/2020
CHEM 1405 – Introductory Chemistry I
Chemical Reactions: Super Slime
Experimental Procedures:
1. Make an aqueous solution of polyvinyl alcohol (PVA): Stir 3 scoops of PVA into
120mL of water in the 250mL glass beaker. Observe the mixture.
-Safety NoteThe glass beaker that you are heating can get very hot.
Use caution!!
If necessary, use an oven mitt or towel when
handling the beaker.
2. Heat the PVA solution using the hot plate.
• Begin by turning the heating knob to the medium setting.
• As the solution heats up, stir the solution with the glass stir rod.
• If necessary, adjust the heating setting either higher or lower to better dissolve
the PVA into solution. NOTE: If heat is turned too high, the solution will boil and
overflow onto the hot plate, resulting in loss of PVA and creating an unnecessary
mess. If heat is too low, the PVA may not go into solution for quite a long time, if
at all.
• Continue stirring until the PVA is completely dissolved, and the solution becomes
clear and transparent. Observe the solution.
• Once completely dissolved, turn off the hot plate off and remove the beaker from
the heat, allowing the solution to cool to room temperature. This may take some
time. Be patient!
• Continue to occasionally stir the PVA solution until it has completely cooled.
Observe the solution.
3. Add food coloring if colored slime is desired (optional).
4. In the 150mL beaker, make an aqueous solution of sodium tetraborate (borax) by
stirring 3 scoops of borax into 120mL of water.
5. Stir borax solution thoroughly. Observe the mixture.
6. Pour all the cooled PVA solution into a zip lock bag and add 10 mL of the borax
solution. Observe this mixture.
7. Zip the bag and knead it until the chemicals are mixed into slime. Again, make
observations of the resulting mixture.
8. Your slime is ready, have fun playing with it!
9. Complete the table on the report sheet.
Waste Disposal:
•
•
Slime can be disposed of in normal waste (garbage), but do not wash it down the
drain.
Clean and dry all glassware and utensils used. Store for use in future lab
experiments.
Page 2 of 2
Dr. Jyoti Gupta and Dr. Prem Adhikari
08/2020
CHEM 1405 – Introductory Chemistry I
Chemical Reactions: Super Slime
Chemical Reactions: Super Slime
Report sheet
Name:_________________________________ Date:__________________________________
Lab Partner(s)__________________________________________________________________
Instructor:______________________________ Section:_______________________________
Pre-lab Questions
1. What is a chemical reaction? ______________________________________________________
____________________________________________________________________________________
____________________________________________________________________________________
2. What kind of chemical reaction is involved in the formation of super slime?
____________________________________________________________________________________
____________________________________________________________________________________
Follow the Experimental
Procedures to complete the observation table below:
Super Slime Observation Table
Experimental
Procedure
step #
Chemical Mixture
1
PVA + water
2
PVA + water
5
6
7
Temperature and
Condition of
Chemical Mixture
room temperature
mixture
Type of
Mixture/Physical State*
heated and stirred
room temperature
mixture
room temperature
PVA(aq) 120mL +
mixture in Ziploc
borax(aq) 10mL
bag
Mixture of PVA and
room temperature
borax
kneaded mixture
*soluble (homogeneous) or insoluble (heterogeneous)
Borax + water
Page 1 of 2
Dr. Jyoti Gupta and Dr. Prem Adhikari
08/2020
CHEM 1405 – Introductory Chemistry I
Chemical Reactions: Super Slime
Questions:
1. Show the chemical reaction involved in the formation of the super slime.
2. Why was the PVA/water mixture heated in Step #2? Was this necessary?
3. Does the super slime act more like a solid or a liquid? Explain.
Page 2 of 2
Dr. Jyoti Gupta and Dr. Prem Adhikari
08/2020
CHEM 1405 – Introductory Chemistry I
Solutions and Solubility
Solutions and Solubility
Required materials provided in the Home Science Tools chemistry kit: safety goggles, 6
small test tubes, metal spatula, 3 transfer pipettes, glass stir rod
Required materials not provided in the Home Science Tools chemistry kit: cell phone (with
camera), salt, sugar, vegetable oil, isopropyl alcohol (IPA), Sharpie or another type of
permanent marker
Objectives:
to identify polar and non-polar solvents and solutes
Introduction:
In chemistry, a mixture contains two or more substances that are not
chemically combined. Mixtures fall
into two different categories:
Homogeneous and Heterogeneous.
A homogeneous mixture is a mixture
in which the components that make
up the mixture are uniformly
distributed throughout the mixture.
