Chapter 3 Mass Relationships Stoichiometry Atomic Weights

Chapter 3 Mass Relationships; Stoichiometry.

Atomic Weights n n weighted average of the masses of the constituent isotopes lower number on periodic chart n How do we know what the values of these numbers are?

Atomic and Molecular Weights The Atomic Mass Scale 1 H weighs 1. 6735 x 10 -24 g and 16 O 2. 6560 x 10 -23 g. We define: mass of 12 C = exactly 12 amu. Using atomic mass units: 1 amu = 1. 66054 x 10 -24 g 1 g = 6. 02214 x 1023 amu

Atomic and Molecular Weights Atomic Masses are determined with a mass spectrometer What is the mass of one atom of 50 V in amu? Experimentally Determined

Mass Spectrometer

Atomic and Molecular Weights The Atomic Mass Scale What is the mass of one atom of 50 V in amu? Experimentally Determined 49. 9472 amu • What is the mass of one atom of 50 V in grams?

Atomic and Molecular Weights The Atomic Mass Scale What is the mass of one atom of 50 V in amu? Experimentally Determined 49. 9472 amu What is the mass of one atom of 50 V in grams?

Atomic and Molecular Weights Average Atomic Mass Relative atomic mass: average masses of isotopes: Naturally occurring C: 98. 892 % 12 C + 1. 108 % 13 C. Average mass of C: (0. 98892)(12 amu) + (0. 0108)(13. 00335) = 12. 011 amu. Atomic weight (AW) is also known as average atomic mass. Atomic weights are listed on the periodic table.

Atomic and Molecular Weights Determine the percent abundance of the two isotopes of Br. 79 Br 78. 9183 amu 81 Br 80. 9163 amu n

Atomic and Molecular Weights Determine the percent abundance of the two isotopes of Br. 79 Br 78. 9183 amu 81 Br 80. 9163 amu n Avg. Atomic Weight = (78. 9183)(x) + (80. 9163)(1 -x) 79. 904 amu = 78. 9183 x + 80. 9163 – 80. 9163 x 1. 9980 x = 1. 0123 79 Br = x = 1. 0123 / 1. 9980 = 0. 50666 50. 67% 81 Br = 1 -x = 0. 4933 49. 33%

Atomic and Molecular Weights Formula weights (FW): sum of AW for atoms in formula. FW (H 2 SO 4) = 2 AW(H) + AW(S) + 4 AW(O) = 2(1. 0 amu) + (32. 0 amu) + 4(16. 0) = 98. 0 amu Molecular weight (MW) is the weight of the molecular formula. MW(C 6 H 12 O 6) = 6(12. 0 amu) + 12(1. 0 amu) + 6(16. 0 amu)

Atomic and Molecular Weights What is the molecular weight of the following molecules? n CH 4 n H 2 SO 4 n Ca(NO 3)2 n Cu. SO 4 • 5 H 2 O n

Atomic and Molecular Weights n What is the molecular weight of the following molecules? n n CH 4 H 2 SO 4 Ca(NO 3)2 n Cu. SO 4 • 5 H 2 O n 12. 011 + 4(1. 008) = 16. 043 amu

Atomic and Molecular Weights n What is the molecular weight of the following molecules? n n CH 4 12. 011 + 4(1. 008) = 16. 043 amu H 2 SO 4 2(1. 008) + 32. 06 + 4(16. 00) = 96. 08 amu Ca(NO 3)2 n Cu. SO 4 • 5 H 2 O n

Atomic and Molecular Weights n What is the molecular weight of the following molecules? n CH 4 12. 011 + 4(1. 008) = 16. 043 amu H 2 SO 4 2(1. 008) + 32. 06 + 4(16. 00) = 96. 08 amu Ca(NO 3)2 40. 08 + 2(14. 007) + 6(16. 00) = 164. 09 amu Cu. SO 4 • 5 H 2 O n n n

Atomic and Molecular Weights n What is the molecular weight of the following molecules? n CH 4 12. 011 + 4(1. 008) = 16. 043 amu H 2 SO 4 2(1. 008) + 32. 06 + 4(16. 00) = 96. 08 amu Ca(NO 3)2 40. 08 + 2(14. 007) + 6(16. 00) = 164. 09 amu Cu. SO 4 • 5 H 2 O 63. 546+32. 06+9(16. 00)+10(1. 008) = 249. 68 amu n n n

Atomic and Molecular Weights Percentage Composition from Formulas Percent composition is the atomic weight for each element divided by the formula weight of the compound multiplied by 100:

Atomic and Molecular Weights n Determine the Mass Percent Composition of nitrogen in NH 4 NO 3?

