Ch.3 Flashcards
Chemical Stoichiometry
Study of the quantities of materials consumed and produced in chemical reactions
Requires understanding the concept of relative atomic masses
Counting Objects by Their Mass
Average mass of objects is required to count the objects by weighing
- For purpose of counting, objects behave as though they are all identical
Sample of matter can contain huge numbers of atoms
-Number of atoms in a sample be determined by finding its mass
Modern System of Atomic Masses
Instituted in 1961
Standard - 12C
- 12C is assigned a mass of exactly 12 atomic mass units (u)
- Masses of all other atoms are given relative to this standard
Atomic Mass
We previously learned that not all atoms of an element have the same mass
-Isotopes
We generally use the average mass of all an element’s atoms found in a sample in calculations
- However, the average must take into account the abundance of each isotope in the sample
We call the average mass the atomic mass
Atomic Mass = ∑(fractional abundance of isotope)n x (mass of isotope)
Mass Spectrometry
Masses and abundance of isotopes are measured with a mass spectrometer
Atoms or molecules are ionized, then accelerated down a tube
- Some molecules into fragments are broken during the ionization process
- These fragments can be used to help determine the structure of the molecule
Their path is bent by a magnetic field, separating them by mass
- Similar to Thomson’s Cathode Ray E.
Mass Spectrum
A mass spectrum is a graph that gives the relative mass and relative abundance of each particle
Relative mass of the particle is plotted in the x-axis
Relative abundance of the particle is plotted in the y-axis
Average Atomic Mass of Carbon
Natural carbon is a mixture of 12C, 13C, and 14C
- Atomic mass of carbon is an average value of these three isotopes
Composition of natural carbon:
12C atoms (mass = 12 u) - 98.89%
13C atoms (mass - 13.003355 u) - 1.11%
Calculation of average atomic mass for natural carbon
98.89% of 12 u + 1.11% of 13.0034 u
= (0.9889)(12 u) + (0.0111)(13.0034 u)
= 12.01 u
For stoichiometric purposes, assume that carbon is composed of only one type of atom with a mass of 12.01
Mole (mol)
Unit of measure established for us in counting atoms
Number equal to the number of carbon atoms in exactly 12 grams of pure 12C
- Techniques such as mass spectrometry determines this number to be 6.02214 x 10^23
Avogadro’s number: one mole of something consists of 6.022 x 10^23 units of that substance
Using the Mole in Chemical Calculations
12 g of 12C contains 6.022 x 10^23 atoms, and 12.01 g sample of natural carbon also contains 6.022 x 10^23 atoms
Ratio of the masses of the samples (12 g/12.01 g) is the same as the ratio of masses of individual components (12 u/12.01 u)
- Both samples contain the same number of atoms (6.022 x 10^23 atoms)
Mole is defined such that a sample of a natural element with a mass equal to the element’s atomic mass expressed in grams contains 1 mole of atoms
Since 6.022 x 10^23 atoms of carbon (each with a mass of 12 u) have a mass of 12 g, then
Relationship can be used to derive the unit factor to convert between atomic mass units and grams
The Mole Concept and Chemical Compounds
Formula mass
- The mass of a formula unit in atomic mass units (u)
Molecular mass
- A formula mass of a molecular compound
Weighted average mass
- Add up the weighted average atomic masses
Exact Mass
-Add up the isotopic masses (see mass spectrometry)
Formula Mass
The mass of an individual molecule or formula unit
- Also known as molecular mass or molecular weight
Sum of the masses of the atoms in a single molecule or formula unit
- Whole = sum of the parts!
Molar Mass of Compounds
The relative masses of molecules can be calculated from atomic masses
Formula Mass = 1 molecule of H2O
= 2(1.01 amu H) + 16.00 amu O = 18.02 amu
1 mole of H2O contains 2 moles of H and 1 mole of O so
Molar mass = 1 mole H2O
= 2(1.01 g H) + 16.00 g O = 18.02 g
So the Molar Mass of H2O is 18.02 g/mol
Molar Mass of a Substance
Mass in grams of one mole of the compound
- Traditionally known as molecular weight
Obtained by summing the masses of the component atoms of a compound
Ionic Substances
Contain simple ions and polyatomic ions
Fundamental unit of an ionic compound is called a formula units
Example - NaCl is the formula unit for sodium chloride
Methods of Approaching a Problem
Pigeonholing method
- Emphasizes memorization
- Involves labeling the problem
- Slotting the problem into the pigeonhole that fits best
Provides steps that one can memorize and store in an appropriate slot for each different problem
Challenge: require a new pigeonhole for every new problem
Conceptual problem solving
- Helps understand the big picture
- Involves looking for a solution within the problem
- Each problem is assumed as a new one
- Involves asking a series of questions
- One uses their knowledge of fundamental principles of chemistry to answer the questions
Conceptual Problem Solving
Method that helps problems in a flexible and creative manner based on the understanding of fundamental concepts of chemistry
Goal of the text
-To help individuals solve new problems on their own
Understanding Conceptual Problem Solving
Where are we going?
