1. Calculations with Chemical Formulas and Equations

2. Mass-Mole Relationship
Topics Featured
Mass-Mole Relationship
Percent Composition
Empirical Formulas
Molecular Formulas
Stoichiometry:
Limiting Reagents
Theoretical Yield
Actual Yield
Percent Yield
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3. Formula Masses
The mass of a carbon atom allows one to determine how many carbon atoms are present in one diamond.
1 x 1021 atoms
© Foto Kormann/zefa/Corbis
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4. Formula Masses
Fig. 6.1
Oranges may be bought in units of mass or units of amount.
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5. Formula Masses
Fig. 6.2
A basic process in chemical laboratory work is determining the mass of a substance.
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6. Mass and Moles of a Substance
Chemistry requires a method for determining the numbers of molecules in a given mass of a substance.
This allows the chemist to carry out “recipes” for compounds based on the relative numbers of atoms involved.
The calculation involving the quantities of reactants and products in a chemical equation is called stoichiometry.
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7. Molecular Weight and Formula Weight
The molecular weight of a substance is the sum of the atomic weights of all the atoms in a molecule of the substance.
For, example, a molecule of H2O contains 2 hydrogen atoms (1.0 amu x 2 atoms), and oxygen atom (16.0 amu), giving a molecular weight of 18.0 amu.
One molecule of water weighs:
2 (1 amu) + 16 amu = 18 amu
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8. Molecular Weight and Formula Weight
The formula weight is the sum of the atomic weights of all the atoms in one formula unit of the compound.
For example, one formula unit of NaCl contains 1 sodium atom (23.0 amu) and one chlorine atom (35.5 amu), giving a formula weight of 58.5 amu.
One molecule of NaCl weighs:
23 amu amu = 58.5 amu
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9. Example
Calculate the mass of a sample that contains 23 nitrogen atoms.
1 N atom weighs 14 amu’s
Thus,
23 x 14 amu = 322 amu
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10. Example #2
How many sodium atoms are there in amu’s?
Since 1 sodium atom weighs amu
amu
22.99 amu/Na atom
= Na atoms
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11. Keeping track of the number of atoms and/or molecules that one mixes together allows one to determine exactly how many molecules of a substance may be produced.
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12. Calculations: Formula Masses, Moles, and Chemical Equations cont’d
Edgar Fahs Smith Collection, University of Pennsylvania
Fig. 6.4
Amadeo Avogadro was the first scientist to distinguish between atoms and molecules.
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13. Atomic and Molecular Masses
Atomic mass - the average mass of an atom
C = amu
Molecular mass - the average mass of a molecule
CHCl3 = 1 C + 1 H + 3 Cl
= [1(12) + 1(1) + 3(35.5)] amu
= amu
However, 1 amu = 1.66 x g ; pg 206
or, x 1023 amu = 1 g
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14. 1 mole of NaCl weighs 58.5 grams
The Mole
The number of carbon atoms in exactly 12 grams of pure 12C.
1 mole of anything = ´ 1023 units of that thing
Thus, x 1023 amu = 1.00 gram
1 mole of H2O weighs 18.0 grams
and
1 mole of NaCl weighs 58.5 grams
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15. Avogadro’s number equals 6.022 ´ 1023 units (denoted as “NA”)

16. Molar Mass
A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound.
CO2 = amu’s x
= grams/mole
6.022 x 1023 /mol
6.022 x 1023 amu/g
Essentially:
formula mass in grams = mass of 1 mole of substance
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17. Mass and Moles of a Substance
The Mole Concept
A mole is defined as the number of atoms in exactly 12 grams of pure carbon–12. (See Figure 3.2)
The number of atoms in a 12-gram sample of carbon–12 is called Avogadro’s number (NA).
The value of Avogadro’s number is 6.02 x 1023.
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18. Mass and Moles of a Substance
The molar mass of a substance is the mass of one mole of a substance.
is numerically equal to the formula weight in atomic mass units.
one mole of any element weighs its atomic mass in grams.
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19. Mass and Moles of a Substance
Mole calculations
Converting the number of moles of substance to its mass, and vice versa
This is known as the mass-mole relationship.
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20. Mass and Moles of a Substance
Mole calculations
Suppose we have grams of iron (Fe). The atomic weight of iron is 55.8 g/mol. How many moles of iron does this represent?
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21. Mass and Moles of a Substance
Mole calculations
Conversely, suppose we have 5.75 moles of magnesium (atomic wt. = 24.3 g/mol). What is its mass?
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22. Mass and Moles of a Substance
Mole calculations
This same method applies to compounds. Suppose we have grams of H2O (molecular weight = 18.0 g/mol). How many moles does this represent?
