Fundamental Observations

The Law of Constant Composition

Suppose we have 10.0g of some gas, which can be decomposed into two constituent elements: element A and element B.

Suppose that the 10.0g of the compound decomposes into 4.67g of element A and 5.33g of element B. This means that there are 1.14 grams of element B for every 1 gram of element A. This ratio holds for any quantity of the compound gas.

In example, if we had 11.5g of the compound now, this would decompose into 5.37g of element A and 6.13g of element B.

Dalton’s Atomic Theory

The Law of Multiple Proportions

This one is a bit different than constant composition.

It states that if we have 2 compounds who are composed of the same elements (lets say the compounds were composed of elements A and B), that the ratio of the ratio of A to B (or B to A) of one compound to the other is a whole number. That was a mouthful.

To put it into context, consider the previous compound where we had 1.14 grams of element B for every gram of element A. Then, let’s say we have another compound is composed of 69.6% element B and 30.4% element A. The ratio of B to A is 2.28 in this scenario, which evenly divides into 1.14 for a ratio of ratios of 2, a whole number. This tells us that the latter compound has twice as many B atoms for every B atom in the former compound.

How can we explain this? One way we can explain this is that the molecules in the former compound are arranged as AB, and the latter compound is arranged as AB2.

And in fact, the values in the example actually correspond to the compound NO and NO2.

Experiments and atom structure

Cathode Ray Tubes were an unlikely place for the next step of chemistry, but it helped us understand the structure of the atom better. JJ Thomson was able to deduce from his experiment involving CRTs the mass-to-charge ratio of the particles in the stream from the CRT. This value was \(-5.69 \times 10^{-9}\text{ g/C}\). He was not able to independently determine the mass or charge though.

Robert Millikan on the other hand was able to determine the charge of an electron, measured to be \(-1.60 \times 10^{-19}\text{ C}\). Now that we know the charge, we can find the mass via the ratio; this comes out to be \(9.10 \times 10^{-31}\text{ kg}\).

These experiments marked the discovery of an electron. We found a subatomic particle, that exhibited the same properties (mass & charge) no matter what electron we sampled.

So.. what’s an atom look like?

We know electrons exist now, but how are they organized in an atom? Well, Thomson proposed the Plum Pudding model. This model is named like this, to depict that an atom mostly contains a diffuse positively charged “pudding” (or spread). In the pudding resides small negative charges (electrons), or the plums. Is this correct? Well only one way to find out..

Rutherford comes around and does this experiment where he takes alpha particles and shoots them at a thin sheet of gold foil. Then he figures out what the path of the particles is after impact with the foil.

The results of this experiment was the discovery of the nucleus. Like in biology, the nucleus refers to the center, in this case of the atom. This is an object with a positive charge that is more massive when compared to the alpha particles.

With these new findings in mind, we present the Nuclear Model of the Atom: where the positively charged, massive part of the atom (nucleus) resides at the center. The electrons in the atom exist somewhere outside of the nucleus.

To keep scale in mind, it’s important to know that the atom itself is MUCH larger than the nucleus. Quantiatively, the nucleus is about 1/20,000th the size of the atom. This means the atom is mostly empty space! This makes sense, because that would explain why the alpha particles were going straight to the foil. Statistically, the nucleus makes up a small part of the atom, so only some were slightly deflected from the foil.

So, what exactly is the nucleus made up of? The nucleus is made up of two fundamental particles: protons and neutrons. Protons are positively charged, and they are about 2000 times more massive than an electron. Neutrons are neutral as their name implies, but their mass is the same as the proton. Nuclei can contain several protons and neutrons, and more importantly the nucleus composition tells us the identity of the atom.

