Chapter 10: Chemical Quantities

external image mole.jpg

Editor: Chris Kelly


Elena Conroy

Nikki Sheehan

Teresa Lynch

This wiki covers Chapter 10 from Prentice Hall's Chemistry. Chapter 10 is split into three sections. These three sections will explain the mole, molar mass, Avogadro's Number, percent composition, empirical formulas, and molecular formulas. These topics are outlined below.



10.1 The Mole: A Measurement of Matter

Co-editor: Elena Conroy
By: Allie Chabot, Bryan Dextradeur, Emily Stewart, and Shannon Leavey

Measuring Matter (PG 287-289)
By: Allie Chabot
· We live in a quantitative world!
· You often measure the amount of something by one of three different methods-by count, by mass, and by volume.
· Some of the units used for measuring indicate a specific number of items. (Ex: a pair= 2, and a dozen=12)
· Knowing how the count, mass, and volume of an object relate, to a specific measure like a pair or a dozen, one is able to convert among those particular units.

Ex: You have 90 apples. 12 apples equal 1 dozen, and 1 dozen equals 2.0 kg apples. What is the mass of 90 apples in kg units?

Mass of apples= 90 apples x 1 dozen apples / 12 apples x 2.0 kg apples / 1dozen apples
90 x 1 x 2.0 / 12 x 1 = Mass of apples = 15 kg

Sample Problem 10.1- Finding Mass From a Count (Elena Conroy)

What is the mass of 90 average-sized apples if 1 dozen of the apples has a mass of 2.0 kg?

1. Analyze-list the knows and unknows

a. Knows

i. Number of apples=90

ii. 12 apples= 1 dozen apples

iii. 1 dozen apples= 2.0 kg apples

b. Unknown

i. Mass of 90 apples=? Kg

2. Caculate-solve for the unknown (using DA)

a. Mass of apples=90 apples x 1 dozen apples x 2.0kg apples = sdfvndskfnerwklgfnrwkg 12 dozen apples 1 dozen apples =15. kg apples

3. Evaluate- Does the result makes sense?

a. Since a dozen apples has a mass of 2.0 kg and 90 apples is less than 10 dozen apples, the mass should be less than 20 kg of apples

What Is A Mole? Pages 290-293

By: Bryan Dextradeur

  • Counting objects such as as apples in a bag is an easy process because they are large size but are small in quantity.
  • On the other hand, counting the particles in a substance is NOT at all an easy process because they are extremely small in size and there is an extremely large quantity of them.
    • To make counting easier, humans have developed specified units to represent other numbers, for example, a dozen eggs equals twelve eggs.
    • Similar to this, Chemists have developed a unit to represent a specified number of particles.
  • A Mole of a substance is 6.02x1023 representative particles of that substance and is the SI unit for measuring the amount of a substance.
  • Avogadro's Number is 6.02x1023, the number of representative particles in a mole.
  • Representative Particle refers to the atoms, molecules, or formula units present in a substance.
    • In most substances, Representative Particles are the atoms in the substance; however, in the Elements that normally exist as Diatomic Molecules (Hydrogen, Nitrogen, Oxygen, Fluorine, Chlorine, Bromine, and Iodine), they are the molecules in the substance and in Ionic Compounds (Such as Calcium Chloride), they are the formula unit of the substance.
  • A Mole of any substance contains Avogadro's Number of Representative Particles, or 6.02x1023 Representative Particles.

Sample Problem 10.2 Converting Number of Particles to Moles

  • The relationship, 1 mol = 6.02x1023 representative particles, is the basis for a conversion factor that you can use to convert numbers of representative particles to moles.

moles= representative particles x 1 mole / 6.02x1023 representative particles

Example: How many moles of magnesium in is 1.25 x 1023 atoms of magnesium?

moles= 1.25 x 1023 atoms Mg x 1 mole Mg / 6.02 x 1023 atoms Mg= 2.08 x 10-1 mol Mg = 0.208 mol Mg

Sample Problem 10.3 Converting Moles to Number of Particles

  • If you know the number of moles in a substance, you can determine the number of representative particles in an element.
    • must first determine how many atoms are in the substance based on its chemical formula.

