CHEM103/ CHEM 103 Module 4 – General
Chemistry I with Lab | Portage Learning | 2026
2027 Latest Exam Questions & Answers | Verified
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Q: Avogadro's number
Answer
6.022 x 10^23 atoms = 1 mole
Q: 1 u = ___ g
Answer
1.661 x 10 ^-24 g
Q: formula mass
Answer
the sum of the average atomic masses of each atom represented in the chemical formula
and is expressed in atomic mass units. --The formula mass of a covalent compound is also called the molecular mass.
Q: The chemical identity of a substance is defined by
Answer
the types and relative numbers of atoms composing its fundamental entities (molecules in
the case of covalent compounds, ions in the case of ionic compounds).
Q: A compound's percent composition provides
Answer
the mass percentage of each element in the compound.
Q: calculate the formula mass of a substance by
Answer
summing the average atomic masses of all the atoms represented in the substance's
formula.
Q: A mole is defined as
Answer
the amount of substance containing the same number of discrete entities (such as atoms,
molecules, and ions) as the number of atoms in a sample of pure 12-C weighing exactly 12 g
Q: percent composition
Answer
the percentage by mass of each element in a compound
i.e. mass X / mass compound x 100%
Q: stoichiometry
Answer
refers to the quantitative relationships between amounts of reactants and products in a
chemical reaction.
Q: Combustion
Answer
a reaction in which an element or compound burns in air or oxygen --The combustion of compounds containing hydrogen and carbon (CxHy) will always
produce two products: water (H2O) and carbon dioxide (CO2). --The same is true for the combustion of compounds containing hydrogen, carbon and
oxygen (CxHyOz)
Q: Chemical formulas provide the
identities of the reactants and products involved in the chemical change, allowing
classification of the reaction
Q: Coefficients provide the
Answer
relative numbers of these chemical species, allowing a quantitative assessment of the
relationships between the amounts of substances consumed and produced by the reaction.
Q: stoichiometric factor
Answer
ratio of coefficients in a balanced chemical equation, used in computations relating
amounts of reactants and products
Q: Limiting Reactant
Answer
A reactant that is totally consumed during a chemical reaction, limits the extent of the
reaction, and determines the amount of product.
Q: theoretical yield
Answer
The amount of product that may be produced by a reaction under specified conditions, as
calculated per the stoichiometry of an appropriate balanced chemical equation, is called the
theoretical yield of the reaction
Q: actual yield
Answer
the amount of product that forms when a reaction is carried out in the laboratory
Q: percent yield
Answer
actual yield/theoretical yield x 100
Q: Mass percent
Answer
Mass % = Mass x / total mass times 100% (assume 1 mole of X)
Q: Use mass percent to calculate molecular formula
Answer
Assume 100 g, not 1 mole
Convert % X into g; ex. 64% N = 64 g of Nitrogen
Convert to moles
Put smaller quantity in the denominator; divide to find ratio of # mol X to # mol Y
Q: To determine an empirical formula, you must
Answer
find the smallest whole-number ratio of atoms in a molecule. --If you begin with the mass of each element in the compound, you must first convert these
masses to moles before determining the ratio.
Q: A compound's percent composition provides the
Answer
mass percentage of each element in the compound, and it is often experimentally
determined and used to derive the compound's empirical formula.
Q: The empirical formula mass of a covalent compound may be compared to
Answer
the compound's molecular or molar mass to derive a molecular formula.
Q: Combustion analysis can be used to
Answer
identify an unknown hydrocarbon because the identity of the products are known to be CO2
and H2O.
Q: Empirical formulas are derived from experimentally measured element masses by
Answer --Deriving the number of moles of each element from its mass --Dividing each element's molar amount by the smallest molar amount to yield subscripts
for a tentative empirical formula --Multiplying all coefficients by an integer, if necessary, to ensure that the smallest whole
number ratio of subscripts is obtained
Q: Empirical formula
Answer
a formula with the lowest whole-number ratio of elements in a compound
Q: Ionization Energy
Answer
Measures how much energy is needed to remove an electron from an atom or how difficult
it is for an atom to form a positive ion.
Q: Ionization Energy (Increase)
Answer
Property increases as you go L --> R across a horizontal row of elements b/c more energy is
needed to remove an electron that's being attracted by a greater nuclear charge.
Q: Ionization Energy (Decrease)
Answer
Property decreases as you go down a vertical column of element b/c the element are
becoming larger and less is needed to remove an electron that's further away from the
attracting nucleus.
Q: Electronegativity
Answer
The attraction an atom has for its outer shell electron or how tightly it holds its valence
electrons.
