Stoichiometry

Chapter 2-6 – 2-12

Chapter 3-1 – 3-8

Key concepts

– Know what a mole is, and how to use it.– Understand the term molar mass and its relationship to

formula weight.– Interconvert between number of particles, moles, and mass.– Understand the term percent composition and know how to

calculate the percent composition of an element in a formula.– Use percentage composition to determine the empirical

formula of a substance.– Understand how to formulate a chemical equation.– Understand how to balance chemical equations by inspection. – Know how to use moles and a chemical formula to determine

quantitative information about chemical reactions.– Understand the terms theoretical yield and limiting reactant.– Introduction to solutions in chemical reactions

The mole

• THE MOLE IS A COUNTING UNIT!!!!– You can have a mole of anything.– 1 dozen = 12 “things”– 1 gross = 144 “things” (a dozen dozen)– 1 mol = 6.0221367 1023 “things”

Molar Mass

• 1 mole = number of atoms in 0.012 kg of carbon-12.

• 1 amu = 1/12 mass of a carbon-12 atom.• As a result, the atomic weight of an element (in

amu) is __________________ to the mass of one mole of that element (in g/mol). This is defined as the molar mass of the element.

• Review: Is molar mass an intensive or extensive property?

Formula weight/molecular weight

• We may use the atomic weight (in amu) of elements in a molecular formula (or formula unit) to determine the molecular weight (or formula weight) of the molecule (unit).

• Examples: Fe2(CO3)3, C11H22O11

)__)(#( atomsofAWMW

Converting from mass to moles

• Just as the atomic weight is numerically equal to the molar mass of an atom, the molecular weight (formula weight) is numerically equal to __________________.

• Now we have a powerful conversion between the mass of a substance and the moles (or the number of particles we have). Why is it so important to have this kind of conversion? (or is it important at all?)

Converting mass moles

• We will need to determine the molar mass of the substance(s) in question.

• Examples: – How many moles of CO2 are in 0.56 g?1.Determine molar mass2.Convert from mass to moles

– What is the mass of 0.45 mol of lithium carbonate?

% composition of a compound

• (we discussed this before, but now we have a better way to work it.)

• Use the chemical formula and molar masses to determine the relative mass of each element in the compound.

• What is the % C in glucose?

%100)__)(#(

% compound

X

MM

XofMMX

Empirical formula

• The relative number of each kind of atom in a substance

• To determine empirical formulas, you must count atoms. Therefore, you must use moles, a counting unit, not grams.

Determining empirical formulas

1. assume 100 g of sample (when % is given). If actual masses are known they may also be used.

2. convert from grams of element to moles of element

3. Ratio all other elements against the element with the smallest number of moles (not mass).

• Example: a compound contains 52.9 % Al and 47.1 % O.

More examples

1. A certain compound is 73.9% Hg and 26.1% Cl by mass. What is the empirical formula?

2. (BLB) A sample of a substance is analyzed and found to contain 5.28 g Sn and 3.37 g F. What is the empirical formula?

3. (BLB) A sample contains 11.66 Fe and 5.01 g oxygen. What is the empirical formula?

• For ionic compounds, the empirical formula is generally the formula unit.

• For molecular compounds, molecular formula is not always the empirical formula, but will always be some whole number multiple of the subscripts of the atom.

• The relative ratio between the atoms must be preserved (if you multiply one subscript by a number, you must multiply all subscripts by the same number) – Law of constant composition.

• What would you need to know in order to find the molecular formula using the empirical formula?

Combustion analysis of compounds

• Done with hydrocarbons and other organic compounds.

• We use deductive reasoning to determine the % composition from the data.

• Example: A 1.000 g sample of an alcohol is burned and produces 1.913 g CO2 and 1.174 g H2O. What is the empirical formula?

Even more fun…

• (BLB) Nicotine contains C, H, and N. A 5.250 mg sample of nicotine was combusted, and 14.242 mg CO2 and 4.083 mg H2O were produced. What is the empirical formula for nicotine?

• The molar mass of nicotine is 160±5 g/mol. What is the molecular formula?

Formula of a hydrate

• A hydrate is a complex where an amount of water is contained inside an ionic salt crystal in a specific ratio.

• We may use the empirical formula calculation process in determining the formula of a hydrate as well.

