1. Chapter 3:Mass Relationships in Chemical Reactions
Chemistry Joanna Sabey

2. Atomic Mass 1H = 1.008 amu 16O = 16.00 amu
Atomic Mass: The mass of the atom in atomic mass units(amu).
Atomic Mass Unit: 1 atom 12C “weighs” 12 amu
On This Scale:
1H = amu
16O = amu

3. Average Atomic Mass
Average Atomic Mass: the weighted average of all of the naturally occurring isotopes of the element.
Average atomic mass of natural carbon=
(0.9890)(12 amu) + (0.0110)( amu) =
12.01 amu
The atomic mass on the periodic table is the average atomic mass.

4. Average Atomic Mass
Calculate the average atomic mass of copper given the following information.
Convert the percent values to decimals.
0.6909
0.3091
Add the contributions together to obtain the average atomic mass.
(0.6909)(62.93amu) + (0.3091) ( amu) =
= amu
Confirm that the average atomic mass matches that in the periodic table.
Isotopes
Mass
Percent
63Cu
62.93
69.09%
65Cu
30.91%

5. Avogadro’s number and the mole
Mole (mol): the amount of a substance that contains as many elementary entities (atoms, molecules, or any other particles) as there are atoms in exactly 12 g of the carbon-12 isotope.
Avogadro’s number: the actual number of atoms in 12 g of carbon This number is X 1023
1 mol of hydrogen atoms contains X H atoms.

6. Molar Mass Molar Mass: the mass of 1 mol of a substance.
The molar mass of an element is equivalent to its atomic mass in amu. Ex: The atomic mass of Sodium, Na is amu, and its molar mass is g/mol.
Conversions from atoms to moles can occur with the knowledge of Avogadro’s number and from mass to moles using the atomic mass.

7. Molar Mass
Helium(He) is a valuable gas used in industry. How many moles of He atoms are in 6.46 g of He?
Find the molar mass of Helium:
He= g/mol
Set up conversion so that the values you are getting rid of cancel:
6.46 g He X 1 mol He g He = 1.61 mol He

8. Mole How many grams of Zn are in 0.356 mole of Zn?
Find the molar mass of Zinc, Zn:
1 mol Zn = g
Set up the conversion so that the values you are getting rid of cancel:
0.356 mol Zn X g Zn 1 mol Zn = 12.6 g Zn

9. Avogadro’s number How many atoms are in 16.3 g of sulfur, S?
Find the molar mass of sulfur, S:
1 mol S = g S
Solve for the moles of sulfur:
16.3 g S X 1 mol S g S = mol S
Use Avogadro’s number to solve for the number of atoms:
1 mol S = X 1023 atoms of S
0.508 mol S X X atoms S 1 mol S = 3.06 X 1023 atoms S

10. Molecular Mass
Molecular mass: molecular weight, is the sum of atomic masses in a molecule.
H2O:
H=1.008 X 2= amu
O=16.00 X 1 = amu
H2O = amu
C8H10N4O2:
C= X 8 = amu
H = X 10 = amu
N= X 4 = amu
O = X 2 = amu
C8H10N4O2 = amu

11. Mole How many moles of CH4 are present in 6.07 g of CH4?
Find the molecular mass of CH4:
C= X 1 = amu
H= X 4 = amu
CH4 = amu
1 mol CH4 = g CH4
Set up the conversion so that the values you are getting rid of cancel:
6.07 g CH4 X (1 mol CH4/ g CH4) = mol CH4

12. Avogadro’s number
How many hydrogen atoms are present in 25.6 g of urea [(NH2)2CO], MW=60.06g?
Find molar mass of urea:
Given: g/mol
Find how many moles of Hydrogen are in 1 mol of [(NH2)2CO]:
4 mole of Hydrogen in 1 mole of [(NH2)2CO]
Set up the conversion so that the values you are getting rid of cancel: X 1024 atoms H

