Chapter 7Chemical Quantities

Section 7.1The Mole: A Measurement of Matter

• OBJECTIVES:– Describe how Avogadro’s number is related to

a mole of any substance.

Section 7.1The Mole: A Measurement of Matter

• OBJECTIVES:– Calculate the mass of a mole of any substance.

What is a Mole?

• You can measure mass, • or volume,• or you can count pieces.• We measure mass in grams.• We measure volume in liters.

• We count pieces in MOLES.

Moles (abbreviated: mol)

• Defined as the number of carbon atoms in exactly 12 grams of carbon-12.

• 1 mole is 6.02 x 1023 particles.

• Treat it like a very large dozen

• 6.02 x 1023 is called Avogadro’s number.

Representative particles

• The smallest pieces of a substance.

– For a molecular compound: it is the molecule.

– For an ionic compound: it is the formula unit

(ions).

– For an element: it is the atom.

• Remember the 7 diatomic

elements (made of molecules)

Types of questions

• How many oxygen atoms in the following?– CaCO3

– Al2(SO4)3

• How many ions in the following?– CaCl2– NaOH– Al2(SO4)3

Types of questions

• 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 ?

• How many moles is 7.78 x 1024 formula units of MgCl2?

Measuring Moles

• Remember relative atomic mass?• The amu was one twelfth the mass of a

carbon-12 atom.• Since the mole is the number of atoms in

12 grams of carbon-12• The mass number on the periodic table is

also the mass of 1 mole of those atoms in grams.

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 pieces as 1.008 grams of H and 55.85 grams of iron.

• We can write this as 12.01 g C = 1 mole C

• We can count things by weighing them.

Examples

• How much would 2.34 moles of carbon weigh?

• How many moles of magnesium is 24.31 g of Mg?

• How many atoms of lithium is 1.00 g of Li?

• How much would 3.45 x 1022 atoms of U weigh?

What about 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

What about compounds?

• 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– this is the mass of one mole of a molecular

compound.

Gram Formula Mass (gfm)

• The mass of one mole of an ionic compound.• Calculated the same way as gmm.

• What is the GFM of Fe2O3?

2 moles of Fe x 55.85 g = 111.70 g

3 moles of O x 16.00 g = 48.00 g

The GFM = 111.70 g + 48.00 g = 159.70 g

Section 7.2Mole-Mass and Mole-Volume Relationships

• OBJECTIVES:– Use the molar mass to convert between mass

and moles of a substance.

Section 7.2Mole-Mass and Mole-Volume Relationships

• OBJECTIVES:– Use the mole to convert among measurements

of mass, volume, and number of particles.

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 mass- just a much broader term.

Examples

• Calculate the molar mass of the following and tell what type it is:

• Na2S

• N2O4

• C

• Ca(NO3)2

• C6H12O6

• (NH4)3PO4

Molar Mass

• The number of grams of 1 mole of atoms, ions, or molecules.

• We can make conversion factors from these.– To change grams of a compound to moles of a

compound.

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

For example

• How many moles is 5.69 g of NaOH?

5 69. g mole

g

need to change grams to moles

For example

• How many moles is 5.69 g of NaOH?

5 69. g mole

g

need to change grams to moles for NaOH

For example

• How many moles is 5.69 g of NaOH?

5 69. g mole

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

For example

• How many moles is 5.69 g of NaOH?

5 69. g mole

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

For example

• How many moles is 5.69 g of NaOH?

5 69. g 1 mole

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

For example

• How many moles is 5.69 g of NaOH?

5 69. g 1 mole

40.00 = 0.142 mol NaOH

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

Examples

• How many moles is 4.56 g of CO2?

• How many grams is 9.87 moles of H2O?

• How many molecules is 6.8 g of CH4?

• 49 molecules of C6H12O6 weighs how much?

Gases

• Many of the chemicals we deal with are gases.–They are difficult to 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.

Standard Temperature and Pressure

• 0ºC and 1 atm pressure• abbreviated 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

Examples

• What is the volume of 4.59 mole of CO2 gas at STP?

• How many moles is 5.67 L of O2 at STP?

• What is the volume of 8.8 g of CH4 gas at STP?

Density of a gas

• 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.

Examples

• Find the density of CO2 at STP.

• Find the density of CH4 at STP.

The other way

• Given the density, we can find the molar mass of the gas.

• Again, pretend you have 1 mole at STP, so V = 22.4 L.

• 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?

• 2.86 g/L?

Summary

• 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.

Section 7.3Percent Composition and Chemical Formulas

• OBJECTIVES:– Calculate the percent composition of a

substance from its chemical formula or experimental data.

Section 7.3Percent Composition and Chemical Formulas

• OBJECTIVES:– Derive the empirical formula and the molecular

formula of a compound from experimental data.

Calculating Percent Composition of a Compound

• Like all percent problems:

part

whole

• Find the mass of each component,• then divide by the total mass.

x 100 %

Example

• Calculate the percent composition of a compound that is 29.0 g of Ag with 4.30 g of S.

Getting it from the formula

• If we know the formula, assume you have 1 mole.

• Then you know the mass of the pieces and the whole.

Examples

• Calculate the percent composittion of C2H4?

• How about Aluminum carbonate?– Sample Problem 7-11, p.191

• We can also use the percent as a conversion factor– Sample Problem 7-12, p.191

The Empirical Formula

• The lowest whole number ratio of elements in a compound.

• The molecular formula = the actual ratio of elements in a compound.

• The two can be the same.

• CH2 is an empirical formula

• C2H4 is a molecular formula

• C3H6 is a molecular formula

• H2O is both empirical & molecular

Calculating Empirical

• Just find the lowest whole number ratio

• C6H12O6

• CH4N

• It 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.

Calculating Empirical

• We 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.

Example

• Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.

• Assume 100 g so• 38.67 g C x 1mol C = 3.220 mole C

12.01 gC • 16.22 g H x 1mol H = 16.09 mole H

1.01 gH• 45.11 g N x 1mol N = 3.219 mole N

14.01 gN

Example

• The ratio is 3.220 mol C = 1 mol C 3.219 molN 1 mol N

• The ratio is 16.09 mol H = 5 mol H 3.219 molN 1 mol N

• = C1H5N1

• A compound is 43.64 % P and 56.36 % O. What is the empirical formula?

• Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?

Empirical to molecular

• 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.• Caffeine has a molar mass of 194 g. what

is its molecular formula?

Example

• 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?

• Sample Problem 7-14, p.194