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Chapter 10 Chemical Quantities
In this chapter we will learn how chemist relate the number of particles to their mass with a unit called the mole
10.1 The Mole: A Measurement of Matter - You couldn't count all the grains of sand in a sand castle, but you could
weigh them and calculate the number of grains of sand. This is what we do in chemistry.
1. Measuring Matter - We often measure the amount of something by one of three different methods—by count, by
mass, and by volume.
For instance, apples can be sold by the count (3 for 2.50), apples can also be sold be mass (1.29/lb), or apples can be sold
by volume ($12.00/bushel)
2. What is a Mole?
Atoms and molecules are much smaller than grains of sand, so in order to quantify them, we need to use mass or
volume.
a mole - is a quantity of 6.02 x 1023 representative particles - this is called Avogadro's number
*the same way a dozen is always 12 "whatevers", a mole is always 6.02 x 1023 "whatevers"
Representative particles - the species present in a substance: usually atoms, molecules, or formulas units.
For instance: we could consider atoms of copper, molecules of water, or a formula unit such as CaCl2 - the quantities of
all of these are represented in moles
A. Converting Number of Particles to Moles - can be accomplished with the following formula
Ex. How many moles is 2.17 X1023 representative particles of bromine
3. The mass of a Mole of an Element
The mass of an element is expressed in amu (atomic mass units), where 1 amu is 1/12 of the mass of a carbon-12 atom
The mass of the hydrogen-1 atom is 1/12 the mass of Carbon-12. This means that 1 mole of the hydrogen atoms weigh
12 times less than 1 mole of carbon atoms.
Atomic Mass - the measurements in the periodic table represent the mass in grams of 1 mole of the atom. Recall that
they are not whole numbers because the mass of each atom is an average of the naturally occurring isotopes for that
atom.
Atomic mass is calculated by multiplying the masses of the various isotopes by their naturally occurring isotopes and
adding them together.
The atomic mass of an element(expressed in grams) is the mass of one mole of the element.
**Another name for the mass of a mole of an element is its molar mass.
4. The Mass of a Mole of a Compound (aka- gram-formula mass or molar mass) In order to calculate gram formula mass, simply add up the masses of each atom in the compound. Consider the gram
formula mass of sulfur trioxide (SO3)
Gram-formula mass is
**This means that 1 mole of SO3 will have a mass of 80.1 grams**
Calculate the gram-formula mass (gfm) of the following compounds (units for gfm are grams/mole or g/mol)
Ex. H2O
CH3Cl
Ex. H2O2
C8H18
Ex. C2H4O2
calcium chloride
Ex. calcium phosphate
magnesium hydroxide
10.2 Mole-Mass and Mole-Volume Relationships
The molar mass of any substance(elements, molecular compounds, and ionic compounds) is the mass in grams of one
mole of the substance.
We can use the molar mass of an element or compound to convert between the mass of a substance and the moles of
a substance.
1. Converting Moles to Mass (grams)
Suppose you need 3.00 moles of NaCl for a chemical reaction. How many grams do you need?
1st - find the gram-formula mass of NaCl, then this number is substituted where it says "mass grams" (directly below)
solution:
176g
=
3.00 moles NaCl
X
58.5 grams/mole NaCl
note: moles of NaCl cancel out and we are left with grams
therefore, you would need 176 grams of NaCl for the reaction requiring 3 moles of NaCl
Convert the following molar quantities to grams
Ex. .5 moles of H2O
Ex. 2.5 moles of NaOH
Ex. 1.2 moles of KClO3
Ex. 9.45 moles of Al2O3
2. Converting Grams to Moles (opposite of the above process) - Table T contains the formula
Suppose a chemical reaction yields 10.0 grams of sodium sulfate (Na2SO4). How many moles is this?
Use the formula below which identical to the mole calculations formula in table T
0.0704 moles Na2SO4 =
10.0 g
Note: grams cancel out and we are left with moles
X
1mole Na2SO4/142.1 g
Try the following gram to mole conversions
Ex. 18.0 grams H2O
Ex. 48.0 g O2
Ex. 15.2 g Ca3(PO4)2
Ex. How many moles of iron(III) oxide (Fe2O3) are contained in 92.2 grams of Fe2O3?
3. Mole-Volume Relationship - The volumes of 1 mole of different solids and liquids are not the same; however, at the
same temperature and pressure, all gases of equal volume contain an equal number of moles (this is called Avogadro's
hypothesis)
At STP, 1 mol or, 6.02  1023 representative particles, of any gas occupies a volume of 22.4 L.
STP stands for standard temperature and pressure (0oC or 273K and 101.3kPs or 1atm)
Mole-Volume Conversions (at STP only)
Using the equivalent measurement for gases at STP: 1 mole = 22.4 liters allows us to easily convert between moles and
liters for gases
Ex. If a reaction generated 0.375 mol O2 at STP, what volume of O2 gas was given off?
8.4 L O2
=
0.375 mole O2
X
22.4L/mol
note: moles cancel and we are left with liters
Ex. How many liters will be produced at STP, when 2.0 moles of H2 gas given off?
Ex. Determine the volume, in liters, of 0.60 moles of SO2 gas at STP.
