2.1 Quantitative Analysis (Stoichiometry)
The assessment for this standard will be comprised of two parts
1. Practical: Quantitative analysis involves collecting primary data from an acid-base titration, and processing both primary and secondary data to solve quantitative problems. 2. Written: solving quantitative problems that may involve more than two steps and use stoichiometric principles such as n = m/M and c = n/V. Answers to calculations must demonstrate correct units and appropriate use of significant figures. The LOWEST score between the two sections will be your final grade. Read the assessment specifications carefully for details of what is expected to achieve at each grade level. |
|
Chapter 6 - The Mole
2.1.1 Relative Mass and Percent composition
Lesson Overview:
Protons and neutrons contribute to the mass of an atom. Because every element has a different number of protons and neutrons, every element will have different atomic mass. The atomic mass can be used to determine the relative composition of a compound by mass. Workbook Questions:
Relative Mass, 6A Percent Composition, 6B
|
|
2.1.2 Mole
Lesson Overview:
Whenever we balance equations, we are making sure the same number of each atom is on each side of the equation. From now on, we will talk about balancing moles of atoms instead of just individual atoms. Atomic mass is, for our purposes, the same as molar mass. The molar mass of an element or molecule, is the mass that one mole of the substance weighs in grams. Definition:
Mole, also spelled mol, in chemistry, a standard scientific unit for measuring large quantities of very small entities such as atoms, molecules, or other specified particles. Description: The mole designates an extremely large number of units, 6.02214179 × 10^23, which is the number of atoms determined experimentally to be found in 12 grams of carbon-12. Carbon-12 was chosen arbitrarily to serve as the reference standard of the mole unit for the International System of Units (SI). The number of units in a mole also bears the name Avogadro’s number, or Avogadro’s constant, in honour of the Italian physicist Amedeo Avogadro (1776–1856). Avogadro proposed that equal volumes of gases under the same conditions contain the same number of molecules, a hypothesis that proved useful in determining atomic and molecular weights and which led to the concept of the mole. (See Avogadro’s law.) The number of atoms or other particles in a mole is the same for all substances. The mole is related to the mass of an element in the following way: one mole of carbon-12 atoms has 6.02214179 × 10^23 atoms and a mass of 12 grams. In comparison, one mole of oxygen consists, by definition, of the same number of atoms as carbon-12, but it has a mass of 16 grams. Oxygen, therefore, has a greater mass than carbon. This reasoning also can be applied to molecular or formula weights. The concept of the mole helps to put quantitative information about what happens in a chemical equation on a macroscopic level. The mole can be used to determine the simplest formula of a compound and to calculate the quantities involved in chemical reactions. When dealing with reactions that take place in solutions, the related concept of molarity is useful. Molarity (M) is defined as the number of moles of a solute in a litre of solution. http://www.britannica.com/science/mole-chemistry Workbook Questions: The Mole, 6C
|
|
2.1.3 Stoichiometry
Lesson Overview:
When carrying out reactions in a lab, you may want to have a reaction go to completion with out having any excess reactants. Or you may want to know how much product you will get by reacting a specific amount of one reactant in excess of another. When making these calculations, use the formula n = m / M and remember to make into account the mole ratios if there is not a 1:1 ratio. Use the tables below to help you stay organized. HOMEWORK Workbook Questions: Using The Mole, 6E
|
|
2.1.4 Water of Crystalization
Lesson Overview:
When many salts crystalize, they trap excess water into their crystal structure. This water can be evaporated off with excess heat. The difference in mass between the hydrated and dehydrated salts can be used to calculate how much water was in the original sample. Use principals of stoichiometry to calculate the water of crystallization for one or more hydrated salts. In Class Activity:
Practical 6.2 (demonstration only) Practical 6.3 Workbook Questions: Water of Crystallization, 6D
Homework: Textbook questions that have been assigned or worked on in class Due will include:
6A, 6B, 6C, 6E and the chapter 6 review questions |
|
Chapter 7 - Solutions
2.1.5 Concentrations by mass and Molarity
Lesson Overview:
Concentration of solutions can be measured either using percent by mass (uncommon for this class) or by molarity (common). Percent by mass is calculated by measuring the mass of solute per 100g of solution. This method is used more commonly in clinical scenarios but is not very useful to most chemists. Molarity is measured in Moles per Liter. This is very useful for chemists as we can tell exactly how many atoms/molecules of a reactant are present in a given volume. The equation for concentration is M = n / V M = Molarity in mol/L (sometimes just "M") n = moles (mol) V = Volume in Liters (L) Workbook Questions:
Concentration by Mass: 7A Concentration in mols per liter: 7B
Homework: Worksheet 'Problems on Concentration'; both pages
|
|
2.1.6 Preparing a Standard Solution
Lesson Overview:
A standard solution is a solution that has been made with an exactly known concentration. See the videos for how a standard solution is prepared. |
|
2.1.7 Dilutions
Lesson Overview
Standard solutions are often made at high concentrations. That solution can then be diluted by adding water to make the correct concentration for a particular experiment. Because the amount of moles will remain constant, the following equation can be used to solve dilution problems: V1 x M1 = V2 x M2 The tables provided in section 2.1.3 may also be useful.
|
|
Chapter 8 - Titrations
2.1.8 Titrations
Lesson Overview:
Titrations are a laboratory procedure used to find the concentration of a solution by comparing it to a solution with a known concentration. This is usually done by reacting an acid with a base, using an indicator to see when the solution becomes neutral. Watch the video for a detailed explanation on how titrations are performed. You will be assessed on how accurately you can perform this procedure, so make sure you pay close attention to each step, as well as the care and precision that is being used. Workbook Questions: Titrations, 8A
|
|
2.1.9 Titration Calculations
Lesson Overview:
Use the table in section 2.1.3 to help keep your calculations organized. Workbook Questions: More complex Titration Problems, 8B
|
|
2.1 Assessment Review
Suggested Review Homework: Make sure you can answer all the questions in the Chapter 7 and Chapter 8 end of chapter reviews
You will not be tested on Empirical Formula or percent composition.
You will be tested on reacting masses, dilutions, titration calculations, converting between mass and concentration
-Read the assessment criteria for full details of what is expected
You will have to remember both formula: M = m/n and c = n/V
It is recommended that you set up a table when solving these problems.
You will not be tested on Empirical Formula or percent composition.
You will be tested on reacting masses, dilutions, titration calculations, converting between mass and concentration
-Read the assessment criteria for full details of what is expected
You will have to remember both formula: M = m/n and c = n/V
It is recommended that you set up a table when solving these problems.