1. 2 the mass of a single atom is too small to measure on a balance. mass of hydrogen atom = 1.673 x...
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The mass of a single atom is too small to measure on a balance.
mass of hydrogen atom = 1.673 x 10-24 g
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This is an
infinitesimal
mass
1.673 x 10-24 g
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• Chemists have chosen a unit for counting atoms.
• That unit is the
• Chemists require a unit for counting which can express large numbers of atoms using simple numbers.
MOLE
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1 mole = 6.02 x 1023 objects
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LARGE6.02 x 1023
is a very
number
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6.02 x 1023
is
number
Avogadro’s Number
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Amadeo (Amedeo) Avogadro
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If 10,000 people started to count Avogadro’s number and counted at the rate of 100 numbers per minute each minute of the day, it would take over 1 trillion years to count the total number.
http://youtu.be/1R7NiIum2TI
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1 mole of any element contains
6.02 x 1023
particles of that substance.
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The atomic weight in grams
of any element23
contains 1 mole of atoms.
12
This is the same number of particles6.02 x 1023
as there are in exactly 12 grams of
C126
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ExamplesExamplesExamplesExamples
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Species
Quantity
Number of H atoms
H
1 mole
6.02 x 1023
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Species
Quantity
Number of H2 molecules
H2
1 mole
6.02 x 1023
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Species
Quantity
Number of Na atoms
Na
1 mole
6.02 x 1023
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Species
Quantity
Number of Fe atoms
Fe
1 mole
6.02 x 1023
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Species
Quantity
Number of C6H6 molecules
C6H6
1 mole
6.02 x 1023
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1 mol of atoms = 6.02 x 1023 atoms
6.02 x 1023 molecules
6.02 x 1023 ions
1 mol of molecules =
1 mol of ions =
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• The mole weight of an element is its atomic weight in grams.
• It contains 6.02 x 1023 atoms (Avogadro’s number) of the element.
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ElementAtomic mass
Mole weightNumber of
atoms
H 1.008 amu 1.008 g 6.02 x 1023
Mg 24.31 amu 24.31 g 6.02 x 1023
Na 22.99 amu 22.99 g 6.02 x 1023
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ProblemsProblemsProblemsProblems
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Convert ToMoles!
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Weight (mass) Numbers
Moles
Periodic Table Avogadro’s Number6.02 × 1023
25
Atomic weight iron = 55.85
How many moles of iron does 25.0 g of iron represent?
Conversion sequence: grams Fe → moles Fe
1 mol Fe(grams Fe)
55.85 g Fe
1 mol Fe(25.0 g Fe)
55.85 g Fe
0.448 mol Fe
Set up the calculation using a conversion factor between moles and grams.
(40.0 g Fe)1 mole Fe55.85 g Fe
236.02 x 10 atoms Fe1 mole Fe
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Atomic weight iron = 55.85Conversion sequence is:
grams Fe → moles Fe → atoms Fe
How many iron atoms are contained in 40.0 grams of iron?
234.31 x 10 atoms Fe
23(3.01 x 10 atoms Na) 23
1 mole Na6.02 x 10 atoms Na
22.99 g Na1 mole Na
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Mole weight Na = 22.99 g
Conversion sequence:
atoms Na → moles Na → grams Na
What is the mass of 3.01 x 1023 atoms of sodium (Na)?
11.5 g Na
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2(2.00 mole O )23
2
2
6.02 x 10 molecules O1 mole O
2
2 atoms O1 molecule O
Conversion sequence:
moles O2 → molecules O2 → atoms O
How many oxygen atoms are present in 2.00 moles of oxygen molecules?
24= 2.41 x10 atoms O
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Mole Weight of Mole Weight of CompoundsCompounds
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The mole weight of a compound can be determined by adding the mole weights of all of the atoms in its formula.
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2 C = 2(12.01 g) = 24.02 g
6 H = 6(1.01 g) = 6.06 g
1 O = 1(16.00 g) = 16.00 g
46.08 g
Calculate the mole weight of C2H6O.
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1 Li = 1(6.94 g) = 6.94 g1 Cl = 1(35.45 g) = 35.45 g4 O = 4(16.00 g) = 64.00 g
106.39 g
Calculate the mole weight of lithium perchlorate.
LiClO4
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Calculate the mole weight of ammonium phosphate.
3 N = 3(14.01 g) = 42.03 g12 H = 12(1.01 g) = 12.12 g
1 P = 1(30.97 g) = 30.97 g4 O = 4(16.00 g) = 64.00 g
149.12 g
(NH4)3PO4
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Water in a hydrate is known as water of hydration or water of crystallization.
Solids that contain water as part of their crystalline structure are known as hydrates.
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Formulas of hydrates are written by first writing the formula for the anhydrous compound and then adding a dot followed by the number of water molecules present.
