1. Chapter 3 Stoichiometry and Chemical Equations

2. Atomic Masses Based on Carbon- 12 as a standard
Carbon- 12 is assigned 12 atomic mass units
Masses can be found using a Mass Spectrometer
This uses a high speed beam of electrons and a magnetic field to deflect ions based on there mass
The amount of deflection of the ion is compared to that of Carbon- 12 to find the mass

3. The Periodic Table of Elements gives the atomic weight of each element
This number is an average of all of the isotopes which we will call the average atomic mass
To find the average atomic mass from percent of each isotope:
Change the percents to decimals
Multiply the decimals by the mass of the isotopes
Add the masses together

4. Solve the following:
An unknown element X has three isotopes and is found to be 84.2% 282X, 10.4% 285X and 5.4% 289X. What is the average atomic mass of element X?
Divide each of the percentages by 100.
84.2/100 = /100 = 0.104
5.4/100 = 0.054

5. Multiply the percentages by the mass of each isotope.
0.842 x 282 amu = amu
0.104 x 285 amu = amu
0.054 x 289 amu = amu
Add the masses together.
amu amu amu = amu
The average mass of element X is amu
Is this answer correct?
Yes! You are do not watch significant figures here because we are taking an average. Therefore we need to keep the whole answer. This is one of the ONLY times you should ignore significant figures.

6. The average atomic mass also allows us to use something know as “counting by weighing”
Solve the following: You now know that the average atomic mass of element X is amu. If you are told to give me a sample of 725 atoms of element X, what would be the quickest way to do it?

7. While you could count out 725 atoms, it is not time efficent or possible when we are talking about atoms, therefore we will count by weighing them:
725 atoms X │ amu = amu
│ 1 atom X
Is this the correct answer?
No! Watch significant figures!
The correct answer is 2.05 x 105 amu

8. The Mole
Defined as the number of carbon atoms in exactly 12 grams of 12C.
Also known as Avogadro’s number.
One mole = x 1023 units of that substance.
One mole of atoms is also defined as a sample of the element equal to the element’s atomic mass expressed in grams.

9. Solve the following:
If you have 537 mg of sulfur, how many sulfur atoms do you have?

10. First you must convert from milligrams to grams.
There are 1000 mg in 1 gram.
Second you must convert from grams to moles.
There are grams of sulfur in 1 mole.
Lastly you must go from moles to atoms.
There are 6.02 x 1023 atoms in a mole.
537 mg │ 1 g │ 1 mole │ 6.02 x 1023 atoms
│1000 mg│ g│ 1 mole
Answer: x 1022 atoms

11. Gram Formula Mass
Also known as the molar mass. This amount is the mass of one mole of any element or compound.
To find the molar mass:
Separate the formula into the elements.
Identify how many atoms of each element there is.
Look up the mass of each element.
Multiply the masses by the number of atoms present for that element.

12. Solve the following:
Please find the gram formula mass for each of the following:
H2O
H2SO4
(NH4)3PO4
Ferric Carbonate
Chloric Acid

13. 2 H and 1 O – 2(1.0079) + 1(15.9994) = 18.015 g 2 H’s, 1 S, 4 O’s
3 N’s, 12 H’s, 1 P, 4 O’s
3(14.01) + 12(1.0079) + 1(30.97) + 4( ) = g
Fe2(CO3)3 = 2 Fe’s, 3 C’s, 9 O’s
2(55.85) + 3(12.01) + 9( ) = g
HClO3 = 1 H, 1 Cl, 3 O’s
1(1.0079) + 1(35.453) + 3( ) = g
For gram formula masses you should keep a few decimal places. These numbers are not measurements so you should not round too soon.

14. Solve the following:
Please calculate the number of atoms of Hydrogen there are present in 250 grams of Ammonium Sulfate.

15. First we must go from grams to moles using the gram formula mass.
There are grams in 1 mole of (NH4)2SO4
Next we must go from moles to molecules.
There are 6.02 x 1023 molecules in 1 mole
Lastly we must go from molecules to atoms.
There are 8 H atoms in every 1 molecule (NH4)2SO4
250 g│1 mole │6.02 x 1023 molecules │ 8 H atoms
│ g│ 1 mole (NH4)2SO4 │1 molecule
Answer: 9.1 x 1024 Hydrogen atoms

16. Percent Composition
While the chemical formula of a compound describes the compound the the number of atoms it contains, we can also describe compounds by the percentage of mass of each element present.
This may be referred to as percent mass or percent weight.

