Topic 3 Mass Relations and - PowerPoint Presentation

Slide1

Topic 3

Mass Relations and

StoichiometrySlide2

Mass and Moles of a Substance

Chemistry requires a method for determining the numbers of molecules in a given mass of a substance

which has lead to the development of the mole (quantity of substance to be discussed later).This allows the chemist to carry out “recipes” for compounds based on the relative numbers of atoms involved.stoichiometry - the calculation involving the quantities of reactants and products in a chemical equation.

2Slide3

Molecular Weight and Formula Weight

The

molecular weight (for molecular substances) 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 (at 1.01 amu each) and 1 oxygen atom (16.00 amu), giving a molecular weight of 18.02 amu.

H: 2 x 1.01 amu = 2.02 amu

O: 1 x 16.00 amu =

16.00 amu

18.02 amu

(mass of 1 molecule of H

2O)

3Slide4

Molecular Weight and Formula Weight

The

formula weight of a substance is the sum of the atomic weights of all the atoms in one formula unit of the compound, whether molecular or not.

Na: 1 x 22.99 amu = 22.99 amu

Cl: 1 x 35.45 amu =

35.45 amu

58.44 amu

(mass of 1 FU NaCl)

4

iron (III)

sulfate,

Fe

2

(SO

4

)

3

Fe: 2 x 55.85 amu = 111.70 amu

S: 3 x 32.07 amu = 96.21 amu

O: 12 x 16.00 amu =

192.00 amu

399.91 amu (mass of 1 FU Fe2(SO4)3)

Glucose, C6 H12 O6

C: 6 x 12.01 amu = 72.06 amu

H: 12 x 1.01 amu = 12.12 amu

O: 6 x 16.00 amu = 96.00 amu

180.18 amu (mass of 1 molecule)

one formula unit (FU) of NaCl:

Note: molecular

wt

and formula

wt

are calculated the same way.Slide5

Mass and Moles of a Substance

The Mole Concept

A mole is defined as the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 grams of carbon–12. 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

6.022 x 10

23.1 mole = 6.022 x 1023 X X = ions, particles, atoms, molecules, items, etc.1 mole Na2CO3  6.022 x 1023 FU Na2CO31 mole CO2  6.022 x 1023 molecules CO2

5Slide6

Mass and Moles of a Substance

The

molar mass (Mm) of a substance is the mass of one mole (6.022 x 1023 FU or molecules) of a substance. This is the term we will use the most in the course and is done the same way as formula wt. except using gram instead of

amu

.

For all substances, molar mass, in grams per mole, is numerically equal to the formula weight in atomic mass units.

That is, one mole of any element weighs its atomic mass in grams.61 molecule of H2O - MW = 18.02 amu  1 mole of H2O - M

m = 18.02 g H2O    

Mm of 1 mole of MgSO4 . 7H2O

Mg: 1 x 24.31 g = 24.31 g

S: 1 x 32.07 g = 32.07 g

O: 4 x 16.00 g = 64.00 g

246.52 g/mol MgSO

4

.

7H

2

O

H:14 x 1.01 g = 14.14 g

O: 7 x 16.00 g =

112.00 g

Note: we can use Mm as a conversion factor between grams and mols.Note: must account for mass of water in hydrates6.022 x 1023 moleculesSlide7

How is it possible that M

m

and FW/MW are the same value but different units?A.)What is the mass in grams of one Cl atom? B.)in one HCl molecule?

7

Let’s convert 1 Na atom

 g/

mol

Notes:

Converting mass

to

mols

or vice versa will require using molar mass.Converting mass to atoms, molecules, ions, etc. or vice versa will require going through mols by using Avogadro’s number.

Avogadro’s number and amu are both based on carbon-12 and inverse values of each other.Slide8

Mass and Moles of a Substance

Mole calculations

Converting the number of moles of a given substance into its mass, and vice versa, is fundamental to understanding the quantitative nature of chemical equations.

58.44 g

NaCl

1 mol NaCl1 mol NaCl 58.44 g NaCl

8   

Mm NaCl = 58.44 g/

mol

Converting mass

to

mols

or vice versa will require using molar mass.

Note: we can use M

m

as a conversion factor between grams and mols with g in numerator or denominator.Slide9

Mass and Moles of a Substance

Mole calculations

Suppose we have 5.75 moles of magnesium. What is its mass?9

Note: we use Mm as a conversion factor between grams and mols with mol in this case being placed in the denominator to cancel with the 5.75 mols we are converting to g.Slide10

Mass and Moles of a Substance

Mole calculations

Suppose we have 100.0 grams of H2O. How many moles does this represent?10

 Slide11

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 by Avogadro’s #:

How many molecules are there in 56 mg HCN?

