MOLES, MOLAR MASS, MOLARITY, PERCENT COMPOSITION, EMPIRICAL FORMULAS & HYDRATES - LECTURE NOTES

VOCABULARY:

Molecular Mass: The mass in grams of one mole of a substance.

Formula Mass: The sum of the atomic masses of the atoms in a formula.

Avogadro's Constant: The number of objects in a mole; 6.02 x 1023

Mole: The Avogadro constant number of objects.

Molar Mass: The mass found by adding the atomic masses of the atoms comprising the molecule.

Molarity: A unit of concentration equal to the number moles of solute in a cubic decimeter (1 liter) of solution.

Percent Composition: The mass of an element in a compound divided by the mass of the compound, multiplied by 100

Hydrate: A compound (crystalline) in which the ions are attached to one or more water molecules.


Lecture Video Link:

H.W.: MOLE WRKST #1


MOLAR RELATIONSHIPS:

How Many (or Much) is a Mole?

The mole is the standard method in chemistry for communicating __________________________ of a substance is present.

Here is how the International Union of Pure and Applied Chemistry (IUPAC) defines "mole:"

The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kg of carbon-12. When the mole is used, the _____________________________ must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.

= 3.74 x 1017 miles !!!

How far is that???

Well,

9.3 x 107 miles = average distance from the _____________________________!

3.7 x 109 miles = average distance from the _____________________________!

5.9 x 1012 miles = __________________________________________________________!!!

So, 6.02 x 1023 straight column of dimes is equal to 3.74 x 1017 miles or ______________________________________________

Therefore, one mole or Avogadro's number, 6.02 x 1023, is HUGE! And hopefully you now understand how _____________________________ an atom is!


THE MOLE AND MOLAR MASS:

MASS OF A MOLE OF ATOMS:

Because 6.02 x 1023 is too large to count one by one, this number of atoms is determined by measuring the _____________________________ of the substance.

This idea makes more sense if you think about a familiar example of _____________________________.

If you ever collected aluminum cans for recycling, you know that the cans are not counted individually - but payment is made per pound. Therefore, if we say that each can weighs approximately 10 grams - and you bring 1000 grams of aluminum cans - how many cans did you bring?

However, counting atoms or molecules is very difficult since they are so small. However, we can "count" atoms or molecules by _____________________________ large amounts of them.

When we weigh one mole of a substance on a balance, this is called the "_____________________________" and has the units g/mol (grams per mole). This idea is very critical because it is used all the time.

Therefore, a molar mass is the mass in grams of 6.02 x 1023 entities.


The molar mass of a substance is the molecular weight (The molecular weight of a substance is the weight in atomic mass units of all the atoms in a given formula) in grams.

Atomic Mass Unit (a.m.u.) = Mass of _____________________________.

All you need to do is calculate the molecular weight and stick the unit "g / mol" after the number and that is the molar mass for the substance in question.

Therefore, use the periodic table to get molar mass - EVERY element has the _____________________________ in a mole.


Let's do a few Molar Conversions to see if you understand…

Find the number of Moles (if Grams is given)

1. How many moles of Magnesium is in a 96.0 gram sample?

 

 

2. How many moles of Nickel is in a 25.0 gram sample?

 

 

3. How many grams of copper are in 2.50 moles of copper?

 

Answers to the Examples:


Lecture Video Link:

H.W.: MOLE WRKST #2


MOLAR MASS OF A COMPOUND:

To find the molar mass of a compound, simply _____________________________ of the elements in the compound. Be sure to account for the proportions of the elements in the compound.

For example, the formula for nitrogen dioxide is NO2. It represents one part Nitrogen and two parts Oxygen.

Using the periodic table, you see that the molar masses (rounded to the nearest one-tenth gram) are 14.0 grams per mole of Nitrogen and 16.0 grams per mole of Oxygen. The molar mass of Nitrogen Dioxide is found this way:

NO2 = One part Nitrogen + Two parts Oxygen

46.0 grams is the mass of one mole of Nitrogen Dioxide.

46.0 grams of Nitrogen Dioxide contains 6.02 x 1023 entities of NO2


Question: What is the molar mass of sucrose, C12H22O11 ?

First, Notice that in one mole of sucrose, there is twelve moles of Carbon, twenty two moles of Hydrogen, and eleven moles of Oxygen. To find the molar mass of sucrose, you need the molar mass of each of these elements.