13. Percent Composition
Percent composition by mass: Percent by mass of each element in a compound. Percent composition of an element = 𝑛 X molar mass of element molar mass of compound X 100 %
n is the number of moles of the element in 1 mole of the compound.
What is the percent composition of Hydrogen and Oxygen in hydrogen peroxide (H2O2)?
Find the molar mass of H2O2:
H2O2 = (1.008 X 2) + (16.00 X 2) = amu
Find the percent composition of H:
1 H has a mass of and we have 2 H atoms
% H = 2 X amu amu 𝑋 100 %= %
Find the percent composition of O:
1 O has a mass of and we have 2 O atoms
% O = 2 X amu amu 𝑋 100 %=94.07 %

14. Percent Composition
Chalcopyrite(CuFeS2) is a principal mineral of copper. Calculate the number of kilograms of Cu in 3.71 X 103 kg of chalcopyrite.
Find the percent composition of Cu in chalcopyrite:
Mass of Cu= amu
Mass of chalcopyrite= (1 X 63.55) + (1 X 55.85)+ (2 X 32.07) = amu
% Cu = amu amu 100 % = %
Use the percent composition to solve for the mass in 3.71 X 103 kg of chalcopyrite:
mass of Cu in CuFeS2 = × (3.71 × 103 kg) = 1.28 × 103 kg

15. Determination of Empirical Formula
Ascorbic acid cures scurvy. It is composed of % carbon, 4.58 % hydrogen and % oxygen by mass. Determine its empirical formula.
Convert to grams and divide by molar mass:
40.92 g C g C mol = mol C
4.58 g H g H/mol =4.54 mol H
54.50 g O g O /mol =3.406 mol O
Divide each mole value by the smallest mole value:
C = mol / mol = 1
H= mol / mol = 1.33
O= mol / mol = 1
Change moles ratios to integer subscripts:
For Hydrogen, find what value multiplied by 1.33 to give a whole number
1.33 X 3 = 3.99 = 4
Multiply all 3 values by 3
C3H4O3

16. Empirical Formula
A g sample of Radium was heated with oxygen to form a g sample of radium oxide. Find the empirical formula of radium oxide.
Step one: determine the mass of oxygen:
Law of Conservation of mass
1.755 g g = g O
Step two: Find the moles of radium and oxygen:
1.640 g radium / (226 g/mol Ra) = mol Ra
0.115 g Oxygen/ (16.00 g/mol O) = mol O
Step Three: Divide the mole number by the smallest mole number:
Ra= mol/ mol = 1
O= mol / mol = 1
Determine the empirical formula by forming integers:
RaO

17. Molecular Formula
A sample contains % nitrogen and % oxygen by mass, the molar mass of the compounds is found to be between 90 g and 95 g. Determine the molecular formula and the accurate molar mass of the compound.
Find the empirical formula:
Determine the moles:
30.46 g N/ (14.01 g / mol N) = 2.17 mol N
69.54 g O /(16.00 g/ mol O) = mol O
Determine the empirical formula:
2.17 mol N / 2.17 mol = 1
4.24 mol O / 2.17 mol = 2
NO2
Find the molecular formula:
Determine the molar mass of the empirical formula:
NO2 = (1 X 14.01) + (2 X 16.00) = amu
Determine the ratio between the molar mass and the empirical molar mass:
Determine the molecular formula
(NO2)2

18. Chemical Equations
Chemical Equation: The use of formulas and symbols to describe a chemical reaction A B  C D
A and B are the reactants, and C and D are the products.
The physical state of the reactants and products are represented by an abbreviation, (s) is solid, (l) is liquid, (g) is gas, and (aq) is aqueous solution A (aq) + B (g)  C (s) + D(aq)

19. Chemical Equations
Catalyst: A substance that speeds up a reaction without being consumed or permanently altered. The presence is indicated by placing its formula above the yield symbol (arrow) S(s) + O2 (g)  SO2 (g)
Symbol
Interpretation of symbol 
Produces, yield, gives +
Reacts with, added to, plus
Heat is the catalyst for the reaction
Iron is the catalyst for the reaction NR
No reaction
(s)
Solid substance or precipitate
(l)
Liquid substance
(g)
Gas substance
(aq)
Aqueous solution Fe

