Unformatted text preview: Chemistry 106 Lecture 7 Topics: Mass and Moles Lectures 6+7+8 = Chapter 3.1 – 3.7 Announcements Ø། Reminder: Third Homework and Tutorial sets are due on the MasteringChemistry website on Friday, September 23rd. Ø། The on line homework is part of your overall course grade. It is NOT opRonal. Exam Reminder Ø། The First Exam will be on September 28th (during normal class <me and loca<on). Ø། Exam #1 will cover Chapters 1 through 3.
Ø། All exams will be mulRple choice. Ø། A pracRce exam is posted on blackboard.syr.edu. Ø། If you have any special needs or issues (such as sports) you must inform me in wri<ng as soon as possible (Before September 21st). Stoichiometry: QuanRtaRve RelaRons in Chemical ReacRons • Stoichiometry is the calculaRon of the quanRRes of reactants and products involved in a chemical reacRon. – It is based on balanced chemical equaRons and on the relaRonship between mass and moles. – Such calculaRons are fundamental to most quanRtaRve work in chemistry. Chemical ReacRons: EquaRons • A chemical equa<on is the symbolic representaRon of a chemical reacRon in terms of chemical formulas. – For example, the reacRon of sodium and chlorine to produce sodium chloride is wriZen: 2Na + Cl2 → 2NaCl
• The reactants are starRng substances in a chemical reacRon. The arrow means “yields”. • The formulas on the right side of the arrow represent the products. Mass and Moles of a Substance • In order to use these balanced equaRons, we need to relate molecules and mass. • 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 relaRve numbers of atoms involved. – These calculaRons are called stoichiometry. Molecular Weight • The molecular weight 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 1.0 amu each) and 1 oxygen atom (16.0 amu), giving a molecular weight of 18.0 amu. 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 of NaCl contains 1 sodium atom (22.99 amu) and 1 chlorine atom (35.45 amu), giving a formula weight of 58.44 amu. CalculaRng the Formula Weight AW = atomic weight, FW = formula weight • Calculate (with 4 signiﬁcant digits) the formula weight of glucose (C6H12O6). • FW = (6 x AW of C) + (12 x AW of H) + (6 x AW of O) AW of C = 12.0107 amu AW of H
= 1.00794 amu AW of O = 15.9994 amu • FW = 180.156 amu FW = 180.2 amu The Mole Concept • A mole (symbol is “mol”) is deﬁned as the quanRty of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 grams of carbon–12. • The number of atoms in a 12 gram sample of carbon–
12 is called Avogadro’s number (to which we give the symbol NA). • The value of Avogadro’s number is 6.02214179 x 1023 , but ocen rounded to 6.02 x 1023. The Mole Concept • A mole of a substance contains Avogadro’s number (6.02 x 1023) of molecules. • Same idea as using the terms “dozen” or “gross” – dozen = 12 – gross = 144 • When using moles, it is ocen important to indicate the formula of the substance. – A mole of oxygen atoms (O) has 6.02 x 1023 O atoms – A mole of oxygen molecules (O2) has 1.20 x 1024 O atoms 1 octanol (C8H17OH) mercury (II) iodide (HgI2) One mole each of various substances. sulfur (S8) methanol (CH3OH) How many sulfur atoms are present in 1.0 mole of Al2(SO4)3? A. 1 sulfur atom B. 3 sulfur atoms C. 4 sulfur atoms D. 6.0 x 1023 sulfur atoms E. 1.8 x 1024 sulfur atoms How many sulfur atoms are present in 1.0 mole of Al2(SO4)3? A. 1 sulfur atom B. 3 sulfur atoms C. 4 sulfur atoms D. 6.0 x 1023 sulfur atoms E. 1.8 x 1024 sulfur atoms Molar Mass • The molar mass 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 mole of C = 12.0107 grams
1 mole of Cl = 35.453 grams
1 mole of Ca = 40.078 grams Mole CalculaRons • ConverRng the number of moles of a given substance into its mass, and vice versa, is fundamental to understanding the quanRtaRve nature of chemical equaRons. mass of X
moles of X =
atomic (or molecular) mass of X Mole CalculaRons • Suppose we have 100.0 grams of iron (Fe). The atomic weight of iron is 55.8 g/mol. How many moles of iron does this represent? 100.0 g Fe 1 mol Fe
moles Fe =
×
55.8 g Fe = 1.79 moles of Fe Mole CalculaRons • Conversely, suppose we have 5.75 moles of magnesium (atomic wt. = 24.3 g/mol). What is its mass? 5.75 mol Mg 24.3 g Mg
mass Mg =
×
1 mol Mg = 140 grams of Mg Mole CalculaRons • This same method applies to compounds. Suppose we have 100.0 grams of H2O (molecular weight = 18.0 g/mol). How many moles does this represent? 100.0 g H 2O 1 mol H 2O
moles H 2O =
×
18.0 g H 2O = 5.56 moles of H 2O Mole CalculaRons • Conversely, suppose we have 3.25 moles of glucose, C6H12O6 (molecular wt. = 180.2 g/mol). What is its mass? 3.25 mol C6 H12O6 180.2 g C6 H12O6
mass C6 H12O6 =
×
1 mol C6 H12O6 = 586 grams of C6 H12O6 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: 1 mole HCl = 6.02 × 10 23 HCl molecules
1 mole Fe = 6.02 × 10 23 Fe atoms
Suppose we have a 3.46g sample of hydrogen chloride,
HCl. How many molecules of HCl does this represent? 1 mole HCl 6.02 x 10 23 HCl molecules
3.46g HCl × 36.5g HCl × 1 mole HCl = 5.71×1022 HCl molecules The formula weight of Na3PO4 is: A.