17. To find the Percent Composition:
Find the gram formula mass for the compound.
Then find the mass of each of the elements in the compound.
REMEMBER – You must multiply the mass of the element by the number of atoms of that element present in the compound.
Divide the mass of each of the elements in the compound by the gram formula mass.
Multiply the decimals by 100.

18. Solve the following:
What is the percent composition of Stannic Phosphate?

19. First write the formula for the compound.
Sn3(PO4)4
Next find the gram formula mass.
3(118.7) + 4(30.97) + 16( ) = g
Then find the mass of each of the elements present in the compound.
Sn = g P = g O = g
Lastly divide each of the element masses by the gram formula mass and multiply by 100.
Sn = g x P = g x O = g x 100
g g g
Sn = 48.4% P = 16.8 % O = 34.8 %

20. Determining Formulas
Empirical Formula- a formula for a compound where the subscripts are as reduced as possible.
Molecular Formula- a formula for a compound that fits a specific molecule and does not have to be reduced.

21. Determining the Empirical Formula
Start with the percent composition (if you do not have the percent composition you must first find it).
Change the percents to decimals.
Assume a 100 grams sample and multiply by the decimals by 100.
Divide the masses by the mass of the element.
Divide each of the numbers by the smallest number found in step 4.
Take the numbers found in step 5 and insert them into the chemical formula.

22. Solve the following:
What is the empirical formula of a compound that is 25 % Hydrogen and 75 % Carbon?
Hydrogen Carbon
25/100 = /100 = 0.75
0.25x 100g = 25 g 0.75x 100g = 75 g
25 g / g = 25 H 75 g/ g = 6.25 C
25 H/ = 4 H C/ 6.25 = 1 C
The empirical formula is CH4

23. Solve the following:
What is the empirical formula for a compound that is 58 grams and contains Carbon and 10 grams of Hydrogen?

24. First we need to find the percent composition.
(10g H/ 58 g)x 100 = 17.24% H
(48g C/ 58 g)x 100 = 82.76% C
Next we need to convert from percents to grams.
(17.24% / 100) x 100 g = g H
(82.76% / 100) x 100 g = g C
Then we need to find the number of atoms of each element in the formula.
17.24 g H/ g = 17.1 H
82.76 g C/ g = 6.89 C

25. Divide each number by the smallest number.
17.1 H/ 6.89 = 2.48 or 2.5 H
6.89 C/ 6.89 = 1.00 H
Place the numbers into the chemical formula.
C1H2.5
What is wrong with this formula?
You cannot have decimal subscripts. (There cannot be half of a Hydrogen atom)
C2H5 (Multiply the formula by 2 to get whole numbers)

26. Determining Molecular Formulas:
This will be the formula that matches the mass of your unknown molecule.
If you have the empirical formula:
Find the mass of the empirical formula.
Divide the mass of the the molecular formula by the mass of the empirical formula.
Take the answer and distribute it into the chemical formula.

27. Solve the following:
What is the molecular formula of a compound that is 58 grams and contains Carbon and 10 grams of Hydrogen?

28. We already found the empirical formula for the compound to be C2H5.
Find the mass of the empirical formula.
2(12.01g) + 5(1.0079g) = g
Divide the molecular mass by the empirical mass.
58 g / g = or 2
Take the answer and distribute it into the formula.
(C2H5)2 = C4H10

29. Solve the following:
Analysis of aspirin shows it is 60.0% C, 4.48% H and 35.5% O. If the molecular mass of this substance is u, what is its molecular formula?

30. First we do not need to find the empirical formula, we can go straight to the molecular formula.
How can we do this?
Instead of assuming a 100 gram sample, we will use the mass of the sample that is given to us for the molecular formula.
(60.0 % / 100) x g = g C
(4.48 % / 100) x g = g H
(35.5 % / 100) x g = g O

31. Next we need to divide the masses by the mass of each element.
g C / g = 9.00 C
8.073 g H / g = 8.01 H
g O / g = 4.00 O
Place the numbers into the chemical formula.
C9H8O4
If we check the gram formula mass:
9(12.01) + 8(1.0079) + 4( ) = g

32. Chemical Changes
A chemical change involves a reorganization of the atoms in one or more substances.
Example: Combustion of Methane (CH4)
CH4(g) + O2 → CO2(g) H2O(g)
reactants products
Notice the atoms are rearranged. Bonds have been broken and new bonds formed.

