Unit 2: (Chapter 7) Chemical Quantities
Section 7.1 The Mole: A Measurement of Matter OBJECTIVES: Describe how Avogadro’s number is related to a mole of any substance. Calculate the mass of a mole of any substance.
What is a Mole? An Animal or What? GHS Honors Chem
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Moles (abbreviated: mol)
What is a Mole? • • • • •
You can measure mass, mass, or volume, volume, or you can count pieces. pieces. We measure mass in grams. grams. We measure volume in liters. liters.
• We count pieces in MOLES MOLES.. GHS Honors Chem
What are Our Representative, or Basic Particles? Particles?
• Defined as the number of carbon atoms in exactly 12 grams of carboncarbon12. • 1 mole is 6.02 x 1023 particles. • Treat it like a very large dozen • 6.02 x 1023 is called Avogadro’s
number. GHS Honors Chem
Let’s Check your Understanding? •
• They are the smallest pieces of a substance. • For a molecular compound: it is the molecule. • For an element: it is the atom. • Remember the 7 diatomic elements (made of molecules) • Br I N Cl H O F GHS Honors Chem
How many oxygen atoms are in each of the following molecules? • • •
•
CaCO3 Al2(SO4)3 O2
How many oxygen atoms in 1 mole of the following? • • •
CaCO3 Al2(SO4)3 O2
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Mole Questions
What’s the Significance of the Mole?
• How many molecules of CO2 are there in 4.56 moles of CO2 ? • How many moles of water is 5.87 x 1022 molecules? • How many atoms of carbon are there in 1.23 moles of C6H12O6 ?
Let’s look at Molar Masses … GHS Honors Chem
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Gram Atomic Mass (gam) • Equals the mass of 1 mole of an element in grams • 12.01 grams of C has the same number of atoms as 1.008 grams of H and 55.85 grams of iron. • Each of these has 6.02 x 1023 atoms • We can write this as 12.01 grams C = 1 mole C • We now have a Factor of 12.01 g/mole of Carbon • Now we can count things by weighing them.
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Is There a Similar Measure for Compounds? • in 1 mole of H2O molecules there are two moles of H atoms and 1 mole of O atoms • To find the mass of one mole of a compound • determine the moles of the elements they have • Find out how much they would weigh • add them up GHS Honors Chem
Examples Using GAM • • • • • •
How much would 2.34 moles of carbon weigh? How much would 1.50 moles of bromine weigh? How many moles of magnesium is 24.31 g of Mg? How many moles are in 56 grams of Nitrogen? How many atoms of lithium is 1.00 g of Li? How much would 3.45 x 1022 atoms of U weigh?
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Gram Molecular Mass (gmm) • What is the mass of one mole of CH4? • 1 mole of C = 12.01 g • 4 mole of H x 1.01 g = 4.04g • 1 mole CH4 = 12.01 + 4.04 = 16.05g
• The Gram Molecular Mass (gmm) of CH4 is 16.05g/mol • this is the mass of one mole of a molecular compound.
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Section 7.2 Mole--Mass and MoleMole Mole-Volume Relationships OBJECTIVES: Use the molar mass to convert between mass and moles of a substance. Use the mole to convert among measurements of mass, volume, and number of particles.
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Another New Term … Molar Mass • Molar mass is the generic term for the mass of one mole of any substance (in grams) • The same as: 1) gram molecular mass, 2) gram formula mass, and 3) gram atomic massmass- just a much broader term.
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Music for the Molar Masses
• • • • • •
Calculate the molar mass of the following: Na2S N2O4 C Ca(NO3)2 C6H12O6 (NH4)3PO4
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Molar Mass The number of grams of 1 mole of atoms, ions, or molecules. Let’s take a closer look at making conversion factors to change grams of a compound to moles of a compound.
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For example
How many moles is 5.69 g of NaOH?
For example
How many moles is 5.69 g of NaOH? 5.69 g
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For example
How many moles is 5.69 g of NaOH?
mole 5.69 g g need to change grams to moles
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For example
How many moles is 5.69 g of NaOH?
mole 5.69 g g need to change grams to moles for NaOH
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For example
How many moles is 5.69 g of NaOH?
mole 5.69 g g need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g GHS Honors Chem
For example
How many moles is 5.69 g of NaOH?
mole 5.69 g g need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g 1 mole NaOH = 40.00 g GHS Honors Chem
For example
How many moles is 5.69 g of NaOH?
1 mole 5.69 g 40.00 g need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g 1 mole NaOH = 40.00 g GHS Honors Chem
For example
How many moles is 5.69 g of NaOH?
