1. Formula Mass the mass of an individual molecule or formula unit
also known as molecular mass or molecular weight (MW)
sum of the masses of the atoms in a single molecule or formula unit
whole = sum of the parts!
mass of 1 molecule of H2O
= 2(1.01 amu H) amu O = amu
Tro, Chemistry: A Molecular Approach

2. Molar Mass of Compounds
the relative masses of molecules can be calculated from atomic masses
Formula Mass = 1 molecule of H2O
= 2(1.01 amu H) amu O = amu
since 1 mole of H2O contains 2 moles of H and 1 mole of O
Molar Mass = 1 mole H2O
= 2(1.01 g H) g O = g
so the Molar Mass of H2O is g/mole
Tro, Chemistry: A Molecular Approach

3. Example
Example – Find the number of CO2 molecules in 10.8 g of dry ice
How many molecules are in 50.0 g of PbO2?
(PbO2 = 239.2)

4. Percent Composition Percentage of each element in a compound
By mass
Can be determined from
the formula of the compound
the experimental mass analysis of the compound
The percentages may not always total to 100% due to rounding
Tro, Chemistry: A Molecular Approach

5. Example Calculate the mass percent composition of calcium chloride
Calculate the mass percent of oxygen in acetic acid (C2H4O2)

6. Mass Percent as a Conversion Factor
the mass percent tells you the mass of a constituent element in 100 g of the compound
the fact that CCl2F2 is 58.64% Cl by mass means that 100 g of CCl2F2 contains g Cl
this can be used as a conversion factor
100 g CCl2F2 : g Cl
Tro, Chemistry: A Molecular Approach

7. Example Find the mass of table salt containing 2.4 g of Na
Benzaldehyde is 79.2% carbon. What mass of benzaldehyde contains 19.8 g of C?

8. Conversion Factors in Chemical Formulas
chemical formulas have inherent in them relationships between numbers of atoms and molecules
or moles of atoms and molecules
these relationships can be used to convert between amounts of constituent elements and molecules
like percent composition
E. g Find the mass of hydrogen in 1.00 gal of water
Tro, Chemistry: A Molecular Approach

9. Grams to Grams conversion
Grams A  Moles A  Moles B  Grams B
How many grams of sodium are in 6.2 g of NaCl? (Na = 22.99; Cl = 35.45)

10. Empirical Formula
simplest, whole-number ratio of the atoms of elements in a compound
can be determined from elemental analysis
masses of elements formed when decompose or react compound
combustion analysis
percent composition
Tro, Chemistry: A Molecular Approach

11. Finding an Empirical Formula
convert the percentages to grams
assume you start with 100 g of the compound
skip if already grams
convert grams to moles
use molar mass of each element
write a pseudoformula using moles as subscripts
divide all by smallest number of moles
if result is within 0.1 of whole number, round to whole number
multiply all mole ratios by number to make all whole numbers
if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3; if ratio 0.25 or 0.75, multiply all by 4; etc.
skip if already whole numbers
Tro, Chemistry: A Molecular Approach

12. Example 3.17
Laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula.
C = 60.00%
H = 4.48%
O = 35.53%
Tro, Chemistry: A Molecular Approach

13. Example
Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70) and the rest fluorine (19.00)
Determine the empirical formula of hematite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00)

14. 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
Tro, Chemistry: A Molecular Approach

15. Example
Benzopyrene has a molar mass of 252 g/mol and an empirical formula of C5H3. What is its molecular formula? (C = 12.01, H=1.01)
A compound has the emperical formular CH and a molar mass of g/mol. Find its molecular formula

16. Combustion Analysis
a common technique for analyzing compounds is to burn a known mass of compound and weigh the amounts of product made
generally used for organic compounds containing C, H, O
by knowing the mass of the product and composition of constituent element in the product, the original amount of constituent element can be determined
all the original C forms CO2, the original H forms H2O, the original mass of O is found by subtraction
once the masses of all the constituent elements in the original compound have been determined, the empirical formula can be found
Tro, Chemistry: A Molecular Approach

17. Combustion Analysis
Tro, Chemistry: A Molecular Approach

18. Combustion Analysis
Unknown formula: CxHyOx (Oxygen can be replaced with other nonmetal)
gCO2  moles CO2  moles C  gC
gH2O  moles H2O  moles H  gH
g O = g sample – (g H + g C)
gO  moles O
Follow steps in determine the empirical formula and molecular formula

19. Example 3.20 Determine the empirical formula of the compound
Combustion of a g sample of a compound containing only carbon, hydrogen, and oxygen produced the following:
CO2 = g
H2O = g
Determine the empirical formula of the compound
Tro, Chemistry: A Molecular Approach