6H2OCoCl2
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Calculate the mole weight of NaC2H3O2 · 3 H2O
1 Na = 1(22.99 g) = 22.99 g 2 C = 2(12.01 g) = 24.02 g
9 H = 9(1.01 g) = 9.09 g 5 O = 5(16.00 g) = 80.00 g
136.10 g
Sodium acetate trihydrate
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Avogadro’s Number of Particles
6.02 x 1023 Particles
Mole Weight
1 MOLE
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1 MOLE Ca
Avogadro’s Number ofCa atoms
6.02 x 1023 Ca atoms
40.078 g Ca
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1 MOLE H2O
Avogadro’s Number of
H2O molecules
6.02 x 1023 H2O
molecules
18.02 g H2O
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In dealing with diatomic elements (H2, O2, N2, F2, Cl2, Br2, and I2), distinguish between one mole of atoms and one mole of molecules.
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Calculate the mole weight of 1 mole of H atoms.
1 H = 1(1.01 g) = 1.01 g
Calculate the mole weight of 1 mole of H2 molecules.
2 H = 2(1.01 g) = 2.02 g
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ProblemsProblemsProblemsProblems
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How many moles of benzene, C6H6, are present in 390.0 grams of benzene?
Conversion sequence: grams C6H6 → moles C6H6
6 6
6 6
78.12 grams C HUse the conversion factor:
1 mole C H
6 6
6 6
1 mole C H 78.12 g C H
6 6(390.0 g C H ) 6 6= 4.992 moles C H
The mole weight of C6H6 is 78.12 g.
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How many grams of (NH4)3PO4 are contained in 2.52 moles of (NH4)3PO4?
Conversion sequence:
moles (NH4)3PO4 → grams (NH4)3PO4
4 3 4
4 3 4
149.12 grams (NH ) POUse the conversion factor:
1 mole (NH ) PO
4 3 4(2.52 mol (NH ) PO )) 4 3 4
4 3 4
149.12 g (NH ) PO1 mol (NH ) PO
4 3 4= 376g (NH ) PO
The mole weight of (NH4)3PO4 is 149.12 g.
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2(56.04 g N ) 2
2
1 mol N28.02 g N
232
2
6.02 x 10 molecules N
1 mol N
56.04 g of N2 contains how many N2 molecules?
The mole weight of N2 is 28.02 g.
Conversion sequence: g N2 → moles N2 → molecules N2
Use the conversion factors
2
2
1 mol N 28.02 g N
232
2
6.02 x 10 molecules N
1 mol N
242= 1.20 x 10 molecules N
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2
2
1 mol N28.02 g N
2(56.04 g N )23
2
2
6.02 x 10 molecules N
1 mol N
56.04 g of N2 contains how many N atoms?
The mole weight of N2 is 28.02 g.
Conversion sequence: g N2 → moles N2 → molecules N2
→ atoms NUse the conversion factors
2
2
1 mol N 28.02 g N
232
2
6.02 x 10 molecules N
1 mol N 2
2 atoms N 1 molecule N
24= 2.41 x 10 atoms N2
2 atoms N1 molecule N
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Percent CompositionPercent Compositionof Compoundsof Compounds
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Percent composition of a compound is the mass percent of each element in the compound.
H2O11.19% H by mass 88.79% O by mass
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Percent Composition Percent Composition From FormulaFrom Formula
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If the formula of a compound is known, a two-step process is needed to calculate the percent composition.
Step 1 Calculate the mole weight of the formula.
Step 2 Divide the total mass of each element in the formula by the mole weight and multiply by 100.
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total mass of the element x 100 = percent of the element
mole weight
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Step 1 Calculate the mole weight of H2S.2 H = 2 (1.01 g) = 2.02 g
1 S = 1 (32.07 g) = 32.07 g34.09 g
Calculate the percent composition of hydrosulfuric acid H2S(aq) .
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Calculate the percent composition of hydrosulfuric acid H2S.
Step 2 Divide the mass of each element by the mole weight and multiply by 100.
H
5.93%
S
94.07%
32.07g SS: (100) 94.07%
34.09g
2.02 g HH: (100) = 5.93%
34.09 g
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Percent Composition Percent Composition From Experimental DataFrom Experimental Data
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Percent composition can be calculated from experimental data without knowing the composition of the compound.
Step 1 Calculate the mass of the compound formed.
Step 2 Divide the mass of each element by the total mass of the compound and multiply by 100.
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Step 1 Calculate the total mass of the compound1.52 g N
3.47 g O
4.99 g
A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition.
= total mass of product
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Step 2 Divide the mass of each element by the total mass of the compound formed.
3.47g O(100) = 69.5%
4.99g
1.52 g N(100) = 30.5%
4.99 g
N
30.5%
O
69.5%
A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition.
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Empirical Formula versus Empirical Formula versus Molecular FormulaMolecular Formula
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• The empirical formula or simplest formula gives the smallest whole-number ratio of the atoms present in a compound.
• The empirical formula gives the relative number of atoms of each element present in the compound.
