Atoms and Molecules Notes

Symbols and Formulas

Each element is represented by a unique name and symbol.
Elemental symbols are based on element properties or derived from:
- Names of famous scientists.
- Places.
- Astronomical bodies.
- Mythological characters.
Elemental symbols:
- Based on the element's name.
- Consist of one capital letter, or a capital letter followed by a lowercase letter.

Compound Formula

Consists of the symbols of the atoms in the molecule.
Each elemental symbol represents one atom of the element.
If more than one atom is present, a subscript follows the elemental symbol.
Example 2.1 - Writing Compound Formulas
- Nitrogen dioxide: One nitrogen (N) atom and two oxygen (O) atoms (NO2)
- Sulfuric acid: Two hydrogen (H) atoms, one sulfur (S) atom, and four oxygen (O) atoms (H2SO4)

Atomic Structure

Atoms are made up of three subatomic particles:
- Protons
- Neutrons
- Electrons
Protons and neutrons:
- Tightly bound together to form the nucleus.
Electrons:
- Located outside the nucleus.
- Move rapidly throughout a relatively large volume of space surrounding the nucleus.
- Electrons move rapidly around a massive nucleus.

Table 2.3 - Characteristics of the Fundamental Subatomic Particles

Characteristic	Particle	Common Symbols	Charge (±)	Mass (g)
Electron	$e^{-}$
$e^{-}$
	1-	$9.07 × 10^{-28}$
1/1836	Outside nucleus
Proton	p, p+, H+
	p, p+, H+
	1+	$1.67 × 10^{-24}$
1	Inside nucleus
Neutron	n	n	0	$1.67 × 10^{-24}$
1	Inside nucleus

Atomic and Mass Numbers

Atomic number of an atom (Z):
- Equal to the number of protons in the nucleus and the number of electrons in an atom.
- In the periodic table, written as a whole number above the element symbol.
Mass number of an atom (A):
- Equal to the sum of the number of protons and neutrons in the nucleus of an atom.
- Isotope Symbol.

Isotopes

Atoms that have the same number of protons but different numbers of neutrons.
- Same atomic number but different mass numbers.
- Example: Isotopes of Cl (Chlorine-35 and Chlorine-37).
All isotopes of the same element have:
- Same number of electrons outside the nucleus.
- Same number of protons in the nucleus.
Example 2.2 - Using the Periodic Table
- a. What are the mass number, atomic number, and isotope symbol for an atom that contains 7 protons and 8 neutrons?
  - A = 7 + 8 = 15
  - Z = 7
  - Element with atomic number 7 is nitrogen (N).
  - Isotope symbol: $\frac{15}{7}N$
- b. How many neutrons are contained in an atom of nickel-60?
  - Nickel (Ni) has an atomic number (Z) of 28.
  - Mass number is 60.
  - Number of neutrons = 60 – 28 = 32 neutrons.
- c. How many protons and how many neutrons are contained in an atom with a mass number of 26 and the symbol Mg?
  - Magnesium (Mg) has an atomic number of 12.
  - Contains 12 protons.
  - Number of neutrons = 26 – 12 = 14 neutrons.

Masses of Atoms and Molecules

Atomic Mass:
- Numbers given beneath the element symbol and name in the periodic table.
- Provide a means of comparing the masses of atoms.
- Atomic weight of elements that occur as mixtures of isotopes is the average relative mass of the atoms in the isotope mixture.
Atomic Mass Unit (u):
- Used to express the relative masses of atoms.
- 1 u = 1/12 the mass of a carbon-12 atom.
- One carbon-12 atom has a relative mass of 12 u.
- Atom with a mass equal to 1/12 the mass of a carbon-12 atom would have a relative mass of 1 u.
- Atom with a mass equal to twice the mass of a carbon-12 atom would have a relative mass of 24 u.
- Two N atoms have a total mass close to the mass of a single Si atom.
Molecular Weight:
- Relative mass of a molecule expressed in atomic mass units.
- Calculated by adding together the atomic weights of the atoms in the molecule.
- Example: Water (H2O)
  - Two atoms of hydrogen (H) and one atom of oxygen (O).
  - $MW = (2 \times \text{atomic weight of H}) + (1 \times \text{atomic weight of O})$
  - $MW = (2 \times 1.01 \text{ u}) + (1 \times 16.00 \text{ u}) = 18.02 \text{ u}$
Example 2.4 - Atomic Weights and Molecular Weights
- Determine the molecular weight of urea, $CH4N2O$

