Chemical Reactions & Writing Chemical Formulas Chapter 15 Section 1.
Chapter 7 “Chemical Formulas and Chemical Compounds” Yes, you will need a calculator for this...
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Transcript of Chapter 7 “Chemical Formulas and Chemical Compounds” Yes, you will need a calculator for this...
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Chapter 7“Chemical Formulas
and Chemical Compounds”
Yes, you will need a calculator for this chapter!
![Page 2: Chapter 7 “Chemical Formulas and Chemical Compounds” Yes, you will need a calculator for this chapter!](https://reader035.fdocuments.in/reader035/viewer/2022062401/5a4d1b077f8b9ab059988e10/html5/thumbnails/2.jpg)
Using Formulas• How many oxygen atoms in
the following?CaCO3
Al2(SO4)3• How many ions in the
following?CaCl2NaOHAl2(SO4)3
3 atoms of oxygen
12 (3 x 4) atoms of oxygen
3 total ions (1 Ca2+ ion and 2 Cl1- ions)
2 total ions (1 Na1+ ion and 1 OH1- ion)
5 total ions (2 Al3+ + 3 SO42- ions)
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Review Problems How many molecules of CO2 are in
4.56 moles of CO2? How many moles of water is 5.87 x
1022 molecules? How many atoms of carbon are in
1.23 moles of C6H12O6? How many moles is 7.78 x 1024
formula units of MgCl2?
2.75 x 1024 molecules
0.0975 mol (or 9.75 x 10-2)
4.44 x 1024 atoms C
12.9 moles
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More Practice How many grams is 2.34 moles of
carbon? How many moles of magnesium is
24.31 g of Mg? How many atoms of lithium is
1.00 g of Li? How many grams is 3.45 x 1022
atoms of U?
28.1 grams C
1.000 mol Mg
8.68 x 1022 atoms Li
13.6 grams U
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What about compounds? in 1 mole of H2O molecules there are
two moles of H atoms and 1 mole of O atoms (think of a compound as a molar ratio)
To find the mass of one mole of a compound – determine the number of moles of the
elements present– Multiply the number times their mass
(from the periodic table)– add them up for the total mass
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Calculating Formula MassCalculate the formula mass of magnesium carbonate, MgCO3.
24.31 g + 12.01 g + 3 x (16.00 g) = 84.32 g
Thus, 84.32 grams is the formula mass for MgCO3.
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Molar Mass Molar mass is the generic term for the
mass of one mole of any substance (expressed in grams/mol)
The same as: 1) Gram Molecular Mass (for
molecules) 2) Gram Formula Mass (ionic compounds)
3) Gram Atomic Mass (for elements)– molar mass is just a much broader
term than these other specific masses
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ExamplesCalculate the molar mass of
the following: Na2S N2O4
C Ca(NO3)2
C6H12O6
(NH4)3PO4
= 78.05 g/mol
= 92.02 g/mol
= 12.01 g/mol
= 164.1 g/mol
= 180.156 g/mol
= 149.096 g/mol
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Since Molar Mass is… The number of grams in 1 mole of
atoms, ions, or molecules, We can make conversion factors
from these.- To change between grams of a compound and moles of a compound.
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For example
How many moles are in 5.69 g of NaOH?
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For example
•How many moles are in 5.69 g of NaOH?5 69. g
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For example
•How many moles are in 5.69 g of NaOH?5 69. g mole
g
We need to change 5.69 grams NaOH to moles
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For example
•How many moles are in 5.69 g of NaOH?5 69. g mole
g
We need to change 5.69 grams NaOH to moles
1mole Na = 22.99 g 1 mol O = 16.00 g 1 mole of H = 1.008 g
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For example
•How many moles are in 5.69 g of NaOH?5 69. g mole
g
We need to change 5.69 grams NaOH to moles
1mole Na = 22.99 g 1 mol O = 16.00 g 1 mole of H = 1.008 g
1 mole NaOH = 39.998 g
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For example
•How many moles are in 5.69 g of NaOH?
We need to change 5.69 grams NaOH to moles
1mole Na = 22.99 g 1 mol O = 16.00 g 1 mole of H = 1.008 g
1 mole NaOH = 39.998 g
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For example
•How many moles are in 5.69 g of NaOH?
We need to change 5.69 grams NaOH to moles
1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.008 g
1 mole NaOH = 39.998 g
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Calculating Percent Composition of a Compound Like all percent problems:
part
whole1)Next, divide by the total mass
of the compound; then x 1002)Find the mass of each of the
components (the elements),
x 100 % = percent
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Example Calculate the percent
composition of a compound that is made of 29.0 grams of Ag with 4.30 grams of S.29.0 g Ag33.3 g total
X 100 = 87.1 % Ag
4.30 g S33.3 g total X 100 = 12.9 % S
Total = 100 %
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Getting it from the formula If we know the formula,
assume you have 1 mole, then you know the mass of
the elements and the whole compound (these values come from the periodic table!).
