CHEMISTRY REVIEW: FINAL EXAM Chapter 7 & 9: Ionic & Metallic Bonding Vocabulary: •
Valence electron – electrons in the highest (outer most) energy level of an atom •
Octet rule – all atoms want to have the e‐ configuration of a noble gas, they want 8 valence e‐ (except for Hydrogen which only want 2, so that it can be like He) •
Cation – a positively charged ion •
Anion – a negatively charged ion •
Ionic bond – a bond between a metal & non‐metal (the metal loses e‐ and the non‐
metal gains the electrons) •
Formula unit – the lowest whole number ration of elements in an ionic bond •
Metallic bonds – bond between a metal & a metal (they all lose electrons, and lost negative e‐ act like ‘glue’ holding the positive metals together) •
Alloy – a mixture between metals, they are not bonded together. How are cations formed? positively charged ions, so they form when an element loses electrons How are anions formed? negatively charged ions, so they form when an element gains electrons How are ionic bonds formed? What keeps them together? Formed between an metal & non‐metal, the metal loses electrons and becomes positive (cation), the non‐metal gains electrons and becomes negative (anion). The anion & cation attract and form an ionic bond (b/c positives & negatives attract) Give four (5) properties of ionic compounds/bonds: 1. between a metal & non‐metal 2. have high melting points 3. solid @ room temp. 4. atoms lose & gain electrons 5. very strong bonds, hard to break How are metallic bonds formed? What keeps them together? Formed with only metals. The metals lose electrons and become positive. The electrons they lost hold them together. Usually described as cations in a sea of moving electrons. For each of the following, give the # of valence e‐, the e‐ dot structure, the amount of e‐ lost/gained to form an ion 1. Ga (3v.e‐; 3 unpaired e‐ in dot struct.; loses 3 e‐; will form a cation w/ charge +3) 2. Xe (nobel gas; 8v.e‐; 8 e‐ (4pairs) in dot struct.; doesn’t lose/gain e‐; won’t form ion) 3. Mg (2v.e‐; 2 unpaired e‐ in dot struct.; loses 2 e‐; will form a cation w/ charge +2) 4. H (1v.e‐; 1 unpaired e‐ in dot struct.; loses 1 e‐; will form a cation w/ charge +1) 5. Rb (1v.e‐; 1 unpaired e‐ in dot struct.; loses 1 e‐; will form a cation w/ charge +1) 6. Br (7v.e‐; 7 unpaired e‐ in dot struct.; gains 1 e‐; will form an anion w/ charge ‐1) Write a formula for the following ionic compounds: 1. nickel (III) sulfide Ni2S3 3. silver acetate AgC2H3O2 2. manganese (II) phosphate Mg3(PO4)2 Name the following ionic compounds: 4. potassium carbonate K2CO3 1. FePO4 Iron(III)Phosphate Or, Ferric Phosphate 3. Cu(OH)2 Copper(II)Hydroxide Or, Cupric Hydroxide 2. K3N Potassium Nitride Chapter 8 & 9: Covalent Bonding Vocabulary: 4. Zn(NO2)2 Zinc Nitrite
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Covalent bond – bond b/w 2 non‐metals •
Molecular compound – a molecule consisting of only non‐metals and covalent bonds •
Diatomic molecule – a molecules consisting of 2 of the same atoms (Br,I,N,Cl,H,O,F) •
Monatomic element – an element that exists alone, doesn’t bond (noble gases) •
Empirical formula – lowest whole number ratio for a covalent bond •