Atomic and Molecular Weights n Determine the Mass Percent Composition of nitrogen in NH 4 NO 3? n Formula weight 80. 043 amu

The Mole n strictly a convenience unit n n amount that is large enough to see and handle in lab mole = number of things dozen = 12 things n mole = 6. 022 x 1023 things n n Avogadro’s number = 6. 022 x 1023 = NA

The Mole n how do you know when you have a mole? count it out n weigh it out n n molar mass - mass in grams equal to the atomic weight H is 1. 00794 g of H atoms = 6. 022 x 1023 atoms n Mg is 24. 3050 g of Mg atoms = 6. 022 x 1023 atoms n

The Mole: convenient measure chemical quantities. 1 mole of something = 6. 0221367 x 1023 of that thing. Experimentally, 1 mole of 12 C has a mass of 12 g. Molar Mass Molar mass: mass in grams of 1 mole of substance (units g/mol, g. mol-1). Mass of 1 mole of 12 C = 12 g.

The Molar Mass Molar mass: sum of the molar masses of the atoms: molar mass of N 2 = 2 x (molar mass of N). Molar masses for elements are found on the periodic table. Formula weights are numerically equal to the molar mass.

Molar Masses of Compounds n n add atomic weights of each atom The molar mass of propane, C 3 H 8, is: One mole of propane is 44. 11 g of propane.

The Mole Interconverting Masses, Moles, and Numbers of Particles

The Mole n Calculate the mass of a single Mg atom, in grams, to 3 significant figures.

The Mole n Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures.

The Mole n How many atoms are contained in 1. 67 moles of Mg?

The Mole n How many moles of Mg atoms are present in 73. 4 g of Mg?

Mole Problems with Compounds n Calculate the number of C 3 H 8 molecules in 74. 6 g of propane.

Mole Problems with Compounds n Calculate the number of O atoms in 26. 5 g of Li 2 CO 3.

1 Mole of Some Common Molecular Substances n 1 Mole n Br 2 n 159. 81 g Contains n n n C 4 H 10 6. 022 x 1023 Br 2 molecules 2(6. 022 x 1023 ) Br atoms

1 Mole of Some Common Molecular Substances n 1 Mole n Br 2 n 159. 81 g Contains n n n C 4 H 10 58. 04 g n n n 6. 022 x 1023 Br 2 molecules 2(6. 022 x 1023 ) Br atoms 6. 022 x 1023 C 4 H 10 molecules 4 (6. 022 x 1023 ) C atoms 10 (6. 022 x 1023 ) H atoms

Formulas from Elemental Composition n empirical formula - simplest molecular formula, shows ratios of elements but not actual numbers of elements n molecular formula - actual numbers of atoms of each element in the compound n determine empirical & molecular formulas of a compound from percent composition n percent composition is determined experimentally

Empirical Formulas Simplest whole number ratio of atoms or ions present in a compound Empirical Formula n H 2 O n C 6 H 12 O 6 n H 2 O 2 n

Empirical Formulas Simplest whole number ratio of atoms or ions present in a compound Empirical Formula n H 2 O n C 6 H 12 O 6 CH 2 O n H 2 O 2 n

Empirical Formulas Simplest whole number ratio of atoms or ions present in a compound Empirical Formula n H 2 O n C 6 H 12 O 6 CH 2 O n H 2 O 2 HO n

Empirical Formulas and Molecular Formula Once we know the empirical formula, we need the MW to find the molecular formula. Subscripts in the molecular formula are always wholenumber multiples of subscripts in the empirical formula.

Empirical Formulas from Analyses Combustion Analysis

Percent Composition mass of that element divided by the mass of the compound x 100% n percent composition of C in C 3 H 8 n

Percent Composition n percent composition of H in C 3 H 8

Percent Composition n Calculate the percent composition of Fe 2(SO 4)3 to 3 sig. fig.

Empirical Formulas from Analyses Start with mass % of elements (i. e. empirical data) and calculate a formula, or Start with the formula and calculate the mass % elements.

