Chemistry Honors - Chapter 5 Class Notes PDF

# Chapter 5 Class Notes ## Formula Mass Formula mass of a substance is the sum of atomic masses of each atom in its chemical formula. For molecular substances formula mass is called as molar mass. **Examples**: Calculate molar mass of C6H12O6. Molar mass= 6( atomic mass of C) + 12 (atomic mass...

# Chapter 5 Class Notes ## Formula Mass Formula mass of a substance is the sum of atomic masses of each atom in its chemical formula. For molecular substances formula mass is called as molar mass. **Examples**: Calculate molar mass of C6H12O6. Molar mass= 6( atomic mass of C) + 12 (atomic mass of H) + 6 (atomic mass of O) = 6(12.01 g) + 12 (1.01g) + 6 (16.00g) =180.18 g **Practice Problems:** 1. Calculate formula mass of Ca(NO3)2 * CaN206 * Ans. 164.10 amu * 1(40.08)+2(14.01) +6 (16.00) = 164.10 amu 2. Calculate formula mass of Al(OH)3 * Al 03 H3 * Ans. 78.01 amu * 1(26.98) + 3(16.00)+3(1.008) = 78.01amu ## Percent Composition Percent composition of a compound is the percent by mass of each element in that substance. **Steps in calculating percent composition**: 1. Calculate molar mass of the compound. 2. Calculate how much of the molar mass comes from each element. 3. Divide each element’s mass by contribution by the total molar mass and multiply by a 100 to convert to percent. **Example**: Calculate % composition of C6H12O6 **Step 1**: Molar mass of C6H12O6 6(12.01)+12(1.008)+6(16.00)=180.16 amu **Step 2**: Find contribution of each element C= 6x12.01=72.06 g H=12. 10g O=96g **Step 3**: Calculate mass percent: C= 72.06 / 180.16 x 100 = 40.00% H=12.10 / 180.16 x 100= 6.72% O= 96 / 180.16 x 100 = 53.29% **Answer**: 40% Carbon, 6.72% hydrogen, 53.29% Oxygen **Some Hints**: * Check answer: All the elements should add up to a 100. * In a binary compound, after calculating the mass % of one element, the other element’s mass percent can be calculated by subtracting from a hundred. **Practice Problems:** 1. Find percent composition of Ba(NO3)2 Ans 52.546% Ba, 10.72% N, 36.73% Ο * BaN206 * 1(137.33)+2(14.01)+6(16.00) = 261.35amu * Ba=137.33g * N= 28.029 * 0=968 * Ba = 137.33 / 261.35 x 100 = 52.546% * N=28.02 / 261.35 x 100=10.72%. * 0=96 / 261.35x100 = 36.73%. 2. Find the mass percent of water in ZnSO4.10 H2O Ans. 52.75% 3. Calculate percent compositon of K3Fe (CN)6 * Ans. K35.62%, Fe 16.96%, С 21.88%, Ν 25.53% ## Empirical Formula Empirical formula is the simplest whole number ratio of atoms in a compound. Ex: For glucose C6H12O6 the empirical formula is CH2O **Calculation of empirical formula of a compound:** **Steps**: 1. Assume that compound weighs exactly 100g. You can therefore convert % to g. 2. Calculate moles of each kind of atoms present. (divide by molar mass) 3. Determine simplest whole number ratio of atoms by dividing the moles of each by smallest calculated mole value. **Mass of elements--→ assume 100g sample-→ g of each element->use atomic masses of elements moles of elements-→ calculate mole ratio -→ empirical formula** **Example:** Phosgene, a poisonous gas used in WW1, contains 12.1% C, 16.2% O and 71.7% Cl by mass. What is the empirical formula of phosgene? **Step 1:** Assuming 100 g of phosgene, C= 12.1 g O= 16.2 g Cl=71.7 g **Step 2:** Calculate moles: Moles of C= 12.1g X 1 mole / 12.01 g =1.01 mole C Moles of O=16.2g x Imole / 16.008 = 1.01 mole O Moles of C1=71.7gx Imole / 35.459 = 2.02 mole Cl **Step 3:** Finding smallest whole number ratio C: O: CI= 1:1:2 Empirical formula of phosgene = COCl2 **Practice Problems:** 1. Determine the empirical formula for a deadly nerve gas that gives following mass percent analysis: C=39.10%, H=7.67%, O=26.11% P-16.82% F= 10.30% Ans: * C6H14O3PF 2. Analysis of a 10.150 g sample of a compound known to contain P and O only, show the P content to be 4.433 g. What is the empirical formula? Hint: mass moles→small whole no. ratio * P2O5 ## Molecular Formula Actual ratio of atoms in a compound. Ex. H2O, C6H12O6 **To determine the molecular formula, divide the molar mass by empirical formula mass. This will give the number of empirical formula units in actual molecule.** **Ex.