Chemistry 101: Stoichiometry

A mole is the number of particles of a substance that must be present in a sample such that the sample’s mass in grams is equal to the substance’s atomic weight in AMU.

One mole is equal to 6.02 x 1023 particles.

4 g

One mole of atoms or molecules is the exact number such that the entire sample will have a mass (in g) of the individual atom of molecule’s mass in AMU.

Atomic weight is the mass, in grams, of one mole of a naturally occurring element.

To calculate atomic weight, one must take into account both the weight of all the naturally-occurring isotopes of that element, and their proportional abundance.

Ex: natural Cl appears as two significant isotopes, 35Cl (75% abundance) and 37Cl (25% abundance). The atomic weight of Cl is therefore:

(0.75 35) + (0.25 37) = 35.5

Atomic mass is the mass, in Atomic Mass Units (AMU), of one atom of a particular isotope of an element.

It also represent the mass in grams of one mole of the isotopes, and can be found by adding together the number of protons and neutrons in that particular isotope.

Ex: the atomic mass of 235U, with 92 protons and 143 neutrons, is 235 AMU.

Molecular weight is the weight of one mole of molecules of a substance.

It can also be calculated by adding together the atomic weights of all the atoms in the molecule.

180.20 AMU

[6 C] + [12 H] + [6 16] =
[6 12.01] + [12 1.01] + [6 16.00]

These 3 masses are worth memorizing. Remember, approximation is almost always good enough, so let C = 12, H = 1, and O = 16, for a total of 180 AMU.

A molecular formula shows the total number and type of atoms in each molecule. This is the full, unsimplified formula.

Ex: The molecular formula of glucose, with 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms, is: C6H12O6

An empirical formula is a simplified ratio of whole numbers for the different elements in a compound.

Ex: The empirical formula of glucose, with 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms, is: CH2O

The molecular formula is the total of all the atoms present in a single molecule of the substance.

The empirical formula is the ratio of the number of atoms in a substance, expressed as the lowest common denominator.

In this case, take the molecular formula, C4H8, and divide both subscripts by 4 to get to the final answer.

1. Find the atomic weight of each element in the molecular formula.

2. Multiply that weight by the subscript (number of that atom present).

3. Add these numbers together to get the total sum.

Ex: 1 mole of H2O has a mass of 18 g.
(1 2) + (16 1) = 18

44 g

To calculate the weight of 1 mole of a substance, add the atomic weights of every atom in the molecular formula. Remember approximation is almost always good enough on most chemistry tests, especially on the AP Chemistry Exam.

Weight (CO2) = 12 + 2(16) = 44 g

1/2 a mole

1 mole of SO2 has a mass of: 32 + 2(16) = 64 g
Thus, 32 g is one half a mole.

Percent composition is a calculation of the proportion of a substance that a particular element makes up, by weight.

53.30%

Disproportionation reactions are a subset of redox reactions, in which one species (in this case, the Hg atom) acts as both the oxidizing agent and the reducing agent.

Oxidation is the process of a chemical species losing electrons.

When a species is oxidized, its oxidation state increases.
Ex: Co2+ (aq) ⇒ Co3+ (aq) + e-

Reduction is the process of a chemical species gaining electrons.

When a species is reduced, its oxidation state decreases.
Ex: Cu2+ (aq) + e- ⇒ Cu+(aq)

Co is the oxidizing agent, Na is the reducing agent.

In a redox reaction, the oxidizing agent is the species which receives electrons and is reduced, while the reducing agent is the species which donates electrons and is oxidized.

1. MnO4-; Oxidizing

2. NaBH4; Reducing

3. Cr2O7; Oxidizing

4. O2; Oxidizing

5. LiAlH4; Reducing

Most common oxidizing agents contain oxygen atoms, particularly in the presence of metal atoms

Most common reducing agents contain hydrogen atoms in the presence of metal atoms.

Oxidation state is the formal charge left on an atom if it is assumed that every bond in a molecule is perfectly ionic.

Ex: in H2O, assume that the more electronegative O atom takes all the electrons in both bonds, giving it an oxidation state of -2, and leaving each H with a +1 oxidation state.

The oxidation state is zero.

Ex: in O2 the oxygen atoms share electrons perfectly, and so each has an oxidation state of zero.

1. 
   oxygen = -2

2. hydrogen = +1

Notable exceptions: oxygen is -1 in peroxides (such as H2O2), hydrogen is -1 in hydrides (such as NaH).