Information about the subatomic particles:

Particle Electric Charge Mass amu
Electron -1.6 \(\times\) 10-19 C 9.11 \(\times\) 10-31 ~0
Proton +1.6 \(\times\) 10-19 C 1.67 \(\times\) 10-27 1
Neutron 0 C 1.67 \(\times\) 10-27 1

Important quantities:

Isotope Notation

A few examples:

  • \(^{23}_{11}\text{ Na}\) – This is sodium (which can be uniquely identified by the atomic number), which has a mass number of 23, and a total of 11 protons. That means its atomic number is also 11, and using subtraction we can determine the number of neutrons is 12. Notice, that the number of neutrons can vary without changing the identity of the atom. In fact, if we hold the number of protons constant and vary the number of neutrons, we effectively can vary the mass of the atom without changing its identity; in other words, we can have isotopes of the same element (isotopes are atoms of the same element that have different masses). We can call this species an atom since it has a neutral charge (the number of electrons equals the number of protons, and we know this since no charge is denoted).
  • \(_{20}\text{ Ca}^{\text{2+}}\) – This is calcium (which can be uniquely identified by the atomic number), which has 20 protons. That means its atomic number is also 20. We can also notice that this ion has a charge of positive 2. An ion is a particle which has a non-neutral charge (# of electrons does not equal # of protons). This means it is a cation, or an ion which has more protons than electron. Conversely, an ion with more electrons than protons is known as an anion.
  • \(^{19}_{9}\text{ F}^{\text{-}}\) – This is Fluorine-19 (notice how we specified the mass after the hyphen, this is a common way to refer to isotopes). It is an ion as it has a charge, and it has 9 protons and 10 neutrons. We can also determine that there are 10 electrons based on the negative charge.

Periodic Table

It is important to know how the periodic table is laid out, so here’s a convenient picture:

Periodic Table

Elements to know by name

Some of these elements’ symbols don’t exactly match up with their name, so listed below are a few of them to keep them in mind:

  • Na – Sodium
  • K – Potassium
  • Mn – Manganese
  • Fe – Iron
  • Cu – Copper
  • Ag – Silver
  • Sn – Tin
  • W – Tungsten
  • Au – Gold
  • Hg – Mercury
  • Pb – Lead

Molecules?

In nature, very few substances exist as just an individual atom. In standard conditions only the noble gasses (that last group in the periodic table) can exist in monatomic, elemental form.

In most cases, we are dealing with substances that exist as molecules or ions.

So what are molecules? Molecules are aggregates of two or more atoms in a definite arrangement held together by chemical forces called bonds.

The definite arrangement explains the fixed ratios described by the Law of Constant Composition.

Some molecules consist of just two atoms, also known as diatomic molecules. Examples include hydrogen gas (\(\text{H}_{2}\)) and oxygen gas (\(\text{O}_{2}\)).

Others can contain more than 2, also known as polyatomic molecules. Examples include water (\(\text{H}_{2}\text{O}\)) and ammonia (\(\text{NH}_{3}\)).

Ion Naming

For anions the suffix becomes -ide.

Some of these ion names just need to be memorized, especially the polyatomic ions..

The monatomic cations/anions are pretty straightforward.

Monatomic transition metal cations to know:

Ionic Compounds

Ionic Compounds are electrically neutral, as they are composed of ionic bonds.

Hydrates

Certain ionic compounds absorb water molecules into their formula structures, known as hydrates. In terms of naming, you just append “hydrate” to the compound’s name.

Bonding

Types of bonding:

  • Covalent bonding (molecular)
  • Ionic bonding
  • Metallic Bonding (not covered)
Covalent Ionic
Electron Sharing Electron Transfer
Distinct Molecules Crystal Lattice Structures
Molecular Formula/Empirical Formula Unit Formula
Nonmetal & Nonmetal Metal & Nonmetal

Formulas:

  • The molecular formula is the actual number of atoms in a type of molecule.
  • The empirical formula is the smallest whole number ratio of atoms.

Chemical Formulas

Distinctions between molecular and empirical formula

Ethanediol (Ethylene Glycol) is a molecule that consists of 2 carbons, 6 hydrogens, and 2 oxygens.

Molecular formula (actual number of atoms):

\(\text{C}_{2}\text{H}_{6}\text{O}_{2}\)

Ethanediol empirical formula (smallest whole number ratio):

\(\text{C}\text{H}_{3}\text{O}\)

Methanol molecular formula:

\(\text{C}\text{H}_{4}\text{O}\)

Methanol empirical formula:

\(\text{C}\text{H}_{4}\text{O}\)

A structural formula shows the connections/bonds between atoms: depict atoms by their elemental symbol, and draw lines between them to denote the bonds.