Q: Electronegativity (Increase)
Answer
Property increases as you got L --> R across a horizontal row of elements and greater
nuclear charge holds the outer shell electrons more tightly
Q: Electronegativity (Decrease)
Answer
Property decreases as you go down a vertical column of elements b/c elements are
becoming larger and the outer shell electrons are more distant from the attracting nucleus.
Atomic Size
Answer
The distance from the center of the nucleus to the outer shell of an atom.
Atomic Size (Increase)
Answer
representation of a molecule that uses chemical symbols to indicate the types of atoms
followed by subscripts to show the actual number of atoms of each type in the molecule
(CH4) or (C_6H_12O_6)
Increases as you go down a vertical column of elements because the element have outer
shells that are further away from the nucleus.
Atomic Size (Decrease)
Answer
Decreases as you go L --> R across the horizontal row of elements because the outer shell is
being attracted by a greater nuclear charge
molecular formula
Answer
structural formula
Answer
gives the same information as its molecular formula (the types and numbers of atoms in the
molecule) but also shows how the atoms are connected in the molecule
Models
Answer
Ball and stick (geometric arrangement, not to scale)
Space Filling (shows relative sizes)
Coefficient vs Subscript
Answer
H = one H atom
2H = Two H atoms
H_2 = one H_2 molecule
2H_2 = Two H_2 molecules
Empirical Formula
Answer
Gives us the relative number of atoms in their whole number ratio (DOES NOT REFLECT
THE STRUCTURE, true nature or anything about the compound itself)
a chemical formula showing the ratio of elements in a compound rather than the total
number of atoms
Answer
Ex: Acetic Acid = CH_3COO = HCH_3COO = HC_2H_3O_2 = C_2H_4_O2
The Empirical Formula is CH_2_O reducted from the final most consolidated form
*NOTE: Molecular Formula is ALWAYS a whole number multiple of the empirical formula
Isomer
Answer
compounds with the same chemical formula but different molecular structures
Formula Mass
Answer
the sum of the average atomic masses (in amu) of all the atoms represented in the formula
of any molecule, formula unit, or ion (the substance's formula)
Formula Mass for Covalent Substances
Covalent Formula = Molecular Formula
Formula Mass = Molecular Mass
Element Quantity * Average Atomic Mass = subtotal (amu)
Mole
The mole provides a link between an easily measured macroscopic property, bulk mass, and
ROUNDING: Same sig figs as the least decimals for addition/substraction
Formula Mass for Ionic Substances
Ionic Formulas DO NOT equal molecular formulas (ionic formula gives the lowest whole
number ration of ions)
the formula for an ionic compound does not represent the composition of a discrete
molecule, so it may not correctly be referred to as the "molecular mass".
Ex: Al_2(SO_4)_3 = 2xAl + 3xS + 12xO amu
an extremely important fundamental property, number of atoms, molecules, and so forth.
A mole of substance is that amount in which there are 6.022 × 10^23 discrete entities
(atoms or molecules)
Avogadro's number (NA)
experimentally determined value of the number of entities comprising 1 mole of substance,
equal to 6.022 × 10^23 mol−1
Molar Mass
the mass in grams of one mole of a substance (g/mol)
- Differs for elements and substances
- Numerically equivalen in grams to its atomic or formula mass in amu
- Enables calculations between moles, molecules, atoms and mass
- Average Atomic Mass C = 12.01 amu - Molar Mass (g/mol) = 12.01 - Atoms/Mole = 6.022 x 10^23
Percent Composition - Chemical formulas represent elemental make up of a compound, percent composition
gives us the relative abundance of any given element in known compound formulas
- Must be determined experimentally
- % = mass of element / mass of compound *100%
Determining Empirical Formulas
Chemical formulas represent relative numbers, not masses, of atoms
Therefore, to derive numbers of atoms from mass, we must convert to moles
Moles can be used to compute who number ratios to derive the empirical formula
These are NOT molecular formulas - can ONLY give Empirical Formula
Empirical Formula Steps
1. Convert mass to moles
2. Divide each element's molar amount by the smallest molar amount found in step 1
3. Multiply all coefficients by an integer, if necessary, to get to smallest whole-number ratio
Empirical Formula From Percent Comp
% = mass of element (g)
(mass of element / mass of compound) 100 = (mass of element (g) / 100 g compound) 100
Deriving the Molecular Formula - must know molecular / molar mass which is experimentally determined
- always a muliple of the empirical fomula, and molecular mass similarly is a multiple of the
