Chapter 3

Calculations Related to Chemical Formulas

Formula MassThe mass of an individual molecule or formula unit

Also known as molecular mass or molecular weight

Sum of the masses of the atoms in a single molecule or formula unit

mass of 1 molecule of H2O

= 2(1.01 amu H) + 16.00 amu O = 18.02 amu

mass of 1 formula unit of MgCl2 = 2(35.45 amu Cl) + 24.30 amu Mg = 95.20 amu

Molar Mass

The relative masses of molecules can be calculated from atomic masses.

Formula Mass = 1 molecule of H2O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu

1 mole of H2O contains 2 moles of H and 1 mole of O.

molar mass = 1 mole H2O = 2(1.01 g H) + 16.00 g O = 18.02 g

so the Molar Mass of H2O is 18.02 g/mole

Molar Mass of Compounds

1. How many moles are in 50.0 g of PbO2? (Pb, 207.2 g/mol; O,16.00 g/mol)

Pb = 1 x 207.2 = 207.200000 O = 2 x 16.00 = 32.000000

PbO2 = 239.20g/mol 1 mol PbO2239.2 g PbO2

239.2 g PbO21 mol PbO2

50.0 g PbO2 x = 0.20903 mol PbO20.209 1 mol PbO2239.2 g PbO2

mol PbO2g PbO2

10.8 g CO2 x x =

1.4778 x 1023 molecules CO2

2. Find the number of CO2 molecules in 10.8 g of CO2.1 mol CO2 = 6.022 x 1023 molecules

C = 1 x 12.011 = 12.011000 O = 2 x 16.00 = 32.000000

CO2 = 44.01g/mol

44.01 g CO21 mol CO2

1 mol CO244.01 g CO2

6.022 x 1023 molecules1 mol CO2

1 mol CO244.01 g CO2

1.48 x 1023

mol CO2 molec CO2g CO2

4.78 x 1024 molec NO2 x x

= 365.207 g NO2

3. What is the mass of 4.78 x 1024 NO2 molecules?

1 mol = 6.022 x 1023 ,1 mol NO2 = 46.01 g

46.01 g NO21 mol NO2

1 mol NO246.01 g NO2

6.022 x 1023 molec NO21 mol NO2

1 mol NO2 6.022 x 1023 molec NO2

1 mol NO2 6.022 x 1023 molec NO2

46.01 g NO21 mol NO2

365 g NO2

mol NO2molec NO2 g NO2

Percent Composition

Percent Composition

Percentage of each element in a compound by mass

Can be determined from

1. the formula of the compound 2. the experimental mass analysis of the compound

4. Find the mass percent of Cl in C2Cl4F2

Mass Percent as a Conversion Factor5. If NaCl is 39% sodium, find the mass of table salt containing 2.4 g of Na.

100 g NaCl39 g Na

39 g Na100 g NaCl

2.4 g Na x = 6.1538 g NaCl 6.2 g NaCl 100 g NaCl39 g Na

g Na g NaCl

Empirical Formulas

Empirical Formula

Simplest, whole-number ratio of the atoms of elements in a compound

Can be determined from elemental analysis

Finding an Empirical Formula from % Composition

1. Convert the percentages to grams

2. Convert grams to moles

3. Write a pseudoformula using moles as subscripts

4. Divide all by smallest number of moles

5. Multiply all mole ratios by number to make all whole numbers

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

Given: C = 60.00%H = 4.48% O = 35.53%

Therefore, in 100 g of aspirin there are 60.00 g C, 4.48 g H, and 35.53 g O

However, ratios of mass are not useful in determining empirical formulas !!

We must convert to a ratio of moles !!

g C g H g O

mol C mol H mol O

pseudoformulaCxHyOz

empirical formulaCxHyOz

Manipulate subscripts to obtain whole-number ratio

Methodology for determining empirical formula:

Calculate the moles of each element

Write a pseudoformula

C4.996H4.44O2.220

Find the mole ratio

C2.25H2.00O1.00Multiply subscripts by factor to give whole number

C9H8O4

(x 4)

÷ 2.220

C4.996H4.44O2.220

6. Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70 g/mol)

and the rest fluorine (19.00 g/mol)

Given: 75.7% Sn, (100 – 75.3) = 24.3% F

g Sn mol Sn

g F mol F

pseudo formula

empirical formula

Element Ratio in Grams Molar Mass Ratio in Moles Ratio in Moles

Sn 75.7g1mol

118.7g0.6377 1

F 24.3g1mol

19.00g1.279 2.005

6. Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70 g/mol)

and the rest fluorine (19.00 g/mol)

SnF2

X

X

÷0.6377

7. Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00)

Given: 72.4% Fe, (100 – 72.4) = 27.6% O

g Fe mol Fe

g O mol O

pseudo formula

empirical formula

7. Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00)

Element Ratio in Grams Molar Mass Ratio in Moles Ratio in Moles Ratio in Moles

Fe 72.4g1mol

55.85g1.296 1 3

O 27.6g1mol

16.00g1.725 1.33 4

Fe3O4

X

X

÷1.296 x 3

Chapter 3 Molecular Formulas

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.

