New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here

Study Guide for Exam 2

Star Star Star Star Star
3 reviews
by: Leslie Pike

Study Guide for Exam 2 CHEM 120

Leslie Pike
GPA 3.9
College Chemistry I
Dr. Darwin Dahl

Almost Ready


These notes were just uploaded, and will be ready to view shortly.

Purchase these notes here, or revisit this page.

Either way, we'll remind you when they're ready :)

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

This exam covers Chapters 3 and 4 only; we are behind yet again. Study guide covers moles, percent composition, specific gravity, stoichiometry, concentrations and parts per million, acids and base...
College Chemistry I
Dr. Darwin Dahl
Study Guide
50 ?




Star Star Star Star Star
2 reviews
Star Star Star Star Star
"It's very helpful."
Joseph McIntyre
Star Star Star Star Star
"Very helpful :)"
Linyue Fan

Popular in College Chemistry I

Popular in Chemistry

This 6 page Study Guide was uploaded by Leslie Pike on Wednesday October 7, 2015. The Study Guide belongs to CHEM 120 at Western Kentucky University taught by Dr. Darwin Dahl in Summer 2015. Since its upload, it has received 121 views. For similar materials see College Chemistry I in Chemistry at Western Kentucky University.


Reviews for Study Guide for Exam 2

Star Star Star Star Star

It's very helpful.

-Joseph McIntyre

Star Star Star Star Star

Very helpful :)

-Linyue Fan

Star Star Star Star Star

-Chad Wilson


Report this Material


What is Karma?


Karma is the currency of StudySoup.