20. Balancing Chemical Equations
To balance an equation, a whole number coefficient is placed in front of each substance. H2 (g) + Cl2 (g)  HCl (g)
Notice the subscript for H and Cl is 2, therefore we have 2 atoms of each substance. In the products, we have HCl, 1 atom of each. We can balance the equation by putting a 2 in front HCl. H2 (g) + Cl2(g)  2 HCl (g)

21. Balancing Chemical Equations
Always make sure that the formula is written correctly before trying to balance an equation, this means subscripts.
Start with the most complex formula first.
Balance polyatomic ions as a single unit unless they breakdown.
The coefficients must be whole numbers.
After balancing an equation, check each symbol with its corresponding number.
Finally, Make sure the coefficients represent the smallest whole number ratio.

22. Balancing Chemical Equations
Al (s) + O2(g) Al2O3 (s)
The most complicated would be O2 . On the left side, there are 2 oxygens, on the right, there are 3. The lowest common multiple is 6, therefore we need to make both sides 6.
Al (s) + 3 O2 (g) 2 Al2O3 (s)
Now, we look at Al, on the left side, we have 1 Al, on the right, we have 4. in order to balance, we must have 4 atoms on both sides.
4 Al (s) + 3O2 (g) 2 Al2O3 (s)

23. Balancing Chemical Equations
Al2(SO4)3(aq) + Ba(NO3)2(aq)  BaSO4(s) +Al(NO3)3(aq)
First look at SO4 as a whole
Al2(SO4)3(aq) + Ba(NO3)2(aq)  3 BaSO4(s) +Al(NO3)3(aq)
Now, what element’s atom number changed?
Al2(SO4)3(aq) + 3 Ba(NO3)2(aq)  3 BaSO4(s) +Al(NO3)3(aq)
Now what element’s atom number changed?
Al2(SO4)3(aq) + 3 Ba(NO3)2(aq)  3 BaSO4(s) +2 Al(NO3)3(aq)
Is everything balanced?

24. Stoichiometry
Stoichiometry: the quantitative study of reactants and products in a chemical reaction.
Mole Method: the stoichiometric coefficients in a chemical equations can be interpreted as the number of moles of each substance.
N2(g) + 3 H2(g)  2 NH3(g) 1mol + 3 mol  2 mol

25. Stoichiometry Write the balanced equation.
Convert the given amount of the reactant to number of moles.
Use the mole ratio from the balance equation to calculate the number of moles of product formed.
Convert the moles of product to desired units of product.

26. Stoichiometry
If 856 g of C6H12O6 is consumed by a person over a certain period, what is the mass of CO2 produced? C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(g)
Convert grams of glucose to moles of glucose:
MW of C6H12O6 = g/mol
856 g C6H12O6 X 1 mol C 6 H 12 O g C 6 H 12 O 6 = mol C6H12O6
Convert moles of glucose to moles of carbon dioxide:
For every 1 mole of glucose, 6 moles of carbon dioxide are produced.
4.750 mol C6H12O6 X 6 mol CO 2 1 mol C 6 H 12 O 6 = mol CO2
Convert moles of carbon dioxide to grams of carbon dioxide:
MW of CO2 = g/mol
28.50 mol CO2 X g CO 2 1 mol CO 2 =1.25 X 103 g CO2

27. Limiting Reagent
Limiting Reagent: the reagent used up first in a reaction. CO(g) + 2 H2(g)  CH3OH(g)
Suppose this reaction takes place with 4 moles of CO and 6 moles of H2 :
Determine the moles of product formed from each reagent:
4 mol CO X (1 mol CH3OH / 1 mol CO) = 4 mol CH3OH
6 moles H2 X (1 mol CH3OH / 2 mol H2) = 3 mol CH3OH
The smaller number of moles of product formed is the limiting reagent:
H2 is the limiting reagent for this reaction.