** Determine the empirical and molecular formula of each of the following: a. Ethylene glycol, the substance used as antifreeze has 38.70% C, 9.70% H and 51.60% O, mm= 62.10 g b. Caffeine, a stimulant in coffee has the following percent composition: 49.50% C, 5.15% H, 28.90% N and 16.50% O, molar mass= 195.00g **MOLECULAR FORMULAS** To determine the molecular formula for a compound: 1) The molecular weight is always a multiple of the empirical formula weight (i.e., M.W. = n x E.F.W.) To determine n, divide the given molecular weight by the empirical formula weight. 2) Multiply all the subscripts in the empirical formula by the answer to the previous step. **Example 2**: If the compound from Example 1 had a molecular weight of 306 g, what would the molecular formula be? **Solution**: The empirical formula was Al2O3. The empirical formula weight is 2 x 27.0 g + 3 x 16.0 g = 102 g The molecular weight is 306 g. 306 + 102 = 3. We multiply the subscripts in the empirical formula by 3 to get the molecular formula Al6O9. **EXERCISES** A. Determine the empirical formula of each compound from its percentage composition by weight: 1) 66.4% Cu, 33.6% S * Cu₂S 2) 79.8% Cu, 20.2% S * Cu₂S 3) 62.6% Ca, 37.4% C * CaC₂ 4) 36.8% Ν. 63.2% O * N2O3 5) 38.9% Cl, 61.2% O * Cl₂O7 6) 39.8% Κ, 27.8% Mn, 32.5% Ο * K₂MnO₄ 7) 32.4% Na, 22.6% S, 45.0% 0 * Na₂SO₄ 8) 52.0% Zn, 9.60% C, 38.4% O * ZnCO₃ 9) 1.90% Η, 67.6% Cl, 30.5% Ο * HClO 10) 60.0% C 13.3% Η, 26.7% Ο * C3H8O **B. Determine the empirical formula of each compound from the given weights:** 1) 7.615 g Ga, 2.622 g O * Ga₂O₃ 2) 0.366 g Na, 0.220 g N, 0.752 g O * NaNO₃ 3) 11.89 g Fe, 5.11g O * Fe₂O₃ 4) 87.3 g Na, 121.5 g S, 91.2 g 0 * Na₂S₂O₃ **C. Determine the molecular formula of each compound from the empirical formula and the molecular weight:** 1) E.F. = NaS2O3, mol. wt. = 270.4 * Na₂S₄O₆ 2) E.F. = C3H2Cl, mol. wt. = 147.0 * C₆H₄Cl₂ 3) E.F. = C2HCl, mol. wt. = 181.4 * C₆H₃Cl₃ 4) E.F. = Na2SiO3, mol. wt. = 732.6 * Na₁₂Si₆O₁₈ 5) E.F. = NaPO3, mol. wt. = 305.9 * Na₃P₃O₉ 6) E.F. = NO2, mol. wt. = 92.0 * N₂O₄ **D. Determine the molecular formula from the percentages by weight and the molecular weight.** 1) 65.45% C, 5.45% H, 29.10% O; mol. wt. = 110 * C₆H₆O₂ 2) 40.0% C, 6.7% H, 53.5% O; mol. wt. = 180 * C₆H₁₂O₆ 3) 7.79% C, 92.21% Cl; mol. wt. = 154 * CCl₄ 4) 10.13% C, 89.87% Cl; mol. wt. = 237 * C₂Cl₆ 5) 25.26% C, 74.74% Cl; mol. wt. = 285 * C₆Cl₆ 6) 11.25% C, 88.75% Cl; mol. wt. = 320 * C₃Cl₈ **SOLUTIONS** A. (1) CuS (2) Cu2S (3) CaC2 (4) N2O3 (5) Cl2O7 (6) K2MnO4 (7) Na2SO4 (8) ZnCO3 (9) HClO (10) C3H8O B. (1) Ga2O3 (2) NaNO3 (3) Fe2O3 (4) Na2S2O3 C. (1) Na2S4O6 (2) C6H4Cl2 (3) C6H3Cl3 (4) Na12Si6O18 (5) Na3P3O9 (6) N2O4 D. (1) C6H6O2 (2) C6H12O6 (3) CCl4 (4) C2Cl6 (5) C6C16 (6) C3Cl8 # Chapter 5 Honors ## LAW OF MULTIPLE PROPORTIONS 1. Benzene, ethane and ethylene are just three of a large number hydrocarbons-compounds that contain only carbon and hydrogen. Show how the following data are consistent with the Law of Multiple Proportions | Compound | Mass of Carbon in 5 sample | |---|---| | Benzene | 4.61 g | | Ethane | 4.00 g | | Ethylene | 4.29 g | 2. In addition to carbon monoxide (CO) and carbon dioxide (CO2), there is a third compound of carbon called carbon suboxide. If a 2.500 g of carbon suboxide contains 1.32 g of carbon and 1.18 g of O how that the law of multiple proportions is followed. 3. The atomic mass of carbon (12.011 amu) is approximately 12 times that of hydrogen (1.008 amu). * (a) Show how you can use this knowledge to calculate possible formulas for benzene, ethane and ethylene. * (b) Show how your answer in part (a) is consistent with the actual formulas of benzene (CH), ethane (C2H6) and ethylene (C2H4) 4. Two elements A and B form two different compound. In the first compound one gram of A is combines with 1.44g of B. In the second compound 0.15g of A is combined with 0.65g of B. Show how these data illustrate the law of multiple proportions. What would the formulas of the two compounds be? 5. Hydrogen and oxygen are known to form two compounds. The hydrogen content in one of these is 5.93% while in the other it is 11.2%. Show that the data illustrates the law of multiple proportions.