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/07/15
Study guide for Exam 2 to be administered October 12th at 8 am Moles An atomic mass unit is onetwelfth of the mass of one atom of carbon12 The masses of atoms are measured in atomic mass units There are 60221023 amus in one gram The molar mass is the mass of one mole of a substance One mole is 60221023 units of a substance A mole is a number like a pair or a dozen or a gross You can have a dozen toothbrushes or a gross of toothbrushes or a mole of toothbrushes technically speaking you would have nowhere to store one mole of toothbrushes but in theory you could have that many The mole is special because one mole of a substance weighs the same number in grams as one molecule weighs in amus This is very convenient when measuring proportions for a chemical reaction For instance diatomic hydrogen weighs 2 amus One mole of diatomic hydrogen weighs 2 grams Diatomic oxygen weighs 32 amus One mole of diatomic oxygen weighs 32 grams In order to make one mole of water you would have to weigh out 2 grams of hydrogen and 16 grams of oxygen Remember hydrogen and oxygen combine in a 21 ratio to make water As a general rule the molecular mass listed on the periodic table is used for the molar mass of the substance EXCEPTIONS TO GENERAL RULE hydrogen nitrogen oxygen uorine chlorine bromine and iodine These seven elements exist in gaseous form at room temperature As gases they form diatomic molecules UNLIKE noble gases which exist as single atoms Therefore a mole of nitrogen gas will weigh 28 grams NOT 14 grams However for percent composition problems you will NOT use the molecular weight you will use the atomic weight Compounds are composed of atoms not molecules Be sure to pay very close attention to which molar mass molecular or atomic the problem asks you for Example mole problem You are performing a chemical reaction which requires two moles of sucrose the scienti c name for ordinary table sugar Obviously you cannot count out 120441023 sucrose molecules You will have to weigh the sucrose to determine the amount you need To determine what one mole of sucrose weighs rst determine what one molecule of sucrose weighs The formula for sucrose is C12H22011 It has twelve carbon atoms 22 hydrogen atoms and 11 oxygen atoms Add up the masses of these atoms to get the weight of the entire molecule 342 amu Therefore one mole of sucrose will weigh 342 grams and two moles will weigh 684 grams Conversely you buy a 5pound bag of table sugar You want to know how many molecules of sucrose you have in the bag You are doing a conversion problem you have pounds and you want to end up with molecules First convert pounds to grams There are 454 grams in one pound so in 5 pounds there are 2270 grams Then convert grams to moles One mole of sucrose weighs 342 grams so divide 2270 grams substance by 342 grams per mole to get the number of moles 664 Now to get the number of molecules There are 60221023 molecules in one mole Therefore in 664 moles you will have about 41024 molecules This problem should be more hideous than anything you will see on a Chem 120 exam If you can understand this problem you39ll be able to work anything Dr Dahl gives you Percent Composition Say you are given an unknown compound that was analyzed in a lab and found to contain 2 hydrogen 33 sulfur and 65 oxygen You will need to nd the moles tomoles ratio of the atoms to identify the compound Convert these percentages to moles To do this assume you have 100 grams of the substance This means that two grams will be hydrogen 33 grams will be sulfur and 65 grams will be oxygen Convert grams to atomic moles NOT molecular moles this is where percent composition problems and stoichiometry problems differ Atomic hydrogen weighs one gram per mole so two grams of hydrogen is two moles Sulfur weighs 32 grams per mole you have 33 grams in our sample giving you just a little more than a mole Atomic oxygen weighs 16 grams per mole you have 65 grams which is just a little more than four moles Our empirical formula would be H2504 which is sulfuric acid This problem is made a little bit easier by the fact that one mole of sulfuric acid weighs 98 grams almost exactly the 100 grams assumed for the calculation however the concept is the same for all percent composition problems Conversely to determine percent composition from given ratios rst determine the molar mass of the compound Then divide the weight of the atom in the compound by the weight of the entire compound For example ironlll oxide weighs 158 gramsmole There are two moles of iron in every ironlll oxide mole lron has a molar mass of 55 gmol so divide 110158 to get 696 iron by weight in ironlll oxide NOTE Due to rounding the percent values may not exactly total 100 and the ratios may not quite be whole numbers If you get close and your answer makes sense you39ve done it correctly One way to determine percent composition of organic compounds compounds containing only carbon hydrogen and possibly oxygen is by combustion analysis Combustion analysis burns an organic compound in an atmosphere full of oxygen The resulting water and carbon dioxide are weighed To calculate the molecular formula of the compound rst take the weight of carbon dioxide and water and convert this into moles Water weighs 18 grams per mole and carbon dioxide weighs 46 grams per mole Then do molestomoles conversion two moles of hydrogen in one mole of water one mole of carbon in one mole of carbon dioxide You now have your ratio of carbon to hydrogen IMPORTANT NOTE YOU CANNOT USE THIS SAME PROCEDURE TO CALCULATE THE PERCENTAGE OF OXYGEN THE COMPOUND IS BURNED IN AN ATMOSPHERE OF OXYGEN THIS MEANS THE PRODUCTS WILL HAVE MORE OXYGEN THAN THE ORIGINAL COMPOUND DID AND YOU HAVE NO MEANS OF TELLING HOW MUCH MORE JUST BY LOOKING AT THE NUMBERS To calculate the amount of oxygen convert your carbon and hydrogen back to grams If this total weight is less than the weight of the original sample the remainder of the weight will be oxygen Convert this weight to moles atomic oxygen weighs 16 grams per mole Stoichiometry Stoichiometry can be de ned as the relationships between reactants and products in a chemical equation