Chemical equation:a visual representation of a chemical reaction

• There are reactants and products in a chemical reaction.

• Reactants:

• Products:

Conservation of matter

• Matter is not created or destroyed in a chemical reaction.

• The number of each type of atom in a chemical reaction must be the same for reactants as it is for products.

• Sandwich example: write a “reaction” for a sandwich made of two slices of bread and one slice or cheese.

Balancing equations by inspection

1. Start with the element found in the fewest chemical formulas on either side of the reaction.

2. Fill out remaining elements, changing coefficients as needed, until the number of each type of atom is same on both sides of the equation.

– When balancing equations, we change the coefficients. we do not change the subscripts. Changing the subscripts means we have changed the molecule involved in the reaction (law of constant composition).

– More balancing examples

Using the chemical equation to determine quantitative information

• A balanced chemical equation gives us the ratio of reactants and products to each other, whether in molecules or moles, but NOT the mass ratio.

• example: combustion of methane• if we know how much we have of one reactant

(or product), we may determine how much of all other reactants (and products) were used (or produced) in the chemical reaction.

In general, comparing substance A and substance B

1. Given mass of A, find moles of A using molar mass of A

2. use ratio of A and B to find moles of B.3. Use molar mass of B to find mass of B.

The ratio (called the stoichiometric ratio or reaction ratio) is the key to solving these types of problems.

• Let’s do some examples.

Limiting reactants.

• limiting reactant—the reactant that limits how much product you can get. The limiting reactant is entirely used up in the reaction.

• Back to the sandwich example: if I have 6 slices of bread and 4 slices of cheese, which is the limiting reactant?

• Note: it doesn’t matter if I start with the bread or the cheese, I’ll still be able to determine the answer if I use the ratio properly.

• Limiting reactants are always determined by counting how many molecules are needed; or, in other words, you must use moles, not mass.

• Lets try some more examples.

• (WGD) How many grams of NH3 can be prepared from 77.3 g N2 and 14.2 g H2?

N2 + H2 NH3

• (WGD) 12.6 g of AgNO3 and 8.4 g of BaCl2 are dissolved in water. A solid, AgCl, forms. How much of this solid, in grams, can form?

AgNO3 + BaCl2 AgCl + Ba(NO3)2

Yield

• The yield is the amount produced in a reaction.

• Theoretical yield:

• Actual yield:

• Percent yield:

Yields

• (WGD) Ethylene glycol (C2H6O2) is formed according to the following rxn:C2H4Cl2 + Na2CO3 + H2O C2H6O2 + 2 NaCl + CO2

(this is balanced)

• When 27.4 g C2H4Cl2 is used in the rxn, 10.3 g ethylene glycol is formed.– What is the theoretical yield?– What is the actual yield?– What is the percent yield?

Sequential reactions

• The same concepts we have been learning in the analysis of single reactions can be applied to multiple reactions. Let’s look at an example…

Sequential reactions

• (WDP) What mass KClO3 is needed to provide enough O2 to react completely with 66.3 g methane?

KClO3 KCl + O2

CH4 + O2 CO2 + H2O

Break problem into steps; plot your course.

Solutions

• A solution is a ____________________.• A Solution contains a solute and a solvent.• Solute:• Solvent:

• We will discuss ways of expressing the concentration of a solute in solution. We’ll talk about the solvation process (process of dissolving in solution) later in the course.

% by mass

%100%'

nsol

solute

m

msolute

This is mass of the solution,not mass of the solvent.

• In very dilute solutions, % solute is an awkward unit to use. Often, parts-per-million (ppm) or parts-per-billion (ppb) are used instead.

6

'

10nsol

solute

m

mppm

9

'

10nsol

solute

m

mppb

Molarity

• Chemists will often use molarity because it is more directly related to the moles of solute (and becomes more useful in analyzing chemical reactions).

nsol

solute

V

nM

'____

Not volume of solvent.

Solution dilution…..

• Suppose we change the volume of a solution, but we leave the amount of solute the same….

2211 VMVM

Solutions in chemical reactions

• Just as we dealt with the mass of reactants and products in a chemical reaction by converting to moles, we may also use the molarity of a solution in analyzing chemical reactivity.

• The key is to always compare moles to moles.