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23. Mass and Moles of a Substance
Mole calculations
Conversely, suppose we have 3.25 moles of glucose, C6H12O6 (molecular wt. = g/mol). What is its mass?
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24. Examples of Calculations using Moles
What is the mass (in grams) of a molecule of formic acid? H C O O H
HCOOH
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25. Calculate the mass of a formic acid molecule.
1 C = 12 amu
2 H = 2 amu
2 O = 32 amu
46 amu = 46 g/mol
Note: This is for 1 mole.
What’s the mass of 1 molecule: (hint: use NA)
_______46 g/mol
6.02 x 1023molecules/mol H C O O H
= 7.6 x10-23 g/molecule
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26. Mass and Moles and Number of Molecules or Atoms
The number of molecules or atoms in a sample is related to the moles of the substance:
Suppose we have a 3.46-g sample of hydrogen chloride, HCl. How many molecules of HCl does this represent?
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27. Mole Calculations Example
How many moles are there in 50 grams of sulfuric acid?
Formula of sulfuric acid: H2SO4
mol. wt. of H2SO4
= (16) = 98 g/mol
# mol = 50 g / (98 g/mol)
= 0.51 mol of H2SO4
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28. Moles to Grams How many grams are in 0.025 mol of sodium hydroxide?
Formula = NaOH
mol. wt. = = 40 g/mol
= (0.025 mol)(40 g/mol)
= 1 g of NaOH
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29. Determining Chemical Formulas
The percent composition of a compound is the mass percentage of each element in the compound.
We define the mass percentage of “A” as the parts of “A” per hundred parts of the total, by mass. That is,
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30. Mass Percentages from Formulas
Let’s calculate the percent composition of butane, C4H10.
First, we need the molecular mass of C4H10.
Now, we can calculate the percents.
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31. Mass Percent using “grams”
Calculate the % of each element in Fe2O3 by mass.
Fe = 58.5 grams/mol; O = 16.0 grams/mol
Formula weight =
2 (58.5) + 3 (16) = 117 g/mol + 48 g/mol
= 165 g/mol
% Fe =
117 g/mol
165g/mol
48 g/mol
165g/mol
%O =
(100%) = 71%
(100%) = 29%
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32. Using Mass Percent
How would one calculate the amount of an element is present in a given mass of sample?
In Fe2O3, iron is 71% by mass in any sample.
How many grams of iron are present in a 258 g sample of iron (III) oxide?
Mass of Fe = (mass of sample) x (percent of Fe/100)
= (258g) x (0.71)
= 183 g of Fe
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33. Determining Chemical Formulas
Determining the formula of a compound from the percent composition.
The percent composition of a compound leads directly to its empirical formula.
An empirical formula (or simplest formula) for a compound is the formula of the substance written with the smallest integer (whole number) subscripts.
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34. Determining Chemical Formulas
Determining the empirical formula from the percent composition.
Benzoic acid is a white, crystalline powder used as a food preservative. The compound contains 68.8% C, 5.0% H, and 26.2% O by mass. What is its empirical formula?
In other words, give the smallest whole-number ratio of the subscripts in the formula
Cx HyOz
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35. Determining Chemical Formulas
Determining the empirical formula from the percent composition.
For the purposes of this calculation, we will assume we have grams of benzoic acid.
Then the mass of each element equals the numerical value of the percentage.
Since x, y, and z in our formula represent mole-mole ratios, we must first convert these masses to moles.
Cx HyOz
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36. Determining Chemical Formulas
Determining the empirical formula from the percent composition.
Our grams of benzoic acid would contain:
This isn’t quite a whole number ratio, but if we divide each number by the smallest of the three, a better ratio might emerge.
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37. Determining Chemical Formulas
Determining the empirical formula from the percent composition.
Our grams of benzoic acid would contain:
now it’s not too difficult to See that the smallest whole number ratio is 7:6:2.
The empirical formula is C7H6O2 .
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38. Determining Chemical Formulas
Determining the molecular formula from the empirical formula.
An empirical formula gives only the smallest whole-number ratio of atoms in a formula.
The molecular formula should be a multiple of the empirical formula (since both have the same percent composition).
To determine the molecular formula, we must know the molecular weight of the compound.
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39. Determining Chemical Formulas
Determining the molecular formula from the empirical formula.
For example, suppose the empirical formula of a compound is CH2O and its molecular weight is 60.0 g/mol.
The molar weight of the empirical formula (the empirical weight) is only 30.0 g/mol.