Empirical Formulas from Analysis n Determine the empirical formula (Crx. Oy) for a compound with the percent composition: 68. 42% Cr, 31. 58% O. 1. ) Assume 100 g n 2. ) Determine mol of each element n 3. ) Divide through by smallest subscript n 4. ) Convert all subscripts to integers n

Empirical Formulas from Analysis n Determine the empirical formula (Crx. Oy) for a compound with the percent composition: 68. 42% Cr, 31. 58% O. n 1. ) Assume 100 g 68. 42 g Cr and 31. 58 g O

Empirical Formulas from Analysis n Determine the empirical formula (Crx. Oy) for a compound with the percent composition: 68. 42% Cr, 31. 58% O. 1. ) Assume 100 g 68. 42 g Cr and 31. 58 g O 2. ) Determine mol of each element n

Empirical Formulas from Analysis n Divide through by smallest whole number 1. 316/1. 316 = 1 Cr 1. 974/1. 316 = 1. 50 O Convert to whole numbers Cr 1 O 1. 5 x 2 = Cr 2 O 3

Empirical Formulas from Analysis n n A compound contains 24. 74% K, 34. 76% Mn, and 40. 50% O by mass. What is its empirical formula? Make the simplifying assumption that you have 100. 0 g of compound. n in 100. 0 g of compound there is n n n 24. 74 g of K 34. 76 g of Mn 40. 50 g of O

Empirical Formulas from Analysis

Empirical Formulas from Analysis n A sample of a compound contains 6. 541 g of Co and 2. 368 g of O. What is empirical formula for this compound?

Chemical Equations n n n Lavoisier: mass is conserved in a chemical reaction. (Law of Conservation of Mass) Chemical equations: symbolic representation of a chemical reaction. Two parts to an equation: reactants and products: 2 H 2 + O 2 2 H 2 O n n n reactants on left side of reaction products on right side of equation relative amounts of each with stoichiometric coefficients

Chemical Equations n attempt to show on paper what is happening at the molecular level

Chemical Equations n look at what an equation tells us reactants yields 1 form. unit 3 mol. 1 mole 3 moles 159. 7 g 84 g products 2 atoms 3 mol. 2 moles 3 moles 111. 7 g 132 g

Chemical Equations n Stoichiometric coefficients: numbers in front of the chemical formulas; give ratio of reactants and products.

Chemical Equations

Law of Conservation of Matter no detectable change in quantity of matter in an ordinary chemical reaction n discovered by Lavoisier n balance chemical reactions to obey this law n propane, C 3 H 8, burns in oxygen to give carbon dioxide and water n C 3 H 8 + 5 O 2 ® 3 CO 2 + 4 H 2 O note that there are equal numbers of atoms of each element on both sides of equation

Law of Conservation of Matter n NH 3 burns in oxygen to form NO & water

Law of Conservation of Matter n C 5 H 12 burns in oxygen to form carbon dioxide & water n balancing equations is a skill acquired only with lots of practice n work many problems

Balance the following Equations CH 4(g) + O 2(g) CO 2(g) + H 2 O(g) n Ca. CO 3(s) Ca. O(s) + CO 2(g) n Ag. NO 3(aq) + Na. I(aq) Na. NO 3(aq) + Ag. I(s) n H 2 SO 4(aq) + KOH K 2 SO 4(aq) + H 2 O(l) n

Balance the following Equations n CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g)

Balance the following Equations CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g) n Ca. CO 3(s) Ca. O(s) + CO 2(g) n

Balance the following Equations CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g) n Ca. CO 3(s) Ca. O(s) + CO 2(g) n Ag. NO 3(aq) + Na. I(aq) Na. NO 3(aq) + Ag. I(s) n

Balance the following Equations CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g) n Ca. CO 3(s) Ca. O(s) + CO 2(g) n Ag. NO 3(aq) + Na. I(aq) Na. NO 3(aq) + Ag. I(s) n H 2 SO 4(aq) + 2 KOH K 2 SO 4(aq) + 2 H 2 O(l) n

Quantitative Information from Balanced Equations Balanced chemical equation gives number of molecules that react to form products. Interpretation: ratio of number of moles of reactant required to give the ratio of number of moles of product. These ratios are called stoichiometric ratios. NB: Stoichiometric ratios are ideal proportions Real ratios of reactants and products in the laboratory need to be measured (in grams and converted to moles).

