Chapter 3 Page 1

CHAPTER THREE: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS

Part One: Mass and Moles of Substance A. Molecular Mass and Formula Mass. (Section 3.1)

1. Just as we can talk about mass of one atom of an element as its Atomic Mass (or

Atomic Weight), we can also extend that to mass of one molecule of a substance, the Molecular Mass (or Molecular Weight).

Molecular Mass = sum of the Atomic Masses in a molecule

Example: Molecular Mass of water is 18.0 amu

2. If the substance is ionic, we talk about its Formula Mass (or Formula Weight).

Formula Mass = sum of the Atomic Masses in its formula

Example: Formula Mass of CaF2 is 78.1 amu

B. The Mole Concept. (Sections 3.2) 1. Atoms combine together most fundamentally by number.

(Their fixed mass-combining ratios are simply a byproduct of this)

When atoms interact, they COUNT each other, they don’t WEIGH each other. 2. Problem: a laboratory sized sample of matter contains an enormously large number of

atoms or molecules. 1 Liter of air contains ~ 3 x 1022 molecules 3. It is inconvenient to handle these large numbers in our chemical calculations, and yet

we must continually keep track of numbers of molecules. 4. Solution: we have adopted the mole concept.

a. Just like a baker uses the dozen to describe 12 of something, or a hardware store

uses a gross to refer to 144 of something.

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b. Better yet the hardware store weighs out nails rather than counting them. Chemists do the same thing, weigh samples but know how to determine the number of atoms or molecules in that mass.

c. Chemist uses the mole to refer to number of atoms in one gram of hydrogen.

(original definition) Now many atoms is that? Turns out to be: 6.02 x 1023 of them. Since Carbon atoms are 12 times the mass of Hydrogen atoms, 12 grams of

carbon would be one mole of Carbon atoms. 16 grams of Oxygen would be one mole of Oxygen atoms, and so forth. d. Modern definition of the mole is based on Carbon and not Hydrogen, because

Carbon is easier to handle. 1 mole = 6.02 x 1023 particles of anything countable.

D. The Mole. (Section 3.2) 1. One mole = amount of any substance that contains as many particles of that substance

as there are atoms of C in 12 g pure Carbon-12. 1 mole = 6.0221367 x 1023 particles = Avogadro’s number

One mole of Fe is 6.022 x 1023 atoms of Fe

One mole of H2O is 6.022 x 1023 molecules of H2O

One mole of NaCl is 6.022 x 1023 formula units of NaCl 2. Abbreviation for mole is mol. 3. Why is mole defined this way? Because now:

1 mole of element X = Atomic Mass of X in grams Examples: 1 mole of C = 12.011 grams C 1 mole of Na = 22.99 grams Na

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4. Atomic Mass (Atomic Weight) of an element = number of grams in 1 mole of that element (55.85 = number of grams in mol of Fe):

26

Fe

55.85

Abbreviate Atomic Mass as AM

5. So now:

6.022 x 1023 atoms of Fe = 1 mol Fe = 55.85g Fe ÷ 6.022 x 1023 x AM of X

atoms of X moles of X grams of X x 6.022 x 1023 ÷ AM of X 6. Problems:

a. How many Na atoms in 10.00g Na?

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10.00gNa × 1mol Na22.99g Na

×

6.022 ×1023atoms1mol

= 2.62 ×1023 Na atoms

b. A reaction used up 5.00g Ni. How many moles Ni were used up?

€

5.00gNi× 1 mol Ni58.69g Ni

= 8.52 ×10−2mol Ni

= 0.0852mol Ni

c. Which weighs more, a mole of C or a mole of H? By what factor? mol of C weighs ≈ 12 times a mol of H

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d. Calculate the mass of a Hydrogen atom in grams.

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1atom H ×1 mol H

6.022 ×1023atoms H

×

1.008g Hmol H

=1.674 ×10−24 grams

B. Molar Mass. (Section 3.2)

1. Molar Mass of a substance = mass of one mole of particles of that substance.

2. Molar Mass of a substance = the number of grams of that substance equal to the sum

of Atomic Masses of elements in the formula. 3. Example: Molar Mass of CaF2.

= 40.078 + 2 x 19.00 = 78.08 grams/mol 4. Example: Molar Mass of NaOH.

= 23.00 + 16.00 + 1.01 = 40.01 grams/mol Therefore: 40.01g NaOH = 1 mol NaOH = 6.022 x 1023 NaOH units = 6.022 x 1023 Na+ ion plus 6.022 x 1023 OH- ions

5. Example: Molar Mass of H2O.

= 2 x 1.008 + 16.00 = 18.02 grams/mol Thus, 18.02g H2O = 1 mole H2O = 6.022 x 1023 H2O molecules. And contains 6.022 x 1023 oxygen atoms 12.04 x 1023 hydrogen atoms 6. Example: Molar Mass of H2

= 2 x 1.008 = 2.016 grams/mol Thus, 2.016 g H2 = 1 mole H2 = 6.022 x 1023 H2 molecules in which there are 12.04 x

1023 H atoms present.

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7. For a molecular substance S (e.g. S = H2, H2O, CH4, etc...) ÷ 6.022 x 1023 x MolarMass

molecules of S moles of S grams of S

x 6.022 x 1023 ÷ MolarMass 8. How many molecules of ozone in 24.0g ozone?