Avogadro’s Number and the Mole Concept

Avogadro’s number:
- Number of atoms or molecules in a specific sample of an element or compound.
Mole (mol):
- Number of particles contained in a sample of an element or compound with a mass in grams equal to the atomic or molecular weight.
- $1 \text{ mol} = 6.022 \times 10^{23}$
- Example: 1 mol S atoms = $6.02 \times 10^{23}$ S atoms = 32.1 g S.
Following factors can be generated for use in factor-unit calculations:
- $1 \text{ mol S atoms} = 6.02 \times 10^{23} \text{ particles S atoms}$
- $6.02 \times 10^{23} \text{ S atoms} = 32.1 \text{ g S}$
- $1 \text{ mol S atoms} = 32.1 \text{ g S}$

The Mole and Chemical Formulas

Chemical formulas represent the numerical relationships that exist among atoms in a compound.
Example: H2O represents a 2:1 fixed ratio of hydrogen atoms to oxygen atoms in a water molecule.
Following relationships can be derived:
- $6.02 \times 10^{23} \text{ H}_2\text{O molecules contain } 12.04 \times 10^{23} \text{ H atoms and } 6.02 \times 10^{23} \text{ O atoms}$
- $1 \text{ mol of H}_2\text{O molecules contains } 2 \text{ mol of H atoms and } 1 \text{ mol of O atoms}$
- $18.0 \text{ g of water contains } 2.0 \text{ g of H and } 16.0 \text{ g of O}$

Mole Calculations: Strategies

Mole Calculation Example (1)

Calculate the number of moles of Ca contained in a 15.84 g sample of Ca
Solution:
- $15.84 \text{ g Ca} \times \frac{1 \text{ mole Ca}}{40.08 \text{ g Ca}} = 0.3952 \text{ moles Ca}$

Example 2.7 - Factor-Unit Calculations for Sulfur

Determine the following using the factor-unit method of calculation and factors obtained from the preceding three relationships given for sulfur (S):
a. The mass in grams of 1.35 mol of S
b. The number of moles of S atoms in 98.6 g of S
c. The number of S atoms in 98.6 g of S
d. The mass in grams of one atom of S

Example 2.7 – Solution

a. Known quantity is 1.35 mol of S, and the unit of the unknown quantity is grams of S
* Factor comes from the relationship 1 mol S atoms = 32.1 g S

*    $1.35 \text{ mol S atoms} \left( \frac{32.1 \text{ g S}}{1 \text{ mol S atoms}} \right) = 43.3 \text{ g S}$

b. Known quantity is 98.6 g of S, and the unit of the unknown quantity is moles of S atoms
* Factor comes from the same relationship used in (a)
* $98.6 \text{ g S} \left( \frac{1 \text{ mol S atoms}}{32.1 \text{ g S}} \right) = 3.07 \text{ mol S atoms}$

c. Known quantity is 98.6 g of S, and the unit of the unknown quantity is the number of S atoms
* Factor comes from the relationship $6.02 \times 10^{23} \text{ S atoms} = 32.1 \text{ g S}$
* $98.6 \text{ g S} \left( \frac{6.02 \times 10^{23} \text{ S atoms}}{32.1 \text{ g S}} \right) = 1.85 \times 10^{24} \text{ S atoms}$

d. Known quantity is one S atom, and the unit of the unknown is grams of S
* Factor comes from the same relationship used in (c), $6.02 \times 10^{23} \text{ S atoms} = 32.1 \text{ g S}$
* $1 \text{ S atom} \left( \frac{32.1 \text{ g S}}{6.02 \times 10^{23} \text{ S atoms}} \right) = 5.33 \times 10^{-23} \text{ g S}$

Note that the factor is the inverse of the one used in (c) even though both came from the same relationship
Thus, we see that each relationship provides two factors

The Mole Concept Applied to Compounds

One mole of any compound is a sample of the compound with a mass in grams equal to the molecular weight of the compound
- $1 \text{ mol CO}2 \text{ molecules} = 6.02 \times 10^{23} \text{ CO}2 \text{ molecules} = 44.0 \text{ g CO}_2$
Following relationships can be used to generate factors for use in factor-unit calculations:
- $1 \text{ mol CO}2 \text{ molecules} = 6.02 \times 10^{23} \text{ CO}2 \text{ molecules}$
- $6.02 \times 10^{23} \text{ CO}2 \text{ molecules} = 44.0 \text{ g CO}2$
- $1 \text{ mol CO}2 \text{ molecules} = 44.0 \text{ g CO}2$
- $1 \text{ mol CO}_2 \text{ molecules contains } 1 \text{ mol C atoms}$
- $1 \text{ mol CO}_2 \text{ molecules contains } 2 \text{ mol O atoms}$
- $1 \text{ mol C atoms} = 12.01 \text{ g C}$

Mole Calculation Example (2)

