TWEED RIVER HIGH SCHOOL - Year 12 Chemistry
TWEED RIVER HIGH SCHOOL
PRELIMINARY CHEMISTRY
Unit 2
Metals
Part 4
For efficient resource use, industrial chemical reactions must use measured amounts of each reactant
• define the mole as the number of atoms in exactly 12g of carbon-12 (Avogadro’s number)
Introduction
Relative Atomic Mass
Isotope: An isotope is an atom of an element that has the same number of protons but a different number of neutrons. For example, Carbon has the following isotopes: 12, 13 and 14.
This means that Carbon 12 has 6 protons and 6 neutrons; Carbon 13 has 6 protons and 7 neutrons and Carbon 14 has 6 protons and 8 neutrons.
The mass that appears on a periodic table is the Relative Atomic Mass. It is the mass of that element relative to the Carbon 12 atom.
For example:
The relative atomic mass of Calcium is 40.08. This means that the Calcium atom is 3.34 (40.08 ( 12) times heavier than Carbon.
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Also it is the ratio of the naturally occurring isotopes of that element.
For Example:
Chlorine has the naturally occurring isotopes 75% 35Cl and 25% 37Cl.
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Therefore, the relative atomic mass of Chlorine is:
(0.75 x 35) + (0.25 x 37) = 35.5
Which indicates that a chlorine atom is 2.96 (35.5 ( 12) times heavier than a carbon atom.
Avogadro’s Number
Avogadro’s number, NA, is the number of atoms in exactly 12 grams of carbon.
Avogadro’s number:
NA = 6.022 x 1023
This means that 12g of carbon has 6.022 x 1023 atoms.
Homework:
Answer the following questions:
1. How many atoms are there in 24.312g of magnesium?
2. How many atoms are there in 20.04g of calcium?
3. How many atoms are there in 127.08g of copper?
The MOLE
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Mole song
A mole of an element is the mass which in grams is numerically equal to the relative atomic mass.
That is:
12g of carbon = 1 mole
40.08g of calcium = 1 mole
16g oxygen = 1 mole
Therefore:
1 mole of any element = 6.022 x 1023 atoms.
Number of moles = number of particles
Number of particles in 1 mole
n = __N_____
NA
• solve problems and analyse information from secondary sources to perform calculations involving Avogadro’s number and the equation for calculating the number of moles of a substance:
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Number of moles = ______mass_______
Mass of 1 mole
n = _m_
M
Therefore:
20.04g of calcium = ½ mole or 3.011 x 1023 atoms
Homework:
Answer the following questions:
1. How many:
a. Atoms
b. Moles
In
- 19.55g of potassium
- 36.048g of Beryllium
- 8.016g of sulfur
Molecular Mass
The molecular mass is the mass of a molecule of the compound relative to the mass of a carbon 12 atom.
To calculate the molecular mass of a substance, add up the mass of all the elements in the compound.
For example:
Methane has the chemical formula CH4 .
This means that there is 1 carbon atom and 4 hydrogen atoms in each molecule. Therefore, methane has a molecular mass of:
C = 12.01g
H = 1.008g
Molecular mass = (12.01 x 1) + (1.008 x 4) = 16.072g
Therefore:
16g methane = 1 mole methane = 6.022 x 1023 molecules
Homework:
Calculate the molecular mass of the following compounds:
1. Carbon dioxide CO2
2. Ethanol C2H5OH
3. Glucose C6H12O6
Formula Mass
Remember: the formula for an ionic compound, is expressed as the ratio of elements that form the ionic lattice.
The formula mass for common table salt, sodium chloride, NaCl is
Atomic mass Na = 22.99
Cl = 35.45
Therefore the formula mass for NaCl = 22.99 + 35.45 = 58.44
Hence, there is 58.44g in 1 mole of NaCl.
Homework:
Calculate the formula mass for the following ionic compounds:
1. magnesium oxide
2. potassium fluoride
3. calcium carbonate
4. sodium sulfate
5. magnesium nitrate
Work through Example 11.5 and Review Exercises 11.4, p176 Chemistry Contexts 1.
Work through Example 11.6 p177 and Review Exercises 11.5, p178 Chemistry Contexts 1.
• recount Avogadro’s law and describe its importance in developing the mole concept
Avogadro’s Law
Avogadro’s Law states that under the same conditions of temperature and pressure, the same volume of gases contain equal numbers of molecules. (Or the same number of moles).
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• distinguish between empirical formulae and molecular formulae
Empirical Formula
The empirical formula of a compound is the formula which expresses the atoms in the compound in the simplest whole number ration. For example:
The empirical formula for magnesium oxide is MgO.
Glucose, has the molecular formula - C6H12O6 and has the empirical formula CH2O
Molecular Formula
The molecular formula is the actual number ratio of the atoms in a compound.
The molecular formula is always a multiple of the empirical formula.