€

24.0g O3 ×1 mol O3

48.00g O3

×

6.022 ×1023 molecules O3

1 mole O3

= 3.01×1023molecules O3

9. How many H atoms are present in 48.0g methane (CH4)?

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48.0g CH4 ×1 mol CH4

16.04g CH4

×

6.022 ×1023 molecules CH4

1 mole CH4

=1.80 ×1024 molecules CH4

4 H for every molecule CH4, so (times 4) = 7.21 x 1024 H atoms. 10. How many moles of NaF are in 147 grams of it? 11. I want to measure out 2.5 moles of NiSO4. How many grams do I measure out?

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Part Two: Determining Chemical Formulas A. Percent Composition from Compound Formulas. (Section 3.3)

1. If formula of compound is known, its percent composition by mass of each element

can be found, and vice versa. ÷ Atomic Masses (AM)

compound percent composition formula in terms of elements

x Atomic Masses (AM) 2. First we’ll go from formula to percent composition. Example: Calculate the percent composition of the elements in ammonium nitrate: given formula = NH4NO3 Formula Mass (FM) = 80.0 g/mol

3 elements present; what is % N, % H, % O?

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% N = mass of Nmass of compound

×100%

=2 × AM of N

FM of compound×100%

=2 ×14.00 g/mol

80.0 g/mol×100% = 35.0% N

€

% H =4 ×AM of H

FM of compound×100%

=4 ×1.008 g/mol

80.0 g/mol×100% = 5.04% H

€

% O =3×AM of O

FM of compound×100%

=3×16.00 g/mol

80.0 g/mol×100% = 60.0% O

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3. Note that the sum of percentages equals 100%. 35.0 % N + 5.04 % H + 60.0 % O = 100%

This makes a good check.

B. Chemical Formula from Percent Composition. 1. Knowing the Percent Composition from experiments, we can determine the Chemical

Formula.

2. Example: A certain acidic substance has the following composition by mass: 2.45 % H, 39.06 % S, and 58.48 % O. Determine the compound formula. a. We want the mole ratio of the elements present, not their masses. b. Convert masses to moles by dividing by atomic mass (AM). c. For convenience, work with 100 g sample of compound and find moles of each

element:

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2.45g H ×1 mol H1.008g H

= 2.43 mol H

39.0g S× 1 mol S32.06g S

=1.218 mol S

58.48g O ×1 mol O16.00g O

= 3.655 mol O

d. Now take this mole ratio and convert it to simplest whole-number ratio; easiest to

first try dividing all 3 numbers by the smallest:

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2.43 mol H1.218

=1.99 mol H ≈ 2H

1.218 mol S1.218

=1 mol S ≈1S

3.655 mol O1.218

= 3.001 mol O ≈ 3O

formula = H2SO3 (sulfurous acid)

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3. This method produces only the simplest or empirical formula of a compound. Illustration:

4. Example: Analysis shows that a compound of Hydrogen and Oxygen is 5.93 % H and

94.07 % O. Calculate it’s formula:

€

5.93g H ×1 mol H1.008g H

= 5.883 mol H

94.07g O ×1 mol O16.00g O

= 5.879 mol O

Now divide through by smallest number:

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5.883 mol H5.879

=1.001 mol H

5.879 mol O5.879

=1 mol O

formula = HO

See, this predicts simplest or empirical formula. Actual chemical formula is H2O2,

hydrogen peroxide. 5. Percent compositions alone give only empirical formula for a compound. 6. This is generally all that is needed for ionic compounds. 7. For molecular substances we also need to know the Molar Mass. 8. Example: In the preceding problem, if we were also given that the Molar Mass of the

compound is 34.0 amu, then we could find the true molecular formula: empirical formula is HO (Formula Mass = 17.0 g/mol) convert to formula H2O2 with Molar mass = 34.0 g/mol

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9. Summary:

percent ÷ Atomic Mass mole ÷ smallest whole- composition ratios of number elements mole ratio empirical true Molar Mass true formula emp Formula Mass molecular formula

10. One more problem: A hydrocarbon has a percent composition of 7.744 % H and 92.27 % C and a MW =

78.1. What is it’s molecular formula? a. First get empirical formula:

€

7.744g H ×1 mol H1.008g H

= 7.683 mol H

92.27g C ×1 mol C12.01g C

= 7.683 mol C

So empirical formula = CH

b. Convert to molecular formula using Molar Mass. Formula Mass (FM) of CH = 13.02 amu 1:1 ratio x 78.1/13.02 = 6:6

molecular formula = C6H6 (benzene) 11. Molecular Formula from Combustion Data: A 2.064 gram sample of an organic compound containing C, H and O is analyzed by

combustion analysis and 5.338 grams of CO2 and 1.093 grams of H2O are produced. In a separate experiment, the molar mass is found to be 136.2 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

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a. All C ends up producing CO2 so every C atom gives rise to a CO2 molecule or every mol of C gives rise to a mole of CO2.

b. Same for H and H2O, except every mol of H gives rise to only half a mole of

H2O.