5.69 g
1 mole = 0.142 mol NaOH 40.00 g
need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g 1 mole NaOH = 40.00 g
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Examples Using Molar Mass • How many moles is 4.56 g of CO2? • 0.104 moles
• How many grams is 9.87 moles of H2O? • 178 grams
• How many molecules is 6.8 g of CH4? • 2.55 x 1023 molecules
Worksheets on Moles, Mass, Avogadro’s Number, & Moles/Molecules/Grams
• 49 molecules of C6H12O6 weighs how much? • 1.47 x 10-20 grams GHS Honors Chem
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What is Standard Temperature and Pressure? Pressure?
What About Gases? • Many of the chemicals we deal with are gases. • They are difficult to weigh weigh..
• Need to know how many moles of gas we have. • Two things effect the volume of a gas • Temperature and pressure
• We need to compare them at the same temperature and pressure. GHS Honors Chem
• 0ºC (or 273 K) and 1 atm pressure is abbreviated as STP • At STP 1 mole of gas occupies 22.4 L • Called the molar volume • 1 mole = 22.4 L of any gas at STP
@ STP: 1 mole/22.4 Liters GHS Honors Chem
Density of a gas
Molar Volume Problems What
is the volume of 4.59 mole of CO2 gas at STP? 103
Liters
How
many moles is 5.67 L of O2 at STP? 0.253
moles
What
is the volume of 8.8 g of CH4 gas at STP? 12.3
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Liters
D=m/V for
a gas the units will be g / L
We
can determine the density of any gas at STP if we know its formula. To find the density we need the mass and the volume. If you assume you have 1 mole, then the mass is the molar mass (from PT) At STP the volume is 22.4 L. GHS Honors Chem
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“Molar Density” Problems Find
the molar density of CO2 at STP.
44 grams / 22.4 L = 1.96 g/L
Find
the molar density of CH4 at STP.
16 grams / 22.4 L = 0.714 g/L
Can we Find the Molar Mass, given the density of 1 Mole of Gas at STP? Pretend you have 1 mole at STP, so V = 22.4 L. Rearranging, m = D x V m is the mass of 1 mole, since you have 22.4 L of the stuff. What is the molar mass of a gas with a density of 1.964 g/L?
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44.0 grams
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Have We Learned Anything??
These four items are all equal: a) 1 mole b) molar mass (in grams) c) 6.02 x 1023 representative particles d) 22.4 L at STP Thus, we can make conversion factors from them.
Grams/mole
Is There an Easy Way to Remember these Mole Conversions?
Introducing Moletown!
6.02 x 1023 molecules/mole
1 mole/22.4 L 6.02 x 1023 molecules/22.4 L GHS Honors Chem
GHS Honors Chem
Mole Town
Molar Mass
The Mole
X Gram Town
X Particle City (molecules)
X
Volume City (L) GHS Honors Chem
Avacado’s Number
22.4 L/mole We Can use Moletown for all of our conversions!
The Mole and Volume, & the Comprehensive Mole (1(1-46) Worksheets GHS Honors Chem
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Section 7.3 Percent Composition and Chemical Formulas
Calculating Percent Composition of a Compound
OBJECTIVES:
Part whole
Calculate the percent composition of a substance from its chemical formula or experimental data. Derive the empirical formula and the molecular formula of a compound from experimental data.
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Like all percent problems:
Find the mass of each component, then divide by the total mass.
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What is the Percent Composition?
x 100 %
Calculate the percent composition of a compound that is 29.0 g of Ag with 4.30 g of S. = 29.0g/(29.0g + 4.30g) x 100% = 87.1 % Ag
What is the Percent Composition?
8.20 grams of Magnesium makes up 60.3% composition of a compound of Mg and Oxygen. What is the weight of Oxygen in the sample?
x
0.603 = 8.20 g/(8.20 g + )
x = 5.4 grams GHS Honors Chem
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Finding the Percent Composition from the Formula
Finding the Percent Composition from the Formula
Use the ratio from the formula, and assume that you have 1 mole of each element.. element Then you would use the Molar Mass of each element to determine the percent composition of each. each. For example: What is the percent composition of hydrogen in Methane, CH4?
What is the percent composition of hydrogen in Methane, CH4?
1 mole of Carbon = 12 grams There’s 1 C, so the mass of C = 12 grams
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1 mole of Hydrogen = 1.01 grams There are 4 H’s, so the total mass of H is 4.04 grams.
% C = [12 g/(16.04 g)] x 100% = 74.8 %
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Percent Composition As a Conversion Factor
More Percent Composition Problems
Calculate the percent composition of Hydrogen & Carbon in C2H4?
C: 85.6 % H: 14.4%
Calculate the percent composition of each element in Aluminum carbonate?