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• The molecular formula is the true formula of a compound.
• The molecular formula represents the total number of atoms of each element present in one molecule of a compound.
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ExamplesExamplesExamplesExamples
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C2H4Molecular Formula
CH2Empirical Formula
C:H 1:2Smallest Whole Number Ratio
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C6H6Molecular Formula
CHEmpirical Formula
C:H 1:1Smallest Whole Number Ratio
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H2O2Molecular Formula
HOEmpirical Formula
H:O 1:1Smallest Whole Number Ratio
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Two compounds can have identical empirical formulas and different molecular formulas.
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CalculatingCalculatingEmpirical FormulasEmpirical Formulas
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The analysis of a compound shows that it contains 27% carbon and 73% oxygen. Calculate the empirical formula for this substance.
C___O___
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Step 1 Assume a definite starting quantity (usually 100.0 g) of the compound, if not given, and express the mass of each element in grams.
Step 2 Convert the grams of each element into moles of each element using each element’s mole weight.
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Step 3 Divide the moles of atoms of each element by the moles of atoms of the element that had the smallest value
– If the numbers obtained are whole numbers, use them as subscripts and write the empirical formula.
– If the numbers obtained are not whole numbers, go on to step 4.
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Step 4 Multiply the values obtained in step 3 by the smallest numbers that will convert them to whole numbers
Use these whole numbers as the subscripts in the empirical formula.
FeO1.5
Fe1 x 2O1.5 x 2 Fe2O3
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• The results of calculations may differ from a whole number.
– If they differ ±0.1 round off to the next nearest whole number.
2.93– Deviations greater than 0.1 unit from a
whole number usually mean that the calculated ratios have to be multiplied by a whole number.
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ProblemsProblemsProblemsProblems
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The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance.Step 1 Express each element in grams. Assume 100
grams of compound.
K = 56.58 g
C = 8.68 g
O = 34.73 g
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The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance.Step 2 Convert the grams of each element to moles.
K: 56.58 g K1 mol K atoms
39.10 g K
1.447 mol K atoms
C: 8.68 g C1 mol C atoms
12.01 g C
0.723 mol C atoms
O: 34.73 g O1 mol O atoms
16.00 g O
2.171 mol O atoms
C has the smallest number of moles
0.723 mol C atoms
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The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance.Step 3 Divide each number of moles by the smallest
value.
1.447 molK = = 2.00
0.723 mol0.723 mol
C: = 1.000.723 mol
2.171 molO = = 3.00
0.723 mol
The simplest ratio of K:C:O is 2:1:3
Empirical formula K2CO3
C has the smallest number of moles
0.723 mol C atoms
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The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.
The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.
Step 1 Express each element in grams. Assume 100 grams of compound.
N = 25.94 g
O = 74.06 g
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Step 2 Convert the grams of each element to moles.
N: 25.94 g N1 mol N atoms
14.01 g N
1.852 mol N atoms
O: 74.06 g O1 mol O atoms
16.00 g O
4.629 mol O atoms
The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.
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Step 3 Divide each number of moles by the smallest value.
1.852 molN = = 1.000
1.852 mol4.629 mol
O: = 2.5001.852 mol
The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.
This is not a ratio of whole numbers.
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Step 4 Multiply each of the values by 2.
The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.
Empirical formula N2O5
N: (1.000)2 = 2.000 O: (2.500)2 = 5.000
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Calculating the Molecular Formula Calculating the Molecular Formula from the Empirical Formulafrom the Empirical Formula
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• The molecular formula can be calculated from the empirical formula if the mole weight is known.
• The molecular formula will be equal to the empirical formula or some multiple n of it.
• To determine the molecular formula evaluate n.
• n is the number of units of the empirical formula contained in the molecular formula.
mole weightn = =
empirical weight formula units number of empirical
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What is the molecular formula of a compound which has an empirical formula of CH2 and a mole weight of 126.2 g?
The molecular formula is (CH2)9 = C9H18
Let n = the number of formula units of CH2.
Calculate the mass of each CH2 unit
1 C = 1(12.01 g) = 12.01g
2 H = 2(1.01 g) = 2.02g
14.03g126.2 g
n 9 (empirical formula units)14.03 g
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A compound made of 30.4% N and 69.6% O has a mole weight of 138 grams/mole. Find the molecular formula.
Old Way:
1mole30.4g 2.17mole of N
14.0g
1mole69.6g 4.35mole of O
16.0g
2.17 24.352.17 2.17
N O NO 46g/ mole
2 3 6
1383units 3(NO ) N O
46
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A compound made of 30.4% N and 69.6% O has a mole weight of 138 grams/mole. Find the molecular formula.
New Way:
N3O6
138gcomp'd1molecomp'd
138gcomp'd1molecomp'd
30.4gN100gcomp'd
69.6gO100gcomp'd
1moleN14.0g
1moleO16.0g
3moleN
6moleO
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