Chemical Formulas and Equations

Chemical Formulas and Equations

Section 3 Chemical Formulas and Equations Mass and Moles of a Substance Chemistry requires a method for determining the numbers of molecules in a given mass of a substance. This allows the chemist to carry out recipes for compounds based on the relative numbers of atoms involved. stoichiometry - the calculation involving the quantities of reactants and products in a chemical equation. 2

Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Molecular Weight and Formula Weight The molecular weight (covalent bonds/nmnm) of a substance is the sum of the atomic weights of all the atoms in a molecule of the substance. For, example, a molecule of H2O contains 2 hydrogen atoms (at 1.01 amu each) and 1 oxygen atom (16.00amu), giving a molecular weight of 18.02 amu. 3 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

Molecular Weight and Formula Weight The formula weight of a substance is the sum of the atomic weights of all the atoms in one formula unit of the compound, whether molecular or not. For example, one formula unit (FU) of NaCl contains 1 sodium atom (22.99 amu) and one chlorine atom (35.45 amu), giving a formula weight of 58.44 amu. 4 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

Mass and Moles of a Substance The Mole Concept A mole is defined as the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 grams of carbon12. The number of atoms in a 12-gram sample of carbon12 is called Avogadros number (Na). The value of Avogadros number is 6.022 x 1023. 1 mole = 6.022 x 1023 ? ions, particles, atoms, molecules, items, etc. 1 mole Na2CO3 6.022 x 1023 FU Na2CO3 1 mole CO2 6.022 x 1023 molecules CO2 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

5 Mass and Moles of a Substance The molar mass (Mm) of a substance is the mass of one mole of a substance. For all substances, molar mass, in grams per mole, is numerically equal to the formula weight in atomic mass units. That is, one mole of any element weighs its atomic mass in grams. 1 molecule of H2O - MW = 18.02 amu 1 mole of H2O - Mm = 18.02 g H2O 1 formula unit of NaCl - FW = 58.44 amu 1 mole of NaCl Mm = 58.44 g NaCl Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

6 How is it possible that Mm and FW/MW are the same value but different units? A.)What is the mass in grams of one Cl atom? B.)in one HCl molecule? 7 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Mass and Moles of a Substance Mole calculations Converting the number of moles of a given substance

into its mass, and vice versa, is fundamental to understanding the quantitative nature of chemical equations. mass of " A" moles of " A" atomic (or molecular) mass of " A" 58.44 g NaCl 1 mole NaCl 1 mole NaCl 58.44 g NaCl

Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 8 Mass and Moles of a Substance Mole calculations Suppose we have 5.75 moles of magnesium. What is its mass? 9 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Mass and Moles of a Substance Mole calculations

Suppose we have 100.0 grams of H2O. How many moles does this represent? 10 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Mass and Moles and Number of Molecules or Atoms The number of molecules or atoms in a sample is related to the moles of the substance: 23 1 mole HCl 6.02 10 HCl molecules 1 mole Fe 6.02 10 23 Fe atoms How many molecules are there in 56 mg

HCN? HW 20-22 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 11 Determining Chemical Formulas The percent composition of a compound is the mass percentage of each element in the compound. We define the mass percentage of A as the parts of A per hundred parts of the total, by mass. That is, mass of " A" in whole mass % " A"

100% mass of the whole 20 g A 20 % A 100 g sample 12 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Mass Percentages from Formulas Lets calculate the percent composition (%C, %H) of one molecule of butane, C4H10. First, we need the molecular wt of C4H10. 4 carbons @ 12.01 amu/atom 48.04 amu 10 hydrogens @ 1.01 amu/atom 10.10 amu

1 molecule of C 4 H10 58.14 amu Now, we can calculate the percents. amu C % C 5848.04 .14 amu total 100% 82.63%C amu H % H 5810.10 .14 amu total 100% 17.37% H Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 13 How many grams of carbon are there in 83.5 g of formaldehyde, CH2O, (40.0% C, 6.73% H, 53.3% O)?

14 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. An unknown acid contains only C, H, O. A 4.24 mg sample of acid is completely burned. It gives 6.21 mg of CO2 and 2.54 mg H2O. What is mass% of each element in the unknown acid? 15 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Determining Chemical Formulas

Determining the formula of a compound from the percent composition. The percent composition of a compound leads directly to its empirical formula. An empirical formula (or simplest formula) for a compound is the formula of the substance written with the smallest integer (whole number) subscripts. 16 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Determining Chemical Formulas Determining the empirical formula from the percent composition.

