Download Section 1.1 Intro

Unit 1: Stoichiometry
Atomic number ‐ the number of protons in an atom or ion
Mass number ‐ the sum of the protons and neutrons in an atom
Isotope ‐ atoms which have the same number of protons and electrons
but different numbers of neutrons
Ex:
https://www.youtube.com/watch?v=xjY5p­1
carbon‐12
(6p+6n)
carbon‐14
(6p+8n)
You can convert an isotope name to isotope notation using the mass number
and the atomic number for the element as shown in Example 1.
Example 1: Write the isotope notation for oxygen‐18 and list the number of
protons, electrons, and neutrons in an atom of this isotope.
Answer: Oxygen ‐ atomic number 8
‐ mass number is 18 (from isotope name)
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Isotope notation: 8O
Since the atomic number is 8, there are 8p+ and 8e‐.
Subtracting the atomic number from the mass number gives the
number of neutrons: 18 ‐ 8 = 10. Thus, oxygen‐18 has 10n°.
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You can convert isotope notation to an isotope name using the mass number
and the symbol for the element as shown in Example 2.
Example 2: Write the isotope name and list the number of protons,
238
electrons, and neutrons in an atom of this isotope: 92U.
Atomic Mass
An atomic mass unit (amu) is defined as exactly 1/12 of the mass of a carbon‐12
atom.
Carbon‐12 has 6p+ and 6n°so each proton or neutron has a mass of 1 amu.
The mass of an electron is insignificant in comparison ‐ about 0.0005 amu.
For example: The atomic mass of chlorine‐35 is 35 amu because its nucleus
contains 17 p+ and 18 n°. In other words, the atomic mass of
chlorine‐35 is based upon the mass unit defined using carbon‐12.
Atomic masses are relative units as opposed to measurable units.
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Average Atomic Mass
There are two naturally occurring isotopes of chlorine:
chlorine‐35 and chlorine‐37.
The atomic mass of this element is a combination of the two isotopes.
The relative abundance of chlorine atoms in nature is 75% chlorine‐35
and 25% chlorine‐37.
Average atomic mass is the weighted average of the atomic masses
of the isotopes of an element.
The average atomic mass is calculated by multiplying the percent
abundance by the mass number for each isotope and then adding the
products.
Exercise 1: Complete the table
Isotope Isotope Atomic # Mass # # protons # neutrons
electrons
Name Notation #
sulfur­3
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67
165
56
85
56
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Exercise 2: Write the isotope notation for each isotope of hydrogen.
hydrogen‐1
hydrogen‐2
hydrogen‐3
The average atomic mass of hydrogen is 1.01. Which isotope of hydrogen do you
think is most abundant in nature? Why?
Read: page 43‐44 Isotopes and atomic mass
Questions:
page 37: # 5 ‐ 7
page 45: # 1 ‐ 4 (#4 hint: Let one isotope be x, the other
will be 1‐x)
page 46: # 1 ‐ 6
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The Mole (Avogadro's Number)
A mole (mol) is the number of carbon‐12 atoms in exactly 12 grams of carbon‐12.
1 mol of a substance contains 6.02 x 1023 particles of the substance.
The constant 6.02 x 1023 mol‐1 is called the Avogardo's constant.
The conversion factor is molar mass (the mass of one mole).
Molar mass is on the periodic table.
1 mole of iron atoms
(6.02 x 1023 atoms)
has a mass of 55.85 g
Calculating Molar Mass
Example: Calculate the molar mass of water.
Molar mass values can be used
to convert mole amounts to
mass and vice versa.
Note: In this course, molar masses
are to be given to 2 decimal places,
regardless of significant figures.
Answer:
Chemical formula: H2O
List the number of atoms of each element: 2 H, 1 O
Multiply each number by the molar mass of the element:
Add the products together:
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Example: Calculate the molar mass of barium hydroxide octahydrate.
Calculate the molar mass of each compound:
a) NaHCO3
b) Sr(NO3)2
c) Al2(SO4)3
d) CuSO4
e) FeSO4·3H2O
f) lead(II) acetate
g) magnesium chloride
Answers:
84.01 g/mol
211.64 g/mol
342.17 g/mol
159.62 g/mol
205.97 g/mol
325.29 g/mol
95.21 g/mol
A note before we do calculations.... Sig Figs!
Zeros:
All zeros to the left of the first non‐zero digit are not significant
Ex: 0.00125 has 3
All zeros in between on‐zero digits are significant
Ex: 2002 has 4
Zeros after a non‐zero digit are not significant unless there is a decimal point
Ex: 800 has 1
800.0 has 4
Add/Subtract: Use the least # of decimal places
Ex: 1.0546 + 2 = 3.0546 = 3
Mulitply/Divide:
Use the least number of significant figures
Ex: 300 x 2.50 = 750 = 800
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Mass‐to‐Mole and Mole‐to‐Mass Conversions
Formula:
n is the number of moles
m is the mass
M is the molar mass
Example: Determine the number of moles in a 56.83 g piece of aluminum.