How many moles of O atoms are contained in 11.57 g of CO2?
- $11.57 \text{ g CO}2 \times \frac{2 \text{ moles O atoms}}{44.01 \text{ g CO}2} = 0.5258 \text{ moles O atoms}$
Note that the factor used was obtained from two of the six quantities given on the previous slide
- $1 \text{ mol CO}2 \text{ molecules} = 44.0 \text{ g CO}2$
- $1 \text{ mol CO}_2 \text{ molecules contains } 2 \text{ mol O atoms}$

Mole Calculation Example (3)

How many CO2 molecules are needed to contain 50.00 g of C?
- $50.00 \text{ g C} \times \frac{1 \text{ mol C atoms}}{12.01 \text{ g C}} \times \frac{6.022 \times 10^{23} \text{ CO}2 \text{ molecules}}{1 \text{ mol C atoms}} = 2.507 \times 10^{24} \text{ CO}2 \text{ molecules}$
Note that the factor used was obtained from three of the six quantities given on a previous slide
- $1 \text{ mol C atoms} = 12.01 \text{ g C}$
- $1 \text{ mol CO}_2 \text{ molecules contains } 1 \text{ mol C atoms}$
- $1 \text{ mol CO}2 \text{ molecules} = 6.02 \times 10^{23} \text{ CO}2 \text{ molecules}$

Mole Calculation Example (4)

What is the mass percentage of C in CO2?
- $\% \text{C} = \frac{\text{mass of C}}{\text{mass of CO}_2} \times 100$
- If a sample consisting of 1 mole of CO2 is used, the mole-based relationships given earlier show that 1 mole CO2 = 44.01 g CO2 = 12.01 g C + 32.00 g O
- Thus, the mass of C in a specific mass of CO2 is known and the problem is solved as follows:
- $\% \text{C} = \frac{12.01 \text{ g C}}{44.01 \text{ g CO}_2} \times 100 = 27.29\%$

Mole Calculation Example (5)

What is the mass percentage of oxygen in CO2?
- $\% \text{O} = \frac{\text{mass of O}}{\text{mass of CO}_2} \times 100$
- Once again, a sample consisting of 1 mole of CO2 is used to take advantage of the mole-based relationships given earlier where:
 - $1 \text{ mole CO}2 = 44.01 \text{g CO}2 = 12.01 \text{ g C} + 32.00\text{g O}$

Example 2.8 - Factor-Unit Calculations for Carbon Dioxide

Determine the following using the factor-unit method of calculation and factors obtained from the preceding three relationships given for carbon dioxide, CO2:
a. The mass in grams of 1.62 mol of CO2
b. The number of moles of CO2 molecules in 63.9 g of CO2
c. The number of CO2 molecules in 63.9 g of CO2
d. The mass in grams of one molecule of CO2

Example 2.8 – Solution

a. The known quantity is 1.62 mol of CO2, and the unit of the unknown quantity is g CO2
* The factor comes from the relationship 1 mol CO2 molecules = 44.0 g CO2

    *    $1.62 \text{ mol CO}_2 \text{ molecules} \left( \frac{44.0 \text{ g CO}_2}{1 \text{ mol CO}_2 \text{ molecules}} \right) = 71.3 \text{ g CO}_2$

b. The known quantity is 63.9 g of CO2, and the unit of the unknown quantity is mol of CO2 molecules
* The factor comes from the same relationship used in (a)
* $63.9 \text{ g CO}2 \left( \frac{1 \text{ mol CO}2 \text{ molecules}}{44.0 \text{ g CO}2} \right) = 1.45 \text{ mol CO}2 \text{ molecules}$

c. The known quantity is, again, 63.9 g of CO2, and the unit of the unknown is the number of CO2 molecules
* The factor comes from the relationship $6.02 \times 10^{23} \text{ CO}2 \text{ molecules} = 44.0 \text{ g CO}2$
* $63.9 \text{ g CO}2 \left( \frac{6.02 \times 10^{23} \text{ CO}2 \text{ molecules}}{44.0 \text{ g CO}2} \right) = 8.74 \times 10^{23} \text{ CO}2 \text{ molecules}$

d. The known quantity is 63.9 g of CO2, and the unit of the unknown quantity is mol of CO2 molecules
* The factor comes from the same relationship used in (c), but the factor is the inverse of the one used in (c)
* $1 \text{ CO}2 \text{ molecule} \left( \frac{44.0 \text{ g CO}2}{6.02 \times 10^{23} \text{ CO}2 \text{ molecules}} \right) = 7.31 \times 10^{-23} \text{ g CO}2$

Example 2.11 - Mass Percentage Calculations

Ammonia (NH3) and ammonium nitrate (NH4NO3) are commonly used agricultural fertilizers
Which one of the two contains the higher mass percentage of nitrogen (N)?