empirical formula mass
- Empirical Formula Mass = sum of average atomic masses of all atoms represented in
formula
= molecular or molar mass (amu or g/mol) / empirical formula mass (amu or g/mol) = n
formula units/molecule
Solvent
In a solution, the substance in which the solute dissolves
(or the medium in which the other components are dispersed)
Solution
A homogeneous mixture of two or more substances
Aqueous Solution
a solution in which water is the solvent
Dilute
small amount of solute
Concentrated
relatively high concentration of solute
Solute
the substance that is dissolved
typically present at lower concentration than the solvent
Concentration
Relative amount of a given solution component
Molarity (M) - concetration unit for solutions
- expressed as the number of moles of solute dissolved in exactly 1 liter of solution
- M = mol of solute / L of solution
Dilution
the process whereby the concentration of a solution is lessened by the addition of solvent
MOLARITY (M) * (L) of Solution = mol of solute (n)
n= ML
Before dilution: n1 = M1*L1
After Dilution: n2 = M2 * L2
NOTE: mole concertration does not change, dilute does not change the amount of solute in
the solution (n1 = n2)
M1 L1 = M2 L2
Dilution Equation
M1 L1 = M2 L2 where
C1 V1 = C2 V2
C = concentration
V = volume
mass percentage
ratio of solute-to-solution mass expressed as a percentage
"percent mass" "percent weight" "weight / weight percent"
symbol: %, % mass, % weight, and (w/w)%
Volume Percentage
Ratio of the volume of a liquid solute to the solution total
VP = volume solute / volume solution * 100%
Symbol: %vol or (v/v)%
Mass-Volume Percentage
Ratio of a solute's mass to the solution's volume expressed as %
Symbol: % mass-volume, (m/v)%
PPM & PPB
Used for very low solute concentations
how many pyruvic acids can be made from 2 sucrose molecules in glycolysis?
8
Can be based on mass, volume, or mixed units or numbers of atoms / molecules
% = pph = parts per hundred
ppm = parts per million
ppb = parts per billion
what % of energy in glucose is saved in glycolysis?
2%
how much CO2 is produced from 2 glucose molecules in human glycolysis only?
0
what steps in glycolysis represents substrate level phosphorylation?
step #7 & #10
what step in glycolysis is oxidation?
step #6
what step in glycolysis is dehydration?
step #9
why is phosphatase found only in the liver?
glycogen in the liver must provide glucose for the entire body
which reaction in the connecting step to Krebs cycle is endothermic?
synthesis
what type of functional group is found at the very end of a coenzyme A molecule?
thioalcohol
what is the high energy bond in acetyl - CoA?
thio ester
what reaction in the connecting step to Krebs cycle is exothermic?
oxidation
anaerobic conditions - without oxygen - humans - lactic acid formed in muscles - yeast - fermentation and ethanol produced
aerobic conditions - has oxygen - krebs cycle - electron transport
phosphate
highest energy compound, cannot move through protein channel
endothermic reaction - needs energy - putting phosphate on
enzyme #1 in glycolysis
glycogen phosphorylase
enzyme #2 in glycolysis
phosphatase (liver only)
substrate-level phosphorylation
making of ATP without O2 or electron transport
staring points of gluconeogenesis
- lactic acid - glycerol - some amino acids
phosphatase - removes phosphate group - found ONLY in the liver bc provides glucose to entire body - converts glucose 6 phosphate to glucose
when glucagon affects liver cell, a phosphate is attached to both glycogen synthetase and
glycogen phosphorylase, which enzyme is activated?
glycogen phosphorylase
Which cells need insulin for glucose entry?
muscle cells, liver, adipose
If glycogen synthetase were allosteric, what effect would ATP have on it?
activate
what happens to most of the glucose 6 phosphate in liver vs muscle tissue? - in muscle it is used for energy, in liver it is converted to glucose to be transported by the
blood - they can both be converted to glycogen if not needed
what can not be converted back to glucose 1 phosphate
fructose 1,6 bisphosphate because step #3 is irreversible
proteins
glucagon, insulin, lactic acid dehydrogenase, and amino acids
Polysaccharides
glycogen, starch, cellulose
can move through cell membrane
glucose and cortisol
can stimulate fatty acid synthesis from acetyl-CoA
insulin
can stimulate the immediate conversion of glycogen to glucose 1 phosphate in the liver
glucagon and epinephrine
Fructose-1,6-bisphosphatase
produces most ATP in glycolysis
can stimulate the conversion of certain amino acids to glucose in gluconeogenesis
cortisol
Anabolism
Metabolic pathways that construct molecules, requiring energy.