Find the molecular formula of butanedione if its empirical formula is C2H3O and its molar mass (MM) is 86.03 g/mol.

Factor of 2Molar Mass (emp. form.) = 2 x (12.01 gC/molC) + 3 x (1.008 gH/molH) + 1 x (16.00 gO/molO) = 43.04 g/mol

Molecular formula = C2H3O x 2 = C4H6O2

Practice – Benzopyrene has a molar mass of 252 g and an empirical formula of C5H3. What is its molecular formula?

(C = 12.01, H=1.01)

Molecular formula = {C5H3} x 4 = C20H12

C5 = 5(12.01 g) = 60.05 gH3 = 3(1.01 g) = 3.03 gC5H3 = 63.08 g

252

?

Combustion Analysis

Combustion Analysis

A known mass of compound is burned in oxygen and the masses of the products formed (CO2 and H2O) are determined.

By knowing the masses of the products and composition of constituent elements in the product, the original amount of

constituent elements can be determined.

It is assumed that all of the carbon in the original sample is converted to carbon dioxide and all of the

hydrogen in the sample is converted to water.

(Generally used for organic compounds containing C, H, O)

Combustion Analysis

Example of Combustion AnalysisCombustion of a 0.8233 g sample of a compound containing only

carbon, hydrogen, and oxygen produced the following:

CO2 = 2.445 g H2O = 0.6003 g

Determine the empirical formula of the compound.

This came from C.This came from H.

gCO2, H2O

molCO2, H2O

molC, H

gC, H

gO

molO

gC, H

pseudoformula empirical formula

molC,H,O

1 mole H = 1.008 g H 1 mole C = 12.01 g C molar masses of elements1 mole O = 16.00 g O

1 mole CO2 = 44.01 g CO2 ⇒1 mole H2O = 18.02 g H2O

1 mole C ⇒ 1 mole CO2 2 mole H ⇒ 1 mole H2O

In the original sample

molar masses of compounds

ratios of compounds formed in combustion

In the original sample

In the original sample

In the original sample

1 mole H = 1.008 g H 1 mole C = 12.01 g C molar masses of elements1 mole O = 16.00 g O

In the original sample

C0.05556H0.06662O0.00556

Pseudo formula

÷ 0.00556

Empiricalformula

8. Combustion of 0.844 g of caproic acid produced 0.784 g of H2O and 1.92 g of CO2.

If the molar mass of caproic acid is 116.2 g/mol,what is the molecular formula of caproic acid?

In the original sample

moles

g

0.0145 0.0870 0.0436

0.232 0.0877 0.524

O H C

In the original sample

Molecular formula = {C3H6O} x 2 = C6H12O2

Chapter 3 Chemical Reactions and

Equations

Chemical Reactions

Reactions involve rearrangement and exchange of atoms to produce new pure substances.

Reactants Products

Chemical Equations

Shorthand way of describing a reaction

Provides information about the reaction

1. formulas of reactants and products 2. states of reactants and products 3. relative numbers of reactant and product molecules

Combustion of MethaneMethane gas reacts with oxygen gas to produce

carbon dioxide gas and gaseous water.

CH4(g) + O2(g) ➜ CO2(g) + H2O(g)

This equation reads “1 molecule of CH4 gas combines with 1 molecule of O2 gas to make 1 molecule of CO2 gas and 1 molecule of

H2O gas.”

+ +

What about conservation of mass ??

+ +

1 C + 4 H + 2 O 1 C + 2 O + 2 H + OX

CH4(g) + O2(g) ➜ CO2(g) + H2O(g)

CH4(g) + O2(g) ➜ CO2(g) + 2 H2O(g)

CH4(g) + 2 O2(g) ➜ CO2(g) + 2 H2O(g)

Combustion of Methane, BalancedTo show the reaction obeys the Law of Conservation of Mass, the

equation must be balanced.

“1 molecule of CH4 gas combines with 2 molecules of O2 gas to make 1 molecule of CO2 gas and 2 molecules

of H2O gas.”

+ +

Symbols used to indicate state after chemical:

(g) = gas; (l) = liquid; (s) = solid (aq) = aqueous = dissolved in water

Energy symbols used above the arrow for conditions for reactions:

Δ = heat hν = light shock = mechanical elec = electrical

Symbols Used in Equations

Steps in Balancing Equations

1. In compounds balance elements other than H and O.

a. Balance elements which occur only once on each side of the equation. b. Start with the elements which occur the most. c. Balance polyatomic ions which do not change in the reaction.

