1. Chapter 3 Atoms: the Building Blocks of Matter

2. 1) Protons (p+) positively charged particle
The parts that make up an atom are called subatomic particles.
1) Protons (p+) positively charged particle
2) Neutron (no) neutral particle (uncharged)
3) Electrons (e-) negatively charged particle
Neutrons and Protons are located in the nucleus of an atom and are called hadrons.
Electrons orbit around the nucleus.

3. Q- How are atoms of different elements distinguished from
one another? In other words, how do we distinguish a
helium atom from a carbon atom?
A- Their number of protons, indicated by the atomic number
Let’s look at helium, He.
It has an atomic number of 2,
which means that is has 2
protons in it’s nucleus.

4. Here are the basics; you need to know these.
Atomic Structure
Here are the basics; you need to know these. 1 H
1.0076
Hydrogen
Atomic Number (Z): the number of protons (p+)
Atomic Mass: the number of protons (p+) + the number of neutrons (n0)
▪ measured in atomic mass units (amu) which is one twelfth
the mass of a carbon-12 atom.
▪ the mass of electrons (1/1860 p+) is negligible.
Number of Neutrons: the atomic mass - the atomic number
Atomic Number
Atomic Symbol
Atomic Mass

5. Lets practice! Find the missing information?
Element
Atomic #
Atomic
Mass
Protons
Electrons
Neutrons Ar 18
amu He 2 O
amu 8

6. The Famous Gold Foil Experiment
This showed us that the atom is made of mostly
empty space.

7. Isotopes Atoms of the same element with different number of neutrons
Because they have the same number of protons, all isotopes of an element have the same chemical properties.

8. Mass Numbers of Hydrogen Isotopes
What would the masses be?

9. Calculating Average Atomic Mass
Look at the mass number of any given element, and you will
notice that it is a rather unruly decimal.

10. The Mole: A Measurement of Matter
At the end of this section, you should
be able to:
Describe how Avogadro’s number is related to a mole of any substance
Calculate the mass of a mole of any substance

11. The Mole (aka Avagadro’s Number):
6.02 x 1023

12. The Mole and Avogadro’s Number
SI unit that measures the amount of substance
1 mole = 6.02 x 1023 representative particles
Representative particles are usually atoms, molecules, or formula units (ions)

13. So, 1 mole of any substance is a set number of
But Why the Mole?
Just as 12 = 1 dozen, or 63,360 inches = 1 mile,
the mole allows us to count microscopic items
(atoms, ion, molecules) on a macroscopic scale.
So, 1 mole of any substance is a set number of
Items, namely: 6.02 x 1023.
Chemistry = awesome

14. Representative Particle Representative Particles in 1.00 mol
Examples:
Substance
Representative Particle
Chemical Formula
Representative Particles in 1.00 mol
Atomic nitrogen
Atom N
6.02 x 1023
Water
Molecule
H2O
Calcium ion
Ion
Ca2+

15. Representative Particles in 1.00 mol
Solve
Substance
Representative
Particle
Formula
Unit
Representative Particles in 1.00 mol
Nitrogen gas N2
Molecule
Calcium Fluoride
CaF2
Sucrose
C12H22O11
Carbon C
Atom

16. All have 6.02 x 1023 representative particles in 1.00 mol
Answers
Nitrogen gas-molecule-N2
Calcium fluoride-molecule-CaF2
Sucrose-molecule-C12H22O11
Carbon-atom-C
All have 6.02 x 1023 representative particles in 1.00 mol

17. How many atoms are in a mole?
Determined from the chemical formula
List the elements and count the atoms
Solve for CO2
C - 1 carbon atom
O oxygen atoms
Add: = 3
Answer: 3 times Avogadro’s number of atoms

18. Solve: How many atoms are in a mole of
1. Carbon monoxide – CO
2. Glucose – C6H12O6
3. Propane – C3H8
4. Water – H2O

19. How many moles of magnesium is 1.25 x 1023 atoms of magnesium?
Divide the number of atoms or molecules given in the example by 6.02 x 1023
Divide (1.25 x 1023) by (6.02 x 1023)
Express in scientific notation
Answer = 2.08 x 10-1 mol Mg

20. Moles and Atoms Conversions
Converting moles to representative particles:
Converting representative particles to moles:

24. Objectives
Use the molar mass to convert between mass and moles of a substance
Use the mole to convert among measurements of mass, volume, and number of particles

25. Molar mass Mass (in grams) of one mole of a substance
Broad term (can be substituted) for gram atomic mass, gram formula mass, and gram molecular mass
Can be unclear: What is the molar mass of oxygen?
O or O2 ? - element O or molecular compound O2 ?

26. Molar Mass
Gram atomic mass (gam) – atomic mass of an element taken from the periodic table
Gram molecular mass (gmm) – mass of one mole of a molecular compound
Gram formula mass (gfm) – mass of one mole of an ionic compound
Can use molar mass instead of gam, gmm, or gfm

27. Calculating the Molar Mass of Compounds (Molecular and Ionic)
1. List the elements
2. Count the atoms
3. Multiply the number of atoms of the element by the atomic mass of the element (atomic mass is on the periodic table)
4. Add the masses of each element
5. Express to tenths place

28. What is the molar mass (gfm) of ammonium carbonate (NH4)2CO3?
N 2 x 14.0 g = g
H 8 x g = g
C 1 x 12.0 g = g
O 3 x 16.0 g = g
Add ________
Answer g

