Transcript Chapter 7 Chemical Formulas and Compounds

What is Avogadro’s Number
Avogadro’s Number is the
number of “things” in 1
mole
6.022 X 1023
Calculating Molar Mass
Molar mass is the mass of 1 mole of a compound.
Calculate the molar mass of magnesium carbonate.
MgCO3
24.31 g + 12.01 g + 3(16.00 g) =
84.32 g/mol
Practice
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Al2S3
NaNO3
Ba(OH)2
Ba(NO3)2
150.17 g/mol
85.00 g/mol
171.35 g/mol
261.35 g/mol
Gram/Mole Conversions
Molar masses are used as conversion
factors.
MgCO3 = 84.32 g/mol
84.32 g
1 mol
or
1 mol
84.32 g
Conversion factors
Mole to Mass Conversions
How many grams are in 2.36 mol of MgCO3?
More Practice!
What is the mass of 1.36 mol of H2O?
What is the mass of 4.36 mol of Ba(OH)2?
Mass to Mole Conversions
How many moles are in 0.37 g of MgCO3?
More Practice!
How many moles are in 4.50 g of H2O?
How many moles are in 471.6 g of Ba(OH)2?
Pause for a Cause #8
Calculate the number of moles in each of the following
masses:
a. 45.0 grams of acetic acid CH3COOH
b. 7.04 grams of lead II nitrate Pb(NO3)2
c. 5000 kg of iron III oxide Fe2O3
Calculate the mass of each of the following amounts:
a. 3.00 mol of selenium oxybromide SeOBr2
b. 488 mol of calcium carbonate CaCO3
c. 0.0091 mol of retonic acid C20H28O2
Pause for a Cause #8
a. 45.0 grams of acetic acid (CH3COOH)
0.749 mol (CH3COOH)
b. 7.04 grams of lead II nitrate Pb(NO3)2
0.0213 mol (Pb(NO3)2
c. 5000 kg of iron III oxide (Fe2O3)
3 .14 X 104 mol (Fe2O3)
a. 3.00 mol of selenium oxybromide (SeOBr2) 764.88g (SeOBr2)
b. 488 mol of calcium carbonate (CaCO3)
c. 0.0091 mol of retonic acid (C20H28O2)
4.88 X 104 g (CaCO3)
2.73g (C20H28O2)
1 mole = molar mass in grams = 6.022 X1023 molecules
Calculate the number of molecules/ formula units
a. 4.27 moles of tungsten oxide (WO3)
b. 0.003 00 moles of strontium nitrate Sr(NO3)2
Calculate the number of molecules/ formula units
a. 285 grams of iron III phosphate Fe(PO4)3
b. 0.0084 grams of C5H5N
Calculate the mass of each of the following quantities
a. 8.39 X 1023 molecules of F2 (fluorine)
b. 6.82 X 1024 formula units of beryllium sulfate BeSO4
Pause for a Cause #8
Calculate the number of molecules/ formula units
a. 4.27 moles of tungsten oxide (WO3)
2.57 X 1024 Molecules
b. 0.003 00 moles of strontium nitrate Sr(NO3)2
1.81 X 1021 Molecules
Calculate the number of molecules/ formula units
5.04 X 1023 Molecules
a. 285 grams of iron III phosphate Fe(PO4)3
6.4 X 1019 Molecules
b. 0.0084 grams of C5H5N
Calculate the mass of each of the following quantities
a. 8.39 X 1023 molecules of F2 (fluorine)
52.9g F2
b. 6.82 X 1024 formula units of beryllium sulfate BeSO4 1.19 X 103
Molecules
What is the mass of 6.022 X 1024
molecules of Iron III Oxide
What is the mass of 6.022 X 1023 atoms of Iron Oxide
6.022 X 1024 atoms Fe2O3 * ____1
mole____ * 159g Fe2O3 = 1590g Fe2O3
6.022 X 1023 atoms
1 mole
Calculating Percentage Composition
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Calculating Percentage Composition
Calculate the percentage composition of barium sulfate.
Step 1: Determine the molar mass.
Step 2: Divide the mass of the element in the
compound by the molar mass, then multiply by 100%.
Page 241
Let’s Practice! #10
• Determine the percent composition of NaCl.
• Na- 1 x 23.0=23.0g (23.0g/58.5g) x 100 = 39.3% Na
Cl- 1 x 35.5=35.5g (35.5g/58.5g) x 100 = 60.7% Cl
23.0g + 35.5g = 58.5 (58.5 is total)
• Determine the percent composition of Ba3(PO4)2
Formulas
Empirical formula: the lowest whole number
ratio of atoms in a compound.
(found experimentally by comparing ratio)
Molecular formula: the true number of
atoms of each element in the formula of a
compound.
(actual compound)
 molecular formula = C6H6
 empirical formula = CH
Khan Academy Video
Formulas (continued)
Formulas for ionic compounds are ALWAYS empirical
(lowest whole number ratio).
Examples:
NaCl
MgCl2
Al2(SO4)3
K2CO3
Formulas (continued)
Formulas for molecular compounds MIGHT be empirical
(lowest whole number ratio).
Molecular:
H2O
C6H12O6
C12H22O11
Empirical:
H2O
CH2O
C12H22O11
Empirical Formula Determination
1. Base calculation on 100 grams of compound.
2. Determine moles of each element in 100 grams
of compound.
3. Divide each value of moles by the smallest of
the values.
4. Convert decimals to fractions and determine
the least common denominator.
C?H?O?
Empirical Formula Determination
Adipic acid contains 49.32% C, 43.84% O,
and 6.85% H by mass. What is the
empirical formula of adipic acid?
C?O?H?
Carbon:
 49.32 g C 1 mol C   4.107 mol C
12.01 g C 
4.107 mol C
 1.50
2.74 mol O
 43.84 g O 1 mol O   2.74 mol O
16.00 g O 
2.74 mol O
 1.00
2.74 mol O
Oxygen:
Hydrogen
 6.85g H 1 mol H   6.78 mol H
1.01 g H 
Carbon: 1.50
x 2
3
Hydrogen: 2.50
x 2
5
Empirical formula:
6.78 mol H
 2.47
2.74 mol O
Oxygen: 1.00
x 2
2
C3H5O2
Pause for a Cause #11
A compound is found to contain 36.48% Na, 25.41%
S, and 38.11% O. Find its empirical formula.
Analysis of a 10.150 g sample of a compound known
to contain only phosphorus and oxygen indicates
a phosphorus content of 4.433 g. What is the
empirical formula of this compound?
Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2. The
molar mass of adipic acid is 146 g/mol. What is the
molecular formula of adipic acid?
Step 1: Find the molar mass of the
empirical formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2. The
molar mass of adipic acid is 146 g/mol. What is the
molecular formula of adipic acid?
Step 2: Divide the molar mass of the compound by
the molar mass given by the empirical formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
146
2
73
Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2. The
molar mass of adipic acid is 146 g/mol. What is the
molecular formula of adipic acid?
3. Multiply the empirical formula by this number to
get the molecular formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
146
2
73
(C3H5O2) x 2 =
C6H10O4
Pause for a Cause #12
What is the molecular formula of a
compound that has an empirical formula
of CH2 and a molar mass of 28 g/mol?
A compound is 81.06% boron and 18.94%
hydrogen. The molar mass was
experimentally determined to be 54 g/mol.
What is the molecular formula of the
compound?
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