5.338 g CO2

€

×1 mol CO2

44 g CO2

= 0.1213 mol CO2 = 0.1213 mol C

1.093 g H2O

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×1 mol H2O18 g H2O

= 0.0607 mol H2O = 0.1214 mol H

Therefore: C and H are present in 1:1 ratio by atoms or moles. c. Now what about oxygen? Must find first what mass oxygen in original sample

by subtracting out masses of C and H.

0.1213 mol C × 12 g/mol = 1.457 g C

0.1214 mol H × 1.008 g/mol = 0.1224 g H

2.064 – 1.457 - 0.1224 g = 0.4846 g oxygen

0.4846 g O

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×1 mol O16 g O

= 0.03028 mol O

mole ratio: 0.1213 mol C; 0.1214 mol H; 0.03028 mol O

divide by smallest 4 : 4 : 1 simplest formula: C4H4O Formula Mass = 68 True Molar Mass = 136 (2x as much) So molecular formula is C8H8O2

12. The Law of Multiple Proportions (revisited) = when two elements, A and B, form

more than one compound together, the amount of B combining with a given amount of A in each compound forms a whole number ratio:

amt B combining w/A in cmpd 1 = ratio of whole numbers amt B combining w/A in cmpd 2

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13. This supplements the Law of Definite Proportions. 14. Example: Remember finding two oxides of hydrogen, with different mass combining ratios: 1.0 g H for every 8.0 g O in one compound 1.0 g H for every 16.0 g O in the other Confirms the Law of Multiple Proportions: for every gram present of hydrogen: 8.0 g O in H2O = 1 (whole number ratio) 16.0 g O in H2O2 2

Part Three: Stoichiometry A. What is Stoichiometry? (Sec 3.6-3.8)

1. Stoichiometry = description of quantitative relationships:

a. among elements within a compound. b. among substances within a chemical reaction.

2. Concerned with:

a. mole ratios which combine together. (balancing chemical equations) b. mass ratios which combine together.

B. Calculations Based on Chemical Equations. (Section 3.6)

1. Here we use mole interpretation of the balanced equation. 2. Balance exists on two levels, mole level and mass level.

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CH4 + 2 O2 → CO2 + 2 H2O mole 1 mol 2 mol 1 mol 2 mol level x MM x MM x MM x MM CH4 O2 CO2 H2O 16 g/mol 32 g/mol 44 g/mol 18 g/ mol mass 16 g 64 g 44 g 36 g level 3. Tells us that 16 g CH4 requires exactly 64 g O2 to completely burn, or a 1:4 ratio by

mass. 4. Problem: What mass of oxygen is required to react completely with 100 g methane?

How many grams CO2 will form?

CH4 + 2 O2 → CO2 + 2 H2O 5. Schematic:

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6. Another problem: How many grams of CH4 and O2 would produce 54 g of H2O?

C. The Limiting Reagent Concept. (Section 3.8) 1. Limiting reagent = reactant that is used up first when reactants are not mixed in their

exact stoichiometric ratios. 2. Example: initially have 4 slices of ham, 6 slices of bread, and wish to make

sandwiches.

2 slices bread + 1 slice ham 1 sandwich a. How many sandwiches can you make? Only 3, until the bread runs out. b. What is limiting reagent? Bread. c. How much ham is left over? 1 slice. d. Which reagent is present in excess? Ham.

3. Problem of Limiting Reagent: How many grams of MgO could be formed by the reaction of 30 g Mg with 16 g O2 in

the following reaction?

2 Mg + O2 → 2 MgO Strategy:

a. Recognize it as a limiting reagent problem. How? b. Convert to moles of reactants first.

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c. Pick one of the reactants. d. Calculate how many moles of the other reactant would be required to completely

use up the one you picked. e. Not enough of the other one? Other must be the limiting reagent. f. The moles of limiting reagent determines how much product could form.

How many grams of MgO could be formed by the reaction of 30 g Mg with 16 g O2 in the following reaction?

2 Mg + O2 → 2 MgO Let’s work it:

O2 is limiting!!

4. Another example problem: 7.0 grams of nitrogen and 3.0 grams of hydrogen are

mixed, and the following reaction takes place. What is the limiting reagent, and how many grams of ammonia would be produced if the reaction goes to completion? N2 + 3 H2 2 NH3

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D. Percent Yields. (Section 3.8) 1. Definition: Percent yield =

2. Theoretical yield is calculated by assuming the reaction goes to completion (all of

limiting reagent used up), no side reactions, and nothing lost in handling. 3. Actual yield is amount of desired product we actually isolated in our lab experiment. 4. Example problem: In preparing Ni(OH)2, 5.0 g of product are isolated after combining 10.0 g NiCl2 and

10.0 g NaOH. What is % yield of Ni(OH)2? % yield = = 70%

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NOTES: