Khalia Simmons
Study Guide-Chapter Three
Mole – the SI unit for amount of substance; the formula mass for an element or compound is measured in grams, an
amount of a chemical substance that contains 6.02x1023 formula units
*One mole of any element is equal to the atomic mass with gram unit*
Using Molar Masses for Conversions
Ex. How many moles of sulfur are there in a 23.5 g sample of sulphur
23.5gS=?molsatomic mass of S is 32.06u=32.06g
Now we use factor-label: 23.5gx(1molS/32.06gS)= 0.733molS
Ex2 We need 0.254mol of iron(III)chloride for an experiment. How many grams de we need?
1mol FeCl3=gram molar mass FeCl3
Atomic mass of Fe=55.85g Cl3=35.45g x 3=106.35
55.85+106.35= 162.20gFeCl3
Now use factor label: 0.254molFeCl3x(162.20gFeCl3/1molFeCl3= 41.2gFeCl3
*The number of particles in a mole is called Avogadro’s Number* 1mole of X=6.02x1023units of X
Ex. How many atoms of copper are there in a piece of pure copper wire with a mass of 14.3 grams?
14.3gCu=?atomsCuatomic mass of Cu is 63.546g
1moleCu=6.02x1023atoms Cu
Grams Cumoles Cuatoms Cu
14.3gCux(1molCu/63.546gCu)x(6.02x1023atomsCu/1molCu)= 1.35x1023atomsCu
Mole to Mole Conversions
In order to perform mole to mole conversions we must find the proper ratio between all atoms within the compound.
Ex. P4O10 4 mol P10 mol 0 or 4 mol P/10 mol O or 10 mol O/4 mol P
1 mol P4O104 mol P or 1 mol P4O10/4 mol P or 4 mol P/1 mol P4O10
1 mol P4O1010 mol O or 1 mol P4O10/10 mol O or 10 mol O/1 mol P4O10
Ex2 In one case, a sample of Calcium Phosphate, 0.864 moles of phosphorus is found. How many moles of Ca3(PO4)2
will that represent?
0.864 mol P=?molCa3(PO4)2if you count how many P atoms are in Ca3(PO4)2 you will come to a
conversion factor of 1molCa3(PO4)2/2molP
0.864molPx(1molCa3(PO4)2/2molP)= 0.432mol Ca3(PO4)2
Percentage Composition
% by mass of element=(mass of element/mass of whole sample)x100%
Ex. A sample of liquid with a mass of 8.657g was decomposed into its elements and gave 5.217g of carbon, 0.9620g of
hydrogen, 2.478g of oxygen. What is the percentage composition of the element?
C=(5.217gC/8.657)x100%=60.26%C H=(0.9620g/8.657)x100%=11.11%H O=(2.478g/8/657)x100%=28.62%O
Sum of Percentages=60.26+11.11+28.62= 99.99%
*If the sum of percentages amount to about 100%, then that means that all calculations are correct*
Using Percent Composition to Find an Unknown Compound
Ex. Do the mass percentages of 25.94% of N and 74.06 of O match the formula N2O5?
N2O5 2 mol N5 mol O=2 mole N/5 mole O or 5 mole O/2 mole N
2N:2molNx(14.01gN/1molN)=28.02gN
5O:5molOx(16.00gO/1moleO)=80.00gO
1molN2O5=28.02+80.00=108.02gN2O5  N=(28.02/108.02)x100%=25.95%N O=(80.00/108.02)x100%=74.06%O
Empirical Formula – expresses the simplest whole number ratio of the atoms of each element in a compound. So if
there is a common factor with all the atoms, then their subscripts can be reduced to the smallest and simplest form
Ex.
Molecular Formula
Empirical Formula
P4O10
common factor of 2
P2O5
Calculating Empirical Formula
A 2.57g sample of a compound of only tin and chlorine was found to contain 1.77g of tin. What is the compounds
empirical formula?
Mass of Cl=2.57g of compound−1.77gSn= 1.40gCl
-now that we know the mass of Cl, we can now calculate how many moles each atom possesses
Sn:1.17gSnx(1molSn/118.71gSn)= 0.00986molSn
Cl:1.40gClx(1molCl/35.45gCl)= 0.0395gCl
Since Sn0.00986Cl0.0395 is not a proper formula, we must now divide each number by the smallest number present, which
in this case is 0.00986
Sn0.00986 / 0.00986Cl0.0395 / 0.00986  Sn1.00Cl4.01 SnCl4
*When using the lowest common divisor, we do not always end up with a whole number. In these cases, we simply
multiply the subscripts until they are close enough to be rounded into a whole number*
Ex. When a 2.448g sample Iron Oxide was analyzed, it was found to have 1.771g of Fe. Calculate the empirical
formula of this compound.
