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Ch 7 Sec 3 Using Chemical Formulas pages 237-244 1 Modern Chemistry Chapter 7 Chemical Formulas and...
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Transcript of Ch 7 Sec 3 Using Chemical Formulas pages 237-244 1 Modern Chemistry Chapter 7 Chemical Formulas and...
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
1
Modern ChemistryModern ChemistryChapter 7Chapter 7
Chemical Formulas and Chemical Formulas and Chemical CompoundsChemical Compounds
Sections 1, 3 & 4Chemical Names and Formulas
Using Chemical FormulasDetermining Chemical Formulas
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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Ch
ap
ter
Vocab
ula
ryMonatomic ionBinary compoundNomenclatureOxyanionSaltFormula Mass Percent CompositionEmpirical Formula
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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Section 3
Using Chemical Formulas
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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Ato
mic
Mass A
nim
ati
on
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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• Formula Mass – – the mass of any molecule, formula
unit or ion.– The sum of all the average atomic
masses of all the atoms represented by the formula
– Unit: a.m.u.
• Molecular mass– Mass of a molecule – in a.m.u.
Definitions
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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• Molar mass – – the mass of one mole of atoms of an
element – The mass of one mole of molecules
or formula units of a compound– Numerically equal to formula mass– Unit: grams/mole (g/mol)
Definitions
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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• List the elements in the compound.• List the number of atoms of each
element.• Multiply each number of atoms by
the – Relative atomic mass (for formula
mass)– Molar mass (for molar mass)
• Add the products.
Calculating Formula Mass or Molar Mass
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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List the elements in the compound.
Calculating Formula Mass or Molar Mass
(NH4)2SO4
N
H
S
O
2
8
1
4
x 14.01 = 28.02
x 1.008 = 8.064
x 32.06 = 32.06
x 16.00 = 64.00
+
132.144List the number of atoms of each
element.
• Multiply each number of atoms by the – Relative atomic mass (for formula mass)– Molar mass (for molar mass)
Add the products…
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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Calculating Formula Mass or Molar Mass
(NH4)2SO4
N
H
S
O
2
8
1
4
x 14.01 = 28.02
x 1.008 = 8.064
x 32.06 = 32.06
x 16.00 = 64.00
132.144From the periodic table4 sig figs
The unit is amu for formula mass. The unit is g/mol for molar mass
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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Pra
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1. Find the formula mass of each of the following:
a. H2SO4
b. Ca(NO3)2
c. PO4 3−
d. MgCl2
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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2. Find the formula mass of each of the following:
a. Al2S3
b. NaNO3
c. Ba(OH)2
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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One mole of …
Nitrogen (balloon)
CdS (yellow)
Water (cylinder)
NaCl (white)
p. 239
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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Mole
An
imati
on
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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The formula could stand for 1 molecule of water with 2 hydrogen atoms and 1 oxygen atom… or …
The formula could stand for 1 mole of water molecule which contain 2 moles of hydrogen atoms and 1 mole of oxygen atoms.
Chemical Formulas & Moles
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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Conversions with Compounds
MASSgrams
AMOUNT
moles
PARTICLESmolecules
MOLAR MASS 6.022 x 10 23
xx
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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Conversions with Compounds
MASSgrams
AMOUNT
moles
PARTICLESatoms
MOLAR MASS 6.022 x 10 23
xx
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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Grams-Moles-Molecules Conversions
p. 240
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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g
mol C
on
vers
ion
An
imati
on
p. 240
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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Dimensional Analysis
# & unit given unit wantedunit givenx
Fill in units first; then numbers !
=unit given
Put in numbers to make the numerator equal to
the denominator
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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Dimensional Analysis
x x x x =
Arrange the units so that all cancel out except the last one, which should be the one you want.
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1. How many moles of compound are there in the following?
a. 6.60 g (NH4)2SO4
b. 4.5 kg Ca(OH)2
2. How many molecules are there in the following?
a. 25.0 g H2SO4
b. 125 g of sugar, C12H22O11
3. What is the mass in grams of 6.25 mol of copper(II) nitrate?
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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% C
om
posit
ion
An
imati
on
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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Find the molar mass of the compound.
Percent Composition(NH4)2SO4
N
H
S
O
2
8
1
4
x 14.01 = 28.02
x 1.008 = 8.064
x 32.06 = 32.06
x 16.00 = 64.00
132.144Divide the subtotal for each element
by the molar mass
132.14 = .2120
132.14 = .06103
132.14 = .2426
132.14 = .4843
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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Multiply each by 100 to get a percent.
Percent Composition(NH4)2SO4
N
H
S
O
28.02
8.064
32.06
64.00
132.144
132.14 = .2120
132.14 = .06103
132.14 = .2426
132.14 = .4843
x100 = 21.20%
x100 = 6.103%
x100 = 24.26%
x100 = 48.43%
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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HydratesSalt which have the ability to bind
water molecules within their crystal lattice.
Anhydrous - salts without waterExamples
CuSO4 5H2O
copper(II) sulfate pentahydrateCoCl2 6H2O
cobalt (II) chloride hexahydrate
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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CuSO4 5H2O
Hydrates
CoCl2 6H2O
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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What percentage of sodium carbonate decahydrate, Na2CO3 10H2O, is sodium
carbonate?1. Find the molar mass of Na2CO3 (the anhydrous compound).
2. Find the mass Na2CO3 10H2O (the hydrate).
3. Divide mass of compound (anhydrous) by mass of the hydrate and multiply by 100.
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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Percent Composition of a HydrateCuSO4 5H2O
Cu
S
O
H
O
1
1
4
10
5
x 63.55 = 63.55
x 32.06 = 32.06
x 16.00 = 64.00
x 1.008 = 10.08
x 16.00 = 80.00
249.69
159.61
90.08
249.69
249.69
x 100 =
63.92%
x 100 =
36.08%
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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What is the percentage of MnCl2 in MnCl2 2H2O?
77.73%
How much water could 100g of anhydrous MnCl2 absorb if the hydrated form is MnCl2 2H2O?
28.64g
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1. Find the percentage compositions of the following:
a. PbCl2 b. Ba(NO3)2
2. Find the mass percentage of water in ZnSO4•7H2O.
3. Magnesium hydroxide is 54.87% oxygen by mass. How many grams of oxygen are in 175 g of the compound? How many moles of oxygen is this?
Ch 7 Sec 3 Using Chemical Formulas pages 237-244
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Section 3 Homework