20. Example
Upon combustion, a compound containing only carbon and hydrogen produced 1.60g CO2 and 0.819g H2O. Find the empirical formula
Tro, Chemistry: A Molecular Approach

21. Chemical Reactions Reactants  Products
Reactions involve chemical changes in matter resulting in new substances
Reactions involve rearrangement and exchange of atoms to produce new molecules
Elements are not transmuted during a reaction
Reactants  Products

22. Chemical Equations Shorthand way of describing a reaction
Provides information about the reaction
Formulas of reactants and products
States of reactants and products
Relative numbers of reactant and product molecules that are required
Can be used to determine weights of reactants used and products that can be made
Tro, Chemistry: A Molecular Approach

23. CH4(g) + O2(g) ® CO2(g) + H2O(g)
Combustion of Methane
methane gas burns to produce carbon dioxide gas and gaseous water
whenever something burns it combines with O2(g)
CH4(g) + O2(g) ® CO2(g) + H2O(g) O C H C + O H O +
1 C + 4 H O
1 C + 2 O H + O
1 C + 2 H + 3 O
Tro, Chemistry: A Molecular Approach

24. Combustion of Methane Balanced
to show the reaction obeys the Law of Conservation of Mass, it must be balanced
CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g) O C + O H + O + H C +
1 C H O
Tro, Chemistry: A Molecular Approach

25. Chemical Equations CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g)
CH4 and O2 are the reactants, and CO2 and H2O are the products
the (g) after the formulas tells us the state of the chemical
the number in front of each substance tells us the numbers of those molecules in the reaction
called the coefficients
Tro, Chemistry: A Molecular Approach

26. Chemical Equations CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g)
this equation is balanced, meaning that there are equal numbers of atoms of each element on the reactant and product sides
to obtain the number of atoms of an element, multiply the subscript by the coefficient
1  C  1
4  H  4
4  O  2 + 2
Tro, Chemistry: A Molecular Approach

27. Symbols Used in Equations
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 decomposition reactions
D = heat
hn = light
shock = mechanical
elec = electrical
Tro, Chemistry: A Molecular Approach

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

29. aluminum(s) + oxygen(g) ® aluminum oxide(s)
Practice
when aluminum metal reacts with air, it produces a white, powdery compound aluminum oxide
reacting with air means reacting with O2
aluminum(s) + oxygen(g) ® aluminum oxide(s)
Al(s) + O2(g) ® Al2O3(s)
Acetic acid reacts with the metal aluminum to make aqueous aluminum acetate and gaseous hydrogen
acids are always aqueous
metals are solid except for mercury
Tro, Chemistry: A Molecular Approach

30. 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 18th century lab
inorganic compounds very difficult to decompose, but able to be synthesized
Tro, Chemistry: A Molecular Approach

31. Modern Classifying 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, and trace amounts of other elements
the main element that is the focus of organic chemistry is carbon
Tro, Chemistry: A Molecular Approach

32. Carbon Bonding carbon atoms bond almost exclusively covalently
compounds with ionic bonding C are generally inorganic
when C bonds, it forms 4 covalent bonds
4 single bonds, 2 double bonds, 1 triple + 1 single, etc.
carbon is unique in that it can form limitless chains of C atoms, both straight and branched, and rings of C atoms
Tro, Chemistry: A Molecular Approach

33. Carbon Bonding
Tro, Chemistry: A Molecular Approach

34. Classifying Organic Compounds
there are two main categories of organic compounds, hydrocarbons and functionalized hydrocarbons
hydrocarbons contain only C and H
most fuels are mixtures of hydrocarbons
Tro, Chemistry: A Molecular Approach

35. Classifying Hydrocarbons
hydrocarbons containing only single bonds are called alkanes
hydrocarbons containing one or more C=C are called alkenes
hydrocarbons containing one or more CC are called alkynes
hydrocarbons containing C6 “benzene” ring are called aromatic
Tro, Chemistry: A Molecular Approach

36. Tro, Chemistry: A Molecular Approach

37. Naming Straight Chain Hydrocarbons
consists of a base name to indicate the number of carbons in the chain, with a suffix to indicate the class and position of multiple bonds
suffix –ane for alkane, –ene for alkene, –yne for alkyne
Base Name
No. of C
meth- 1
hex- 6
eth- 2
hept- 7
prop- 3
oct- 8
but- 4
non- 9
pent- 5
dec- 10
Tro, Chemistry: A Molecular Approach

38. Functionalized Hydrocarbons
functional groups are non-carbon groups that are on the molecule
substitute one or more functional groups replacing H’s on the hydrocarbon chain
generally, the chemical reactions of the compound are determined by the kinds of functional groups on the molecule
Tro, Chemistry: A Molecular Approach

39. Functional Groups
Tro, Chemistry: A Molecular Approach