HW

20-22

11

code: molesSlide12

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,

12

If given the %A by mass, it is useful to put the percentage in ratio form for dimensional analysis calculations.Slide13

Mass Percentages from Formulas

Let’s calculate the percent composition (%C, %H) of

one molecule of butane, C4H10.

First, we need the molecular

wt

of C

4H10.

Now, we can calculate the

percents

(basically, part/whole).

13

What if we wanted the %C & %H for

1

mol

of butane?

Same calculation and results except we would use molar mass, g/

mol

, instead of amu/atom.Note: % is a unit like g, etc. Don’t use the % button on calculator.Slide14

How many grams of carbon are there in 83.5 g of formaldehyde, CH

2

O, (40.0% C, 6.73% H, 53.3% O)?14

rewrite % into ratio

= 33.4 g CSlide15

An unknown acid contains only C, H, O. A 4.24 mg sample of acid is completely burned. It gives 6.21 mg of CO

2

and 2.54 mg H2O. What is mass% of each element in the unknown acid?15

%C, H, O in sample

unknown acid

 CO

2

+ H2O

O

2

has C

, H, O

4.24 mg

6.21 mg 2.54 mg

C:

H:  

O:

Total mass = 4.24 mg = mass C + mass H + mass O = 1.69 mg + 0.285 mg + mass O

mass O = 4.24 mg – 1.69 mg – 0.285 mg = 2.26 mg O

%C:

%H:

%O:

100% - 39.9% - 6.72% = 53.4% O

goal of problem

Note: We did the same calculation for H as we did for C except we did it in terms of

mg

and

mmol

to demonstrate how taking advantage of the prefix can simplify your work by adding

m

to the

mols

and

g

of the molar mass conversion factor (

mmol

/mg

) and

mol

to

mol

ratio (

mmol

/

mmol

). This eliminates the useless conversion from

mg

to

g

and back again.

mol

to

mol

ratio needed to convert from one species to another

Alternatively, we could have just subtracted the percent C and H from 100% to determine the %O.Slide16

Determining Chemical Formulas

We can determine 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.

From the empirical formula, we can find the

molecular formula

of a substance which will be a multiple of the empirical formula based on its molar mass.

16Slide17

Determining Chemical Formulas

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 based on mols (C:H:O)

C

x

H

yOz17

molsSlide18

Determining Chemical Formulas

For the purposes of calculations of this type, we will

assume we have 100.0 grams of sample; therefore, benzoic acid at 68.8% C means

which means 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 HyOz18 

  Slide19

Determining Chemical Formulas

This isn’t quite a whole number ratio, but if we divide each number by the smallest of the three, a better ratio will emerge

. We are dividing by the smallest number to get the species all on the same scale – “normalizing” the values.

Our

assumed 100.0

grams of benzoic acid (68.8%C, 5.0%H, 26.2%O) would contain:

19

Note: 68.8g + 5.0 g + 26.2 g =

assumed 100.0

g sampleSlide20

Determining Chemical Formulas

Next step is to get all species on the same scale, basically normalizing to moles of smallest species (in this case, 1.63

mols of O) for the purpose of getting values that can easily be converted to whole number ratios.

now it’s not too difficult to see that the smallest whole number ratio is 7:6:2 (multiple everything by 2 to get

whole

number.

The empirical formula is

C

7

H

6

O

2

.

2 x

 C7H6O220

C3.5 H3 O1

This is the empirical as well as molecular formula of benzoic acid because the molar mass of the empirical formula is the same as the molecular formula of benzoic acid.Slide21

Determining Chemical Formulas

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).

C2H3O2 empirical formula (lowest whole # ratio) C4H6O4 molecular formula (multiple of 2 x emp) C8H12O8 molecular formula (multiple of 4 x

emp)Which is not an empirical formula meaning not lowest whole #? CH4 CH4O C2H4O2 C2H6OTo determine the molecular formula, we must know the molecular weight (molar mass) of the compound.21

C

2

H

4

O

2

is divisible

by 2; CH

2O would be its empirical formulaSlide22

Determining Chemical Formulas

Determining the molecular formula from the empirical formula.

molar mass = n x empirical formula mass where n is the multiple factorn = molar mass

empirical mass

22Slide23

Acetic Acid contains 39.9% C, 6.7% H, 53.4% O. Determine empirical formula. The molecular mass of acetic acid is

60.0 g/mol

. What is the molecular formula?