Al: 23.1 %
We can also use the percent as a conversion factor:
For Example … Sulfur makes up 26.7 % of the mass of NaHSO4. What is the mass of the sulfur in 16.8 grams of NaHSO4? Mass of Sulfur = 16.8 grams NaHCO3 x (26.7 grams S/100 grams NaHCO3) Mass of Sulfur = 4.49 grams
C: 15.4 % O: 61.5 %
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Percent Composition As a Conversion Factor A bit more difficult:
Calculate the mass of carbon in 82.0 grams of C3H8? The Solution: % Composition of C in C3H8 = (3x12g)/[(3x12g)+(8x1g)] % Composition of C in C3H8 = 81.8 % Mass of Carbon = 82.o grams C3H8 x (81.8 grams C/100 grams C3H8) Mass of Carbon = 67.1 grams GHS Honors Chem
And the Last Tidbit in Chapter 7: The Empirical Formula vs. the Molecular Formula GHS Honors Chem
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Percent Composition Problems for those times when you have difficulty getting to sleep
Please try these problems in the textbook: Page 191, practice problems 31, 32, 33, & 34
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The Empirical Formula is the lowest whole number ratio of elements in a compound
CH2
vs. The Molecular Formula is the actual ratio of elements in a compound
C2H4 C8H16
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The Molecular Formula and Empirical Formula can be the same
H2O CO 2
CO
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How Can we Calculate the Empirical Formula? The
Empirical Formula is not just the ratio of atoms, it is also the ratio of moles of atoms. 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. GHS Honors Chem
Calculating the Empirical Formula can get a ratio from the percent composition. Assume you have a 100 g. The percentages become grams. Convert grams to moles. Find lowest whole number ratio by dividing by the smallest.
Calculating the Empirical Formula
We
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Confused ? Let’s try an example … GHS Honors Chem
Here’s an Example
Here’s an Example
The Problem: Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.
The Problem: Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.
1. Assume 100 g so • 38.67 g C • 16.22 g H • 45.11 g N
Now, convert to moles … GHS Honors Chem
1. 38.67 g C x 1mol C = 3.220 mole C 12.01 g C 2. 16.22 g H x 1mol H = 16.09 mole H 1.01 g H 3. 45.11 g N x 1mol N = 3.219 mole N 14.01 g N GHS Honors Chem
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Here’s an Example 1. 38.67 g C x 1mol C = 3.220 mole C 12.01 g C 2. 16.22 g H x 1mol H = 16.09 mole H 1.01 g H 3. 45.11 g N x 1mol N = 3.219 mole N 14.01 g N Find the lowest whole number ratio: 1. Mole C = 3.220 / 3.219 = 1 2. Mole H = 16.09 / 3.219 = 5 3. Mole N = 3.219 / 3.219 = 1
C 1H 5N 1
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More Examples A compound is 43.64 % P and 56.36 % O. What is the empirical formula? Moles of P = 43.64 g/ 31 g/mol = 1.40 mol Moles of O = 56.36 g / 16 g/mol = 3.50 mol Dividing by 1.40, we get P1O2.5 Multiply by 2 to get P2O5 GHS Honors Chem
More Examples Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula? Moles of C = 4.12 Moles of H = 5.10 • Moles of N = 2.06 • Moles of O = 1.03 •
A Slightly Different Twist? What is the Empirical Formula of a compound that has 1.04g K, 0.70g Cr, and 0.86g O? Hint: Convert to % first, then follow previous example …
•
C 4H 5N 2O GHS Honors Chem
Can I Go from Empirical to Molecular Formulas?
Absolutely GHS Honors Chem
Ans. K2CrO4 GHS Honors Chem
Can I Go from Empirical to Molecular Formulas? Since
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. GHS Honors Chem
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Can I Go from Empirical to Molecular Formulas? Caffeine
has a molar mass of 194 g. what is its molecular formula? The
Empirical Formula for Caffeine is C4 H5 N2 O The empirical formula mass is 97 grams. 194 g / 97 g = 2 The molecular formula is 2 x the Empirical formula, or C8H10N4O2. GHS Honors Chem
Here’s Another …
A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) to be 98.96 g. What is its molecular formula? Moles
of Cl = 71.65 g / 35.45 g/mol = 2 of C = 24.27 g / 12 g/mol = 2 Moles of H = 4.07 g / 1 g/mol = 4 Empirical Formula = CH2Cl Empirical Molar Mass = 49.45 g / mol 98.96 g / 49.45 g = 2 Molecular Formula = C2H4Cl2 Moles
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Percent Composition & Molecular Formula Worksheet Chapter 7 Practice Problems Worksheet GHS Honors Chem
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