Carbon ( C ) 12.0g x 12 atoms = 144.0g

Hydrogen (H) 1.0g x 22 atoms = 22.0g

Oxygen ( O ) 16.0g x 11 atoms = 176.0g

Total = 342.0g

Or

342.0 grams is the mass of one mole of sucrose.

342.0 grams of sucrose contains 6.02 x 1023 entities of C12H22O11.


PRACTICE PROBLEMS:

Calculate the mass of one mole of each of these substances. ( name each )

1. Mg(C2H3O2)2

 

2. Al2(C2O4)3

 

3. calcium oxalate

 

4. iron(III) sulfide

Answers to the Examples:


MOLAR CONVERSIONS FOR COMPOUNDS:

Converting the number of moles to number of molecules (using ratios)

Given the question:

How many molecules are in 5.0 moles of H2SO4 ?

First, ask yourself - what am I looking for? The number of molecules. therefore, set the equation as follows:


How many molecules are in 3.98 moles of aluminum hydroxide, Al(OH)3?

 

 

Answers to the Examples:


Converting the number of molecules to number of Atoms (using ratios)

Given the question:

How many atoms of hydrogen are in 2.50 moles Ammonia, NH3 ?

First, start with what you know. You have 2.50 moles of NH3 and then ask yourself - how many hydrogen atoms are in one molecule of ammonia (NH3). After you have figured those components of the equation begin:

= 4.52 x 1024 Hydrogen atoms in 2.50 moles of Ammonia


How many atoms of chlorine are there in 0.205 moles of phosphorus trichloride?

 

 

 

Answers to the Examples:


Lecture Video Link:

H.W.: 1a - 1f, & 2a - 2e


CONCENTRATION:

Many chemical compounds are stored, measured, and used as solutions. Medicines are commonly prepared by dissolving them in water so they may enter the bloodstream more quickly.

Household cleaners like bleach, ammonia, and vinegar contain compounds dissolved in water so they can be used and measured more easily.

Solutions can range from very _____________________________ to very _____________________________. Therefore, in using a solution a chemist needs to know how much of a certain substance the solution contains.

This is the measure of its _____________________________. There are various ways you can express concentration. The way that is chosen depends upon how the concentration is being used, on convenience, and clarity.

ppm and ppb:

The letters ppm represent a common means of concentration measurement - _____________________________.

Conversely, ppb represents _____________________________.

The solutions of some solutions are best expressed in parts per million, ppm. You may ask yourself if such a dilute concentration can be significant. Remember that an atom is very small, so even a very dilute solution contains a very large number of atoms.

For example, many communities add sodium fluoride to its water supply, between 0.7 to 1.0 ppm NaF, to prevent tooth decay. Maintaining a concentration this small is essential because larger concentrations of fluoride can cause mottling of the tooth enamel.


MOLARITY:

Percentage, ppm, and ppb have many applications in science, but chemists find it most useful to describe the concentrations of solutions used in the lab by indicating the number of moles of a substance dissolved in _____________________________ of solution. The amount of solution is commonly described in _____________________________ and is measured in a graduated cylinder.

Moles in a Given Volume:

Molarity is the ratio of the Moles of solute per Liter solution.

M = Molarity

4.90 M NaCl, represents that there are 4.90 moles of NaCl in one Liter of solution.

*** DO NOT confuse the molarity of a solution with the _____________________________ of a substance in a particular sample of solution.

Example: How many moles of HCl are contained in 1.45 L of a 2.25 M solution?

therefore, set up equation

A great rule to remember, if you KNOW the volume and molarity, ALWAYS start with the volume !!!


Practice Problems:

1. How many moles of H2SO4 are contained in 3.50 L of a 6.50 M solution?

 

 

 

2. What is the molarity of a solution that has 75.0 grams of calcium hydroxide dissolved in enough water so that the total volume is 3.50 liters?

 

 

Answers to the Examples:


PREPARING SOLUTIONS:

The preparation of one liter of a 1.00 M sugar solution involves the same process anywhere in the world.

Molar solutions contain a known amount of the dissolved solution in a given volume of solution. For example, a 1.00 M solution of sugar, C12H22O11, is prepared by measuring 342 grams of sugar (one mole) and then adding _____________________________ so that the _____________________________ of the solution is 1.00 liter when the sugar is completely dissolved.

Example:

Give directions for preparation of a 2.50 L of a 1.34 M sodium chloride solution.

? grams of NaCl needed to prepare 2.50 L of 1.34 M NaCl solution.

First, Always find molar mass first, NaCl = (23g) + (35.5g) = 58.5 g / mol

Then, you need to find the number of moles:

Finally, find the number of grams of NaCl:

Or you can place this whole process in ONE equation:

Directions: Measure 195g of NaCl and add enough water so that the final volume of the solution is 2.50 liters solution process is complete.


Practice Problem:

1.Give Directions for the preparation of 3.00 L of a 1.50 M CuSO4 solution.

 

 

Answers to the Examples:


HOMEWORK PROBLEMS:

1a. How many moles of H2SO4 are dissolved in 4.95 liters of a 2.33 M H2SO4 solution?

 

 

1b. How many moles of sodium chloride are dissolved in 1.75 liters of a 1.40 M NaCl solution?

 

 

1c. How many grams are in 1.25 liters of a 1.64 M copper(II) sulfate solution?

 

 

1d. How many grams of HCl are in 1.56 liters of a 9.32 M HCl solution?

 

 

1e. What is the molarity of a NaOH solution where 10.3 g of sodium hydroxide is dissolved in a 300. mL volume?

 

 

1f. What is the molarity of a solution where 23.4 g of nickel(II) carbonate is dissolved in a solvent with 1.72 liters of volume?

 

Answers to the Homework:


PERCENT COMPOSITION:

Percent means "_____________________________." In this case, percent indicates the score you would have gotten if the test had been graded on a 100-point basis.

In a similar fashion, if a 36 gram sample of iron oxide produces 28 grams of iron and 8 grams of oxygen, while a 160 gram sample of iron oxide produces 112 grams of iron and 48 grams of oxygen, you can determine whether the two samples are the same compound or different oxides or iron by using percent of iron and oxygen as a basis of comparison.

Example:

The following data are obtained from _____________________________ two iron oxide samples. Are the samples the same compound?

A 36 gram sample contains 28 grams of Fe and 8 grams of oxygen.

A 160 gram sample contains 112 grams of Fe and 48 grams of oxygen.

*****

Using the definition of percent, set up the equation that you need in order to find the percentages of iron or oxygen in each sample.

Substitute the data into the equation and solve for the percentage of iron in each sample:

Sample 1:

Sample 2:

These two samples are NOT the same...


Practice Problem:

Determine the percent composition for each of the elements in magnesium nitrate

 

Answers to the Examples:


HOMEWORK PROBLEMS: Determine the percent composition for each of the elements in the compound.

2a. aluminum sulfide

 

 

2b. nickel(II) iodide

 

 

2c. calcium cyanide, Ca(CN)2

 

Determine if the following samples are the same compound.

2d. Sample #1) 45.0 gram sample containing 35.1 grams of iron and 9.90 grams of oxygen and sample #2) 215 gram sample containing 168 grams of iron and 47.3 grams of oxygen.

 

 

 

2e. Sample #1) 75.0 gram sample containing 20.5 grams of carbon and 54.5 grams of oxygen and sample #2) 157 gram sample containing 67.0 grams of carbon and 90.0 grams of oxygen.

 

 

Answers to the Homework:


Lecture Video Link:

H.W.: 3a - 3d, 4a - 4b, & 5a - 5e


EMPIRICAL FORMULAS:

You already learned that every compound can be represented by a chemical formula.

The formula that represents the _____________________________ of the various types of atoms in a compound is called the empirical formula.

To determine the empirical formula for an unknown compound, the compound is analyzed to find the _____________________________ of each element in the compound.

Using this information, the percent composition by _____________________________ of each element in sample is determined. The below example illustrates how the empirical formula can be calculated.

Example:

What is the empirical formula for an iron oxide compound having the following composition?

A 36 gram sample that is 77.7% iron.

- when finding the formula of a compound from percentage data, use the fact that percent means "per hundred parts" to determine how many grams of iron and oxygen there are in each sample.

In a 100 gram sample, there would be 77.7 parts out of 100, or 77.7 grams of iron. There would be 22.3 grams of oxygen (100-77.7). Using the molar masses of each element, calculate the number of moles present to determine the mole ratio in the compound.

Using 77.7 grams of iron and 22.3 grams of oxygen, calculate the moles of iron and oxygen.

The mole ratio is 1 to 1 which means the formula must be FeO for the sample.

Another Way to work the problem:

If there is 77.7% iron in a 36 gram sample, then there is (0.777)*(36 grams) = 27.972 grams of Fe,

And (36 grams) - (27.972 grams) = 8.028 grams of oxygen. Finding the number of moles of Fe and O, you would get:

Doing the problem this way gives the same RATIO. Notice that the number of moles are NOT the same, but the ratios are the same - that is empirical formulas !!!


Practice Problem:

1. Determine the Empirical formula for the 400 gram sample containing 31.5% cobalt, 34.28% sulfur, and 34.22% oxygen.

 

 

 

 

2. What is the Empirical formula for a 425.0 gram sample containing 69.94% of iron and 30.06% of oxygen.

 

 

 

Answers to the Examples:


HOMEWORK PROBLEMS:

What are the Empirical Formulas for the two sulfur oxide compounds (in parts 3a and 3b) having the following compositions? VIDEO: (Homework Questions 3a & 3b)

3a.) a 27.0 g sample that is 50% sulfur by mass.

 

3b.) a 78.0 g sample that is 60% oxygen.

 

3c.) Calculate the Empirical formula of a compound that contains 1.67 grams of cerium, Ce, and 4.54 grams of iodine.

 

 

3d.) A synthetic rubber contains 0.556 grams of carbon and 0.0933 grams of hydrogen. Determine its Empirical Formula.

 

Answers to the Homework:


EMPIRICAL FORMULA (CONTINUED):

To determine the empirical formula for a compound, it is not always necessary to know the percent composition.

If you know _____________________________ of each element in the compound you can find the number of moles and determine the empirical formula.

Below are the steps for finding the empirical formula of a compound:

1. The mass (or mass percent) of each element in a sample of the compound is determined.

2. The mass (or mass percent) of each element is divided by its molar mass to determine the number of moles of each element in the sample of the compound.

3. The number of moles of each element is divided by the smallest number of moles to give the ratio of atoms in the compound.


Example: Charcoal is mixed with 15.53 grams of rust (an oxide of iron) and heated in a covered crucible to keep out air until all of the oxygen atoms in the rust combine with carbon. When this process is complete, a pellet of pure iron with a mass of 10.87 grams remains. What is the empirical formula for rust?

Step One: Begin by writing an equation that summarizes what you know about the reaction.

C(s) + Fe?O? (s) ® Fe (s) + CO? (g)

Step Two: Since you know the mass of iron oxide (Fe?O?) and the mass of iron (Fe), the mass of the oxygen can be found by subtraction. Using the molar masses of iron and oxygen atoms, calculate the number of moles of each element.

15.53 grams (Fe?O?) - 10.87 grams (Fe) = 4.66 grams (oxygen in rust)

Step Three: Calculate the moles for each element.

Step Four: Divide by the smallest number of moles to get the simplest ratio of atoms.

There are 1.49 moles of oxygen atoms for every mole of iron atoms in the compound. The formula could be written as Fe1O1.49.

However, because atoms combine in _____________________________, multiply both numbers by some whole-number. Multiplying by 2 gives 2.00 moles of iron and 2.98 moles of oxygen. Because of the uncertainty involved in the experimental measurements, you can assume that 2.98 is the same as 3. The ratio expressed in whole numbers becomes Fe2O3 for rust.

To check the percent composition of iron in the empirical formula is the same as the percent iron in the compound.

The comparison of the empirical formula agrees with the experimental data.


PRACTICE PROBLEMS:

1. When iron(III) oxide, Fe2O3, reacts with 18.94 grams of aluminum metal, iron is produced along with 35.74 grams of aluminum oxide. What is the empirical formula for aluminum oxide?

 

 

 

 

2. Propane is a gas commonly used for cooking and heating. It contains the elements carbon and hydrogen. When 72.06 grams of carbon reacts with hydrogen gas, 88.06 grams of propane is produced. What is the empirical formula of propane?

 

 

 

 

Answers to the Examples:


HOMEWORK PROBLEMS:

4a. Calculate the Empirical formula of a compound that contains 2.40 grams of carbon, 0.333 grams of hydrogen, and 3.20 grams of oxygen

 

 

 

4b. Methyl p-hydroxybenzoate is a mold inhibitor. Calculate the Empirical formula if 3.84 grams of carbon, 0.32 grams of hydrogen, and 1.92 grams of oxygen.

 

 

Answers to the Homework:


Lecture Video Link:

H.W.: 5a - 5e

H.W.: EMPIRICAL, MOLECULAR and HYDRATE FORMULAS


DETERMINING MOLECULAR FORMULAS:

Rust, Fe2O3, is an ionic compound that forms crystals of various sizes rather than molecules.

In ionic compounds, the empirical formula indicates the proportions of elements in the compound. This is what is called a _____________________________ of the ionic compound.

The formula for molecular compound is the molecular formula. The molecular formula of ethylene is C2H4.

The empirical formula of ethylene is CH2.

Remember, the empirical formula gives only the simplest whole number ratio of atoms. In the case of ethylene, the empirical formula tells you only that there are twice as many hydrogen atoms as carbon atoms in the molecule so CH2 is the empirical formula.

molecular formula = (empirical formula)n

To find n, or the multiple of the empirical formula, compare the molecular formula mass to the empirical formula mass. The empirical formula and molecular masses are the sum of the molar masses of the atoms in the formulas.

This ratio tells you how many empirical formula units are the actual molecule. From this information, you can determine the molecular formula of the compound, as shown in the below example.


EXAMPLE: A compound composed of hydrogen and oxygen is analyzed and a sample of the compound yields 0.59 grams of hydrogen and 9.40 grams of oxygen. The molecular mass of this compound is 34.0 grams/mole. Find the empirical formula and the molecular formula for the compound.

-- Determine the empirical formula by finding the number of moles of each element and the ratio of atoms in the compound, as you did in the previous problems. Then compare each empirical formula mass to the molecular formula mass to find the molecular formula.

Step 1: Determine the number of moles of each element.

The ratio of hydrogen atoms to oxygen atoms is 1:1. The empirical formula is HO. The empirical formula mass is

1.0 grams H + 16.0 grams O = 17.0 grams

Step 2: Solve for n.

The molecular formula for the compound is (HO)2, which is commonly written as H2O2.


PRACTICE PROBLEMS:

1. The empirical formula for a common drying agent is P2O5. The molecule has a molar mass of 283.88 g/mol. Find the molecular formula of the compound.

 

 

 

2. Hydrazine is a widely used compound. It can be used to treat waste water from chemical plants removing ions that may be hazardous to the environment; it can be used in rocket fuels; and it can help prevent corrosion in the pipes of electric plants. In a 32.0 gram sample of hydrazine, there are 28.0 grams of nitrogen and 4.0 grams of hydrogen. The molar mass of the molecule is 32.0 g/mol. What is the empirical formula of hydrazine?

 

 

Answers to the Examples:


HOMEWORK PROBLEMS:

5a. The molecular mass of benzene, an important industrial solvent, is 78.0 g/mol and its empirical formula is CH. What is the molecular formula for benzene?

 

 

5b. What is the molecular formula of dichloroacetic acid, if the empirical formula is CHOCl and the molecular mass of the acid is 129 g/mol?

 

 

5c.What is the molecular formula of cyanuric chloride if the empirical formula is CClN and the molecular mass is 184.5 g/mol?

 

 

5d. Asorbic acid, vitamin C, has a percentage composition of 40.9% carbon, 4.58% hydrogen, and 54.5% oxygen. Its molecular mass is 176.1 g/mol. What is the molecular formula?

 

 

5e. Aspirin contains 60.0% carbon, 4.48% hydrogen, and 35.5% oxygen. It has a molecular mass of 180 g/mol. What is the its empirical and molecular formulas?

 

Answers to the Homework:


HYDRATES:

There are many compounds that crystallize from a water solution with water molecules adhering to the ions or molecules and becoming part of the crystal. These hydrates, as they are called, usually contain a _____________________________ to compound.

Calculating for hydrates is similar to calculating molecular formulas.

EXAMPLE: A 344 gram sample of hydrated calcium sulfate and the sample is heated to evaporate the water. The dry sample of calcium sulfate has a mass of 272 grams. What is the mole ratio between the calcium sulfate, CaSO4 and water, H2O? What is the formula of the hydrate?

Step 1: Calculate the difference between the hydrated sample and the dry sample.

344 grams (hydrate) - 272 grams (dry sample) = 72.0 grams of water

Step 2: Convert mass to moles for each sample.

Dividing both mole values by 2 gives a ratio of 1 to 2. The formula for the hydrate is written as CaSO4 * 2H2O

PRACTICE PROBLEMS:

1. What is the formula for a hydrate that is 90.7% SrC2O4 and 9.30% H2O?

 

 

 

2. What is the formula for a hydrate that is 433.5 grams of Mo2S5 and 66.5 grams of H2O?

 

Answers to the Examples: