Empirical and Molecular Formulas Answers

Answers Honors Chemistry Assignment Sheet- Unit 7 Honors Assign #1-% Comp, Empirical Formulas, & Molecular Formulas ● Read each section in your book (shown below), see sample problems, & use reliable internet resources Percent Composition 1. Determine the percent composition of each element of the following compounds: a. NaCl b. AgNO3 Total :Na= 22.99 + Cl= 35.45 = 58.44g Na= 22.99 = 39.34% 58.44 Cl= 35.45 = 60.66% 58.44 c. Mg(OH)2 total: Mg= 24.31 + O 2 = 32 + H 2 = 2 = 58.31 g d. What is the mass of carbon present in 635.45 grams of glucose (C6H12O6)? Molar mass of C6H12O6 = 180.16 g/mol. Empirical Formula & Molecular Formulas Write the empirical formula for each of the following molecular formulas: 2. a. N2O4 b. NO2 c. C2H6 NO2 Empirical formula = Molecular because it is already reduced. d. C3H9 e. H2SO4 ** CH3 f. Hg2(NO3)2 3. Determine the empirical formula of a compound containing 63.50% silver, 8.25% nitrogen, and the remainder oxygen. Step 1: Turn % comp to moles Ag= 63.50g Step 2: Convert moles to smallest whole # ratios 1mol 107.87 N= 8.25g 1mol 14.01 O= 28.25g 1mol 16g 4. A sample is analyzed as containing 24.09 grams of potassium, 0.308 moles of manganese, & 7.42 x 1023 atoms of oxygen. What is the empirical formula? 5. a. What is the molecular formula of the molecule that has an empirical formula of CH2O and a molar mass of 120.12 g/mol? Step 1: find the multiplying factor between molecular & empirical Molecular formula mass = Empirical formula mass x Empirical mass = C + H2 + O = 30.02 (plug into equation) 120.12g = 4 🡪 Multiply all the subscripts by 4 -🡪 4(CH2O) = C4H8O4 30.02g b. (Almost same question as above) What is the molecular formula of the molecule that has an empirical formula of CH2O and a molar mass of 60.04 g/mol? 60.04 = 2 🡪 2(CH2O) = C2H4O2 30.02 This is the same as in the above question, however, this molecular formula is only 2 x’s more than the empirical. c. (Almost same question as above) What is the molecular formula of the molecule that has an empirical formula of CH2O and a molar mass of 30.02 g/mol? Since the Empirical formula mass and the Molecular formula mass are equal, this means CH2O is both the Molecular and Empirical formulas. 6. A sample compound with a formula mass of 34.00 amu is found to consist of .44 g H and 6.92 g O. Find its molecular formula. (Hint: even though it’s asking for the molecular formula, you will still need to find the empirical formula first. In the easier problems, like #5, they give you the empirical formula, but here, you’ll have to determine it first) 7. The molar mass of a compound is 92 g/mol. Analysis of a sample of the compound indicates that it contains 31.1 % N and 68.9% O. (error here!) Find its molecular formula. (again, you must find the empirical formula first in order to solve) Step 1: Empirical Forumla & then mass N= 31.1 1mol 14.01g O= 68.9 1mol 16 g Step 2: Determine multiplying factor & multiply by the empirical formula 92g = 2 🡪 2(NO2) = ** N2O4 ** 46 8. If 4.04 g of N combine with 11.46 g O to produce a compound with a formula mass of 108.0 amu, what is the molecular formula of this compound? (this one’s a bit tricky!) Honors Assign #2-Oxidation/ Reduction Part I: Oxidation States (reviewed) ● Assign oxidation numbers to each element in the compounds/ Ions below: a. HF H = +1 F= -1 b. CI4 c. H2O d. PI3 e. CS2 f. Na2O2 C= +4 S= -2 i. SO4-2 h. NO2- g. H2CO3 Na= +1 O= -1 H= +1 C= +4 O= -2 j. ClO2- k. N2 N= +3 O= -2 l. Fe 3+ Part II: Oxidation & Reduction 1. What does it mean when an element has been oxidized? Reduced? Oxidation= Increase in oxidation state (or loss of electrons thus a mone + charge) Reduction= Decrease in oxid state ( or gain of electrons, thus a mone – charge) 2. Explain the helpful pneumonic devices “LEO goes GER” & “OIL RIG” for the oxidation/ reduction reactions. “Oil rig” = Oxidation is loss of electrons Reduction is gain “leo says GER” = Lose electrons is oxidation , Gain electrons is reductions #3 a. reduction a. Oxidation b. Oxidation c. Reduction d. Reduction e. Oxidation f. Reduction g. Reduction #4 a. Redox reaction b. Redox reaction c. Not redox d. Redox reaction e. Not redox #5 a. O (-2) is oxidized to O2 (zero) N (+5) is reduced to N (+1) b. H2 (zero) is oxidzed to H (+1) Cu2+ (+2) is reduced to Cu (zero) d. H2 (zero) is oxidzed to H (+1) Cl2 (zero) is reduced to Cl (-1)