33. Conservation of Atoms
In an ordinary chemical reaction, atoms are neither created nor destroyed.
All atoms present in the reactants must be
accounted for among the products.

34. Balancing Chemical Equations
CH4(g) + O2 (g) → CO2(g) + H2O(g)
Equation is not balanced!
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Equation is balanced!

35. Physical State Symbols
Solid
(s)
Liquid
(l)
Gas
(g)
Dissolved in water
(in aqueous solution)
(aq)

36. Balancing Chemical Equations
Atoms must be conserved in a chemical equation.
The same number of each type of atom must be found among the reactants and products.
The formulas of the compounds in an equation must never be changed in balancing a chemical equation.

37. Steps to Balancing Equations I
1. Determine what reaction is occurring. What are the reactants, the products, and the physical states involved?
2. Write the unbalanced equation that summarizes the reaction described in step 1.
3. Balance the equation by inspection, starting with the most complicated molecule(s). Determine what coefficients are necessary so that the same number of each type of atom appears on both reactant and product sides.

38. Steps to Balancing Equations II
1. Balance metals.
2. Balance polyatomic ions.
3. Balance nonmetals other than H and O.
4. Balance H and O
DO NOT CHANGE ANY FORMULAS!

39. Decomposition of Ammonium Dichromate

40. Decomposition of Ammonium Dichromate
When ignited, solid ammonium dichromate decomposes into chromium (III) oxide, nitrogen gas, and water vapor.
The formula for the reactant is (NH4)2Cr2O7(s).
The formula for chromium (III) oxide can be determined by recognizing the Roman numeral III means Cr3+ ions. Since each oxide ion is 2-, the formula is Cr2O3.
The unbalanced equation is:
(NH4)2Cr2O7(s)→ Cr2O3(s) + N2(g) + H2O(g)

41. (NH4)2Cr2O7(s)→ Cr2O3(s) + N2(g) + H2O(g)
Note that the nitrogen and chromium are balanced but hydrogen and oxygen are not.
A coefficient of 4 for H2O balances the hydrogen atoms. In balancing the hydrogen atoms, the oxygen atoms are also balanced.
(NH4)2Cr2O7(s)→ Cr2O3(s) + N2(g) + 4H2O(g)
The equation is balanced.

42. Solve the following:
At 1000oC, ammonia gas, NH3(g), reacts with oxygen gas to form gaseous nitric oxide, NO(g), and water vapor.

43. NH3(g) + O2(g) → NO(g) + H2O(g)
The unbalanced equation for the reaction is:
NH3(g) + O2(g) → NO(g) + H2O(g)
A coefficient of 2 for NH3 and a coefficient of 3 for H2O give six atoms of hydrogen on both sides:
2NH3(g) + O2(g) → NO(g) + 3H2O(g)
The nitrogen can be balanced with a coefficient of 2 for NO:
2NH3(g) + O2(g) → 2NO(g) + 3H2O(g)
There are 2 atoms of left and 5 atoms on the right. It can be balanced with a coefficient of 5/2 for O2:
2NH3(g) + 5/2O2(g) → 2NO(g) + 3H2O(g)

44. Can we leave the the equation like this?
No! We can NEVER have fractions as coefficents!
Multiply the equation by 2 to get whole-numbers coefficients.
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

45. Balancing Equations Practice
Cu(s) + AgNO3(aq) → Ag(s) + Cu(NO3)2 (aq)
AgNO3(aq) + H2SO4(aq) → Ag2SO4(s) + HNO3(aq)
Solid iron (III) sulfide reacts with gaseous hydrogen chloride to form solid iron (III) chloride and hydrogen sulfide gas.

46. Cu(s) + 2AgNO3(aq) → 2Ag(s) + Cu(NO3)2 (aq)
2AgNO3(aq) + H2SO4(aq) → Ag2SO4(s) + 2HNO3(aq)
Fe2S3 (s) + 6HCl (g) → 2FeCl3 (s) + 3H2S (g)

47. Stoichiometric Calculations: Amounts of Reactants and Products
Remember chemical equations represent numbers of molecules, not masses of molecules.
Counting is always done by weighing.
Remember, before doing any calculations involving a chemical reaction, be sure the equation for the reaction is balanced.