Quantitative Information from Balanced Equations The ratio of grams of reactant cannot be directly related to the grams of product.

Calculations Based on Chemical Equations n n can do this in moles, formula units, etc. most often work in grams (or kg or pounds or tons) How many CO molecules are required to react with 25 formula units of Fe 2 O 3?

Calculations Based on Chemical Equations n How many iron atoms can be produced by the reaction of 2. 50 x 105 formula units of iron (III) oxide with excess carbon monoxide?

Calculations Based on Chemical Equations What mass of CO is required to react with 146 g of iron (III) oxide?

Calculations Based on Chemical Equations n What mass of carbon dioxide can be produced by the reaction of 0. 540 mole of iron (III) oxide with excess carbon monoxide?

Calculations Based on Chemical Equations n What mass of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8. 65 grams?

Calculations Based on Chemical Equations

Calculations Based on Chemical Equations n How many pounds of carbon monoxide would react with 125 pounds of iron (III) oxide?

Calculations Based on Chemical Equations 159. 6 g 84. 0 g 111. 6 g 132. 0 g or 159. 6 lb 84. 0 lb 111. 6 lb 132. 0 lb

Other Interpretations of Chemical Formulas n What mass of ammonium phosphate, (NH 4)3 PO 4, would contain 15. 0 g of N?

Patterns of Chemical Reactivity Using the Periodic Table n Properties of compounds vary systematically because of good ordering in the periodic table. 2 K(s) + 2 H 2 O(l) 2 KOH(aq) + H 2(g) 2 M(s) + 2 H 2 O(l) 2 MOH(aq) + H 2(g) Combustion in Air n Combustion is the burning of a substance in oxygen from air: C 3 H 8(g) + 5 O 2(g) 3 CO 2(g) + 4 H 2 O(l)

Patterns of Chemical Reactivity

Patterns of Chemical Reactivity Combination and Decomposition Reactions 2 Mg(s) + O 2(g) 2 Mg. O(s) There are fewer products than reactants; the Mg has combined with O 2 to form Mg. O. 2 Na. N 3(s) 2 Na(s) + 3 N 2(g) (the reaction that occurs in an air bag) There are more products than reactants; the sodium azide has decomposed into Na and nitrogen gas.

Patterns of Chemical Reactivity Combination and Decomposition Reactions Combination reactions: fewer reactants than products. Decomposition reactions: more products than reactants.

Limiting Reactant Concept If the reactants are not present in stoichiometric amounts, at end of reaction some reactants are still present (in excess). Limiting Reactant: one reactant that is consumed.

Limiting Reactant Concept n What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95. 6 g of carbon disulfide with 110 g of oxygen?

Limiting Reactant Concept What do we do next?

Limiting Reactant Concept n What can we conclude from our data now? Which is limiting reactant? What is maximum mass of sulfur dioxide? n n

Limiting Reactant Concept n n limiting reactant is O 2 maximum amount of SO 2 is 147 g

Theoretical yield The amount of product predicted from stoichiometry taking into account limiting reagents is called theoretical yield. The percent yield relates the actual yield (amount of material recovered in the laboratory) to theoretical yield:

Percent Yields from Reactions n n theoretical yield is what we have been calculating actual yield is what you have in your flask A 10. 0 g sample of ethanol, C 2 H 5 OH, was boiled with excess acetic acid, CH 3 COOH, to produce 14. 8 g of ethyl acetate, CH 3 COOC 2 H 5. What is the percent yield?

Percent Yields from Reactions

Synthesis Problem n In 1986, Bednorz and Muller succeeded in making the first of a series of chemical compounds that were superconducting at relatively high temperatures. This first compound was La 2 Cu. O 4 which superconducts at 35 K. In their initial experiments, Bednorz and Muller made only a few mg of this material. How many La atoms are present in 3. 56 mg of La 2 Cu. O 4?

Synthesis Problem

Group Activity n Within a year after Bednorz and Muller’s initial discovery of high temperature superconductors, Wu and Chu had discovered a new compound, YBa 2 Cu 3 O 7, that began to superconduct at 100 K. If we wished to make 1. 00 pound of YBa 2 Cu 3 O 7, how many grams of yttrium must we buy?