Catabolism
Metabolic pathways that break down molecules, releasing energy.
glucose
simple sugar, produces least ATP in glycolysis
in anaerobic exercise, the pH of the muscles will ...
decrease
processes that need a lot of ATP
muscle activity and synthesizing protein from amino acids (active transport, all anabolic
reactions)
starting point of glycolysis in a red blood cell
glucose
starting point of glycolysis in a muscle cell
glycogen
purpose of converting pyruvic acid to lactic acid is to regenerate
NAD
cannot move through cell membrane - glycogen - glucose-1-phosphate - protein hormones - epinephrine - polypeptides
cAMP
secondary messenger, used for protein hormones
Digestion is the process of
breaking down food into molecules the body can use, hydrolysis
diabetes - occurs when there are not enough insulin receptor sites - will occur when not enough insulin is secreted b the pancreas - kidney will excrete large amounts of water in order to remove excess sugar from the blood
covalent modification
when a phosphate is added to an enzyme to either activate or deactivate it
Glucagon
A protein hormone secreted by pancreatic endocrine cells that raises blood glucose levels;
an antagonistic hormone to insulin - famine hormone
Where is galactose converted to glucose?
liver
epinephrine
A substance secreted from the adrenal glands in response to a threat, strenuous exercise or
stress - reaches receptor sites on muscle/liver and stimulates glycogenolysis
where does digestion start?
carbohydrates - mouth
proteins - stomach
lipids - small intestine
the enzyme in step 4 of the Krebs cycle is an allosteric enzyme with sites for ATP, ADP and
NADH, which will active the enzyme?
ADP, needs more
what produces the most CO2?
Krebs cycle
electron transport - uses electrons from the Krebs cycle to change ADP to ATP - uses O2 - converts FADH2 to FAD - oxidative phosphorylation
Produce the most ATP per glucose
glucose metabolism in a typical cell under aerobic conditions
high energy
more hydrogens and less oxygens
percent of energy stored as ATP in aerobic oxidation of glucose
40%
NADH
highest energy compound
mitochondria - a cell which must do a lot of work will have more than one that doesn't work - a cell can produce more mitochondria if it needs to do more work
blood glucose
primary source of energy in red blood cells and brain during healthy eating
ketone bodies
the product of the incomplete breakdown of fat when glucose is not available in the cells - primary source of energy in the brain during starvation - too much in blood can cause ketosis - since 2/3 ketone bodies are acids it can lead to acidosis - synthesized by liver
fatty acids
long carbon chains that make up most lipids - primary source of energy in resting muscle and liver
glycogen
primary source of energy in muscle during strenuous activity
if no oxygen present, which processes occur?
Krebs cycle, glycolysis, and glycogenolysis
Pyruvate to Acetyl CoA
1. decarboxylation
2. oxidation
3. synthesis
how many COs are produced from the complete aerobic oxidation of 4 acetyl-CoA
8
how many ATP are produced from anaerobic oxidation of 9 glucose molecules?
18
complete aerobic oxidation of 5 pyruvic kids will produce how many CO2?
15
For 6 glucose molecules going to pyruvic acid, how many NAD molecules are used up?
12
fermentation
Process by which cells release energy in the absence of oxygen - pyruvic acid to ethanol
metabolism of fatty acids - long chains are broken down into groups of carbons
body temperature
energy from glucose metabolism is not stored as ATP and is used for maintaining
step #1 Krebs cycle - enzyme = synthetase - substrates - acetyl CoA and oxaloacetic acid - product - citric acid
ketone body that is not acidic
acetone
CHEM103 / CHEM 103 Module 3 Exam
Questions & Answers | Latest 2026–2027
Update | General Chemistry I with Lab |
Portage Learning | Verified Solutions
...
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Grade A
2026 / 2027 Academic Year
Q: Heat temp change = qtemp change = m x c x ∆t
Answer -(40.5 x 4.184 x (∆t - 85.7)) = 36.8 x 4.184 x (∆t - 26.3) -(169.452 x (∆t - 85.7)) = 153.9712 x (∆t - 26.3) -(169.452∆t - 14522.0364) = 153.9712∆t - 4049.44256 -169.452∆t + 14522.0364 = 153.9712∆t - 4049.44256
18571.479 = 323.423∆t
57.4 C = ∆t
1. Show the calculation of the final temperature of the mixture when a 40.5 gram sample of
water at 85.7C is added to a 36.8 gram sample of water at 26.3C in a coffee cup calorimeter.
c (water) = 4.184 J/g C
Q: qs↔l = mass x Heat of Fusion = m x ∆Hfusion
Answer
120 x 0.334 = 40.08 kJ
2. Show the calculation of the energy involved in melting 120 grams of ice at 0oC if the Heat
of Fusion for water is 0.334 kJ/g.
