Why is Knowledge of Composition Important?

Because everything in nature is either chemically or physically combined with other substances it is necessary to have knowledge of chemical composition. To know the amount of a material in a sample, you need to know what fraction of the sample it is.

• Examples that show the need to know about chemical composition:– the amount of sodium in sodium chloride for

dietary need– the amount of iron in iron ore for steel production– the amount of hydrogen in water for hydrogen

fuel– the amount of chlorine in Freon to estimate

ozone depletion

Counting Atoms by Moles• If we can find the mass of a particular number of atoms, we can use this

information to convert the mass of an element sample to the number of atoms in the sample.

• The number of atoms we will use is 6.022 x 1023 and we call this a mole– 1 mole = 6.022 x 1023 things, like 1 dozen = 12 things– The dozen and the mole make it easier to talk about large quantities– A mole can refer to anything, but we usually use it to talk about atoms,

molecules, and ions

• mole = number of particles equal to the number of atoms in 12 g of C-12

– 1 atom of C-12 weighs exactly 12 amu

– 1 mole of C-12 weighs exactly 12 g

• The number of particles in 1 mole is called Avogadro’s Number = 6.0221421 x 1023 (remember Avogadro’s # to 4 Sig. Figs.)

– 1 mole of C atoms weighs 12.01 g and has 6.022 x 1023 atoms

• the average mass of a C atom is 12.01 amu

Mass Number is Not the Same as Atomic Mass

• the atomic mass is an experimental number determined from all naturally occurring isotopes

• the mass number refers to the number of protons + neutrons in one isotope

– natural or man-made

Calculating Atomic Mass

Gallium has two naturally occurring isotopes: Ga-69 with mass 68.9256 amu and a natural abundance of 60.11% and Ga-71 with mass 70.9247 amu and a natural abundance of 39.89%. Calculate the atomic mass of gallium.

Solution: Convert the percent natural abundance into decimal form.

Ga-69 0.6011Ga-71 0.3989

Determine the Formula to Use Atomic Mass = (abundance1)∙(mass 1) + (abundance2)∙(mass 2) + ...Apply the Formula: Atomic Mass = 0.6011 (68.9256 amu) + 0.3989 (70.9247 amu) = 69.72

amu

Relationship Between Moles and Mass

• The mass of one mole of atoms is called the molar mass.

• The molar mass of an element, in grams, is numerically equal to the element’s atomic mass, in amu.

One mole

Converting Between Moles and Number of Atoms

A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?

• Write down the given quantity and its units. Given: 1.1 x 1022 Ag atoms

• Write down the quantity to find and/or its units. Wanted: ? Moles

• Collect Needed Conversion Factors: 1 mole Ag atoms = 6.022 x 1023 Ag atoms

= 1.8266 x 10-2 moles Ag

= 1.8 x 10-2 moles Ag

moles atoms Ag 106.022

Ag mole 1atoms Ag 101.1

2322

Mole Relationships in Chemical Formulas

• since we count atoms and molecules in mole units, we can find the number of moles of a constituent element if we know the number of moles of the compound

Moles of Compound Moles of Constituents

1 mol NaCl 1 mole Na, 1 mole Cl

1 mol H2O 2 mol H, 1 mole O

1 mol CaCO3 1 mol Ca, 1 mol C, 3 mol O

1 mol C6H12O6 6 mol C, 12 mol H, 6 mol O

Masses of different substances that equal to one mole

Converting from Grams to Moles and Moles to Grams

How do you calculate the number of moles of sulfur in 57.8 g of sulfur?

S moles S g 2.063

S mole 1S g 57.8

= 1.80287 moles S = 1.80 moles S

Converting Between Grams and Number of Atoms

How many aluminum atoms are in an aluminum can with a mass of 16.2 g?

= 3.6159 x 1023 atoms Al

Sig. Figs. & Round: = 3.62 x 1023 atoms Al

Al mole 1

Al atoms 10022.6

Al g 26.98

Al mole 1Al g 6.21

23

Converting Between Grams and Moles of Compound

Calculate the mass (in grams) of 1.75 mol of water

OH g OH mol 1

OH g 18.02OH mol 1.75 2

2

22

= 31.535 g H2O

Sig. Figs. & Round: = 31.5 g H2O

Converting Between Grams and Number of Molecules

Find the mass of 4.78 x 1024 NO2 molecules?

= 365.21 g NO2

Sig. Figs. & Round: = 365 g NO2

2

2

223

22

24

NO mole 1

NO g 6.014

NO molec 10022.6

NO mole 1NO molec 1078.4

Converting Between Grams of a Compound and Grams of a Constituent ElementL-Carvone, (C10H14O), is found in spearmint oil. It has a pleasant odor and mint flavor. It is often added to chewing gum, liquors, soaps and perfumes. Find the mass of carbon in 55.4 g of carvone.

