Chapter 3. Calculations Related to Chemical Formulas
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1 Chapter 3 Calculations Related to Chemical Formulas
2 Formula Mass The mass of an individual molecule or formula unit Also known as molecular mass or molecular weight Sum of the masses of the atoms in a single molecule or formula unit mass of 1 molecule of H2O = 2(1.01 amu H) amu O = amu mass of 1 formula unit of MgCl2 = 2(35.45 amu Cl) amu Mg = amu
3 Molar Mass
4 Molar Mass of Compounds The relative masses of molecules can be calculated from atomic masses. Formula Mass = 1 molecule of H2O = 2(1.01 amu H) amu O = amu 1 mole of H2O contains 2 moles of H and 1 mole of O. molar mass = 1 mole H2O = 2(1.01 g H) g O = g so the Molar Mass of H2O is g/mole
5 1. How many moles are in 50.0 g of PbO2? (Pb, g/mol; O,16.00 g/mol) g PbO2 mol PbO2 Pb = 1 x = O = 2 x = g PbO2 1 mol PbO2 PbO2 = g/mol 1 mol PbO g PbO2 1 mol PbO g PbO g PbO2 x = mol PbO2
6 2. Find the number of CO2 molecules in 10.8 g of CO2. 1 mol CO2 = x molecules g CO2 mol CO2 molec CO2 C = 1 x = O = 2 x = CO2 = 44.01g/mol g CO2 1 mol CO2 1 mol CO g CO g CO2 x 1 mol CO2 x x molecules = g CO2 1 mol CO x 10 x molecules CO2
7 3. What is the mass of 4.78 x NO2 molecules? 1 mol = x 10 23,1 mol NO2 = g molec NO2 mol NO2 g NO x molec NO2 1 mol NO2 1 mol NO x molec NO g NO2 1 mol NO2 1 mol NO g NO2 1 mol NO x molec NO x molec NO2 x x g NO2 1 mol NO2 = g g NO2 NO2
8 Percent Composition
9 Percent Composition Percentage of each element in a compound by mass Can be determined from 1. the formula of the compound 2. the experimental mass analysis of the compound
10 4. Find the mass percent of Cl in C2Cl4F2
11 Mass Percent as a Conversion Factor 5. If NaCl is 39% sodium, find the mass of table salt containing 2.4 g of Na. g Na g NaCl 100 g NaCl 39 g Na 39 g Na 100 g NaCl 100 g NaCl 2.4 g Na x = g NaCl 39 g Na
12 Empirical Formulas
13 Empirical Formula Simplest, whole-number ratio of the atoms of elements in a compound Can be determined from elemental analysis
14 Finding an Empirical Formula from % Composition 1. Convert the percentages to grams 2. Convert grams to moles 3. Write a pseudoformula using moles as subscripts 4. Divide all by smallest number of moles 5. Multiply all mole ratios by number to make all whole numbers
15 Example: Find the empirical formula of aspirin with the given mass percent composition Given: C = 60.00% H = 4.48% O = 35.53% Therefore, in 100 g of aspirin there are g C, 4.48 g H, and g O However, ratios of mass are not useful in determining empirical formulas!! We must convert to a ratio of moles!!
16 Methodology for determining empirical formula: g C g H g O mol C mol H mol O pseudoformula CxHyOz empirical formula CxHyOz Manipulate subscripts to obtain whole-number ratio
17 Calculate the moles of each element Write a pseudoformula C4.996H4.44O2.220
18 C4.996H4.44O2.220 Find the mole ratio C2.25H2.00O1.00 Multiply subscripts by factor to give whole number (x 4) C9H8O4
19 6. Determine the empirical formula of stannous fluoride, which contains 75.7% Sn ( g/mol) and the rest fluorine (19.00 g/mol) Given: 75.7% Sn, ( ) = 24.3% F g Sn g F mol Sn mol F pseudo formula empirical formula
20 6. Determine the empirical formula of stannous fluoride, which contains 75.7% Sn ( g/mol) and the rest fluorine (19.00 g/mol) Element Ratio in Grams Molar Mass Ratio in Moles Ratio in Moles Sn 75.7g X 1mol 118.7g F 24.3g X 1mol 19.00g SnF2
21 7. Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00) Given: 72.4% Fe, ( ) = 27.6% O g Fe g O mol Fe mol O pseudo formula empirical formula
22 7. Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00) x 3 Element Ratio in Grams Molar Mass Ratio in Moles Ratio in Moles Ratio in Moles Fe 72.4g X 1mol 55.85g O 27.6g X 1mol 16.00g Fe3O4
23 Chapter 3 Molecular Formulas
24 Molecular Formulas The molecular formula is a multiple of the empirical formula. To determine the molecular formula you need to know the empirical formula and the molar mass of the compound.