Q: moles = grams/molecular weight
Answer
moles (S) = 42.8/32.07 = 1.335 mols S
= ΔHrx x new moles / original moles
q = -792 x ( 1.335/2 )= -526.7 kJ
3. Sulfur undergoes combustion to yield sulfur trioxide by the following reaction equation:
2 S + 3 O2 → 2 SO3 ΔH = - 792 kJ
If 42.8 g of S is reacted with excess O2, what will be the amount of heat given off?
Q: moles = grams/molecular weight
Answer
moles (H2S ) = 26.2/34.086 = 0.7686 mols H2S
ΔHrx = q / (new moles / original moles)
q reaction = - 58,791 J + (-295,365 J) = - 354,156 J = - 354,156 J x 1 kJ / 1000 J = - 354.16
kJ -431.8 / ( 0.7686/2) = -1123.6 kJ
4. Hydrosulfuric acid (H2S) undergoes combustion to yield sulfur dioxide and water by the
following reaction equation:
2 H2S + 3 O2 → 2 SO2 + 2 H2O
What is the ΔH of the reaction if 26.2 g of H2S reacts with excess O2 to yield 431.8 kJ?
1)
q water = s (specific heat of water) x mass x Δt = 4.18 J / g / K x 500 g x (53.13 C - 25.00 C)
= - 58,791 J
q calorimeter = heat capacity x Δt = (10.5 kJ/C) x (53.13 C - 25.00 C) = - 295.365 kJ x 1000
J/1 kJ= - 295,365 J
(new) moles C6H6 = 7.05 g / 78 = 0.0904 mole C6H6
ΔH = q / (new moles / original moles)
ΔH = - 354.16 / (0.09048 / 2) = - 7828 kJ / mole
(2) Isolated system (bomb calorimeter)
(3) Exothermic (temperature of water rises due to heat given off by combustion reaction)
5. A sample of benzene (C6H6), weighing 7.05 g underwent combustion in a bomb
calorimeter by the following reaction:
2 C6H6 (l) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (l)
Q: The heat given off was absorbed by 500 g of water and caused the temperature of the
water and the calorimeter to rise from 25.00 to 53.13 C. The heat capacity of water = 4.18
J/g/C and the heat capacity of the calorimeter = 10.5 kJ/C. (1) what is the ΔH of the
reaction? Using the definitions at the beginning of the module describe (2) the calorimeter
+ contents, (3) the type of process.
Answer
q water = s (specific heat of water) x mass x Δt = 4.18 J / g / K x 100 g x 5.14 K = 2148.52 J
q calorimeter = heat capacity x Δt = 150 x 5.14 K = 771 J
q reaction = 771 J + 2148.52 J = + 2919.52 J = + 2919.52 J x 1 kJ / 1000 J = 2.92 kJ
(1) ΔH reaction = 2.92 kJ / 14.5 g NaHCO3 / 84 g NaHCO3 / mol NaHCO3 = + 16.9 kJ /
mol
(2) The calorimeter + contents (no lid) = open system, (3) Endothermic process
6. A sample of 14.5 g of sodium bicarbonate (NaHCO3) was dissolved in 100 ml of water in
a coffee-cup calorimeter with no lid by the following reaction
Answer
NaHCO3 (s) → Na+ (aq) + HCO3- (aq)
Q: If the temperature of the water and the calorimeter (heat capacity of calorimeter = 150
J/C) decreases from 25.00oC to 19.86C, (1) what is the ΔH of the reaction? Using the
definitions at the beginning of the module describe (2) the calorimeter + contents, (3) the
type of process.
Answer
2 (2 NH3 (g) → 3 H2 (g) + 2 N2 (g) ΔH = + 91.8 kJ)
3 (2 H2 (g) + O2 (g) → 2 H2O (g) ΔH = - 483.7 kJ)
2 (N2 (g) + O2 (g) → 2 NO (g) ΔH = 180.6 kJ)
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g) ΔHrxn = - 906.3 kJ
ΔHrxn = 2 (+ 91.8) + 3 (-483.7) + 2 (180.6) = - 906.3 kJ
7. The combustion of ammonia by the following reaction yields nitric oxide and water
Answer
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g)
Determine the heat of reaction (ΔHrxn) for this reaction by using the following
thermochemical data:
N2 (g) + O2 (g) → 2 NO (g) ΔH = 180.6 kJ
N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔH = - 91.8 kJ
2 H2 (g) + O2 (g) → 2 H2O (g) ΔH = - 483.7 kJ
Q: C2H5OH (l) → 2 C (graphite) + 3 H2 (g) + 1/2 O2 (g) ΔHf0 (C2H5OH) = -
Exam
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