A condensed structural formula is similar to a structural formula, but it depicts the structure without any lines as contiguous text (\(\text{C}\text{H}_{2}\text{O}\text{H}\text{C}\text{H}_{2}\text{O}\text{H}\)).

The line angle formula shows carbon groups. Each line’s ends/angles represent a carbon, and the carbon’s other bonds are hydrogen unless explicitly written out.

Dimethyl peroxide is \(\text{C}_{2}\text{H}_{6}\text{O}_{2}\). Notice it is the same formula as ethanediol, but these are different molecules. If two molecules have the same molecular formula, but different structural formulas, then they are called isomers.

Ionic formulas must use unit structure. Consider a NaCl crystal. The crystal lattice might be composed of many sodium and chlorine atoms, but when we use unit structure we describe the atoms in their smallest whole number ratio.

Covalent Compounds

Covalent compounds are always between a non-metal and a non-metal, and their bonds share electrons. In terms of naming, first element keeps its full name, and second element has an “ide” ending. You also use prefixes to denote the number. i.e. CO refers to carbon monoxide and \(\text{CO}_{2}\) refers to carbon dioxide.

Acids

Acids have a hydrogen atom as the first element. 3 categories:

Introduction to Organic compounds

Number Formula Condensed Formula Prefix name
? \(C_{n}H_{2n+2}\) \(C_{n}H_{2n+2}\) Alk- Alkane
1 \(CH_{4}\) \(CH_{4}\) Meth- Methane
2 \(C_{2}H_{6}\) \(CH_{3}CH_{3}\) Eth- Ethane
3 \(C_{3}H_{8}\) \(CH_{3}CH_{2}CH_{3}\) Prop- Propane
4 \(C_{4}H_{10}\) \(CH_{3}CH_{2}CH_{2}CH_{3}\) But- Butane

Organic Nomenclature – Functional Groups

Three examples:

  • Hydroxyl Group: -\(OH\) alcohol
  • Amine Group: -\(NH_{2}\) amines
  • Carboxyl Group: -\(COOH\) carboxylic acid

Examples of the functional groups:

  • \(CH_{3}CH_{3}\) – Ethane (Alkane, no functional group)
  • \(CH_{3}CH_{2}OH\) – Ethanol (Hydroxyl group)
  • \(CH_{3}CH_{2}NH_{2}\) – Ethylamine (Amine group)
  • \(CH_{3}COOH\) – acetic acid (Carboxyl group)

Atomic Mass

Atomic mass is not the mass number of an element.

Example: Chlorine has two naturally occurring isotopes:

These mass numbers aren’t exactly the mass:

We also need the relative abundances

To calculate the atomic mass, we take a weighted average with the abundances acting as the weights:

weighted_cl35 <- 34.97 * 0.7578
weighted_cl37 <- 36.86 * 0.2422
weighted_cl35 + weighted_cl37
## [1] 35.42776

Moles

Mole: \(6.022 \times 10^{23}\) elementary units per mole (Avogadro’s number: \(N_{A}\))

Definition: 1 mole of \(^{12} C\) atoms has a mass of 12 grams exactly.

\(1\text{ g} = 6.022 \times 10^{23}\text{ amu}\)

1.27 grams of Carbon to moles?

\[ \frac{1.27\text{ g}}{} * \frac{1\text{ mol}}{12.011\text{ g}} = 0.106\text{ mol of Carbon} \]

Moles to atoms?

\[ 0.10573\text{ mol} * \frac{6.022 \times 10^{23}\text{ atoms}}{\text{mol}} = 6.37 \times 10^{22}\text{ atoms} \]

5 skills of Mass Composition

  1. Calculate molar mass of a substance from its formula
  2. Convert between grams, moles, molecules, atoms
  3. Determine percent composition by mass
  4. Determine empirical formula from:
    1. Percent composition
    2. Combustion analysis data
  5. Finding a molecular formula from an empirical formula and molar mass

Example: Molar Mass of a substance

  1. What is the molar mass of methanol (\(CH_{3}OH\))?
Elements Atoms Atomic Mass Mass contribution of each element
C 1 12.011 g/mol 12.011 g/mol
H 4 1.00794 g/mol 4.03176 g/mol
O 1 15.9994 g/mol 15.9994 g/mol

Sum the mass contribution for each element: \(32.04216\text{ g/mol}\). This is the atomic mass of methanol.

