CHEMISTRY CP Name: KEY Period: UNIT 5: TEST REVIEW SHEET MOLE CONVERSIONS, EMPIRICAL/MOLECULAR FORMULA, STOICHIOMETRY

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1 CHEMISTRY CP Name: KEY Period: UNIT 5: TEST REVIEW SHEET MOLE CONVERSIONS, EMPIRICAL/MOLECULAR FORMULA, STOICHIOMETRY MOLES AND MOLE CONVERSIONS 1. Things to Know: a. What is a mole? Avogadro s number of particles; 1Mole = particles = gram formula mass of an element/compound b. What is Avogadro s number? c. Be able to determine the empirical formula when given the molecular formula. What is the empirical formula of C6H6 and N2O4? Molecular Formula: C6H6 Molecular Formula: N2O4 Empirical Formula: CH Empirical Formula: NO2 d. The relationship between the weight of one atom of an element and the weight of atoms of the element. If a single molecule of H2O has a mass of 18 a.m.u., what is the mass of molecules? They have the same numerical value, but one atom will have a mass in a.m.u. While atoms () will have a mass in grams 1 molecule/atom of H2O = 18.0 amu atoms = = 18.0 g e. The relationship between empirical formula mass and molecular mass. The molecular mass is a whole number multiple of the empirical formula mass. The empirical formula is always in lowest or simplest terms 2. Calculate the formula mass in a.m.u. of the following: a. CaCO3 Ca: = 40.1 amu CaCO3 = 40.1 amu amu amu = amu C: = 12.0 amu O: = 48.0 amu b. KNO3 K: = 39.1 amu KNO3 = 39.1 amu amu amu = amu N: = 14.0 amu O: = 48.0 amu 3. Calculate the gram formula mass (formula mss in grams) of the following: a. Na2CO3 Na: = 46.0 g Na2CO3 = 46.0 g g g = 106 g/m C: = 12.0 g O: = 48.0 g b. AlPO4 Al: = 27.0 g AlPO4 = 27.0 g g g = g/m P: = 31.0 g O: = 64.0 g 4. Make the following conversions. a. What is the number of moles of CaS in 144 grams of the compound? Ca: 40.1 g S: g CaS = 72.2 g/m 144 grams CaS 72.2 grams CaS = 2.0 M

2 b. What is the mass in grams of molecules of CO2? C: 12.0 g 1 = 12.0 g O: 16.0 g 2 = 32.0 g CO2 = 44.0 g/m mol 44.0g/M mol = 132g c. How many molecules of CO2 are in.25 moles? 0.25 M CO mol 1 M = molecules d. How many ions of Cl - are in 1 mole of CaCl2? CaCl2 (Ratio of Ca to Cl - is 1 : 2) CaCl2 2 Cl - = 2 Moles of Cl - 2 Moles Cl mol = M molecules EMPIRICAL/MOLECULAR FORMULA/PERCENT COMPOSITION 5. Find the empirical formula of a compound which is 49.5% Fe, and 50.5% F. Step 1: Convert to grams Fe: 49.5 grams F: 50.5 grams Step 2: Convert grams to Moles Fe: 49.5 grams = M F: 50.5 grams = 2.66 M 55.8 grams 19.0 grams Step 3: Divide by Smallest Moles Fe: 0.887M = 1 F: 2.66M = 3 EF: FeF M 0.887M 6. The empirical formula of a compound is CH and its molecular mass is 78 a.m.u. What is the molecular formula? C: amu n = Molar Mass = 78 amu = 6 MF: C6H6 H: amu EF Molecular Mass 13.0 amu CH = 13.0 amu 7. A sample of a compound is composed of 63.0 grams of Rb and 5.9 grams of Oxygen. What is the empirical formula of the compound? Step 1: Convert grams to Moles Rb: 63.0 grams = M O: 5.9 grams = M grams 16.0 grams Step 3: Divide by Smallest Moles Rb: = 2 O: = 1 EF: Rb2O A compound has an empirical formula of C2H8N and a molar mass of 46 grams per mole. What is the molecular formula n = 1, MF = C2H8N