Example:
Diamonds are pure carbon. Calculate the number of moles of carbon
atoms in a 1.52 g diamond.
Example: Calculate the number of moles in a 48.12 g sample of cerium nitrate,
Ce(NO3)3?
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Example: A sample of neon gas contains 2.00 mol of atoms. Calculate the mass
of the sample.
Example:
Calculate the mass of 2.40 mol of aluminum nitrate Al(NO3)3.
Molar Volume
The SI unit of volume is the litre.
Chemists compare gas volumes at 0°C and 101.3 kPa. These
conditions are known as Standard Temperature and Pressure or STP.
Molar volume:
1 mol of any gas at STP has a volume of 22.4 L.
Mole to Volume Conversions
n is moles
Little v is volume
Big V is molar volume (22.4 L/mol)
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Example: Calculate the number of moles of neon in a 6.81 L sample at STP
conditions.
Example: Calculate the volume of 0.833 mol of sulfur trioxide gas at STP
conditions.
Homework
1. Calculate the mass of 88.6 mol copper
2. Calculate the number of moles in 369 L of oxygen gas at
STP conditions.
3. Calculate the volume of 0.295 mol of carbon monoxide at
STP conditions.
Page 59 #'s 20‐26
Page 73 #'s 38‐43
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Mole to Number of Particles Conversions
Formula: # Particles = nA
n is the number moles
A is Avogadro's number
The particles could be atoms (for an element), molecules (for a molecular
compound), or formula units (for ionic compounds)
Example: How many particles are there in each sample?
a) 0.25 mol of helium (atoms)
b) 0.750 mol of carbon dioxide (molecules)
c) 1.20 mol of sodium sulfide (formula units)
Example:
Convert these numbers of particles to mole amounts.
a) 1.204 x 1024 argon atoms
b) 1.35 x 1022 sodium chloride formula units
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Example:
a) What is the volume of 2.7 g of hydrogen gas
(assume H2 or diatomic hydrogen) at STP conditions?
b) How many molecules are there in this sample?
1. Calculate the mass of each sample.
a) 0.0346 mol silver
b) 0.037 mol tin(IV) silicate
c) 4.23 mol Ba(BrO3)2
d) 65.72 mol diphosphorus pentaoxide
2. Calculate the number of moles in each sample of gas at
STP conditions.
a) 65 L of helium
b) 67 L of carbon dioxide
3. Calculate the volume of each amount of gas at STP
conditions.
a) 2.7 mol of krypton
b) 3.726 mol of dinitrogen monoxide
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4. Which sample in each pair contains the greater number of particles (i.e. greater
number of moles)?
a) 25.0 g of aluminum or 25.0 g of copper
b) 10.0 L of oxygen gas or 10.0 L of nitrogen gas (both at STP).
c) 15.0 L of hydrogen gas (at STP) or 98.77 g of calcium chloride
Qualitative Analysis versus Quantitative Analysis
Qualitative analysis is carried out to identify the composition of a substance or a
mixture. The focus is on determining which chemical species are present in a
sample.
Quantitative analysis involves identifying the quantity of each species is present in
a sample.
Percentage Composition is a type of quantitative analysis used to find the percent
by mass of each element in a substance and is a key step in the determination of
the compound's chemical formula.
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Calculating Percentage Composition
Example: AlBr3
Example: Calculate the percentage composition of ethanol, C2H5OH.
Two kinds of chemical formulas:
1. Empirical formula is a lowest whole number ratio of the elements in a
compound.
Example: A prospector finds a rock that may contain rutile ‐ an ore of titanium. A
crystal from the sample is analyzed and found to be 59.94% titanium
and 40.06% oxygen by mass. Determine the empirical formula of the
Remember
crystalline substance.
. . .
Percent to mass
Mass to mole
Divide by small
Multiply 'til whole
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2. Molecular formula shows the number of atoms of element in the
compound; it is not necessarily a lowest whole number ratio.
Example: An unknown was subjected to a combustion analysis and found to be
85.60% carbon and 14.40% hydrogen. Mass spectrometry indicated that
its molar mass was 28.06 g/mol. Determine the molecular formula of
the compound.
Practice Problems...
.
1.
What is the percent composition of epsom salts, MgSO4 7H2O ?
2.
Determine the empirical formula of a compound that is 74.39% gallium and 25.61% oxygen.
3.
Vanillin has a molar mass of 152.16 g/mol and the following percent composition: C: 63.14%, H: 5.31%, O: 31.55%. Determine the molecular formula of vanillin.
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3.
Vanillin has a molar mass of 152.16 g/mol and the following percent composition: C: 63.14%, H: 5.31%, O: 31.55%. Determine the molecular formula of vanillin.
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