Molecular formula – gives the actual number of atoms in a covalent bond What are the diatomic molecules? Br,I,N,Cl,H,O,F Name six monatomic elements: He, Ne, Ar, Kr, Xe & Rn What is shared in a single covalent bond? Double covalent bond? Triple covalent bond? Single: 2 e‐ are shared; Double: 4 e‐ are shared; Triple: 6 e‐ are shared Give five (5) properties of covalent molecules/bonds: 1. weaker than a ionic bond 2. exist usually as a gas or liquid 3. have low melting points 4. consist of 2 non‐metals 5. e‐ are shared between the atoms What type of bond would each of the following sets of atoms form? (ionic/metallic/covalent) 1. Na & Mg (M) 2. Al & O (I) 3. Na & S (I) 4. B & Cl (C) 5. Ca & I (I) 6. F & H (C)
Classify each of the following as an atom or a molecule: 1. Be ‐ atom 2. CO ‐ molecule 3. N2 ‐ molecule 4. H2O ‐ molecule 5. Ne ‐ atom Draw the structure of each of the following covalent molecules. They may contain single, double, or triple bonds: See Answer Key in Class for Diagrams
What is a polar covalent bond? A covalent bond between atoms of different electronegativities Would a bond between the following elements be polar or non‐polar? 1. C & I Non‐polar 2. Si & N polar Write formulas for the following covalent compounds: 3. H & H Non‐polar
1. Hexaboron nonasilicide B6Si9 2. Chlorine dioxide ClO2 3. Hydrogen diiodide HI2 4. Iodine pentafluoride IF5 5. Carbon monosulfide CS 6. Dinitrogen heptoxide N2O7 Write the names for the following covalent compounds: 1. P4S5 tetraphosphorous pentasulfide 2. SeF6 Selenium hexaflouride 3. B2Si7 Diboron heptasilicide 4. NF7 Nitrogen heptaflouride Chapter 10: Chemical Quantities Vocabulary: •
Mole – a unit used to measure the amount of matter, 1 mole contains 6.02 x 1023 particles/atoms/molecules •
Avogadro’s number ‐ 6.02 x 1023 •
Molar mass – the mass (weight) of 1 mole of a particle/atom/molecule How many liters are in a mole of hydrogen? 22.4L How many grams are in a mole of hydrogen? 1.01g How many atoms are in a mole of hydrogen? 6.02 x 1023 atoms Conversions using dimensional analysis: 1. How many molecules are in.78 moles of iron? .78 moles Fe 6.02 x 1023 molecules 1 mole Fe = 4.70 x 1023 molecules 2. Calculate the volume of 5.0 moles of Al2O3. 5 moles Al2O3 22.4 liters 1 mole Al2O3 = 112 liters 3. How many atoms of helium do you need to fill a 5.50 liter balloon? 5.5 liters 1 mole 6.02 x 1023 atoms 22.4 liters 1 mole = 1.48x1023 atoms What does percent composition tell you? The % of each element in a compound Percent Comp. Calculations: 1. What is the percent composition of a sample containing 1.03g of N and 5.29 g of O? 1.03 + 5.29 = 6.32 N: 1.03/6.32 (x100) = 16% O: 5.29/6.32 (x100) = 84% 2. Find the % comp of Pb(NO3)3. Pb: 207.20 N: 14.01 x 3 = 42.03 O: 16 x 9 = 144 144 + 42.03 + 207.20 = 393.23 Pb: 207.20/393.23 (x100) = 53% N: 42.03/393.23 (x100) = 11% O: 144/393.23 (x100) = 37% 3. Find the empirical formula of a molecule with 40.0% C, 6.67% H, and 53.33% O. (hint: convert % to grams, grams to moles, divide by the smallest amount, and find the smallest ratio) C – 40% = 40g / 12.01 = 3.33 / 3.33 = 1 H – 6.67% = 6.67g / 1.01 = 6.60 / 3.33 = 2 O – 53.33% = 53.33g / 16 = 3.33 / 3.33 = 1 CH2O 4. Now, calculate the molecular formula for the compound above, if the molar mass is 90g/mol. CH2O = 12.01 + 1.01x2 +16 = 30.03 30.03/90 = 3 (round to whole #) C3H6O3 Chapter 11: Chemical Reactions Vocabulary: •
Product – what is made in a reaction •