Representative Particles= moles x 6.02 x 1023 Representative Particles / 1 mole

Example: How many atoms are in 2.12 moles of C3H8?

2.12 mol C3H8 x 6.02 x 1023 molecules C3H8 / 1 mole C3H8 x 11 Atoms C3H8 / 1 molecule C3H8 = 1.4039 x 1025 atoms = 1.40 x 1025 atoms C3H8.
    • Since there are 3 atoms of C and 8 atoms of H, there is a total of 11 atoms.

  • Everyone, I'm very sorry about the math problems because I know they're very confusing... when I tried to make a fraction bar for the division the formatting got all messed up so I had to end up using a backslash instead... if you want to see what the problems normally look like when set up they're on pages 291 and 292 of the textbook.

Above Images: Bryan Dextradeur

The Mass of a Mole of an Element - p293
By: Emily Stewart
-The atomic mass of an element is expressed in amu and is relative to the mass of carbon-12.

-The atomic mass of an element expressed in grams is the mass of a mole of the element, which is its molar mass.

-The molar masses of any two elements must contain the same number of atoms: 6.02 x 10^23 atoms

-Molar mass is the mass of 1 mol of atoms of any element.

*one mole of a substance contains 6.02 x 10^23 molecules of the substance, with the molar mass expressed in grams, NOT amu*

Use DA to help solve problems - Identify whether you need to deal with molecules or particles to reach the answer. If not, it isn't necessary to include 6.02 x 10^23 in the equation.

The Mass of a Mole of a Compound (pg 295)
By: Shannon Leavey
- in order to find this, you must know the formula of the compound
-calculate mass of molecule by adding atomic masses of the atoms
-then substitue the unit grams for atomic mass units to find the molar mass of SO3.
-this applies to any compund
-example: SO3
32.1(sulfur atom)amu + 16.0(oxygen atom)amu + 16.0(oxygen atom)amu + 16.0(oxygen atom)amu = 80.1 amu

Group 2: Mole-Mass and Mole-Volume Relationships

Co-Editor: Nikki Sheehan
By: Lindsey Chou, Lauren Murphy, Hannah Kumlin, Jessen Foster, Nikki Sheehan

The Mole-Mass Relationship (pgs. 297-298)
By Lindsey Chou
· Molar mass = the mass (g) of one mole of a substance
· You can use the molar mass of an element or compound to convert between the mass and moles of a substance:
Mass (grams) = number of moles × (mass in grams/1 mole)
Practice Problems:
1) What is the mass of 3.7 moles of iron phosphate?
2) What is the mass of 41.2 moles of NaOH?

(Answers: )
1) 498.34 g FePO3
2) 1648 g NaOH

Lauren Murphy
p. 299

· Use a conversion factor based on the molar mass to convert moles to mass
· OR use a conversion factor based on the grams to convert molar mass to moles
Sample problem:

How many moles of iron (III) oxide are contained in 92.2 g of pure Fe2O3?
Known: mass = 92.2 g Fe2O3

Unknown: number of moles= ? g mol Fe2O3

Solve for the unknown: determine the molar mass of Fe2O3: 1 mol = 159.6 g Fe2O3
Multiply the given mass by the conversion factor relating mass of Fe2O3 to moles of Fe2O3

Hannah Kumlin
Page 300

The Mole-Volume Relationship

Hannah Kumlin

- Notice that in the figure above the volumes of one mole of different solid and liquid substances are not the same. For instance, the volumes of one mole of glucose and one mole of paradichlorobenzene are much larger than the volume of one mole of water. *Unlike liquids and solids, the volumes of moles of gases, measured under the same physical conditions, are much more predictable.