Volume-Mole Conversions (at STP only) - the opposite of the above process. (just divide by the volume by 22.4 L/mole
Ex. A reaction generated 2.5 liters of H2 gas at STP? How many moles of H2? How many grams of H2(g)?
The Mole Road Map
Section Review Questions
1. Describe how to convert between the mass and the number of moles of a substance.
2. What is the volume of one mole of any gas at STP?
3. How many grams are in 5.66 mol of CaCO3?
4. Find the number of moles in 508 g of ethanol (C2H6O).
5. What is Avogadro's hypothesis?
6. Which gas sample at STP has the same total number of molecules as 2.0 liters of CO2 at STP?
(1) 5.0 L of CO2(g)
(2) 2.0 L of Cl2(g)
(3) 3.0 L H2S(g)
(4) 6.0 L of He(g)
7. Three balloons filled with three different gaseous compounds each have a volume of 22.4 L at STP. Would these
balloons have the same mass or contain the same number of molecules? Explain
10.3 Percent Composition and Chemical Formulas - In this section, you will learn how the percents of the elements in a
compound are important.
1. The Percent Composition of a Compound (Formula: Table T) - determined by comparing the "mass of the part", to
the "mass of the whole."
The percent composition 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%.
Ex.
K2CrO4 gram formula mass (39.1+39.1+52.0+ 16.0+16.0+16.0+16.0) = 194.2g/mole
%K =
39.1+39.1/194.2
% Cr = 52.0/194.2
X 100 = 40.3 %
X 100 = 26.8 %
% O = 16+16+16+16/194.2 X 100 = 33.0
Note: At times, the percents may only add up to 99.9 or 100.1 due to rounding
Ex. Find the % mass of each element in propane gas - C3H8
%C=
%H=
Ex. Find the % composition of each element in Mg(NO3)2
% Mg =
%N=
%O=
2. Percent Composition as a Conversion Factor
In the above example, we determined that propane (C3H8) is 81.7 % carbon and 18.3 % hydrogen. This means that if you
had 100.0 grams of propane, 81.7 grams of it is carbon and 18.3 % is hydrogen.
If you had only 55.0 grams of propane, how many grams of element would you have
Solution: grams of carbon = 55.0 g C3H8 X 81.7 g C/100 g C3H8 = 44.9 g C
grams of hydrogen = 55.0 g C3H8 X 18.3 g H/100 g C3H8 = 10.1 g H
Ex. Calculate the % mass of each element in NaCl, and determine the mass of each element in 146.3 g NaCl sample?
3. Empirical Formulas
empirical formula - gives the lowest whole number ratio of the atoms of the elements in a compound.
For instance ethyne (C2H2) is a gas used in welders torches, its empirical formula is CH
styrene (C8H8) is a solid used to make polystyrene (styrofoam), its empirical formula is also CH
Both compounds have the same empirical formulas, but different molecular formulas (the molecular formula shows
the actual numbers of each kind of atom present in a molecule or compound)
All formulas can be looked at in two ways: macroscopically and microscopically. For instance consider the formula CO2
microscopic interpretation: it contains 1 C atom and 2 O atoms
macroscopic interpretation: it contains 1 mole of carbon atoms and 2 moles of oxygen atoms
note: the empirical formula of CO2 is the same as the molecular formula
Determining empirical formulas from percent data
Ex. A compound is analyzed and found to contain 25.9 % nitrogen and 74.1 % oxygen. What is the empirical formula?
Solution: 1. Consider the percent data as mass data( in grams)
2. Convert the grams to moles
3. Divide the larger number of moles by the smaller number of moles in order to get the whole
numbers for each element.
moles of N =
25.9 g N
X
1mol N/14.0 g N = 1.85 mol N
moles of O =
74.1 g O
X
1 mol O/16.0 g O = 4.63 mol O
N will have no subscript in the empirical formula because it has the lower number of moles
Oxygen's subscript will be equal to 4.63/1.85, which is equal to 2.5
This yields the formula NO2.5; however we cannot have 2.5 atoms of oxygen in the formula, so we multiply the entire
formula by 2 in a distributive fashion: 2 (NO2.5) = N2O5
N2O5 is the empirical formula, which has a gram formula mass of 2(14.0) + 5(16.0) = 108 g/mol
If data suggested that the gram formula mass of the molecular formula was 216 g/mol, the molecular formula would be
N4O10
Calculate the empirical formulas for the following sets of data
Ex. 94.1 % O, 5.9% H
Ex. 67.6 % Hg, 10.8 % S, 21.6 % O
4. Molecular Formulas
The molecular formula of a compound is either the same as its experimentally determined empirical formula, or it is a
simple whole-number multiple of its empirical formula.
Consider the table below
Finding the molecular formula when we are given a empirical formula and molar mass.
Ex. Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N
solution: 1 calculate the gram formula mass of the empirical formula
2. Divide the given molar mass by the gram formula mass calculated above
3. multiply the result from step 2 by the empirical formula (in a distributive fashion)
1. gfm CH4N = 30.0 g/mol
2. 60.0 g/mol/30.0 g/mol = 2
3. 2 (CH4N) = C2H8N2 = molecular formula
Try the following problem
Ex. Find the molecular formula of ethylene glycol, which is used in antifreeze. The molar mass is 62 g/mol and the
empirical formula is CH3O