Stoichiometry problems start with a chemical equation usually it will not be balanced and you will be expected to balance it For example the decomposition of potassium chlorate into potassium chloride and oxygen gas KCO3 I KCI 02 This equation is unbalanced there are three oxygen moles on the left and two on the right Dr Dahl did not explain this in class but there is a shortcut to balancing equations and because the exam is multiple choice you will not be penalized for using the shortcut KCO3 I KCI 3202 Tada there are now three oxygen moles on both sides Some professors do not like fractions some are ne with them To be safe multiply everything by the denominator to remove the fraction Now you have dealt with the equation On to the rest of the problem Stoichiometry problems give you the mass of one reactant and ask you how many grams of product can be produced from this reactant Alternatively you may be given the mass of the product and be asked how many grams of reactant are needed If you know how to convert grams to moles good If you don39t go back to the rst section of this study guide and read it until you know how to convert grams to moles This is important knowledge I will be very surprised if it DOESN T show up on the exam Sometimes the product will be in aqueous solution and you will be dealing with molars instead of grams Regardless of which you are given CONVERT IT TO MOLES Converting molars to moles is actually simpler than converting grams to moles Multiply the molarity of the solution by the liters of solution The volume will cancel leaving you with the moles You may be given the volume in milliliters instead of liters In this case just move the decimal point three places to the left No big deal Speci c Gravity Speci c gravity is measured in units of grams per cubic centimeter or grams per mL By de nition water has a speci c gravity of 1 Everything else is based off of that mass of solution g S 39 G 39 pea c rawzy volumeof soluti0nmL Calculating molarity from speci c gravity is yet another factorlabel problem like most other chemistry problems Dr Dahl has gone over so far Sample problem what is the molarity of a 35 hydrochloric acid solution with a speci c gravity of 12 Hint in a 100 g sample that is 35 HCI by weight 35 g will be HCI and 100 g will be the weight of the entire solution 12 g solution 35 g HCl mL solution mol HCl 100 g solution 3645 g HCl L 1000 mL 115MHCl Parts per Million ppm The concentration of very dilute solutions is measured in parts per million ppm De nition as follows mass of solute mg m pp mass of solution kg Note the solute weighs so little that quotmass of solventquot and quotmass of solutionquot can be used interchangeably here If the solvent is water we can take a shortcut because by de nition water weighs one kilogram per liter mass of solute mg m pp volume of solution L Sample problem You need to prepare a 300 mL solution of 50ppm Cl All you have available is barium chloride and water How much barium chloride in grams do you need 50quotI 3 Cl 03 L Cl 1 8 1000 mg Cl 3454 g Cl 2mol Cl mol Ba Cl2 mol Cl mol Ba Cl2 2079 g Ba Cl2 20044 g Ba Cl2 This gives you the weight of barium chloride Dilutions The relationship of dilutions is de ned as follows MiViszVf This is a plugandchug type of problem Plug in the knowns and use algebra to solve for the unknown Acids Bases Salts and Precipitates An acid is a molecule that can lose a proton A base is a molecule that can gain a proton or lose a hydroxyl group Strong acids and bases dissociate completely in water weak acids and bases only dissociate a small percent in water The six strong acids are Hydrochloric acid HCI Hydrobromic acid HBr Hydroiodic acid HI Sulfuric acid H2504 Nitric acid HNO3 Perchloric acid HCO4 Most bases are strong with the exception of ammonia NH3 Note AMMONIA is different from AMMONIUM The latter has four hydrogens and a charge of 1 An acidbase reaction involves a proton transfer A neutralization reaction turns an acid and a base into water and a salt A precipitation reaction as seen in chem lab involves the exchange of anions and the formation of a nonsoluble product which precipitates out of solution Most compounds containing silver lead and mercury are insoluble All compounds containing alkali metals alkali earth metals and nitrates are soluble The types of equations we have seen so far such as the potassium chlorate one in a previous section are molecular equations lonic equations are just like molecular equations but with all aqueous compounds unless they are weak electrolytes broken into ions A weak electrolyte is an acid base or salt that does not dissociate fully in water A strong electrolyte is an acid base or salt that dissociates fully in water A net ionic equation is an ionic equation with the spectator ions removed They can be computed by working quotbackwardsquot in many cases nd the product that is a solid liquid gas or weak electrolyte and remove everything else from the equation except the cation and anion that form the nondissolved product For neutralization reactions the net ionic equation will almost always be the combining of a proton with hydroxide to form liquid water Oxidationreduction reactions These reactions involve transfer of electrons When an element is oxidized it loses electrons When an element is reduced it gains electrons Remember this with the term quotoil rigquot oxidation is losing reduction is gaining When an element is oxidized the element taking its electrons is called the oxidizing agent Makes logical sense Same is true for reduction


Buy Material

Are you sure you want to buy this material for

50 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Jim McGreen Ohio University

"Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

Jennifer McGill UCSF Med School

"Selling my MCAT study guides and notes has been a great source of side revenue while I'm in school. Some months I'm making over $500! Plus, it makes me happy knowing that I'm helping future med students with their MCAT."

Steve Martinelli UC Los Angeles

"There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

Parker Thompson 500 Startups

"It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.