In 1811, Amedeo Avogadro proposed a groundbreaking explanation for this.

Avagadro's hypothesis states that equal volumes of gases at the same temperature and pressure contain equal numbers of particles.

The particles that make up different gases are not the same size. Large expanses of space exist between individual particles of gas as you can see in the image below.

Hannah Kumlin

-If you buy a party balloon filled with helium and take it home on a cold day, you might notice the balloon shrinks while it's outside.
This is because the volume of a gas varies with a change in temperature.
The volume of a gas also varies with a change of pressure.

Because of these variations in temperature and pressure, the volume of a gas is usually measured at a standard temperature in pressure.
*Standard temperature and pressure (STP) means a temperature of 0°C and a pressure of 101.3kPa, or 1 atmosphere (atm).

At STP, 1 mol or 6.02 x 10^23 representative particles, of any gas occupies a volume of 22.4 L.

The quantity, 22.4L, is called the molar volume of a gas.

Jessen Foster
group 2: page 301

Calculating Volume at STP
-molar volume is used to convert a known number of moles to the volume of the gas
volume of gas = moles of gas X 22.4 L
1 mol -the opposite conversion uses the same relationship -sample problem Sulfur dioxide is gas produced by burning coal. It is an air pollutant and one of the causes of acid rain. Determine the volume, in liters, of 0.60 mol. SO2 gas at STP.

      1. analyze
        use the relationship to write the conversion factor to convert moles to volume known: moles = 0.60 mol SO2 1 mol SO2 = 22.4 L SO2

volume = ? L SO2 conversion factor: 22.4 L SO2 1 mol SO2

      1. calculate
        volume = 0.60 mol SO2 X 22.4 L SO2 = 13 L SO2
        1 mol SO2
3. evaluate
because one mole of any gas at STP has a volume of 22.4 L, 0.60 mol should have a volume slightly larger than one half of a mole or 11.2 L. The answer should have two significant figures.

Calculating Molar Mass from Density (302-303)

Nikki Sheehan

-Different gasses have different densities
-Density of a gass is usually measured in grams per liter -> g/L
-The density of a gas at STP and the molar volume at STP can be used to calculate the molar mass -> 22.4L/mol

molar mass=density at STP*molar volume at STP

Try it: Find the molar mass in g/mol when:
1. density=1.96g/L
1 mol=22.4L
2. density=6.93g/L
1 mol=22.4L
3. density=122.43g/L
1 mol=22.4L
4. density=34.64g/L
1 mol=22.4L

Group 3: Percent Composition and Formulas

Co-editor: Teresa Lynch

Ellie Kawa, Kayla Anghinetti, Neal McGovern, and David Monti

David Monti

Percent Composition – The percent by mass of each element in a compound
- A percent value for each different element in a compound
- The percents of each element add up to 100
- The percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound, multiplied by 100%

% mass of element = mass of element X 100%
mass of compound

Percent composition from mass data

There is a simple 3 step process to finding % composition from Mass data:
1. List the knowns and the unknows. This will usually include mass data from each element of the compound or the compound all together. The unknown would be the % composition of each element
2. Solve for the unknown. This will be the percent composition of each element in the compound
3. Check your answer. Add up the percent compositions and see if they add up to 100. If they do the answer is correct.

Neal McGovern Page 307
Percent Composition from the Chemical Formula

  • You can find the percent composition through many methods, one of which is through the chemical formula
  • To obtain the percent composition from the formula, first you need to find the mass of each type of element in the compound and the molar mass of the compound
  • Next you divide the mass of each type of atom into the molar mass and multiply it by 100%
  • Here is the complete formula
% composition= mass of one type of element in compound x 100%
molar mass of compound
Example Problem

H2O external image small-arrow.gif
 external image 350px-water_molecule-svg.png?w=262(Neal McGovern)
H atomic mass= 2.02g
O atomic mass= 16.00g
molar mass of H2O= 18.02g

%H= 2.02g x 100%
%H= 0.1121x 100%
%H= 11.21%

%O= 16g x 100%
%O= 0.8879 x 100%
%O= 88.79%


Sample Problems
Find the % composition


Kayla Anghinetti

p. 308
Percent Composition as a Conversion Factor
I. You can use percent composition to calculate the number of grams in a specific mass of a compound
a. Multiply mass of compound by a conversion factor based on the percent composition of the element.