2. Be prepared to rebalance if something changes!!!

3. Balance H.

4. Balance O.

5. Balance elements which appear in their “elemental” forms.

1. When aluminum metal reacts with air, it produces a white, powdery compound, aluminum oxide.

aluminum(s) + oxygen(g) ➜ aluminum oxide(s)

Al(s) + O2(g) ➜ Al2O3(s)

2 Al(s) + O2(g) ➜ Al2O3(s)

2 Al(s) + 3 O2(g) ➜ 2 Al2O3(s)

4 Al(s) + 3 O2(g) ➜ 2 Al2O3(s)

2. Solid phosphorous (P4) reacts with hydrogen gas to produce phosphorous trihydride.

P4 (s) + H2 (g) ----------> PH3 (g)

P4 (s) + H2 (g) ----------> 4 PH3 (g)

P4 (s) + 6 H2 (g) ----------> 4 PH3 (g)

3. Solid potassium chlorate decomposes to produce oxygen gas and potassium chloride.

KClO3 (s) -------------------> O2 (g) + KCl (s) 2 KClO3 (s) -------------------> 3 O2 (g) + KCl (s)

2 KClO3 (s) -------------------> 3 O2 (g) + 2 KCl (s)

H2SO4 (aq) + NaCN (s) ------> Na2SO4 (aq) + HCN(g)

H2SO4 (aq) + 2 NaCN (s) ------> Na2SO4 (aq) + HCN(g)

H2SO4 (aq) + 2 NaCN (s) ------> Na2SO4 (aq) + 2 HCN(g)

4. Aqueous sulfuric acid reacts with solid sodium cyanide to produce aqueous sodium sulfate and hydrogen cyanide gas.

K3PO4 (aq) + Ca(NO3)2 (aq) ----------> Ca3(PO4)2 (s) + KNO3 (aq)

K3PO4 (aq) + 3 Ca(NO3)2 (aq) ----------> Ca3(PO4)2 (s) + KNO3 (aq)

K3PO4 (aq) + 3 Ca(NO3)2 (aq) ----------> Ca3(PO4)2 (s) + 6 KNO3 (aq)

2 K3PO4 (aq) + 3 Ca(NO3)2 (aq) ----------> Ca3(PO4)2 (s) + 6 KNO3 (aq)

5. Aqueous potassium phosphate reacts with aqueous calcium nitrate to produce solid calcium phosphate and aqueous potassium nitrate.

Organic Chemistry

Classifying Compounds Organic vs. Inorganic

In the18th century, compounds from living things were called organic; compounds from the nonliving

environment were called inorganic.

Organic compounds easily decomposed and could not be made in the 18th-century lab.

Inorganic compounds were very difficult to decompose, but could be synthesized.

Modern Classification of Compounds Organic vs. Inorganic

Today we commonly make organic compounds in the lab and find them all around us.

Organic compounds are mainly made of C and H, sometimes with O, N, P, S, halogens, and

trace amounts of other elements.

The main element that is the focus of organic chemistry is carbon.

Write a balanced equation for the combustion of butane, C4H10.

C4H10 (g) + O2 (g) ➝ CO2 (g) + H2O (g)

C4H10 (g) + O2 (g) ➝ 4 CO2 (g) + H2O (g)

C4H10 (g) + O2 (g) ➝ 4 CO2 (g) + 5 H2O (g)

C4H10 (g) + 13/2 O2 (g) ➝ 4 CO2 (g) + 5 H2O (g)

2 C4H10 (g) + 13 O2 (g) ➝ 8 CO2 (g) + 10 H2O (g)

Carbon Bonding in Organic Compounds

Carbon atoms bond almost exclusively covalently in organic compounds.

When C bonds, it forms four covalent bonds.

Carbon is unique in that it can form limitless chains of C atoms, both straight and branched,

and rings of C atoms.

Classifying Organic CompoundsThere are two main categories of organic compounds,

hydrocarbons and functionalized hydrocarbons.

All chemical compounds

Inorganic Organic

Hydrocarbons Functionalized Hydrocarbons

Name Molecular formula Structure Use

methane CH4 Natural gas

propane C3H8 LP gas

n-butane C4H10 Butane lighters

n-pentane C5H12 Gasolines

ethene C2H4 Polymers

ethyne C2H2 Welding

Hydrocarbons-Contain only Carbon and Hydrogen

Alkanes Aldehydes

Alkenes Ketones

Alkynes Carboxylic Acids

Aromatics Esters

Alcohols Amines

Ethers Amides

Families of Organic Compounds

C C

C C

C C

CC

CC

C

C

C O

H

C OC

CH

O

CC

O

C

CO

H

O

C

O

OC

N

CN

O