Example: Conversions for mass

Sample of methanol has a mass of 79.18 g. How many atoms are in the sample?

  • Total atoms
  • Carbon atoms
  • Hydrogen atoms

Let’s convert from mass to moles. Notice that we have already calculated the molar mass in the previous example: \(32.04216\text{ g/mol}\).

\[ \frac{79.18\text{ g}}{} * \frac{1\text{ mol}}{32.042\text{ g}} = 2.47\text{ mol of methanol} \]

Moles to molecules

\[ \frac{2.4711364\text{ mol}}{} * \frac{6.022 \times 10^{23}\text{ molecules}}{\text{ mol}} = 1.488 \times 10^{24}\text{ molecules} \]

Atoms

\[ \frac{1.488 \times 10^{24}\text{ molecules}}{} * \frac{6\text{ atom}}{\text{molecule}} = 8.928 \times 10^{24}\text{ atoms} \]

Carbon atoms

\[ \frac{1.488 \times 10^{24}\text{ molecules}}{} * \frac{1\text{ C atom}}{\text{molecule}} = 1.488 \times 10^{24}\text{ atoms} \]

Can do the same for hydrogen

Percent Composition for Chemical Formula

Divide each mass contribution for each element by the total molar mass. This gives you the percent composition.

atomic_masses <- c(12.011, 4.03176, 15.9994)
total_mass <- sum(atomic_masses)
((atomic_masses) / total_mass) * 100
## [1] 37.48499 12.58267 49.93234

Determine empirical formula from mass percent

For a certain compound, the mass percent is:

Element Percent Mass
C 24.27%
Cl 71.65%
H 4.07%

What is the empirical formula of this compound?

We can think of this question as:

Convert to moles, then divide by smallest value:

masses <- c(24.27, 71.65, 4.07)
atomic_masses <- c(12.011, 35.4527, 1.00794)
# Moles conversion
moles <- masses / atomic_masses
# Get smallest value of mole for the empirical formula
smallest_value <- min(moles)
moles / smallest_value
## [1] 1.000000 1.000176 1.998339
round(moles / smallest_value)
## [1] 1 1 2

Determine empirical formula from mass percent (second example)

Element Percent Mass
Na 17.55%
Cr 39.70%
O 42.75% 
masses <- c(17.55, 39.70, 42.75)
atomic_masses <- c(22.99, 52.00, 16.00)
# Moles conversion
moles <- masses / atomic_masses
# Get smallest value of mole for the empirical formula
smallest_value <- min(moles)
moles / smallest_value
## [1] 1.000000 1.000113 3.500080

We can double these to get our whole number ratio

masses <- c(17.55, 39.70, 42.75)
atomic_masses <- c(22.99, 52.00, 16.00)
# Moles conversion
moles <- masses / atomic_masses
# Get smallest value of mole for the empirical formula
smallest_value <- min(moles)
2 * (moles / smallest_value)
## [1] 2.000000 2.000226 7.000160

Combustion Analysis

Combustion analysis of a C, H, and N compound.

A sample of 0.1156g yields 0.1638g \(CO_{2}\) and 0.1676g \(H_{2}O\).

  1. Find moles of \(CO_{2}\)
  2. Find moles of \(C\)
  3. Find moles of \(H_{2}O\)
  4. Find moles of \(H\)
  5. Find mass of \(C\)
  6. Find mass of \(H\)
  7. Find mass of \(N\)
  8. Find moles of \(N\)
  9. Find smallest whole number ratio
total_sample_mass <- 0.1156
carbondioxide_sample <- 0.1638
water_sample <- 0.1676

carbondioxide_molarmass <- 44.01
carbon_molarmass <- 12.011
water_molarmass <- 18.01
hydrogen_molarmass <- 1.00794
nitrogen_molarmass <- 14.00674

carbondioxide_moles <- carbondioxide_sample / carbondioxide_molarmass
carbon_moles <- carbondioxide_moles
carbon_mass <- carbon_moles * carbon_molarmass

water_moles <- water_sample / water_molarmass
# For every mole of H2O, there are 2 moles of H
hydrogen_moles <- 2 * water_moles
hydrogen_mass <- hydrogen_moles * hydrogen_molarmass