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Examples Calculate the percent
composition of C2H4.
How about aluminum carbonate?
85.63% C, 14.37 % H
23.1%Al, 15.4% C, and 61.5 % O
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Formulas
• Example: molecular formula for Example: molecular formula for benzene is Cbenzene is C66HH66 (note that (note that everything is divisible by 6) everything is divisible by 6) • Therefore, the empirical formula = Therefore, the empirical formula = CH
(the lowest (the lowest wholewhole number ratio) number ratio)
Empirical formula: the lowest whole number ratio of atoms in a compound.Molecular formula: the true number of atoms of each element in the formula of a compound.
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Formulas (continued)
Formulas for ionic compounds are ALWAYS empirical (the lowest whole number ratio = cannot be reduced).
Examples:
NaCl MgCl2 Al2(SO4)3 K2CO3
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Formulas (continued)Formulas for molecular compounds might be empirical (lowest whole number ratio).
Molecular:
H2O
C6H12O6 C12H22O11
Empirical:
H2O
CH2O C12H22O11
(Correct formula)
(Lowest whole number ratio)
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Calculating EmpiricalJust find the lowest whole number ratioC6H12O6 CH4N
A formula is not just the ratio of atoms, it is also the ratio of moles.
In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen.
In one molecule of CO2 there is 1 atom of C and 2 atoms of O.
= CH2O
= this is already the lowest ratio.
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Calculating Empirical We can get a ratio from the
percent composition.1)Assume you have a 100 g sample
- the percentage becomes grams (75.1% = 75.1 grams)
2)Convert grams to moles. 3)Find lowest whole number ratio by
dividing each number of moles by the smallest value.
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ExampleCalculate the empirical formula of
a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.
Assume 100 g sample, so38.67 g C x 1mol C = 3.220 mol
C 12.01 g C 16.22 g H x 1mol H = 16.09 mol
H 1.008 g H45.11 g N x 1mol N = 3.220 mol
N 14.01 g N
Now divide each value by the smallest value
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ExampleThe ratio is 3.220 mol N = 1 mol
N 3.220 mol NThe ratio is 3.220 mol C = 1 mol
C 3.220 mol N The ratio is 16.09 mol H = 5 mol H
3.220 mol N = C1H5N1 which is =
CH5N
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PracticePracticeA compound is 43.64 % P and 56.36
% O. What is the empirical formula? PP22OO55
Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?
CC44HH55NN22OO
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Empirical to molecularSince the empirical formula is the
lowest ratio, the actual molecule would weigh more.By a whole number multiple.
Divide the actual molar mass by the empirical formula mass – you get a whole number to increase each coefficient in the empirical formula
Caffeine has a molar mass of 194 g. what is its molecular formula?
= C8H10N4O2
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HydratesHydrates• Hydrate: a compound that has a Hydrate: a compound that has a
specific number of water molecules specific number of water molecules bound to its atomsbound to its atoms
• You can determine the formula of a You can determine the formula of a hydrate by heating it until the water hydrate by heating it until the water is gone, and analyzing the datais gone, and analyzing the data
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BaClBaCl22 ?H ?H22OO
• Mass of the compound before heating Mass of the compound before heating (hydrate): 5.00 g(hydrate): 5.00 g
• Mass after heating (anhydrous): 4.26 gMass after heating (anhydrous): 4.26 g• Find the mass of the water lost by Find the mass of the water lost by
subtracting 5.00g – 4.26 g = 0.74 g subtracting 5.00g – 4.26 g = 0.74 g HH22OO
• The mass of just the BaClThe mass of just the BaCl22 is 4.26 g is 4.26 g
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ContinuedContinued• Convert the mass of BaClConvert the mass of BaCl22 and the mass and the mass
of Hof H22O to molO to mol• 4.26 g BaCl4.26 g BaCl221 mol BaCl1 mol BaCl22 = 0.02046 mol = 0.02046 mol 208.2 g BaCl208.2 g BaCl22
• 0.74 g H0.74 g H22O O 1 mol H1 mol H22O = 0.04107 molO = 0.04107 mol 18.016 g H18.016 g H22OO• Divide by the smallest molar valueDivide by the smallest molar value
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ContinuedContinued
• 0.02046 mol BaCl0.02046 mol BaCl22 = 1 = 1 0.02046 mol BaCl0.02046 mol BaCl22
• 0.04107 mol H0.04107 mol H22O = 2O = 2 0.02046 mol BaCl0.02046 mol BaCl22
• Formula of the hydrate: BaClFormula of the hydrate: BaCl22 2H 2H22OO
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Another PracticeAnother Practice
• FePOFePO44 ? H ? H22OO• Mass before heating: 7.61 gMass before heating: 7.61 g• Mass after heating: 5.15 gMass after heating: 5.15 g• Mass of water: 2.46 gMass of water: 2.46 g• Formula of the hydrate: FePOFormula of the hydrate: FePO44 4H 4H22OO