A heterogeneous mixture is a
mixture in which the components of
the mixture are not uniform.
A solution is a type of homogeneous
Figure 1. Types of Mixtures
mixture. In a solution, one
substance, called a solute, is dissolved in another substance, known as a solvent. The
component of a solution with the greater amount is the solvent, and the component with
the lesser amount is the solute. Because of its polar nature, water acts a dominant,
universal solvent and can dissolve many solutes such as common table salt (sodium
chloride, NaCl) or the sugar one puts in coffee (sucrose, C12H22O11). Solutes, as well as
solvents, can be either solid, liquid, or gas. The resulting solution can also be in the form
of a solid, liquid, or gas.
Page 1 of 5
Dr. Prem Adhikari
08/2020
CHEM 1405 – Introductory Chemistry I
Solutions and Solubility
The polarity of the component molecules (solute and solvent) determines the nature of the
resulting solution or mixture. The molecular polarity of the solute, as well as the solvent,
is determined by the differences in the electronegativity of its bond pairs and its molecular
geometry, or shape.
a) A nonpolar covalent compound is a chemical compound wherein the atoms
comprising the molecule equally share their bond electrons.
b) A polar covalent compound is a molecule in which the distribution of electrons
between the covalently bonded atoms is not even.
c) An ionic compound is a chemical compound composed of ions held together
by electrostatic forces termed ionic bonding. The compound is neutral overall but
consists of positively charged ions called cations and negatively charged ions called
anions.
Figure 2. Types of Molecular Bonding
Polar molecules have both
centers of positive and
negative charge that are not in
the same location in the
molecule. Non-polar molecules
have the center of positive and
negative charge in the same
place.
If a molecule has more than
two atoms, its shape
(molecular geometry) can
affect the polarity in a crucial
way. Water has a bent
structure, making it a polar
molecule. Carbon dioxide’s
linear molecular shape cancels
its bond polarities resulting in
a nonpolar molecule.
Figure 3. Molecular Shape Affecting Polarity
Page 2 of 5
Dr. Prem Adhikari
08/2020
CHEM 1405 – Introductory Chemistry I
Solutions and Solubility
Weak, intermolecular forces can also contribute to solutes dissolving better in a solvent.
Dispersion forces, also known as London dispersion forces or Van der Waals forces, are
attractive forces between the atoms of molecules and are present in all molecules. Their
presence is due to temporary dipoles created by the momentary differences in the
distribution of molecular electron clouds. A dipole-dipole force is caused by a permanent
dipole present in the polar ends of a compound. Ion-dipole forces occur where the cations
or anions of an ionic compound interact with the end of another molecular dipole.
Figure 4. Intermolecular Forces
As seen in the flowchart below, if a solvent and a solute share the same polarity, they are
likely to result in a solution. The ability of the solute to dissolve in a solvent is known as
solubility. The greater the similarity in polarity between the solute and solvent, the
higher the solubility – the phrase, “like dissolves like” is helpful to remember when
predicting the solubility of a mixture.
Page 3 of 5
Dr. Prem Adhikari
08/2020
CHEM 1405 – Introductory Chemistry I
Solutions and Solubility
Flowchart – “Matter: Elements, Compounds, and Mixtures”
Page 4 of 5
Dr. Prem Adhikari
08/2020
CHEM 1405 – Introductory Chemistry I
Solutions and Solubility
Experimental Procedures:
*** A VIDEO RECORDING must be produced of the student
executing the following steps of the lab experiment as the steps are
performed. ***
[This video will either be (1) attached as a file to the report sheet for submission or
(2) provided as a link on the report sheet to a YouTube video.]
1. Using the permanent marker, label six test tubes 1 through 6.
2. Add a small amount (enough to fully cover the bottom of the tube) of salt to test
tube #1 and to test tube #4.
3. Add about the same amount of sugar to test tube #2 and to test tube #5
4. Using a transfer pipette, add a small amount (enough to fully cover the bottom of
the tube) of vegetable oil to test tube #3 and to test tube #6.
5. Using a different transfer pipette, add 2-3 squirts of isopropyl alcohol (CH3OH) to test tubes #1,
#2, and #3.
6. Using a third pipette (different from both the one used for vegetable oil and the one used for the
alcohol) add 2-3 squirts of water (H2O) to test tubes #4, #5, and #6.
7. Closely examine the six test tubes and determine the solubility of each mixture.
8. Record the results in the Solutions and Solubility table on the report sheet.
Waste Disposal:
Clean all six test tubes by rinsing them out with water and allowing them to air dry for use
in other labs.