HW

23

23

We assume 100.0 g sample meaning we have 39.9 g C, 6.7 g H, 53.4 g O and calculate the mols of each.  

empirical formula = CH2O

Mm= 30.03 g/mol

 2 x CH

2O  C2H4O2Next, we get them on same scale by dividing each by the smallest mols, 3.33 mols

code: formulaSlide24

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 (molar mass is an important concept here; g  mol, mol  g

) and

mol

to

mol ratios.Such calculations are fundamental to most quantitative work in chemistry.24

What we are discussing deals with mol to mol ratios (very important concept) and is the basis of many calculations.Slide25

The balanced chemical equation can be interpreted in numbers of molecules, but generally chemists interpret equations as “

mole-to-mole

” relationships. Lets look at the following reaction:

1 molecule N

2

+ 3 molecules H

2

2 molecules NH

328.02 g N2 + 3(2.02 g) H2  2 (17.04 g) NH328.02 g + 6.06 g  34.08 g

34.08 g = 34.08 g

Molar Interpretation of a Chemical Equation

25

If the number of each particular atom is the same on both sides, the Law of Conservation of Mass will be preserved and the mass of reactants will be equal to the mass of the products.Slide26

Molar Interpretation of a Chemical Equation

Suppose we wished to determine the number of moles of NH

3 we could obtain from 4.8 mol H2. We will assume N2 is in excess which means H2

is limited and will dictate the amount of product that will be formed; once the H

2

is gone, the reaction will stop.

26

 We should realize that we have conversion factors that exist from the mol to mol ratios available from the balanced equation to go from one species to another.Slide27

Mass Relationships in Chemical Equations

How many grams of HCl are required to react with 5.00 grams MnO

2 according to this equation?

27

? g

5.00 g

Note: To get from one substance to another requires the use of

mol

to

mol

ratios.Slide28

Mass Relationships in Chemical Equations

How many grams of CO

2 gas can be produced from 1.00 kg Fe2O3?Fe2O3 (s) + 3 CO (g)

 2 Fe

(s)

+ 3 CO

2 (g)

HW 2428

1.00 kg assume excess

? g

code:

stoichSlide29

Limiting Reagent

The

limiting reactant (or limiting reagent) is the reactant that is entirely consumed when the reaction goes to completion. The limiting reagent ultimately determines how much product can be obtained.For example, bicycles require one frame and two wheels. If you have 20 wheels but only 5 frames, how many bicycles can be made?

o

o

o o o o o o o o o o o o o o o o o o 20 wheels

1 frame + 2 wheels  1 bike29

R + R  P

10 wheels in excess; therefore, frames are limiting factor

Smaller

value is correct answer or you can determine which is the limiting reagent and only calculate that value.

Reaction stops when one component , limiting reagent, is exhausted

Basically, we must find out which species will run out first because it will dictate how much product will form.Slide30

Limiting Reagent

If 0.30

mol Zn is added to hydrochloric acid containing 0.52 mol

HCl, how many grams of H

2

are produced?

30

0.30 mol 0.52 mol ? g

The maximum

product possible is based

on the

limiting reagent; smaller

mols

of product is maximum possible based on the limiting reagent

which is

HCl

in this case.

limiting reagent HCl   

First, we calculate how many mols of H2 is possible for each reactant.We use the mols of H2 possible from the limiting reagent, HCl, to determine the mass of H2.Note: The limiting reagent is not necessarily the species with the smaller mass or #mols

; you must account for the mol-to-mol ratio as demonstrated in this problem.Slide31

If 7.36 g Zn was heated with 6.45 g sulfur, what amount of

ZnS

was produced?

31

7.36 g 6.45 g ? g

smaller mols; limiting reagent

This is known as the theoretical yield.

mols

produced by each reactant:Slide32

Theoretical and Percent Yield

The

theoretical yield of product is the maximum amount of product that can be obtained from given amounts of reactants.

The

percentage yield

is the actual yield (experimentally determined) expressed as a percentage of the theoretical yield (calculated like in previous example).

32Once again, % is a unit.Slide33

Theoretical and Percent Yield

To illustrate the calculation of percentage yield, recall that the theoretical yield of

ZnS in the previous example was 11.0 g ZnS.If the actual yield experimentally for the reaction is 9.32 g ZnS, what is the %yield?

33

experimental value

theoretical value;

calculated value

% is unit, not button on calculatorSlide34

If 11.0 g CH

3

OH are mixed with 10.0 g CO, what is the theoretical yield of HC2H3O2 in the following reaction? If the actual yield was 19.1 g, what is the %yield of HC2H

3

O

2

? CH3OH (l) + CO (g)  HC2H3O2 (l)34

11.0 g 10.0 g ? g  

smaller mols; limiting reagent

theoretical yield

experimental value

theoretical value;

calculated value

HW 25

mols produced by each reactant:

code: limiting