IIIIII II. Formula Calculations Ch. 10 – The Mole.
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Transcript of IIIIII II. Formula Calculations Ch. 10 – The Mole.
![Page 1: IIIIII II. Formula Calculations Ch. 10 – The Mole.](https://reader030.fdocuments.in/reader030/viewer/2022032612/56649efe5503460f94c130c7/html5/thumbnails/1.jpg)
I II III
II. Formula Calculations
Ch. 10 – The Mole
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Hydrates
Some compounds contain H2O in their structure. These compounds are called hydrates.
This is different from (aq) because the H2O is part of the molecule (not just surrounding it).
The H2O can usually be removed if heated.
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Hydrates
A dot separates water in the formula CuSO4•5H2O
When naming a Greek prefix indicates the # of H2O groups.
copper(II) sulfate pentahydrate
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Naming Hydrates
Na2SO4•10H2O
NiSO4•6H2O
CoCl2•6H2O
MgSO4•7H2O
Sodium sulfate decahydrate
Nickel (II) sulfate hexahydrate
Cobalt (II) chloride hexahydrate
Magnesium sulfate heptahydrate
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Writing Formulas of Hydrates sodium carbonate
monohydrate barium chloride
dihydrate Tin (IV) chloride
pentahydrate Barium hydroxide
octahydrate
Na2CO3•H2O
BaCl2•2H2O
SnCl4•5H2O
Ba(OH)2•8H2O
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What is Percent Composition?
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A. Percent Composition
the percentage by mass of each element in a compound
100compound of mass total
element of mass totalelement% mass of
100compound of mass molar
compound mol 1 in element of masselement% mass of
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%Fe =28 g
36 g 100 = 78% Fe
%O =8.0 g
36 g 100 = 22% O
Find the percent composition of a sample that is 28 g Fe and 8.0 g O.
A. Percent Composition
Known: Mass of Fe = 28 g Mass of O = 8.0 g Total Mass = 28 + 8.0 g = 36 g
Unknown: % Fe = ? % O = ?
Check:78% + 22%
= 100%
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100 =
A. Percent Composition Find the % composition of Cu2S. Use this formula when finding % composition from a
chemical formula:
%Cu =127.10 g Cu
159.17 g Cu2S 100 =
%S =32.07 g S
159.17 g Cu2S
79.85% Cu
20.15% S
100compound of mass molar
compound mol 1 in element of masselement% mass of
Known: Mass of Cu in 1 mol Cu2S =
2(63.55g) = 127.10 g Cu Mass of S in 1 mol Cu2S = 32.07 g S Molar Mass = 127.10 g + 32.07 g = 159.17 g/mol
Unknown: % Cu = ? % S = ?
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How many grams of copper are in a 38.0-gram sample of Cu2S?
Use answer from last question as a conversion factor.
Cu2S is 79.85% Cu =
A. Percent Composition
38.0 g Cu2S 79.85 g Cu
100 g Cu2S = 30.3 g Cu
79.85 g Cu
100 g Cu2S
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Find the percent composition of Cu2SO4.
A. Percent Composition
Known: Mass of Cu in 1 mol Cu2SO4 = 2(63.55 g) = 127.10 g Cu Mass of S in 1 mol Cu2SO4 = 32.07 g S Mass of 0 in 1 mol Cu2SO4 = 4(16.00 g) = 64.00 g O Molar Mass = 127.10 g + 32.07 g + 64.00 g = 223.17 g/mol
Unknown: % Cu = ? % S = ? % O = ?
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Find the percent composition of Cu2SO4.
A. Percent Composition
%Cu =127.10 g
223.17 g 100 = 56.95% Cu
%S =32.07 g
223.17 g 100 = 14.37% S
Check:56.95% + 14.37% + 28.68% = 100%
%O =64.00 g
223.17 g 100 = 28.68% O
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100 =%H2O = 36.04 g/mol
147.02 g/mol 24.51%
H2O
Find the mass percentage of water in calcium chloride dihydrate, CaCl2•2H2O.
A. Percent Composition
Known: Mass of H2O in 1 mol compound =
2(2(1.01g) + 16.00 g) = 36.04 g H2O Molar Mass = 40.08 g + 2(35.45 g) +
36.04 g = 147.02 g
Unknown: % H2O = ?
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B. Empirical Formula
C2H6
CH3
reduce subscripts
Smallest whole number ratio of atoms in a compound
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B. Empirical Formula1. Find mass (or %) of each element.
2. Find moles of each element.
3. Divide moles by the smallest # to find subscripts.
4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.
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B. Empirical Formula
Find the empirical formula for a sample of 25.9% N and 74.1% O.
25.9 g N 1 mol N
14.01 g N = 1.85 mol N
74.1 g O 1 mol O
16.00 g O = 4.63 mol O
1.85 mol
1.85 mol
= 1 N
= 2.5 O
N1.85O4.63
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B. Empirical Formula
N1O2.5Need to make the subscripts whole
numbers multiply by 2
N2O5
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B. Empirical Formula
Find the empirical formula for a sample of 94.1% O and 5.9% H.
94.1 g O 1 mol O
16.00 g O = 5.88 mol O
5.9 g H 1 mol H
1.01 g H = 5.84 mol H
5.84 mol
5.84 mol
= 1 O
= 1 H
EF = OH
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C. Molecular Formula “True Formula” - the actual number of atoms
in a compound, either the same as or a whole-number multiple of the empirical formula
CH3
C2H6
empiricalformula
molecularformula
?
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C. Molecular Formula1. Find the empirical formula.2. Find the empirical formula mass.3. Divide the molar mass by the empirical formula mass.4. Multiply each subscript in your EF by the answer from step 3.
nmass EF
mass molar nEF
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C. Molecular Formula The empirical formula for ethylene is CH2. Find the
molecular formula if the molar mass is 28.1 g/mol?
28.1 g/mol
14.03 g/mol = 2.00
empirical formula mass = 14.03 g/mol
(EF)n = (CH2)2 C2H4
n =
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D. Put it all together!! 1,6-diaminohexane is 62.1% C, 13.8% H,
and 24.1% N. What is the empirical formula? If the molar mass is 116.21 g/mol, what is the molecular formula?
62.1 g C 1 mol C
12.01 g C
= 5.17 mol C
13.8 g H 1 mol H
1.01 g H
= 13.7 mol H
= 3 C
= 8 H
24.1 g N= 1 N
1 mol N
14.01 g N
= 1.72 mol N
1.72 mol
1.72 mol
1.72 mol EF = C3H8N
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C3H8N
116.21 g/mol
58.12 g/mol = 2.00
empirical formula mass = 58.12 g/mol
(C3H8N)2
n =
C6H16N2
D. Put it all together!!