Benzoic acid is a white, crystalline powder used as a food preservative. The compound contains 68.8% C, 5.0% H, and 26.2% O by mass. What is its empirical formula? In other words, give the smallest whole-number ratio of the subscripts in the formula Cx HyOz 17 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Determining Chemical Formulas For the purposes of this calculation, we will assume we have 100.0 grams of benzoic acid. Then the mass of each element equals the

numerical value of the percentage. Since x, y, and z in our formula represent molemole ratios, we must first convert these masses to moles. Cx HyOz 18 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Determining Chemical Formulas Determining the empirical formula from the percent composition. Our 100.0 grams of benzoic acid would contain: 1 mol C

68.8 g C 5.72(8) mol C 12.01 g 1 mol H 5.0 g H 4.9(5) mol H 1.01 g 1 mol O 26.2 g O 1.63(7)mol O 16.00 g Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. This isnt quite a whole number ratio,

but if we divide each number by the smallest of the three, a better ratio might emerge. 19 Determining Chemical Formulas Determining the empirical formula from the percent composition. Our 100.0 grams of benzoic acid would contain: 5.728 mol C 1.63(7) 3.50 4.95 mol H 1.63(7) 3.0

1.63(7) mol O 1.63(7) 1.00 2x C3.5H3O1 C7H6O2 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. now its not too difficult to see that the smallest whole number ratio is 7:6:2 (multiple everything by 2 to get whole number. The empirical

formula is C7H6O2 . 20 Determining Chemical Formulas An empirical formula gives only the smallest whole-number ratio of atoms in a formula. The molecular formula should be a multiple of the empirical formula (since both have the same percent composition). C2H3O2 empirical formula (lowest whole #) C4H6O4 molecular formula C8H12O8

molecular formula Which is not an empirical formula? CH4 CH4O C2H4O2 C2H6O To determine the molecular formula, we must know the molecular weight (molar mass) of the compound. 21 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

Determining Chemical Formulas Determining the molecular formula from the empirical formula. Molecular weight = n x empirical formula wt. where n is the multiple factor n = molecular wt empirical wt 22 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Acetic Acid contains 39.9% C, 6.7% H, 53.4% O. Determine empirical formula. The molecular mass of acetic acid is 60.0 g/mol. What is the molecular

formula? HW 23 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 23 Stoichiometry: Quantitative Relations in Chemical Reactions Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction. It is based on the balanced chemical equation and on the relationship between mass and moles. Such calculations are fundamental to most

quantitative work in chemistry. 24 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Molar Interpretation of a Chemical Equation The balanced chemical equation can be interpreted in numbers of molecules, but generally chemists interpret equations as mole-to-mole relationships. N 2 (g)

1 molecule N2 1 mol N 2 + 28.02 g N2 + 28.02 g + 3H 2 (g)

3 molecules H2 3 mol H 2 2 NH 3 (g ) 2 molecules NH3 2 mol NH 3 3(2.02 g) H2 2 (17.04 g) NH3 6.06 g 34.08 g

34.08 g = 34.08 g Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 25 Molar Interpretation of a Chemical Equation Suppose we wished to determine the number of moles of NH3 we could obtain from 4.8 mol H2 (assume N2 in excess). N 2 (g) 3H 2 (g) 2 NH 3 (g ) 26

Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Mass Relationships in Chemical Equations How many grams of HCl are required to react with 5.00 grams MnO2 according to this equation? 4 HCl(aq) MnO 2 (s ) 2 H 2O(l) MnCl 2 (aq) Cl 2 (g ) 27 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Mass Relationships in Chemical Equations

How many grams of CO2 gas can be produced from 1.00 kg Fe2O3? Fe2O3 (s) + 3 CO (g) 2 Fe (s) + 3 CO2 (g) HW 24 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 28 Limiting Reagent The limiting reactant (or limiting reagent) is the reactant that is entirely consumed when the reaction goes to completion. The limiting reagent ultimately determines how much product can be obtained.

For example, bicycles require one frame and two wheels. If you have 20 wheels but only 5 frames, how many bicycles can be made? oo oo oo oo oo oo oo oo oo oo 20 wheels 1 frame + 2 wheels 1 bike 29 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Limiting Reagent Zn(s) 2 HCl(aq) ZnCl 2 (aq) H 2 (g ) If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, how many moles of H2 are produced?

30 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 8 Zn(s) S8 8ZnS If 7.36 g Zn was heated with 6.45 g sulfur, what amount of ZnS was produced? 31 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Theoretical and Percent Yield The theoretical yield of product is the

maximum amount of product that can be obtained from given amounts of reactants. The percentage yield is the actual yield (experimentally determined) expressed as a percentage of the theoretical yield (calculated). actual yield (exp) %Yield 100% theoretical yield (calc) 32 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Theoretical and Percent Yield To illustrate the calculation of percentage yield, recall

that the theoretical yield of ZnS in the previous example was 11.0 g ZnS. If the actual yield of the reaction had been 9.32 g ZnS, what is the %yield? 9.32 g ZnS %Yield 100% 84.7% 11.0 g ZnS HW 25 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 33 If 11.0 g CH3OH are mixed with 10.0 g CO, what is the theoretical

yield of HC2H3O2 in the following reaction? If the actual yield was 19.1 g, what is the %yield of HC2H3O2? CH3OH (l) + CO (g) HC2H3O2 (l) 34 Material was developed by combining Janusas material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

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