Find how much carbon and hydrogen are contained in 82.0g of propane. (C3H8)
1. Find the percent composition of carbon and hydrogen
a. To do this, find the mass of carbon, divide it by the mass of propane, then multiply by 100. Do the same for hydrogen.
2. The percent composition of C= 81.8%. The percent composition of H=18%
a. This means that in 100g of propane there is 81.8g of carbon and 18g of hydrogen.
3. Once you have found the percent composition, you can use dimensional analysis.
4. To calculate the mass of carbon contained in 82.0g propane, use the ratio 81.8g C/100g C3H8
external image chem003.jpg

5. Using the ratio 18g H/100g C3H8 you can calculate the mass of hydrogen.
external image chem002.jpg
6. To check your answer, add the two masses together to see if it equals the sample size. In this case, 82g of propane. (67.1gC + 15gH = 82g C3H8)

Ellie Kawa
Pgs 309-310

Empirical Formulas
· formulas for some compounds show a basic ratio of elements
o multiplying that ratio by any factor can produce the formulas for other compounds
o the percent composition of the newly synthesized compound is the data you need to calculate the basic ratio of the elements contained in the compound
· empirical formula- basic ratio
o gives the lowest whole-number ratio of the atoms of the elements in a compound
§example: a compound may have the empirical formula CO2; empirical formula shows the kinds and the lowest relative count of atoms or moles of atoms in molecules or formula units of a compound
o an empirical formula may or may not be the same as a molecular formula
· the empirical formula of a compound shows the smallest whole-number ratio of atoms in the compound
o molecular formula tells the actual number of each kind of atom present in a molecule of the compound
Sample Problem
Determining the Empirical Formula of a Compound
A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula for the compound?
Analyze: List the knowns and the unknowns
Ellie Kawa
Ellie Kawa

· percent of nitrogen= 25.9% N
· percent of oxygen= 74.1% O
· empirical formula= N?O?
Calculate: Solve for the unknown
25.9g N*1 mol N/14.0g N= 1.85 mol N74.1g O*1 mol O/16.0g O= 4.63 mol O
1.85 mol N/1.85= 1mol N; 4.63 mole O/1.85= 2.50 mol O
1 mol N*2= 2 mol N
2.5 mol O*2= 5 mol O
The empirical formula is N2O5.

Teresa Lynch
Molecular Formulas
- The molecular formula of a compound is either the same as its experimentally determined empirical formula, or it is a sample whole-number multiple of its empirical formula.

-You must know the compound's molar mass to determine the molecular formula after finding the empirical formula.

-The mass of a compound is the amount of deflection its particles experience (particles are deflected by a magnetic field that ions travel through).

-Once you have the formula, you can find the efm, or empirical formula mass (molar mass represented by empirical number).

-If you then divide the efm into the determined molar mass you can use that fraction as a multiplier to get the molecular formula.

ex: Empirical formula for hydrogen peroxide is HO. Its efm is 17.0 g/mol. The molar mass of H2O2 is 34.0 g/mol.

34.0 g/mol / 17.0 g/mol = 2

-The molecular formula of hydrogen peroxide is found: from its empirical formula, by multiplying subscripts in empirical formula by 2. (HO) X 2 = H2O2.

ethanoic_molecule.gif glucose_molecule.jpg methanol_molecule.jpg

Ethanoic molecule Glucose molecule Methanol molecule

Description of image: Ethanoic acid, glucose, and methanal have same empirical formula. By: Teresa Lynch