Ch.3

Question 1

Chemical Stoichiometry

Answer 1

Study of the quantities of materials consumed and produced in chemical reactions

Requires understanding the concept of relative atomic masses

Question 2

Counting Objects by Their Mass

Answer 2

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

Question 3

Modern System of Atomic Masses

Answer 3

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

Question 4

Atomic Mass

Answer 4

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)

Question 5

Mass Spectrometry

Answer 5

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.

Question 6

Mass Spectrum

Answer 6

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

Question 7

Average Atomic Mass of Carbon

Answer 7

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

Question 8

Mole (mol)

Answer 8

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

Question 9

Using the Mole in Chemical Calculations

Answer 9

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

Question 10

The Mole Concept and Chemical Compounds

Answer 10

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)

Question 11

Formula Mass

Answer 11

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!

Question 12

Molar Mass of Compounds

Answer 12

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

Question 13

Molar Mass of a Substance

Answer 13

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

Question 14

Ionic Substances

Answer 14

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

Question 15

Methods of Approaching a Problem

Answer 15

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

Question 15

Conceptual Problem Solving

Answer 15

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

Question 16

Understanding Conceptual Problem Solving

Answer 16

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

Question 17

Method’s of Describing a Compound’s Competition

Answer 17

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%

Question 18

Determining the Formula of a Compound: Process

Answer 18

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

Question 19

Empirical Formula

Answer 19

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

Question 20

Chemical Change

Answer 20

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

Question 20

Problem Solving Strategy: Empirical Formula Determination

Answer 20

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

Question 21

Balancing a Chemical Equation

Answer 21

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

Question 22

Information Provided by Chemical Equations

Answer 22

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

Question 23

Problem-Solving Strategy: Writing and Balancing the Equation for a Chemical Reaction

Answer 23

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

Question 24

Problem Solving Strategy: Calculating Masses of Reactants and Products in Reactions

Answer 24

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

Question 25

Stoichiometry Mixture

Answer 25

A type of mixture that contains relative amounts of reactants that match the numbers in the balanced equation

3H2(g) + N2(g) → 2NH3(g)

Question 26

Limiting Reactant

Answer 26

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

Question 27

Determining of the Limiting Reactant Using Reactant Quantities

Answer 27

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

Question 28

Determination of Limiting Reactants Using Quantities of Products Formed

Answer 28

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

Question 29

Concept of Yield

Answer 29

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

Question 30

Problem Solving Strategy

Answer 30

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