Reactant – what a reaction is begun with •
Balanced chemical equations – follows the rule of conservation of mass/matter; has the same amount of atoms on each side •
Combination reaction – 2 elements come together to form 1 compound •
Decomposition reaction – 1 compound breaks apart into 2 elements •
Single displacement reaction – a more reactive element switches with a less reactive element in a compound; 1 element + 1 compound make 1 element + 1 compound •
Double displacement reaction – the cations & anions in two compounds switch places; 1 2 compounds make 2 compounds •
Combustion reaction – a hydrocarbon + oxygen gas make carbon dioxide + water •
Precipitate ‐ when a solid forms in a liquid Writing Chemical Equations: 1. Write the balanced equation for hydrogen and oxygen reacting to form water. What type of reaction is this? 2 H2 + O2 → 2 H2O 2. Write the balanced equation for the following : Aqueous silver nitrate + coper metal → silver metal + aqueous copper nitrate (ignore the word ‘aqueous’ – sorry!) AgNO3 + Cu → Ag + Cu(NO3)2 Balance the following reactions & identify the reaction typed. Then, predict the products. 1. ___ Br2 + ___ NaCl → No rxn; would be single displacement, but Br is less reactive than Cl 2. _2_ HF → H2 + F2 decomposition 3. ___ Zn(NO3)2 + ___ SnCl2 → ZnCl2 + Sn(NO3)2 double dis. 4. ___ CH2O + ___ O2 → H2O + CO2 combustion (it’s already balanced!) 5. _2_ Al + _3_ H2 → 2AlH3 For any of the double displacement reactions above, copy the full reaction and label any ppt’s: Ppt’s only occur in double dis. reactions. There was only 1 double dis. reaction above: Zn(NO3)2 + SnCl2 → ZnCl2 + Sn(NO3)2 There will be NO ppt’s b/c zinc chloride is soluble (rule #4) & tin nitrate is also soluble (rule #2) Chapter 12: Stoichiometry & % Yield Vocabulary: •
Stoichiometry ‐ using dimensional analysis & balanced equations to calculate the amounts of a product or reactant •
Molar ratio – the ratio of products to reactants in moles •
Theoretical yield ‐ the amount of product that could be produced in a reaction if all goes perfectly •
Actual yield ‐ the amount of product that is actually produce in a reaction •
Percent yield – the % of product made out of the total possible product •
Error – problems that can occur in the lab which lower the actual yield •
Percent yield formula – Actual/Theoretical x 100 = % Stoichiometry: __2__ C2H6 + __7__ O2 → __4__ CO2 + __6__ H2O 1. If you were given 3 grams of C2H6 how many grams of water would you produce? ***correction: change the question to C2H6 *** 3g C2H6 1 mole C2H6 30.08g 6 moles H2O 18.02g 2 mole C2H6 1 mole H2O =5.39g H2O 2. If you were given 4.56 liters of O2 how many milli‐grams of carbon dioxide would you produce? 4.56L O2 1 mole O2 4 moles CO2 44.01g CO2 1000 mg 22.4L O2 7 moles O2 1 mole CO2 1 g = 5119.53 mg Percent Yield: __8__ NH3 + __1__ N2SO4 → __3__ (NH4)2SO4 + __2__ N2 *** correction: add N2 as second product in equation*** 1. I began this reaction with 200 grams of NH3 and excess N2SO4 what is my theoretical yield of (NH4)2SO4 ? 200g NH3 1 mole NH3 3 mole (NH4)2SO4 132.17g (NH4)2SO4 17.04g NH3 8 mole NH3 1 mole (NH4)2SO4 = 581.73g (NH4)2SO4 2. If I produced 20 grams of the product in the lab, what is my percent yield? 20/581.73 x 100 = 3% 3. Instead, if I produced a 15% yield, what was my actual yield? 581.73 x .15 = 87.26g