Mass of O=2.448g of compound−1.771gFe= 0.677gO
Fe:1.771gFex(1molFe/55.845gFe)= 0.03171molFe
O:0.677gOx(1molO/16.00gO)= 0.0423molO
Fe0.03171O0.0423Fe0.03171 / 0.03171O0.0423 / 0.03171Fe1.00O1.33
*Since Fe1.00O1.33 can not be used as a proper formula and 1.33 is not close enough to be rounded to a whole number,
we must multiply both subscripts until they are both whole numbers*
First try to multiply Fe1.00O1.33 by 2Fe(1.00 x 2)O(1.33 x 2)=Fe2.00O2.66
-since 2.66 is not close enough to round up to a whole number, multiply the original formula by 3
Fe(1.00 x 3)O(1.33 x 3)=Fe3.00O3.993.99 is close enough to be rounded up to 4, therefore the empirical formula
is Fe3O4
When Balancing an Equation You Should:
1) Balance all elements before H and O
2) Single elements should be balanced last
3) If there are polyatomic ions, balance them as a group
Ex. Zn+HCl→ZnCl2+H2  Zn+2HCl→ZnCl2+H2  2Zn+4HCl→2ZnCl2+2H2
*Balancing equations is important because it provides mole to mole relationships*
Ex. 2Zn+4HCl→2ZnCl2+2H2
2 molecule Zn  4 molecule HCl
2 molecule Zn  2 molecule H2
4 molecule HCl  2 molecule H2
2 molecule Zn  2 molecule ZnCl2 4 molecule HCl  2 molecule ZnCl2 2 molecue ZnCl2  2 molecule H2
Limited and Excess Reactants
Gold(III) hydroxide is made by the following reaction:
2KAuCl4(aq) + 3Na2CO3(aq) + 3H2O → 2Au(OH)3(aq) + 6NaCl(aq) + 2KCl(aq) + 3CO2(g)
To prepare a fresh supply of Au(OH)3, a chemist has mixed 20.00g of KAuCl4 with 25.00g of Na2CO3. What is the
maximum number of grams of Au(OH)3 that can form?
2 mol KAuCl4  3 mol Na2CO3
20.00gKAuCl4x(1molKAuCl4/377.88g KAuCl4)x(3mol Na2CO3/2molKAuCl4)x(105.99gNa2CO3/1molNa2CO3)
= 8.415g Na2CO3
-according to these calculation, 20.00g of KAuCl4 needs 8.415g of Na2CO3. Since the provided 25.00g of Na2CO3 is
more than enough to react with the 20.00g of KAuCl4, we can conclude that Na2CO3 is the excess reactant and KAuCl4
is the limiting reactant
-now that we know what the limited reactant is, we can move onto the next step, which is converting the mass of the
limited reactant(KAuCl4) to the mass of the product(Au(OH)3)
1 mol KAuCl4  1 mol Au(OH)3
20.00g KAuCl4x(1molKAuCl4/377.88gKAuCl4)x(1molAu(OH)3/1molKAuCl4)x(247.99gAu(OH)3/1molAu(OH)3)
= 13.13gAu(OH)3
Percentage Yield
-The actual yield of desired product is how much is isolated and the theoretical yield is what must be obtained if there
is no loss
Percentage Yield=(actual yield/theoretical yield)x100%
Ex. A chemist set up a synthesis of phosphorus trichloride by mixing 12.0g of phosphorus with 35.0g of chlorine gas
and obtained 42.4g of liquid phosphorus trichloride. Calculate the percentage yield of this compound.
-First we must find the limited reactant
2P(s) + 3Cl2(g) → 2PCl3(l)
12.0gPx(1molP/30.97gP)x(3molCl2/2molP2)x(70.90gCl2/1molCl2)= 41.2gCl2
-according to these calculation, the limited reactant is Cl2. When trying to find the theoretical yield, we always convert
the mass of the limited reactant(Cl2) to the mass of the product(PCl3)
35.0gCl2x(1molCl2/70.90gCl2)x(2molPCl3/3molCl2)x(137.32gPCl3/1molPCl3)= 45.2gPCl3
Percentage Yield=(42.4gPCl3/45.2gPCl3)x100%= 93.8% of the theoretical yield was obtained