Chemical Composition:

Molar Mass and Formulas

Chem 101

Lectures 12 and 13

What can we tell from a

balanced equation?

C + O2 CO2

CH4 + 2O2 CO2 + 2H2O

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

What else can we figure out

from a balanced equation?

• If you know how many C atoms at start

– how many O2 molecules you will need

– how many CO2 molecules you can make

• Count the atoms

• Weigh the carbon and calculate the

number of atoms.

• Atomic masses allow us to convert

weights into numbers of atoms.

Moles

• The mass of 1.0 mole of an element is equal to the atomic mass in grams.

• One mole element = 6.022 x 1023 atoms. This number is called Avogadro’s number.

• Example: 1 mole of C atoms weighs 12.0 g and has 6.02 x 1023 atoms.

A 1-mol sample of graphite

(a form of carbon) weighs 12.01 g.

Hmco Photo File

Equalities

• For any element:

• 1 mole = 6.02 x 1023 atoms = Atomic mass

• 1 mole of calcium = – 6.02 x 1023 atoms of calcium

– 40.078 grams of calcium

• 1 mole of oxygen (O) atoms = – 6.02 x 1023 atoms of oxygen

– 15.999 grams of oxygen

• 1 mole of oxygen gas (O2) = – 12.04 x 1023 atoms of oxygen

– 31.998 grams of oxygen

All of these examples of

pure elements contain the same number

(a mole) of atoms: 6.02 x 1023 atoms.

Hmco Photo File

One-mole samples of iron (nails),

iodine crystals, liquid mercury, and powdered

sulfur.

Hmco Photo File

How many moles and atoms are

in 10.0 g of Al?

1.What are the equalities?

2.Use the atomic mass as a conversion

factor for grams-to-moles.

3.Use Avogadro’s number as a conversion

factor for moles-to-atoms.

1. What are the equalities?

2. Use Avogadro’s number as a conversion

factor for atoms-to-moles.

3. Use atomic mass as a conversion factor

for moles-to-grams.

How many moles and grams

are in 2.23 x 1023 atoms of Al.

Various numbers of methane molecules

and their constituent atoms.

One mole of SO2 contains

1 mole of S and 2 moles of O.

Molar Mass

• Molar mass: the mass in grams of one mole of a compound

• The relative weights of molecules can be calculated from atomic masses

Water = H2O = 2(1.008 g) + 16.00 g = 18.02 g

• 1 mole of H2O will weigh 18.02 g, therefore the molar mass of H2O is 18.02 g

• 1 mole of H2O will contain 16.00 g of oxygen and 2.02 g of hydrogen

Molar Mass

• For any molecule:

• 1 mole = 6.02 x 1023 molecules = molar mass

• 1 mole of hydrogen gas = 1 mole of H2

– 6.02 x 1023 molecules of H2

– 2.016 grams of H2

– 2 x 6.02 x 1023 atoms of H for a total of

12.04 x 1023 atoms or 1.204 x 1024 atoms

Molar Mass

• For any molecule:

• 1 mole = 6.02 x 1023 molecules = molar mass

• 1 mole of water = 1 mole of H2O

– 6.02 x 1023 molecules of H2O

– 18.02 grams of H2O

– How many atoms?

• 1 mole of hydrogen gas = 1 mole of H2

– 6.02 x 1023 molecules of H2

– 2.016 grams of H2

– How many atoms?

Practice Mole Problems

1. How many moles are in 5.5 x 1025 atoms of S?

2. What is the mass in g of 1.505 x 1024 atoms of Cl?

3. How many moles of Ne are in 25.0 g of Ne?

4. How many atoms of Na are in 75.0 g of Na?

5. Which contains more atoms: 50.0 g of Al or

50.0 g of Fe?

6. Which contains more Ni: 20.0g

2.85 x 1023 atoms

0.45 mol?

7. How many molecules are in 0.5 moles of CaCl2?

Chemical Composition: Part 2

• Percent composition

• Empirical formulas

• Molecular formulas

What is Percent Composition?

• Percentage by mass of each element in a

compound

• Can be determined from the formula of the

compound or by experimental mass analysis of

the compound

• The percentages may not always total to 100%

due to rounding.

100%whole

part Percentage

• Determine the mass of each element in 1

mole of the compound.

• Determine the molar mass of the compound

by adding the masses of the elements.

• Divide the mass of each element by the molar

mass of the compound and multiply by 100%

How to determine the percent composition

for each element in a formula

What is the percent composition for

each element in Ethanol? C2H5OH

Percent composition for elements in C2H5OH

• Determine the mass of each element in 1 mole of the compound C2H5OH. • 2 moles C = 2( g) = g

• 6 moles H = 6( g) = g

• 1 moles O = 1( g) = g

• Determine the molar mass of the compound by adding the masses of the elements. • 1 mole C2H5OH = g

• Divide the mass of each element by the molar mass of the compound and multiply by 100% • C

• H

• O

Empirical Formulas

• Empirical formula: the simplest, whole-number ratio of atoms in a molecule

– Can be determined from percent composition or by combining masses

– Different substances can have the same empirical formula

• Molecular formula: a multiple of the empirical formula

Same Empirical Formula

CH2O

Formaldehyde Acetic acid Dextrose

Benzopyrene, C20H12

• Benzopyrene is found in nature from the eruption of volcanoes and forest fires. It is also produced by burning plants, wood, coal, and operating cars, trucks and other vehicles.

• The major indoor sources of benzopyrene in the air are wood-burning fireplaces and stoves, and tobacco smoking.

• Benzopyrene can be found in surface water, tap water, rainwater, groundwater, wastewater and sewage sludge.

• No known industry production or use.

• Skin and eye irritant, Carcinogenic

• Find the greatest common factor (GCF) of the

subscripts.

factors of 20 = (10 x 2), (5 x 4)

factors of 12 = (6 x 2), (4 x 3) GCF = 4

• Divide each subscript by the GCF to get the

empirical formula.

C20H12 = (C5H3)4

Empirical Formula = C5H3

Determine the empirical formula

of benzopyrene, C20H12

Determine the empirical formula of acetic

anhydride by percent composition: 47% carbon, 47% oxygen, and 6.0% hydrogen.

Convert the percentages to grams by assuming you have

100 g of the compound.

Convert the grams to moles

Divide by the smallest number of moles

If any of the ratios is not a whole number, multiply all the

ratios by a factor to make it a whole number.

Use the ratios as the subscripts in the empirical formula.

Molecular Formulas

• The molecular formula is a multiple of

the empirical formula.

• To determine the molecular formula you

need to know the empirical formula and

the molar mass of the compound.

Determine the molecular formula of

benzopyrene if it has a molar mass of 252 g

and an empirical formula of C5H3

• What is the empirical formula: C5H3

• Determine the molar mass of C5H3

• 5 C =_____g, 3 H =_____g, so C5H3 =______g

• Divide the given molar mass of the compound by the molar mass of the empirical formula

• Round to the nearest whole number

• Multiply the empirical formula by the calculated factor to give the molecular formula

(C5H3)__ = C H

Empirical and Molecular

Formulas

• Different compounds may have the same

empirical formula

• The molecular formula and structure

identifies the different compounds