28. Limiting Reagent
In a reaction g of NH3 are treated with 1142 g of CO2. Which of the two reactants is the limiting reagent and what is the mass of (NH2)2CO formed in the reaction? NH3(g) + CO2(g)  (NH2)2CO(aq) + H2O (l)
Find the moles of (NH2)2CO produced from each reactant:
MM NH3: g/mol
637.3 g NH3 X 1 mol NH g NH 3 X 1 mol (NH 2 ) 2 CO 2 mol NH 3 = mol (NH2)2CO
MM CO2: g/mol
1142 g CO2 X 1 mol CO g CO 2 X 1 mol (NH 2 ) 2 CO 1 mol CO 2 =25.95 mol (NH2)2CO
Determine the limiting reactant:
NH3 is the limiting reagent
Using the limiting reactant, determine the mass of (NH2)2CO formed:
MM (NH2)2CO =60.06 g/mol
18.71 mol (NH2)2CO X g (NH 2 ) 2 CO 1 mol (NH 2 ) 2 CO = 1124 g (NH2)2CO

29. Limiting Reagent
How many grams of CH3Br and LiC4H9 will be needed to carry out the preceding reaction with 10.0 g of CH3OH? CH3OH +CH3Br + LiC4H9  CH3OCH3 + LiBr + C4H10
Find the grams of CH3Br needed with CH3OH:
MM CH3OH = g/mol
10.0 g CH3OH X 1 mol CH 3 OH g CH 3 OH X 1 mol CH 3 Br 1 mol CH 3 OH = mol CH3Br
MM CH3Br=94.93 g /mol
0.312 mol CH3Br X g CH 3 Br 1 mol CH 3 Br = 29.6 g CH3Br
Find the grams of LiC4H9 needed with CH3OH:
MM CH3OH =32.04 g/mol
10.0 g CH3OH X 1 mol CH 3 OH g CH 3 OH X 1 mol LiC 4 H 9 1 mol CH 3 OH = mol LiC4H9
MM LiC4H9 = g/mol
0.312 mol LiC4H9 X g LiC 4 H 9 1 mol LiC 4 H 9 = 20.0 g LiC4H9

30. Reaction Yield % Yield =
Theoretical Yield: the amount of product that would result if all the limiting reagent reacted.
Actual Yield: The amount of product actually obtained from a reaction.
% Yield =
Actual Yield
Theoretical Yield
x 100%

31. Reaction Yield
In a certain industrial operation, 3.54 X 107 g of TiCl4 are reacted with 1.13 X 107 g of Mg. A) Calculate the theoretical yield of Ti in grams. B) Calculate the percent yield if 7.91 X 106 g of Ti are actually obtained. TiCl4(g) + 2 Mg(l)  Ti(s) + 2 MgCl2(l)
Determine the moles of Ti formed from TiCl4:
MM TiCl4= g /mol
3.54 X 107 g TiCl4 X 1 mol TiCl g TiCl 4 X 1 mol Ti 1 mol TiCl 4 = 1.87 X 105 mol Ti
Determine the moles of Ti formed from Mg:
MM Mg=24.3 g /mol
1.13 X 107 g Mg X 1 mol Mg 24.3 g Mg X 1 mol Ti 2 mol Mg = 2.33 X 105 mol Ti
Determine the limiting reagent:
TiCl4 is the limiting reagent
Determine the theoretical yield, mass, of Ti formed:
MM= 47.9 g/mol
1.87 X 105 mol Ti X g Ti 1 mol Ti = 8.96 X 106 g Ti

32. Reaction Yield Use the formula to solve for percent yield:
Theoretical yield = 8.96 X 106 g Ti
Actual Yield = 7.91 X 106 g Ti
% yield = X 10 6 g Ti X g Ti X 100 % = % Ti