Last week…

Review of ionic and net ionic equations. Review of water solubilities.

Table of Water Solubilities

i= nearly insoluble

ss=slightly soluble

s=soluble

d=decomposes

n=not isolatedacetate

bromide

carbonate

chloride

hydroxide

iodide

nitrate

oxide

phosphate

sulfate

sulfide

sodium s s s s s s s d s s S

silver ss i i i n i s i i ss i

Being able to use this table is essential for both writing ionic equations and ranking the solubility of compounds.

Ionic and Net Ionic Equations

i= nearly insoluble

ss=slightly soluble

s=soluble

d=decomposes

n=not isolatedacetate

bromide

carbonate

chloride

hydroxide

iodide

nitrate

oxide

phosphate

sulfate

sulfide

sodium s s s s s s s d s s S

silver ss i i i n i s i i ss i

Last week we mixed solution #2 (NaCl) with solution #4 (AgNO3).

This equation is written as:NaCl + AgNO3 NaNO3 + AgCl

Ionic and Net Ionic Equations

i= nearly insoluble

ss=slightly soluble

s=soluble

d=decomposes

n=not isolatedacetate

bromide

carbonate

chloride

hydroxide

iodide

nitrate

oxide

phosphate

sulfate

sulfide

sodium s s s s s s s d s s S

silver ss i i i n i s i i ss i

To write an ionic equation from the given equation:NaCl + AgNO3 NaNO3 + AgCl

We first need to check the table to see which compounds disassociate in water…

Ionic and Net Ionic Equations

i= nearly insoluble

ss=slightly soluble

s=soluble

d=decomposes

n=not isolatedacetate

bromide

carbonate

chloride

hydroxide

iodide

nitrate

oxide

phosphate

sulfate

sulfide

sodium s s s s s s s d s s S

silver ss i i i n i s i i ss i

Since the compounds in red disassociate, we need to split them up:

NaCl + AgNO3 NaNO3 + AgCl

becomes

Na+ + Cl- + Ag+ + NO3- Na+ + NO3

- + AgCl

Notice the insoluble AgCl stays together.

Ionic and Net Ionic Equations

i= nearly insoluble

ss=slightly soluble

s=soluble

d=decomposes

n=not isolatedacetate

bromide

carbonate

chloride

hydroxide

iodide

nitrate

oxide

phosphate

sulfate

sulfide

sodium s s s s s s s d s s S

silver ss i i i n i s i i ss i

To write a net ionic equation from the newly created ionic equation:

Na+ + Cl- + Ag+ + NO3- Na+ + NO3

- + AgCl

Simply eliminate any ion that appears on both the reactant and product side of the equation (shown in orange).

This leaves us with: Ag+ + Cl- AgCl

Table of Water Solubilities

i= nearly insoluble

ss=slightly soluble

s=soluble

d=decomposes

n=not isolatedacetate

bromide

carbonate

chloride

hydroxide

iodide

nitrate

oxide

phosphate

sulfate

sulfide

sodium s s s s s s s d s s S

silver ss i i i n i s i i ss i

To rank solubility of compounds, simply use the table.

For example, in order of increasing solubility:

silver chloride (i) silver sulfate (ss) silver nitrate (s)

Questions?

If anything from the last two labs is still unclear, be sure to see me today BEFORE you go take the quiz!

Lab #6Stoichiometry

Chemistry 108

Instructor:Kristine Cooper

Stoichiometry

Chemical math

The branch of chemistry that quantifies the substances in a chemical reaction

The art of figuring how much stuff you'll make in a chemical reaction from the amount of each reagent you start with

Atomic weight

Atomic weight of an element (found on the periodic table) is given in AMUs.

For example, the atomic weight of carbon is 12 amu.

This is the mass of ONE atom of carbon.

Gram atomic weight

The gram atomic weight of an element is the same NUMBER as the element’s atomic weight, the UNITS change.

For example, the gram atomic weight of carbon is 12 g.

This is the mass of 6.02x1023 atoms of carbon (an Avagardo’s number or one mole).

Formula weight / Molecular weight

Formula weight is the combined weight of all atoms in a compound.

Example:– Propanol CH3(CH2)2OH

What is the formula weight?

Count the number of each atom, multiply by it’s atomic weight of that atom, then total.

Percent composition

Calculated by determining the percent (by mass) of each individual atom of the formula weight of the compound.

Example:– Propanol CH3(CH2)2OH

We determined the formula weight earlier to be 60 amu.

Now determine what % of 60 amu the total mass of each atom comprises.

Use the total mass of each type of atom determined before, then divide by the

formula weight.

Simplest empirical formula

Example:

A compound was found to contain 88.8% oxygen and 11.2% hydrogen, what is the empirical formula?

1. Set the amount of the compound as 100g.In a 100g sample, there would be 88.8g oxygen and 11.2g hydrogen.

2. Determine the moles of each kind of atom.

88.8 g O / 16 g/mol=5.55 mol O

11.2 g H / 1 g/mol = 11.1 mol H

3. Set these numbers as subscripts, then simplify the ratio (divide by the smallest subscript present).

H11.1O5.5 H2O

True molecular formula

Simple empirical formulas only tell us the ratio of atoms present in a compound.

True molecular formulas tell us the true number of each atom present.

THESE ARE ALWAYS EVEN MULTIPLES OF THE EMPIRICAL FORMULA!

True molecular formula

For example, we determined water’s empirical formula to be H2O, which has a formula weight of 18g for each water molecule.

What is the true molecular formula for 90g of water?

True molecular formula

First, divide 90g by the molecular weight of water (18g). 90g/18g= 5

Next, multiply each subscript in the empirical formula (H2O) by this number (5).

H=2x5=10 and O=1x5=5, so the true molecular formula is H10O5.

Law of Definite Composition

The chemical composition of a substance is fixed, it never varies with sample size.

Simply stated, water is still H2O whether you have a mL vs. 5L, etc.

Stoichiometry practice

Calculate the molecular weight of methane (CH4).

Stoichiometry practice

12 moles of a certain compound weighs 702 g. What is the molecular weight of the compound?

Stoichiometry practice

Calculate the percent composition of each element in hydrogen peroxide (H2O2).

Stoichiometry practice

Calculate the mass (in grams) of 1.15 moles of vandium sulfide (VS2).

Stoichiometry practice

You want to manufacture sodium monoxide following the process:2Na + O2 → 2Na2O.

If you begin with 11.5g Na, how much O2 will be necessary to ensure all the Na has reacted?

Today in Lab

Given an unknown oxygen-containing potassium salt, we will determine its chemical formula.

By removing the oxygen from the compound and then calculating the difference in weight with and without the oxygen, we can determine the original composition.

Check out the demo compounds, compare volumes.