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Chemistry 111 Study guide (Dr. patton)

by: Jamisha Evans

Chemistry 111 Study guide (Dr. patton) Chem 111

Marketplace > Eastern Kentucky University > Chemistry > Chem 111 > Chemistry 111 Study guide Dr patton
Jamisha Evans
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These notes include all Exam 2 content
General Chemistry 1
Dr. James Patton
Study Guide
general, Chem, 111
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This 11 page Study Guide was uploaded by Jamisha Evans on Friday October 7, 2016. The Study Guide belongs to Chem 111 at Eastern Kentucky University taught by Dr. James Patton in Fall 2016. Since its upload, it has received 7 views. For similar materials see General Chemistry 1 in Chemistry at Eastern Kentucky University.


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Date Created: 10/07/16
Chemistry 111 Exam 2 Study guide (Dr patton) HIGHLIGHT: Important HIGHLIGHT: Summary of section 2.7-Naming Compounds  Acids  Definition: Substance that yields H when dissolved in water +  EXAMPLE: HCL is an acid. It contains Hydrogen and yields H when dissolved in water. HCL H2O H + Cl -  Acids have several groups  Simple acids: Contains hydrogen and an anion EXAMPLE: HCL (Hydrochloric acid)  Oxoacids: Contains hydrogen, a central atom and oxygen EXAMPLE: HNO (Nit3ic acid)  Oxo-acids often have variable numbers of oxygen EXAMPLE: HNO (Nitr3c acid) HNO (2itrous acid)  Polyprotic acids: Contains more than one ionizable hydrogen EXAMPLE: H SO 2Sul4uric acid) **The prefix oxo- means oxygen**  Bases  Definition: Substance that produces OH ions when dissolved in water  EXAMPLE: NaOH NaOH H2O Na + OH -  Hydrates  Definition: Hydrates are formed from water molecules in the crystal of an ionic compound. They aren’t specifically associated with a cation or anion and they can be removed by heating.  EXAMPLE: FeCL ∙ 6H3O (I2on (III) chloride heptahydrate) Know the definition of an acid and base Know the acids and bases highlighted (pg. 4.2 in text book) Remember the prefixes when naming a hydrate 3.1- Atomic Mass  Atomic mass  Definition: The mass of an atom of a chemical element. AKA atomic weight. (reported in amu)  Carbon-12 isotope is the reference used for the atomic mass unit. 1 atom of carbon-12 is 12 amu  Mass of atoms are really small. EXAMPLE: There are two isotopes an element has, one having an atomic mass and percent natural abundance of 11.0093 amu (80.22%) and 10.0129amu (19.78%) Add the product of the (amu) (percentage in decimal form) of each isotope. But make sure you divide the percentages by 100 in order to make it a decimal.  (11.0093) (.8022) + (10.0129) (.1978) =10.81 amu *This element is Boron because when you look at the periodic table you will see that the atomic mass of boron is 10.81. * Remember, atomic mass is reported in amu Remember to change the percent natural abundance of the isotopes to decimals Know how to identify an element when only given the percentages and mass of its isotopes 3.2- Avogadro’s number and the molar mass of an element  Avogadro’s number  Definition: A counting number is used to represent very large numbers of atoms, molecules, or ions  The unit used to count large numbers of small particles is the mole  1 mole: 6.022 x 10 (Avogadro’s number; N ) Areported in atoms)  Mol is the SI unit for amount of substance  The mass of a single atom (in amu) is numerically equal to the mass (in grams) of 1 mol of that element. EXAMPLE: 1 atom K has a mass of 39.0983 amu 1 mol of atoms of K has a mass of 39.0983 g  Convert mass to atoms  Mass to moles to atoms  Convert atoms to mass  Atoms to moles to mass  Molar mass  Definition: The mass of 1 mole of atoms of that element (reported in units of grams/mol)  EXAMPLE: Grams of gold in 15.3 moles of gold (Au)? 1. State the given first- 15.3 mol of Au 2. Convert moles of gold to mass of gold 3 15.3 mol Au x 197 g of Au/1 mol of Au = 3.01 x 10 g of Au Know the Avogadro’s number Know how to convert mass to atoms and atoms to mass Remember, molar mass is reported in grams /mol 3.3- Molecular Mass and formula mass  Molecular mass  Definition: Mass of a molecule (reported in amu)  EXAMPLE: Molecular mass of CCl (Add4mass of C and mass of Cl ) 4 C: 12.01 x 1= 12.01 Cl4: 35.45 x 4= 141.8 153.8 amu  Formula mass  Definition: Sum of atomic weights of the atoms in the empirical formula of the compound. AKA formula weight (reported in grams)  EXAMPLE: Formula mass of NaBr Na: 22.99 x 1=22.99 Br: 79.90 