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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  

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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

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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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