Molar Mass C10H14O = 10(atomic mass C) + 14(atomic mass H) + 1(atomic mass O)

= 10(12.01) + 14(1.01) + (16.00) = 150.2 g/mo

1 mole C10H14O = 150.2 g C10H14O

1 mole C10H14O 10 mol C

1 mole C = 12.01 g C

Solution Strategy “converting units” :

gC10H14O

C mol 1

C g 0112.OHC g 50.21

OHC mol 1

1410

1410

molC10H14O

molC

gC

OHC mol 1

C mol 10

1410

C mole 1

C g 2.011

OHC mol 1

C mol 10

OHC g 50.21

OHC mole 1OHC g 5.45

14101410

14101410

= 44.2979 g C

= 44.3 g C

Percent Composition• Percentage of each element in a compound by mass can be determined

from:1. the formula of the compound2. the experimental mass analysis of the compound

• The percentages may not always total to 100% due to rounding

100%wholepart

Percentage The mass percent tells you the mass of a constituent element in 100 g of the compound.The fact that NaCl is 39% Na by mass means that 100 g of NaCl contains 39 g Na.This can be used as a conversion factor:

100 g NaCl 39 g Na

Na g NaCl g 100

Na g 39 NaCl g NaCl g

Na g 39

NaCl g 100 Na g or

Example - Percent Composition from the Formula C2H5OH

1. Determine the mass of each element in 1 mole of the compound:Mass of Carbon = 2 moles C x (12.01 g) = 24.02 g

Mass of Hydrogen = 6 moles H x (1.008 g) = 6.048 g Mass of Oxygen = 1 mol O x (16.00 g) = 16.00 g

2. Determine the molar mass of the compound by adding the masses of the elements:

1 mole C2H5OH = 46.07 g

3. Divide the mass of each element by the molar mass of the compound and multiply by 100%:

52.14%C100%46.07g

24.02g 13.13%H100%

46.07g

6.048g

34.73%O100%46.07g

16.00g

Example – Percent Composition of Carvone if a 30.0 g sample contains 24.0 g of C, 3.2 g O and the rest H?

1. Determine the masses of all the elements in the sample

C = 24.0 g, O = 3.2 g

H = 30.0 g – (24.0 g + 3.2 g) = 2.8 g

2. Divide the mass of each element by the total mass of the sample then multiply by 100% to give its percentage

80.0%C100%30.0g

24.0g 11%O100%

30.0g

3.2g

9.3%H100%30.0g

2.8g

Hydrates

Hydrates are compounds that include water molecules as part of their structure

Examples include:

Na2CO3 • 10 H2O

NiCl2 • 6 H2O

What does this have to do with percent composition?

Percent Composition of Water

Let’s say we take 1.20 g of NiCl2 • x H2O (a hydrate)

With lots of heat from a Bunsen burner, we remove all the water and are left with 0.65 g of NiCl2. What is the percent composition of water in the hydrate?

(1.20 g of NiCl2 • 6 H2O) - (0.65 g of NiCl2) =

= 0.55 g H2O

% Comp. H2O = 0.55 g H2O (part) x 100% = 45% 1.20 g NiCl2 • x H2O (whole)

*We didn’t need to know the exact composition of the NiCl2 hydrate

% Composition and Formula

We find a compound and determine that it is composed of 31.2% nitrogen and 68.8% oxygen. What is the chemical formula of the compound?

For every 100. grams of material, there is 30.9 g N and 69.1 g O. We first convert to moles.

30.9 g N 1 mole N = 2.21 mole N 14.01 g

69.1 g O 1 mole O = 4.32 mole O 16.0 g

Examples: Empirical Formulas

Hydrogen PeroxideMolecular Formula = H2O2

Empirical Formula = HO

BenzeneMolecular Formula = C6H6

Empirical Formula = CH

GlucoseMolecular Formula = C6H12O6

Empirical Formula = CH2O

Finding an Empirical Formulafrom Experimental Data

A laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula.

C = 60.00%H = 4.48%

O = 35.53%

• Collect Needed Conversion Factors:

1 mole C = 12.01 g C

1 mole H = 1.01 g H

1 mole O = 16.00 g O

– calculate the moles of each element

Information

Given: 60.00 g C, 4.48 g H, 35.53 g O

Find: empirical formula, CxHyOz

Example:Find the empirical formula of aspirin with the given mass

percent composition.

C mol 996.4C g 12.01

C mol 1C g 00.60

H mol 44.4H g 1.01

H mol 1H g .484

O mol 220.2O g 16.00

O mol 1O g 5.523

This gives the formula ofC4.996H4.44O2.220. But, the formulamust have the atoms in whole number ratios!

Find the mole ratio by dividing by the smallest number of moles!

122.25

2.220

2.220

2.220

4.44

2.220

4.996

OHC

OHC

Multiplying the subscripts by 4 gives the final ratio!

C9H8O4

All these molecules have the same Empirical Formula. How are the molecules different?

Name Molecular

Formula

Empirical

Formula

glyceraldehyde C3H6O3 CH2O

erythrose C4H8O4 CH2O

arabinose C5H10O5 CH2O

glucose C6H12O6 CH2O

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

Molar Massreal formula

Molar Massempirical formula

= factor used to multiply subscripts

Determine the Molecular Formula of Cadinene if it has a molar mass of 204 g/mol and an empirical formula of C5H8

Determine the empirical formula.Determine the molar mass of the empirical formula:

5 C = 60.05 g, 8 H = 8.064 g

C5H8 = 68.11 g

Divide the given molar mass of the compound by the molar mass of the empirical formula and round to the nearest whole number:

3g 11.68

g 204

Multiply the empirical formula by the factor above (3) to give the molecular formula:

(C5H8)3 = C15H24

Cadinenes: Sesquiterpenes found in junipers and cedars (oil of cade)