x 1=79.90 102.89g Remember that molecular mass, like atomic mass, is reported in amu Remember, formula is reported in grams Know how to find the molecular mass and formula mass of a molecule 3.5 Percent Composition of compounds  Percent Composition  Definition: The percentage by mass contributed by each element in the substance  Formula: %element= (n)(molar mass of element)/(molar mass of the compound) x 100  n: Integer, subscript/moles of a particular element EXAMPLE: Percent composition of sulfuric acid (H S2 )4 1. Calculate the molar mass of the compound 2 H: 2.016g Sum 4 O: 64.00g of this is 1 S: 32.07g 98.08g 2. Calculate the percent by mass of each element in formula… I will only show how to calculate the percent by mass for hydrogen but the same steps are followed for oxygen and sulfur. %H= (2) (1.008)/(98.09) x 100= 2.055% **The sum of all the percentages should equal 100%**  The percent composition by mass is the same for the molecular formula and the empirical formula of a compound. This means that the empirical formula can be used to calculate the empirical formula of a compound  Empirical formula from percent by mass composition  EXAMPLE: Calculate the empirical formula of this unknown compound using percent composition by mass composition. Given C:10.4% S:27.8% Cl:61.7% Steps 1. Convert each of our data into grams assuming we have 100g of sample C:10.4g S:27.8g Cl:61.7g 2. Determine number of moles by using molar mass as a conversion factor C:10.4gC x 1molC/12.01gC=.866 mol C S:27.8gS x 1 molS/32.06 g S= .866 mol S Cl:61.7gCl x 1 molCl/34.45g Cl = 1.74 mol Cl 3. Find whole number containing ratio by dividing by the smallest amt of mol C:.866/.866 =1 S:.866/.866= 1 Cl:1.74/.866= 2 4. The least electronegative element is listed first and the most electronegative is listed last. (more about this is Ch. 8) Final answer: CSCl 2  Molecular formula from empirical formula EXAMPLE: Empirical formula for a compound is CH and its molar mass is 30.1 g/mol. 3 What is the molecular formula of the unknown?  Use this formula  n= (molar mass of compound)/(empirical formula’s molar mass) n= (30.1g)/(15.03) **15.03 is the mass of CH **3 n=2 …Now multiply each subscript by 2 C 1 x 2= 3 x 2= Final answer:C 2 6  Mass of element from percent by mass composition EXAMPLE: Mass of aluminum (grams) in 371g of aluminum oxide? Mass of Al 2 3 moles of Al2O 3 moles Al mass of Al Final answer: 196 g Al Know the formula used to find percent by mass composition Know how to find empirical formula when given mass by percentages Know the formula used to find empirical formula from molecular formula 3.7 Chemical Reactions and Chemical Equations  Chemical reaction  Definition: process in which a substance or substances are changed into one or more new substances  Chemical equation  Definition: Uses chemical symbols to represent this process  Chemical equations must be balanced  The physical state of all the reactants and products are represented  Gas: (g)  Liquid: (l)  Solid: (s)  Aqueous: (aq)  Make sure to balance the equation. Change only the coefficient NOT the subscripts EXAMPLE: write a chemical equation for the formation of ammonia from hydrogen and nitrogen gas 3H 2g) + N (2) 2NH 3(g) Product Reactant H:6 H:6 N:2 N:2 Remember to represent the state of matter of the elements and balance the equation 3.8- Amounts of reactants and products  Stoichiometry  Definition: Qualitative study of reactants and products in a chemical reaction in a chemical reaction  EXAMPLE: 2 H (g) + O (g) 2H O(g) How many moles of water can be 2 2 2 formed if… a) 1 mol of O reacts completely with H 2 2 Begin with given (1 mol O ) … 2 1 mol O (2 mol H O/1 mol O ) = 2 mol H O 2 2 2 2 b) 0.41 mole of O react completely with H 2 2 Begin with given (0.41 mol of O ) 2 0.41 mol O 22 mol H O/2 mol O ) =.22 mol H O 2  ANOTHER EXAMPLE: 2 H (g) + O (2) 2 2 H 2 (g) How many grams of water can be formed if…  3.65 grams of H r2act completely with O 2 3.65 g H 21 mol H /22016 g H ) (2 mol H O/22mol H ) =1.21 mol H O 2 Make sure you know how to set up stoichiometry problem and know how to find what you are asked to find. 3.9- Limiting Reagents  Definition: The reagent that is consumed completely and limits the maximum amount product formed. It runs out first.  