B.
C.
D. 70 grams/mole 164 grams/mole 265 grams/mole 116 grams/mole From the Periodic Table Na = 22.989770 amu P = 30.973761 amu O = 15.9994 amu The formula weight of Na3PO4 is: A.
B.
C.
D. 70 grams/mole 164 grams/mole 265 grams/mole 116 grams/mole From the Periodic Table Na = 22.989770 amu P = 30.973761 amu O = 15.9994 amu Determining Chemical Formulas • The percent composi<on of a compound is the mass percentage of each element in the compound. – We deﬁne the mass percentage of “X” as the parts of “X” per hundred parts of the total, by mass. That is, mass of X in whole
mass % X =
× 100%
mass of the whole Mass Percentages from Formulas • Let’s calculate the percent composiRon of butane, C4H10. First, we need the molecular mass of C4H10.
4 carbons @ 12.0 amu/atom = 48.0 amu
10 hydrogens @ 1.00 amu/atom = 10.0 amu
1 molecule of C4 H10 = 58.0 amu
Now, we can calculate the percentages.
4 amu C
% C = 58.8.0mu total × 100% = 82.8%C
0a
1 amu H
% H = 58.0.0 mu total × 100% = 17.2%H
0a CombusRon Method for Determining the Percentages of Carbon and Hydrogen in a Compound If you have an unknown compound (unknown
formula), elemental analysis is an experiment that
can tell you the percentage composition.
Sample + O2 → CO2 + H2O 100 g butane 17.2 g H 82.8 g C Determining Chemical Formulas • Determining the formula of a compound from the percent composiRon. – The percent composiRon of a compound leads directly to its empirical formula. – An empirical formula (or simplest formula) for a compound is the formula of the substance wriZen with the smallest integer (whole number) subscripts. Determining Chemical Formulas • Determining the empirical formula from the percent composiRon. – Benzoic acid is a white, crystalline powder used as a food preservaRve. 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 raRo of the subscripts in the formula: Cx HyOz Determining the Empirical Formula from the Percent ComposiRon • For the purposes of this calculaRon, we will assume we have 100.0 grams of benzoic acid. Remember: The compound contains 68.8% C, 5.0% H, and 26.2% O by mass. • Then the mass of each element equals the numerical value of the percentage. • Since x, y, and z in our formula represent mole mole raRos, we must ﬁrst convert these masses to moles. Cx HyOz Determining the Empirical Formula from the Percent ComposiRon • Our 100.0 grams of benzoic acid would
contain: C5.733H5.000O1.637 1 mol C
68.8 g C ×
= 5.733 mol C
12.0 g
1 mol H
5.0 g H ×
= 5.000 mol H
1.0 g
1 mol O
26.2 g O ×
= 1.637 mol O
16.0 g This isn’t quite a
whole number ratio,
but if we divide
each number by the
smallest of the
three, the correct
ratio will be found. Determining the Empirical Formula from the Percent ComposiRon • Our 100.0 grams of benzoic acid would contain: 5.73 mol C ÷1.637 = 3.50 5.0 mol H ÷1.637 = 3.0
1.637 mol O ÷1.637 = 1.00
Now you can see that the smallest whole number raRo is: 7:6:2 The empirical formula is: C7H6O2 Determining the Molecular Formula from the Empirical Formula • An empirical formula gives only the smallest whole number raRo of atoms in a formula. • The molecular formula should be a mulRple of the empirical formula (since both have the same percent composiRon). • To determine the molecular formula, we must know the molecular weight of the compound. Molecular Formula vs. Empirical Formula Determining the Molecular Formula from the Empirical Formula • For example, suppose the empirical formula of a compound is CH2O and its molecular weight is 60.0 g/mol. • The molar weight of the empirical formula (the empirical weight) is only 30.0 g/mol. molecular weight
Number of empirical formula units (n) =
empirical formula weight • This indicates that the molecular formula is actually the empirical formula doubled (n=2), or: C2H4O2 Next Lecture • Chap 3 conRnues ...
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 Fall '08
 Freedman
 Chemistry, Mole, Mass, Molecule

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