Page 5 of 5
Dr. Prem Adhikari
08/2020
Moles from Masses, Masses from Moles
Goal
•
•
to apply the mole conversion factors to convert grams to moles and moles to
grams.
to determine experimentally the simplest (empirical) formula of an oxide of
magnesium.
Introduction
A. Finding the Simplest Formula
As previously learned, the chemical formula of a compound shows its composition
using the symbols and relative numbers of its constituent elements. The simplest
form of a chemical formula, known as its empirical formula, is based on the simplest
whole-number ratio of its constituent elements. The molecular formula indicates the
exact numbers of atoms present in a molecule of a compound. In most experiments,
determining the empirical formula is the first step in determining the molecular
formula of a compound.
A combination reaction is a chemical reaction between two or more elements
or compounds to produce a more complex compound. For example, metals combine
with oxygen to produce their oxides. The reaction observed in this laboratory
experiment is based on the combination reaction between magnesium and oxygen
and represented by the unbalanced chemical equation:
∆
Mg(s) +O2(g) →
MgxOy(s)
In this experiment, magnesium metal is heated to a high temperature until it reacts
with the oxygen (O2) in the air. In the equation above, “x” and “y” are the simplest
whole numbers that we are looking for. We start with a known amount of
magnesium and heat strongly to convert it into its oxide (residue). As the
magnesium burns in the air, it may also combine with nitrogen (N2) in the air.
∆
3Mg(s) + N2(g)
→ Mg3N2(s)
To remove any nitride product, water is added and the product is reheated. Any
nitride product is converted to magnesium oxide and ammonia.
Mg3N2(s) + 3H2O(l)
∆
→
3MgO(s) + 2NH3(g)
To begin the experiment, a piece of magnesium metal will be obtained and the mass
of the magnesium measured using a digital scale. At the end of the experiment, the
mass of the oxide product will be measured. The mass of oxygen that combined
with the magnesium in the reaction is the difference between the mass of the oxide
product and the original mass of the magnesium.
grams of the oxide product – grams of Mg = grams of O
The simplest formula of the magnesium oxide product is determined by calculating
the moles of magnesium and the moles of oxygen using their respective molar
masses.
moles of Mg = grams of Mg(s) ribbon ×
1 𝑚𝑜𝑙𝑒 𝑀𝑔
𝑔
24.3𝑚𝑜𝑙𝑀𝑔
moles of O = grams of O(g) combined ×
1 𝑚𝑜𝑙𝑒 𝑂
𝑔
16.0𝑚𝑜𝑙𝑂
B. Molar Mass
Molar mass is defined as the mass in grams of one mole of a substance. Another
way to say this is, the molar mass is the mass of a given chemical element
or chemical compound (g) divided by the amount of substance (mol). The units
of molar mass are grams per mole, abbreviated as g/mol.
So, the calculations for this experiment involve first, measuring the mass of
magnesium and the magnesium oxide product, determining the mass of oxygen
involved in the reaction, and converting the masses of reactants into moles. From
these values, the simplest whole-number ratio between the two-mole amounts gives
the empirical, or simplest formula. This “whole-number ratio” is obtained by
dividing the number of moles of each reactant by the number of mole of the element
with the fewest number of moles. This simplest formula is then reported as the
chemical formula of magnesium oxide.
To illustrate, suppose in a similar experiment, it was determined that 0.040
mole of Zn had combined with 0.080 mole of Cl to form a compound. Proceeding as
follows:
1. Ratio of moles as 0.040 mole Zn : 0.080 mole Cl
2. Determine the smaller number of moles (0.040 mole Zn).
3. Divide the moles of each element by the smaller number of moles and
round to the nearest whole number.
0.080 𝑚𝑜𝑙 𝐶𝑙
0.040
= 2 moles Cl
0.040 𝑚𝑜𝑙 𝑍𝑛
0.040
= 1 mole Zn
4. Use the whole numbers as subscripts to write the formula of the
compound.
Zn1Cl2 or ZnCl2
Experimental Procedures
•
•
•
•
Complete the Pre-Lab Study Questions on the report sheet.
Watch the video, “Finding the Empirical Formula for MgO.wmv”, followed by
the video, “Finding the Empirical Formula for Magnesium Oxide part 2.wmv”
both found in the online folder, “Moles from Masses, Masses from Moles”.
Use the data from these videos to make calculations and to complete the
report sheet.
Submit your completed report sheet.