25 Find the molecular formula of butanedione if its empirical formula is C2H3O and its molar mass (MM) is g/mol. Molar Mass (emp. form.) Factor of 2 = 2 x (12.01 gc/molc) + 3 x (1.008 gh/molh) + 1 x (16.00 go/molo) = g/mol Molecular formula = C2H3O x 2 = C4H6O2
26 Practice Benzopyrene has a molar mass of 252 g and an empirical formula of C5H3. What is its molecular formula? (C = 12.01, H=1.01) C5 = 5(12.01 g) = g H3 = 3(1.01 g) = 3.03 g = g C5H3? 252 Molecular formula = {C5H3} x 4 = C20H12
27 Combustion Analysis
28 Combustion Analysis (Generally used for organic compounds containing C, H, O) A known mass of compound is burned in oxygen and the masses of the products formed (CO2 and H2O) are determined. By knowing the masses of the products and composition of constituent elements in the product, the original amount of constituent elements can be determined. It is assumed that all of the carbon in the original sample is converted to carbon dioxide and all of the hydrogen in the sample is converted to water.
29 Combustion Analysis
30 Example of Combustion Analysis Combustion of a g sample of a compound containing only carbon, hydrogen, and oxygen produced the following: CO2 = g H2O = g This came from C. This came from H. Determine the empirical formula of the compound. g CO2, H2O mol CO2, H2O mol C, H g C, H g C, H g O mol O mol C,H,O pseudoformula empirical formula
31 1 mole H = g H 1 mole C = g C molar masses of elements 1 mole O = g O 1 mole CO2 = g CO2 molar masses of compounds 1 mole H2O = g H2O 1 mole C 1 mole CO2 2 mole H 1 mole H2O ratios of compounds formed in combustion In the original sample
32 1 mole H = g H 1 mole C = g C molar masses of elements 1 mole O = g O In the original sample In the original sample In the original sample In the original sample
33 Pseudo formula C H O Empirical formula
34 8. Combustion of g of caproic acid produced g of H2O and 1.92 g of CO2. If the molar mass of caproic acid is g/mol, what is the molecular formula of caproic acid? In the original sample
35 C H O g moles In the original sample
36 Molecular formula = {C3H6O} x 2 = C6H12O2
37 Chapter 3 Chemical Reactions and Equations
38
39 Chemical Reactions Reactions involve rearrangement and exchange of atoms to produce new pure substances. Reactants Products Chemical Equations Shorthand way of describing a reaction Provides information about the reaction 1. formulas of reactants and products 2. states of reactants and products 3. relative numbers of reactant and product molecules
40 Combustion of Methane Methane gas reacts with oxygen gas to produce carbon dioxide gas and gaseous water. CH4(g) + O2(g) CO2(g) + H2O(g) This equation reads 1 molecule of CH4 gas combines with 1 molecule of O2 gas to make 1 molecule of CO2 gas and 1 molecule of H2O gas. + +
41 What about conservation of mass?? + + X 1 C + 4 H + 2 O 1 C + 2 O + 2 H + O
42 Combustion of Methane, Balanced To show the reaction obeys the Law of Conservation of Mass, the equation must be balanced. CH4(g) + O2(g) CO2(g) + H2O(g) CH4(g) + O2(g) CO2(g) + 2 H2O(g) CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) molecule of CH4 gas combines with 2 molecules of O2 gas to make 1 molecule of CO2 gas and 2 molecules of H2O gas.
43 Symbols Used in Equations Symbols used to indicate state after chemical: (g) = gas; (l) = liquid; (s) = solid (aq) = aqueous = dissolved in water Energy symbols used above the arrow for conditions for reactions: Δ = heat hν = light shock = mechanical elec = electrical
44 Steps in Balancing Equations 1. In compounds balance elements other than H and O. a. Balance elements which occur only once on each side of the equation. b. Start with the elements which occur the most. c. Balance polyatomic ions which do not change in the reaction. 2. Be prepared to rebalance if something changes!!! 3. Balance H. 4. Balance O. 5. Balance elements which appear in their elemental forms.