29. Practice Problems 1. How many grams are in 9.45 mol
of dinitrogen trioxide (N2O3) ?
a. Calculate the grams in one mole
b. Multiply the grams by the number
of moles
2. Find the number of moles in 92.2 g
of iron(III) oxide (Fe2O3).
b. Divide the given grams by the grams in one mole

30. Answers 1. 718 g N2O3 (one mole is 76.0g)
mol Fe2O3 (one mole is g)

31. Volume of a Mole of Gas
Varies with a change in temperature or a change in pressure
At STP, 1 mole of any gas occupies a volume of 22.4 L
Standard temperature is 0°C
Standard pressure is kPa (kilopascals), or 1 atmosphere (atm)
22.4 L is known as the molar volume

32. 22. 4 L of any gas at STP contains 6
22.4 L of any gas at STP contains 6.02 x 1023 representative particles of that gas
One mole of a gaseous element and one mole of a gaseous compound both occupy a volume of 22.4 L at STP (Masses may differ)
Study Figure 7.13 on page 186
Molar mass (g/mol) = Density (g/L) x Molar Volume (L/mol)

33. Objectives Define the terms
Calculate the percent composition of a substance from its chemical formula or experimental data
Derive the empirical formula and the molecular formula of a compound from experimental data

34. Terms to Know
Percent composition – relative amounts of each element in a compound
Empirical formula – lowest whole- number ratio of the atoms of an element in a compound

35. An 8. 20 g piece of magnesium combines completely with 5
An 8.20 g piece of magnesium combines completely with 5.40 g of oxygen to form a compound. What is the percent composition of this compound?
1. Calculate the total mass
Divide each given by the total mass
and then multiply by 100%
Check your answer: The
percentages should total 100%

36. Answer The total mass is 8.20 g + 5.40 g = 13.60 g
Divide 8.2 g by 13.6 g and then multiply by 100% = = 60.3%
Divide 5.4 g by 13.6 g and then multiply by 100% = = 39.7%
Check your answer: 60.3% % = 100%

37. 1) Find the percent composition of
Aluminum Oxide (Al3O2)
2) How much of a 5-g piece of Iron
Bromide (FeBr3) is iron?

38. Calculate the percent composition of propane (C3H8)
1. List the elements
2. Count the atoms
3. Multiply the number of atoms of the element by the atomic mass of the element (atomic mass is on the periodic table)
4. Express each element as a percentage of the total molar mass
5. Check your answer

39. Answer
Total molar mass = 44.0 g/mol
36.0 g C = 81.8%
8.0 g H = 18.2%

40. Calculate the mass of carbon in 52.0 g of propane (C3H8)
Calculate the percent composition using the formula (See previous problem)
2. Determine 81.8% of 82.0 g
Move decimal two places to the
left (.818 x 82 g)
3. Answer = 67.1 g

41. Calculating Empirical Formulas
Microscopic – atoms
Macroscopic – moles of atoms
Lowest whole-number ratio may not be the same as the compound formula
Example: The empirical formula of hydrogen peroxide (H2O2) is HO

42. Empirical Formulas
The first step is to find the mole-to-mole ratio of the elements in the compound
If the numbers are both whole numbers, these will be the subscripts of the elements in the formula
If the whole numbers are identical, substitute the number 1
Example: C2H2 and C8H8 have an empirical formula of CH
If either or both numbers are not whole numbers, numbers in the ratio must be multiplied by the same number to yield whole number subscripts

43. What is the empirical formula of a compound that is 25
What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen?
1. Assume 100 g of the compound, so that
there are 25.9 g N and 74.1 g O
2. Convert to mole-to-mole ratio:
Divide each by mass of one mole
25.9 g divided by 14.0 g = 1.85 mol N
74.1 g divided by 16.0 g = 4.63 mol O
3. Divide both molar quantities by the
smaller number of moles

44. /1.85 = 1 mol N
4.63/1.85 = 2.5 mol O
5. Multiply by a number that converts each to a whole number (In this case, the number is 2 because 2 x 2.5 = 5, which is the smallest whole number )
2 x 1 mol N = 2
2 x 2.5 mol O = 5
Answer: The empirical formula is N2O5

45. Determine the Empirical Formulas
1. H2O2
2. CO2
3. N2H4
4. C6H12O6
5. What is the empirical formula of a compound that is 3.7% H, 44.4% C, and 51.9% N?

46. Answers Compound Empirical Formula 1. H2O2 HO 2. CO2 CO2 3. N2H4 NH2
4. C6H12O CH2O
5. HCN

47. Calculating Molecular Formulas
The molar mass of a compound is a simple whole-number multiple of the molar mass of the empirical formula
The molecular formula may or may not be the same as the empirical formula

48. Calculate the molecular formula of the compound whose molar mass is 60
Calculate the molecular formula of the compound whose molar mass is 60.0 g and empirical formula is CH4N.
1. Using the empirical formula, calculate the empirical formula mass (efm)
(Use the same procedure used to calculate molar mass.)
2. Divide the known molar mass by the efm
3. Multiply the formula subscripts by this value to get the molecular formula

49. Answer Molar mass (efm) is 30.0 g 60.0 g divided by 30.0 g = 2
Answer: C2H8N2

50. Practice Problems
1) What is the empirical formula of a compounds that is 25.9% nitrogen and 74.1% oxygen?
2) Calculate the empirical formula of a compound that is 32.00% C, 42.66% O, 18.67% N, and 6.67% H.
3) Calculate the empirical formula of a compound that is 42.9% C and 57.1% O.

51. Practice Problems 4) What is the molecular formula for each compound:
a) CH2O, 90 g/mol
b) HgCl, g/mol
c) C3H5O2, 146 g/mol