3 9. What is the percentage composition of CaCO3? CaCO3 Ca: = 40.1 g % Ca: (40.1 g / g) 100 = 40.0% C: = 12.0 g % C: (12.0 g / g) 100 = 12.0% O: = 48.0 g_ % O: (48.0 g / g) 100 = 48.0% CaCO3 = g 10. What is the percentage by mass of oxygen in Ammonium nitrate, NH4NO3? NH4NO3 N: = 28.0 g % O: (48.0 g / 80 g) 100 = 60.0% H: = 4.0 g O: = 48.0 g_ NH4NO3 = 80.0 g MIXED STOICHIOMETRY PROBLEMS 11. Given the equation: Zn + 2 HCl ZnCl2 + H2 a. How many moles of hydrogen will be formed when 4 moles of HCl are reacted? Mole Ratio: H2 = 1 = x Moles H2 x = 2 Moles H2 HCl 2 4 Moles HCl b. How many moles of Zn will be reacted to form 3 moles of ZnCl2? Mole Ratio: Zn = 1 = x Moles Zn x = 3 Moles Zn ZnCl2 1 3 Moles ZnCl2 c. How many moles of HCl will be needed to react completely with 3 moles of Zn? Mole Ratio: Zn = 1 = 3 Moles Zn x = 6 Moles HCl HCl 2 x Moles HCl 12. According to the following reaction, calculate the number of grams of sodium hydroxide that will be produced if grams of sodium are reacted with excess water. 2 Na + 2 H2O 2 NaOH + H2 Mass to Mass Problem with Balanced Equation Step 1: Convert grams to Moles g Na Na = Moles Na 23.0 g Na Step 2: Mole Ratio Na = 2 = Moles Na = x = Moles NaOH NaOH 2 x Moles NaOH Moles NaOH 40 g NaOH = x = grams NaOH

4 13. According to the following reaction, if 35 grams of C6H12O6 react with 40 grams of O2 how many grams of CO2 will be produced? What is the limiting reactant? C6H12O6 + 6 O2 6 CO2 + 6 H2O Step 1: Balance the equation Step 2: Convert grams of reactants into Moles and compare 35 g C6H12O6 C6H12O6 = Moles C6H12O6 **Limiting Reactant** 180 g C6H12O6 40 g O2 O2 = 1.25 Moles O2 32 g O2 Step 3: Use Limiting Reactant to solve for Moles of CO2 C6H12O6 = 1 = Moles = x = Moles CO2 CO2 6 x Moles Step 4: Convert Moles to grams Moles CO2 44 g CO2 = x = 51.2 grams CO2 14. The amount of product in a given reaction turns out to be only 2.34 grams although the theoretical amount was calculated to be 3.12 grams. What is the percentage yield? Actual yield 100 Theoretical yield 2.34 grams 100 = 75% 3.12 grams 15. In the reaction of 10 grams of NaHCO3 with excess HCl, 4.5 grams of NaCl is actually produced. What is the percentage yield from this reaction? NaHCO3 + HCl NaCl + H2O + CO2 Step 1: Convert grams of reactant into Moles 10 g NaHCO3 NaHCO3 = Moles NaHCO3 84 g NaHCO3 Step 2: Set Up Mole Ratio and Solve NaHCO3 = 1 = Moles = x = Moles NaCl NaCl 1 x Moles Moles NaCl 58.5g NaCl = x = 6.96 grams NaCl Step 4: Find Percent Yield 4.5 grams 100 = 64.7% 6.96 grams

5 16. Excess copper reacts with 60.5 grams of silver nitrate to produce copper (II) nitrate and silver. Determine the number of grams of Cu(NO3)2 produced. The reaction is: Cu + 2 AgNO3 Cu(NO3)2 + 2 Ag Mass to Mass Problem with Balanced Equation Step 1: Convert grams to Moles 60.5 g AgNO3 AgNO3 = Moles AgNO g AgNO3 Step 2: Mole Ratio AgNO3 = 2 = Moles AgNO3 = Moles Cu(NO3)2 Cu(NO3)2 1 x Moles Cu(NO3) Moles Cu(NO3) g Cu(NO3)2 = 33.5 grams Cu(NO3)2

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