45 1. When aluminum metal reacts with air, it produces a white, powdery compound, aluminum oxide. aluminum(s) + oxygen(g) aluminum oxide(s) Al(s) + O2(g) Al2O3(s) 2 Al(s) + O2(g) Al2O3(s) 2 Al(s) + 3 O2(g) 2 Al2O3(s) 4 Al(s) + 3 O2(g) 2 Al2O3(s)
46 2. Solid phosphorous (P4) reacts with hydrogen gas to produce phosphorous trihydride. P4 (s) + H2 (g) > PH3 (g) P4 (s) + H2 (g) > 4 PH3 (g) P4 (s) + 6 H2 (g) > 4 PH3 (g)
47 3. Solid potassium chlorate decomposes to produce oxygen gas and potassium chloride. KClO3 (s) > O2 (g) + KCl (s) 2 KClO3 (s) > 3 O2 (g) + KCl (s) 2 KClO3 (s) > 3 O2 (g) + 2 KCl (s)
48 4. Aqueous sulfuric acid reacts with solid sodium cyanide to produce aqueous sodium sulfate and hydrogen cyanide gas. H2SO4 (aq) + NaCN (s) > Na2SO4 (aq) + HCN(g) H2SO4 (aq) + 2 NaCN (s) > Na2SO4 (aq) + HCN(g) H2SO4 (aq) + 2 NaCN (s) > Na2SO4 (aq) + 2 HCN(g)
49 5. Aqueous potassium phosphate reacts with aqueous calcium nitrate to produce solid calcium phosphate and aqueous potassium nitrate. K3PO4 (aq) + Ca(NO3)2 (aq) > Ca3(PO4)2 (s) + KNO3 (aq) K3PO4 (aq) + 3 Ca(NO3)2 (aq) > Ca3(PO4)2 (s) + KNO3 (aq) K3PO4 (aq) + 3 Ca(NO3)2 (aq) > Ca3(PO4)2 (s) + 6 KNO3 (aq) 2 K3PO4 (aq) + 3 Ca(NO3)2 (aq) > Ca3(PO4)2 (s) + 6 KNO3 (aq)
50 Organic Chemistry
51 Classifying Compounds Organic vs. Inorganic In the18 th century, compounds from living things were called organic; compounds from the nonliving environment were called inorganic. Organic compounds easily decomposed and could not be made in the 18 th -century lab. Inorganic compounds were very difficult to decompose, but could be synthesized.
52 Modern Classification of Compounds Organic vs. Inorganic Today we commonly make organic compounds in the lab and find them all around us. Organic compounds are mainly made of C and H, sometimes with O, N, P, S, halogens, and trace amounts of other elements. The main element that is the focus of organic chemistry is carbon.
53 Write a balanced equation for the combustion of butane, C4H10. C4H10 (g) + O2 (g) CO2 (g) + H2O (g) C4H10 (g) + O2 (g) 4 CO2 (g) + H2O (g) C4H10 (g) + O2 (g) 4 CO2 (g) + 5 H2O (g) C4H10 (g) + 13/2 O2 (g) 4 CO2 (g) + 5 H2O (g) 2 C4H10 (g) + 13 O2 (g) 8 CO2 (g) + 10 H2O (g)
54 Carbon Bonding in Organic Compounds Carbon atoms bond almost exclusively covalently in organic compounds. When C bonds, it forms four covalent bonds. Carbon is unique in that it can form limitless chains of C atoms, both straight and branched, and rings of C atoms.
55 Classifying Organic Compounds There are two main categories of organic compounds, hydrocarbons and functionalized hydrocarbons. All chemical compounds Inorganic Organic Hydrocarbons Functionalized Hydrocarbons
56 Hydrocarbons-Contain only Carbon and Hydrogen Name Molecular formula Structure Use methane CH4 Natural gas propane C3H8 LP gas n-butane C4H10 Butane lighters n-pentane C5H12 Gasolines ethene C2H4 Polymers ethyne C2H2 Welding
57 Families of Organic Compounds C C Alkanes O C H Aldehydes O C C Alkenes C C C Ketones C C Alkynes O C O H Carboxylic Acids C C C C C C Aromatics O C O C Esters C O H Alcohols N Amines O C O C Ethers C N Amides
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