48. C3H5(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Write the equation:
C3H5(g) + O2(g) → CO2(g) + H2O(g)
The balanced equation:
C3H5(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
This equation means that 1 mole of C3H5 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O.

49. Solve the following:
What mass of O2 will react with 96.1 grams of propane (C3H8)?

50. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
1. Find the number of moles of the propane.
96.1 g C3H8| 1 mol C3H8| = 2.18 mol C3H8
44.1 g C3H5
2. Use the balanced equation to set up a mole ratio to calculate the number of moles of the desired reactant or product.
2.18 mol C3H8| 5 mol O2| = 10.9 mol O2
1 mol C3H 5
3. Convert from moles to units required by the problem.
10.9 mol O2| 32 g O2| = 349 g O2
1 mole O2

51. Summary of Steps in Calculating Quantities of Reactants and Products in Chemical Reactions.
1. Balance the equation.
2. Convert mass to moles.
3. Set up mole ratios.
4. Use mole ratios to calculate moles of desired substance.
5. Convert moles to units asked for in the problem.

52. Practice Problems.
Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment by forming solid lithium carbonate and liquid water. What mass of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide?
Answer: g of CO2

53. First write the equation: LiOH(s) + CO2 (g)  Li2CO3 (s) + H2O (g)
Balance the equation:
2LiOH(s) + CO2 (g)  Li2CO3 (s) + H2O (g)
Work out the stoichiometry:
1.00 kg LiOH| 1000 g | 1 mole LiOH = | 1 kg | g LiOH moles
moles LiOH | 1 mole CO2 | 44 gCO2 =
| 2 mole LiOH | 1 mole CO2
Answer: grams or 919 grams CO2

54. NaHCO3 (s) + HCl (aq) → NaCl (aq) + H2O (l) + CO2(aq)
Baking soda (NaHCO3) is often used as an antacid. It neutralizes excess hydrochloric acid secreted by the stomach:
NaHCO3 (s) + HCl (aq) → NaCl (aq) + H2O (l) + CO2(aq)
Milk of magnesia, which is an aqueous suspension of Mg(OH)2, is also used as an antacid:
Mg(OH)2(aq) + 2HCl(aq)→2H2O(l) + MgCl2(aq)
Which is the more effective antacid per gram, NaHCO3, or Mg(OH)2?

55. The Milk of magnesia is a better antacid. How do we know?
By looking at the balanced equations we can see that for every one mole of baking soda we will neutralize one mole of hydrochloric acid.
For every one mole of magnesia we will neutralize 2 moles of hydrochloric acid.
Therefore the milk of magnesia is twice as effective because it will neutralize twice as much hydrochloric acid.

56. Calculations Involving a Limiting Reactant
Limiting reaction problems are where the quantities of two (rarely more than two) reactants are stated.
The limiting reactant controls the amount of product that can form.
The problem allows the assumption of 100% reaction for one of the substances.
It is then necessary to determine which of the two would be used up ... the “limiting” reactant.
The limiting reactant is then used with the “mole ratio” to compute the amount of product produced.

57. Solving Problems with a Limiting Reactant
Write the balanced equation for the reaction.
For each of the reactants, convert to moles of product (it must be the same product).
The reactant that gives you the least moles of product is the limiting reactant.
Use the limiting reactant to find the amount of product(s) that the problem asks for.

58. Solve the following:
If you burn 225 grams of Butane (C4H10) in a container with only 225 L of O2, how many grams of each product will be produced?

59. Write the balanced equation:
2C4H10 (l) + 13O2 (g)  8CO2 (g) + 10H2O (g)
Find the moles of one of the products for each of the reactants.
225 g C4H10 | 1 mole C4H10 | 8 mole CO2 = mol CO2
| 58 g C4H10 | 2 mole C4H10
225 L O2 | 1 mole O2 | 8 mole CO2 = 6.18 mol CO2
| 22.4 L | 13 mole O2
Therefore O2 is the limiting reactant.

60. Use the limiting reactant to solve for both products.
6.18 mol CO2 | 44 g CO2 = 272 g CO2
|1 mole CO2
225 L O2 | 1 mole O2 | 10 mole H2O | 18 g H2O = 139 g H2O
| 22.4 L O2 | 13 mole O2 |1 mole H2O

61. Percent Yield
This will compare the actual yield you found in an experiment (in the lab) to the theoretical value that you calculated.
Actual Yield (experimental value) x 100
Theoretical Yield (calculated value)

62. End of Chapter!!