The oxidation state of the nitrogen is +5

Add up the total oxidation for all the molecule’s known atoms: (-2) * 3 = -6
Subtract that amount from the molecule’s net charge: (-1) - (-6) = +5
The remaining amount is the unknown atom’s oxidation state.

1. Alkali Metals = +1

2. Alkali Earth Metals = +2

3. Halogens = -1

Elements in these groups will typically give up or accept their standard number of electrons when making bonds in compounds.

• In O2, O = 0

The oxidation number for every element in its standard state is zero.

• In H2O, O = -2

This is oxygen's typical oxidation state in compounds.

• In H2O2, O = -1

Peroxides are the one compound where oxygen has this oxidation state.

• In NO, N = +2

Oxygen's oxidation state is -2, and the molecule is neutrally charged, so the nitrogen must be +2.

• In NO2, N = +4

Oxygen's oxidation state is -2 (total of -4 for both), and the molecule is neutrally charged, so the nitrogen must be +4.

• In HNO3, N = +5

Oxygen's oxidation state is -2 (total of -6 for all 3), hydrogen's oxidation state is +1, and the molecule is neutrally charged, so the nitrogen must be +5.

• In H2S, S = -2

H2=2(+1)=2, S must be -2 to compensate.

• In S8, S = 0

This is sulfur's standard state, all atoms are zero oxidation in their standard state.

• In SO2, S = +4

The two oxygens (-2 each) have a total charge of -4, S must be +4 to compensate.

H2SO4 + 2 NaOH →
Na2SO4 + 2 H2O

The first step in balancing any reaction is finding an atom which exists in a single molecule on each side (Na, in this case), and changing coefficients to have equal numbers on each side. H2O also needed to be doubled, to equate the total number of H's and O's on both sides.

2 AgNO3 + Cu →
Cu(NO3)2 + 2 Ag

The first step to balancing this reaction was to balance the NO3 groups by doubling AgNO3 on the left. But doing so unbalanced the equation in Ag, making it necessary to double Ag on the right to restore balance.

4 CrO3 →
2 Cr2O3 + 3 O2

The first step in this reaction was to balance Cr atoms by doubling the CrO3. To balance O atoms at that point would have required a coefficient of 3/2 in front of the O2. The rules of equation balancing require all coefficients to be whole numbers though. Doubling every coefficient, however, allows you to balance the equation using whole numbers.

The limiting reactant is the reactant which is totally consumed when a chemical reaction is complete.

The amount of limiting reactant present defines the amount of product which can be created.

AgNO3 is the limiting reactant.

1 mole of AgNO3 will react completely with 1 mole of NaCl, since the equation shows that they react in a 1:1 ratio. Thus, there will be 1 mole of NaCl left over (in excess).

1. If necessary, balance the reaction.

2. If given amounts of molecules in grams, convert to moles.

3. Take the number of moles of the substance, and divide into that the coefficient in front of that substance in the balanced equation.

4. Do this for all reactants.

5. Whichever reactant has the lowest final value from that calculation is the limiting reactant.

HCl is the limiting reactant.

After converting grams to moles: there is one mole of HCl and 1.5 moles of NaOH. Since these combine in the ratio 1:1, when all the HCl has been consumed, there will be 0.5 moles of excess NaOH remaining.

O2 is the limiting reactant.

Converting grams to moles: there are 2 moles of O2 and 1.5 moles of CH4. Due to the 2:1 ratio in the equation, two moles of O2 are needed for every mole of CH4 consumed.

The 2 moles of O2 will react completely with 1 mole CH4, leaving 0.5 moles excess CH4 remaining.

Theoretical yield (or expected product) is the maximum amount of product that the given amount of reactants are capable of producing.

This amount can be in number (moles) or weight (grams) and assumes ideal reactivity, with 100% efficiency and no experimental error.

The theoretical yield is 1 mole of NaCl.

The 1.5 moles NaOH could support the formation of 1.5 moles NaCl. The 1 mole HCl could support the formation of 1.0 moles NaCl.

The limiting reagent is HCl though, hence 1 mole is the maximum product that is possible from this system.

Heat added

The Δ sign indicates that heat is necessary for the reaction to occur. In the case of this reaction, one could find in a table that 172.5 kJ per mole are required.

The two-sided arrow ⇔ implies a reaction which reaches an equlibrium between reactants and products.

The single-sided arrow ⇒ indicates a reaction which goes to completion, leaving only products.