In class work through example 11.8, p180 Chemistry Contexts 1
Do exercises 10a and 13 p196 Chemistry Contexts 1
• describe the contribution of Gay-Lussac to the understanding of gaseous reactions and apply this to an understanding of the mole concept
Gay-Lussac’s Law of Combining Gas Volumes.
The ratios of the volumes of gases involved in a reaction, if measured at the same temperature and pressure, are expressed by small, whole number ratios.
In class work through example 11.9, p183 Chemistry Contexts 1
Do exercises 11.8 and 13 p183 Chemistry Contexts 1
• process information from secondary sources to interpret balanced chemical equations in terms of mole ratios
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Stoichiometric Calculations
- A chemical equation tells us the ratios by weight in which substances react or are formed in a reaction.
- Chemical equations indicate the number of moles if reactants and products in a chemical reaction.
- Stoichiometric calculations involve the calculation of the masses and volumes of gases of products formed in chemical reactions.
Mass – Mass calculations
In carrying out stoichiometric calculations it is useful to remember the following:
n = m from n = m
M M
equation
For example:
If 26.13g of barium nitrate is reacted with excess potassium sulfate, what is the mass of barium sulfate formed?
Step 1:
Write a balanced chemical equation:
Ba(NO3)2 + K2SO4 ( BaSO4 + 2KNO3
1 1 1 2
This shows that 1 mole of barium nitrate reacts with 1 mole of potassium sulfate to produce 1 mole of barium sulfate and 2 moles of potassium nitrate.
Calculations:
1. Mass of known = 26.13g
2. Moles of known = n = m
M
M = formula (molecular mass)
m = mass of reactant
Formula mass = sum of atomic masses of the elements in the compound
Barium = 137.3
Nitrogen = 14
Oxygen = 16
Therefore for the compound Ba(NO3)2
the formula mass = 137.3 + (14 x 2) + (16 x 3 x 2) = 261.3g
(
n = m = 26.13 = 0.1 mole
M 261.3
Moles of unknown from stoichiometric ratio = 0.1 mole
M (formula mass of barium sulfate) BaSO4 =
Barium = 137.3g
Sulfur = 32.1g
O = 16g
Therefore for the compound BaSO4
the formula mass = 137.3 + 32.1 + (16 x 4) = 233.4g
Mass of unknown
n = m
M
← m = n x M
= 0.1 x 233.4
= 23.34g
Example 2:
What mass of sodium carbonate is needed to react with 14.1g of nitric acid, and how much sodium nitrate is formed?
Follow the processes as in example 1.
Attempt the following exercises:
Questions 5, 6 and 7 p187 Chemistry Contexts 1
Questions 17 and 18 p197 Chemistry Contexts 1
• process information from secondary sources to investigate the relationship between the volumes of gases involved in reactions involving a metal and relate this to an understanding of the mole
Mass - Volume Calculations
The calculations for mass – volume calculations follows the same processes as in mass – mass calculations.
n = m from n = m
M M
equation
Restating Avogadro’s Law in terms of volumes of gases:
Equal numbers of molecules of different gases occupy the same volume, at the same temperature and pressure.
Thus one mole of any gas at the same temperature and pressure occupies the same volume.
When calculating volumes of gases standard conditions are used. These are:
One mole of any gas occupies:
1. At Standard Temperature and Pressure, (STP), 0(C and 101.3 kPa
1 mole of any gas occupies = 22.4 Lmol-1
2. At Standard Laboratory Conditions, (SLC), 25(C and 101.3 kPa
1 mole of any gas occupies = 24.5 Lmol-1
For example:
1. What is the volume of hydrogen gas produced, at STP, when 4.86g of magnesium is reacted with excess Hydrochloric acid.
Step 1:
Write a balanced chemical equation:
Mg + 2HCl ( MgCl2 + H2
1 2 1 1
This shows that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of magnesium chloride and 1 mole of hydrogen.
Calculations:
2. Mass of known = 4.86 g
Moles of known = n = m
M
M = formula (molecular mass)
m = mass of reactant
Atomic mass magnesium =
Magnesium = 24.3g
( Number of moles of magnesium=
n = m = 4.86 = 0.2 mole
M 24.3
Moles of hydrogen produced from stoichiometric ratio = 0.2 mole
When calculating volume of gas produced the following equation is used:
Volume of gas produced = number of moles x molar volume (MV)
V = n x MV (In this example molar volume is at STP = 22.4 Lmol-1)
V(hydrogen) = 0.2 x 22.4 (mol x Lmol-1)
= 4.48 L
Volumes of Gases where reactants and products are gases:
Example 11.9 p183 Chemistry Contexts 1
• compare mass changes in samples of metals when they combine with oxygen
• perform a first-hand investigation to measure and identify the mass ratios of metal to non-metal(s) in a common compound and calculate its empirical formula
• perform a first-hand investigation to measure and identify the mass ratios of metal to non-metal(s) in a common compound and calculate its empirical formula
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Mass
of Unknown
Moles
of
Unknown
Mass
of
Known
Moles
of
Known
Volume
of Unknown
Moles
of
Unknown
Mass
of
Known
Moles
of
Known
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