Think of the sandwich analogy…  If you make a sandwich that requires 2 pieces of bread and a single piece of cheese. How many sandwiches can you make if you have 10 pieces of bread and 6 slices of cheese? Which is the limiting reactant? ANSWER: You can make 5 sandwiches. The bread is the limiting reactant  EXAMPLE: 2H (g) +2O (g) 2 2H 2 (g) (A Little more complex) React 3 moles of oxygen gas with 4 moles of hydrogen gas to form water… a) How many moles of water can you form? From oxygen... 3mol O (2 mol H O/ 1 mol O ) = 6 mol H O 2 2 2 2 From H .2. 4 mol H (2mol H O/22mol H )= 4 2ol H O 2 b) Which reagent is our limiting reagent? H 2 Because it produces the least amount of water c) How many moles of excess reagent remains? Use the limiting reagent to find how many moles of excess reagent remains Hint: O 2s the excess reagent… 4 mol H 21 mol O / 2 mol H )= 2 mol O 2 **If mass of reactant is given you may need to convert it to moles in order to find moles of product** Know how to identify the liming reactant and how to find the number of moles of excess reagent that remains. 3.10 Reaction Yield  Definition: Criteria used to report processing efficiency. AKA percent yield  Formula  % yield= (actual yield/theoretical yield) (100) EXAMPLE: if 10.0 grams of H react2with excess N accordin2 to the equation below, what is the reaction yield if 45.8 grams of NH ar3 collected? 3H (g) + N (g) 2NH (g) 2 2 3 …From the reaction we know that H react2 completely and N is the l2miting reactant. We also know that 45.8 is the actual yield. Find moles of NH 1030g H x (12molH /2.016 2 H )(2 mol 2H /3 mol H )317.034g 2 NH /3 mol NH )=53.3 g NH 3 …56.3 is the theoretical yield Fill in the information we have into our formula % yield = (45.8g/56.3g)(100)=81.3% FINAL ANSWER: Our Percent yield is 81.3% Know the formula used to find the percent yield. Remember to convert moles to mass. Chapter 4: Reactions in Aqueous Solutions 4.1-General Properties of Aqueous Solutions  Solutions  Definition: Homogenous mixture of 2 or more pure substances that may occur as a solid liquid or gas. To create a solution, the solute dissolves in the solvent.  Solvent definition: present in greatest abundance  Solute definition: all other substances present in smaller amount EXAMPLE: If salt was poured into water, the salt (solute) would dissolve in the water (solvent) creating a solution. When the solvent is water, it is called an aqueous solution.  Aqueous solutions may or may not conduct electricity  Electrolytes…  Conduct electricity.  They dissociate (break up) when dissolved in water  Most commonly made up of a metal and non-metal (ionic compound) EXAMPLE: NaCl(aq) Na (aq)+Cl (aq) ***The single arrow mean the ions will not recombine to reform NaCl***  Non-electrolytes…  Do not conduct electricity  They don’t dissociate. They retain their molecular structure when dissolved in H 2 EXAMPLE: Sugars like sucrose, and alcohols like ethanol  Electrolytes can be further classified into weak or strong electrolytes  Strong electrolytes: Strong acids and strong bases. Dissociate completely. Strong Acids EXAMPLES: HCl, HBr, HNO HCLO , 3,SO 4 2 4 Strong Bases EXAMPLES: LiOH, NaOH, KOH, Ba (OH) 2  Weak electrolyte: weak acids and weak bases. Only dissociate partially. Weak acids EXAMPLES: CH COOH 3 Weak bases EXAMPLES: H O, N2 4 Know how to distinguish between an electrolyte and non-electrolyte. Know how to identify a base (strong and weak) and acid (strong and weak) Remember, a single arrow is used to represent a strong acid or base reaction whole a double arrow is used to represent a weak acid or base 4.2- Precipitation reactions  Definition: involves the formation of a solid from a homogenous solution  Common type of precipitation reaction is double displacement  It is important to know solubility rules in order to determine whether or not a precipitate will form  Solubility: Maximum amount of solute that will dissolve in a given quantity of solvent at a certain temperature (solubility rules on page 101, table 4.2) EXAMPLE: Write the net ionic equation for the molecular equation below Molecular Equation: NaCl(aq)+AgNO (aq) 3 AgCl(s)+ NaNO (aq)3 Ionic Equation: Na (aq)+ Cl (aq) +Ag (aq) +NO (aq) _ AgCl(s)+ Na (aq) +NO (aq) - 3 3 ***Cross out spectator ions. These are ions that don’t do anything in the solution. rSpectator ions are on both the product and reactant side*** - + Net Ionic Equation: Cl (aq) + Ag (aq) AgCl(s) Know how to write a molecular equation, ionic equation, and net ionic equation. Know how to write the products of an equation when given the reactants Remember to cross out spectator ions Remember the solubility rules Know how to apply the double displacement


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