nitrogen_mass <- total_sample_mass - (carbon_mass + hydrogen_mass)
nitrogen_moles <- nitrogen_mass / nitrogen_molarmass

smallest <- min(carbon_moles, hydrogen_moles, nitrogen_moles)
result <- c(carbon_moles, hydrogen_moles, nitrogen_moles) / smallest
paste0("Carbon moles ", result[1])
## [1] "Carbon moles 1"
paste0("Hydrogen moles ", result[2])
## [1] "Hydrogen moles 5.00066507617868"
paste0("Nitrogen moles ", result[3])
## [1] "Nitrogen moles 1.00010352749297"

So, our empirical formula is: \(CH_{5}N\)

Molecular Formula from Empirical Formula and Molar Mass

Say we have this compound: \(CH_{2}Cl\)

With this molecular mass: 98.95 g/mol

  1. Find molar mass of empirical formula
carbon_mass <- 12.011
hydrogen_mass <- 1.00794
chlorine_mass <- 35.4527

molar_mass <- 1 * carbon_mass + 2 * hydrogen_mass + 1 * chlorine_mass
molar_mass
## [1] 49.47958
  1. Divide molecular mass by molar mass of empirical formula
round(98.95 / 49.47968)
## [1] 2
  1. This represents the scaling factor that we apply to the empirical formula. Apply scale to empirical formula.
paste0("Carbon scaled up to ", 2 * 1)
## [1] "Carbon scaled up to 2"
paste0("Hydrogen scaled up to ", 2 * 2)
## [1] "Hydrogen scaled up to 4"
paste0("Chlorine scaled up to ", 2 * 1)
## [1] "Chlorine scaled up to 2"

So, the molecular formula is \(C_{2}H_{4}Cl_{2}\).

5 Skills of Quantifying Substances in Chemical Reactions

  1. Writing and balancing chemical reactions
  2. Stoichiometry (quantifying substances in chemical rxns)
  3. Limiting reagents (reactants that bottleneck the reaction)
  4. Excess reagents (reactants leftover due to the bottleneck)
  5. Yields (actual yield / theoretical yield is the percent yield, where theoretical is actual if all of the excess could be used)

3 Parts to a chemical equation (expression)

\[ A + B \rightarrow C + D \]

A and B are reactants (reagents) (everything on the left hand side of the arrow).

C and D are products (everything on the right hand side of the arrow).

Rxn arrow indicates that a chemical reaction takes place.

  • \(\rightarrow\) indicates forward rxn only (at least for this semester)
  • bidirectional arrow indicates equilibrium rxn (not this semester)

Products appear on the right hand side, and represent the results of a rxn.

Balancing experssions

A couple of heuristics:

  1. Look for elements that only appear once on each side, and balance those first.
  2. If an element appears by itself, balance last.

Stoichiometry

  1. Find moles of reactant, or the limiting reactant if there is one
  2. Ensure that is it not in excess (otherwise, find the limiting reactant)
  3. Use mole ratio to find the product
  4. Can find mass using molar mass

Deriving Chemical Formulas

Problem statement:

Write a balanced equation for the chemical reaction when an ionic compound is formed from 0.636 g of oxygen and 0.689 g of chromium.

Moles of oxygen:

oxygen_present <- 0.636
oxygen_molar_mass <- 32
oxygen_present / oxygen_molar_mass
## [1] 0.019875

Moles of chromium:

chromium_present <- 0.689
chromium_molar_mass <- 52
chromium_present / chromium_molar_mass
## [1] 0.01325

Ratio of moles of oxygen gas to chromium:

oxygen_moles <- 0.019876
chromium_moles <- 0.013251
oxygen_moles / chromium_moles
## [1] 1.499962

This means for every 3 moles of oxygen gas we have 2 moles of chromium (1.5 mole ratio).

Our chemical formula is then:

\[ 2 Cr + 3 O_{2} \rightarrow 2 Cr O_{3} \]

Notice the 2 present in the chromium compound in the products – we had to balance.

State symbols in chemical formula

In chemical formulas, you may see the state of matter appended to the substance

  • \(\text{(s)}\) for solids
  • \(\text{(l)}\) for liquids
  • \(\text{(g)}\) for gasses/vapor
  • \(\text{(aq)}\) for aqeuous

Limiting and Excess Reagents

Suppose we have a recipe that calls for:

These ingredients are to make 4 servings of pastry creme.