The ratio of the formula mass to empirical mass determines the whole number multiple.
molecular
empirical
C2H4O2
whole # = = n
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40. Stoichiometry: Quantitative Relations in Chemical Reactions
Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction.
It is based on the balanced chemical equation and on the relationship between mass and moles.
Stoichiometric calculations involve mole-to-mole and mass-mole relationships from balanced equations.
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41. Molar Interpretation of a Chemical Equation
The balanced chemical equation represents molecule-to-molecule, or “mole-to-mole” relationships.
For example, the Haber process for producing ammonia involves the reaction of hydrogen and nitrogen.
reactants
products
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42. Molar Interpretation of a Chemical Equation
This balanced chemical equation shows that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
1 molecule N molecules H molecules NH3
Because moles can be converted to mass, one can determine the mass of reactants needed, and mass of products produced.
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43. Molar Interpretation of a Chemical Equation
Suppose we wished to determine the number of moles of NH3 we could obtain from 4.8 mol H2.
Because the coefficients in the balanced equation represent mole-to-mole ratios, the calculation is simple.
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44. Mass Relationships in Chemical Equations
Amounts of substances in a chemical reaction by mass.
How many grams of HCl are required to react with 5.00 grams manganese (IV) oxide according to this equation?
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45. Mass Relationships in Chemical Equations
First, you write what is given (5.00 g MnO2) and convert this to moles.
Then convert to moles of what is desired.(mol HCl)
Finally, you convert this to mass (g HCl)
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46. Example 2
Solid lithium hydroxide is used by space vehicles to remove exhaled carbon dioxide from the living environment.
The products of the reaction are solid lithium carbonate and liquid water.
What mass of gaseous carbon dioxide can 1 kg of lithium hydroxide absorb?
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47. Example 2 Write down the equation for the reaction and balance it:
LiOH + CO Li2CO3 + H2O
Next, convert the initial mass of LiOH to moles
1 kg = 41.8 mol of LiOH
Use the mole ratio of CO2 to LiOH to calculate the moles of CO2 needed.
41.8 mol = 20.9 mol CO2 2
1mol
24g/mol
1 mol CO2
2 mol LiOH
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48. Example 2 Finally, Convert moles of CO2 to grams of CO2
20.9 mol CO = 920 g of CO2
Thus, 1 kg of LiOH can absorb 920 g of CO2
44 g CO2
1 mol CO2
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49. Limiting Reagent
The limiting reactant (or limiting reagent) is the reactant that is entirely consumed when the reaction goes to completion. (See Figure 3.14)
The limiting reagent determines how much product can be produced.
For example, bicycles require one frame and two wheels.
If you have 20 wheels but only 5 frames, how many bicycles can be made?
How many wheels are left?
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50. Limiting Reagent
Zinc metal reacts with hydrochloric acid by the following reaction.
If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, how many moles of H2 are produced?
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51. Limiting Reagent
From each reactant one can calculate the amount of product to expect.
The reactant that gives the smaller amount is the limiting reagent.
Since HCl is the limiting reagent, the theoretical yield of H2 is 0.26 mol. (How many grams of HCl?)
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52. Theoretical and Percent Yield
The theoretical yield refers to the maximum amount of product can be produced from a given amounts of reactants.
The percentage yield is the actual yield (experimental amount) divided by the theoretical yield (ie., the calculated yield) x 100%.
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53. Theoretical and Percent Yield
In the previous example the theoretical yield of H2 was 0.26 mol (or 0.52 g) H2.
An experiment was performed, and the amount of H2 obtained was only 0.22 g H2.
Calculate the %Yield of H2 = (A.Y./T.Y.) x 100%
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54. Operational Skills Calculating the formula weight from a formula.
Calculating the mass of an atom or molecule.
Converting moles of substance to grams and vice versa.
Calculating the number of molecules in a given mass.
Calculating the percentage composition from the formula.
Calculating the mass of an element in a given mass of compound.
Calculating the percentages C and H by combustion.
Determining the empirical formula from percentage composition.
Determining the true molecular formula.
Relating quantities in a chemical equation.
Calculations with a limiting reagent, actual yield, theoretical yield and percent yield.
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55. End of Chapter 3 Practice Problems
6, 8, 10, 12 (balance them), 14, 18, 20, 27, 28, 30, 34, 40, 46 (balance them), 51, 52, 56, 62, 63, 83, 97.
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56. Figure 3. 2: One mole each of various substances
Figure 3.2: One mole each of various substances. Photo courtesy of American Color.
Return to Slide 5
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57. Figure 3.14: Limiting reactant analogy using cheese sandwiches.
Return to Slide 28
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