- Read the problem and decide on the final goal
- Sort through the given facts and focus on the key words
- Draw a diagram of the problem
- This stage involves a simple, visual analysis of the problem
How do we get there?
- Involves working backwards from the final goal in order to decide the starting point
- Understanding of the fundamental principles can help lead to the final solution
Reality check
-Involves checking if the answer makes sense and whether the answer is reasonable
Method’s of Describing a Compound’s Competition
In terms of the numbers of the compound’s constituent atoms
In terms of mass percent (weight percent)
Mass % = (mass of an element in 1 mol of the compound / mass of 1 mol of the compound ) x 100%
Determining the Formula of a Compound: Process
Weigh the sample of the compound by:
- Decomposing the samples into its constituent elements
- Reacting it with oxygen
Resulting substances are then collected and weighed
- This type of analysis is done by a combustion device
- Results of such analyses proved the mass of each type of element in the compound which can be used to determine the mass percent of each element
Empirical Formula
Any molecule that can be represented as (CH5N)n has the empirical formula CH5N
- n = integer
- Molecular formula: exact formula of the molecules present in a substance
Molecular formula = (empirical formula)n
Requires the knowledge of the molar mass
Chemical Change
Involves the reorganization of atoms in one or more substances
- Atoms are neither created nor destroyed
Represented by a chemical equation
- Reactants: presented on the left side of an arrow
- Products: presented on the right side of an arrow
Problem Solving Strategy: Empirical Formula Determination
Base the calculation on 100 g of compound as mass percentage gives the number of grams of a particular element per 100 g of compound
- Each percent will then present the mass in grams if the element
Determine the number of moles of each element present in 100g of compound using the atomic masses of the elements present
Divide each value of the number of moles by the smallest of the values
- If each resulting number
Balancing a Chemical Equation
All atoms present in the reactants must be accounted for among the products
- Same number of each type of atoms should be present on the products side and on the reactants side of the arrow
Information Provided by Chemical Equations
Nature of the reactants and products
- Compounds involved in the reaction
- Physical states of the reactants and products
Relative numbers of reactants and products
- Indicated by coefficients in a balanced equation
Mass remains constant
-Atoms are conserved in a chemical reaction
Problem-Solving Strategy: Writing and Balancing the Equation for a Chemical Reaction
1) Determine what reaction is occurring
a) Determine the reactants, the products, and the physical states involved
2) Write the unbalanced equation that summarizes the reaction
3) Balance the equation by inspection, starting with the most complicated molecule(s)
a) Determine what coefficients are necessary
i) Same number of each type of atom needs to appear on both reactant and product sides
b) Do not change the formulas of any of the reactants or products
Problem Solving Strategy: Calculating Masses of Reactants and Products in Reactions
Balance the equation for the reaction
Convert the known mass of the reactant or product to moles of that substance
Use the balanced equation to set up the appropriate mole ratios
Use the appropriate mole ratios to calculate the number of moles of the desired reactant or product
Convert from moles back to grams if required by the problem
Stoichiometry Mixture
A type of mixture that contains relative amounts of reactants that match the numbers in the balanced equation
3H2(g) + N2(g) → 2NH3(g)
Limiting Reactant
One that runs out first and thus limits the amount of product that can form
- To determine how much product can be formed from a given mixtures of reactants, one must look for the reactant that is limiting
Some mixtures can be stoichiometric
- All reactants run out at the same time
- Requires determining which reactant is limiting
Determining of the Limiting Reactant Using Reactant Quantities
Compare the moles of reactants to ascertain which runs out first
- Use moles of molecules instead of individual molecules
- Use the balanced equation to determine the limiting reactant
- Determine the amount of limiting reactant formed
Use the amount of limiting reactant formed to compute the quantity of the product
Determination of Limiting Reactants Using Quantities of Products Formed
Consider the amounts of products that can be formed by completely consuming each reactant
- Reactant that produces the smallest amount of product must run out first and thus be the limiting reactant
Concept of Yield
Theoretical yield: amount of product formed when the limiting reactant is entirely consumed
- Amount of product predicted is rarely obtained b/c of side reactions and other complications
Percent yield: actual yield of product
- Often provided as a percentage of the theoretical yield
Actual Yield/Theoretical Yield x 100% = percent yield
Problem Solving Strategy
Solving a stoichiometry involving masses of reactants and products
1) Write and balance the equation for the reaction
2) Convert the known masses of substances to moles
3) Determine which reactant is limiting
4) Using the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product
5) Convert from moles to grams, using the molar mass