Now suppose we have readily available:

The limiting reagent is the eggs: they cap the recipe to be made 9 times, whereas the other ingredients can make 10 or more. The other ingredients are what we call the excess reagents.

There are three important questions for limiting/excess reagents questions:

Example 1

For the combustion rxn between 35.00 g \(C_{4}H_{10}\) (butane) and 45.00 g \(O_{2}\)

  • Limiting reagent?
  • Amount of product?
    • in moles?
    • in grams?
  • Amount of excess reagent?
    • in moles?
    • in grams?

Balance equation:

\[ 2 C_{4}H_{10} + 13O_{2} \rightarrow 8 CO_{2} + 10 H_{2}O \]

Moles of butane:

butane_present <- 35.00
butane_molar_mass <- (12.011 * 4) + (1.00794 * 10)
butane_present / butane_molar_mass
## [1] 0.6021671

Moles of oxygen gas:

o2_present <- 45.00
o2_molar_mass <- 15.9994 * 2
o2_present / o2_molar_mass
## [1] 1.406303
butane_moles <- 0.6021671
o2_moles <- 1.406303
o2_moles / butane_moles
## [1] 2.335403

Notice that this ratio of \(\approx 2.3\) is less than \(\frac{13}{2}\) (the mole ratio between \(O_{2}\) and \(C_{4}H_{10}\)). This means that the oxygen gas has to be the limiting reagent, since the supply of oxygen gas is not high enough to meet the mole ratio.

Now use the limiting reagent moles to find the product moles via stoichiometry:

Moles of \(CO_{2}\)

\[ \frac{1.406303}{} * \frac{8}{13} = 0.8652 \text{ mol} \]

This can then be converted into grams using the molar mass of carbon dioxide:

\[ 0.8652 * 44.01 = 38.08 \]

Moles of \(H_{2}O\)

\[ \frac{1.406303}{} * \frac{10}{13} = 1.08177 \text{ mol} \]

Grams of water:

\[ 1.08177 * 18.01 = 19.49 \]

Now, how much of the excess reagents remain (in this case, butane)?

Find used amount of moles of butane:

\[ 1.406303 * \frac{2}{13} = 0.2163 \]

We had \(0.6022\) moles of butane originally, so that means what we have remaining is:

\[ 0.6022 - 0.2163 = 0.3859 \]

We can then convert to grams:

\[ 0.3859 * 58.124 = 22.43 \]

So we have 22.43 grams of butane left after rxn is complete.

Example 2

Suppose we have the following chemical formula:

\[ 2 NH_{3} + 3O_{2} = 2 CH_{4} \rightarrow 2 HCN + 6 H_{2}O \]

Following are the molar masses:

Suppose we start with:

  • 11.5g of ammonia
  • 12.0g of oxygen gas
  • 10.5g of methane

Questions to ask:

  • Limiting reagent?
  • Amount of product (\(HCN\))?
    • in moles?
    • in grams?
  • Amount of excess reagent?
    • in moles?
    • in grams?

Ammonia moles:

\[ 11.5 * 17.03 = 0.675 \]

Oxygen moles:

\[ 12.0 * 32.00 = 0.375 \]

Methane moles:

\[ 10.5 * 16.04 = 0.655 \]

Notice that the number of oxygen moles is the smallest, yet we must have 3 moles of oxygen gas compared to the other two reactants which require only 2 moles. So, oxygen is the limiting reagent.

We can then use the number of moles of oxygen gas to find the moles of HCN (This comes out to be 0.250 moles). This can then be converted to grams.

Then, we can find the used moles of the excess reagents via stoichiometry with the limiting reagent, and then subtract to find the excess moles. We can then convert these into grams.

Percent Yield

In the case of a limiting reagent, it’s fair to say that the rxn did not go to its full potential. That is to say, there is a theoretical max potential of the rxn, and what actually happened. The actual yield is what we actually have left, which may or may not have suffered the consequences of a limiting reagent. The theoretical yield is what we would have had if the excess reagents could have been used in full. The ratio is the percent yield, or a description of the utilization of the reagents.