Topic 3

Mass Relations and Stoichiometry

Mass and Moles of a SubstanceChemistry 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.

2

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 amuO: 1 x 16.00 amu = 16.00 amu

18.02 amu (mass of 1 molecule of H2O)3

Molecular Weight and Formula WeightThe 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 amu58.44 amu (mass of 1 FU NaCl)

4

iron (III) sulfate, Fe2(SO4)3 Fe: 2 x 55.85 amu = 111.70 amu S: 3 x 32.07 amu = 96.21 amuO: 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 amu180.18 amu (mass of 1 molecule)

one formula unit (FU) of NaCl:

Note: molecular wt and formula wt are calculated the same way.

Mass and Moles of a SubstanceThe 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 1023.

1 mole = 6.022 x 1023 X X = ions, particles, atoms, molecules,

items, etc.1 mole Na2CO3 6.022 x 1023 FU Na2CO3

1 mole CO2 6.022 x 1023 molecules CO2

5

Mass and Moles of a SubstanceThe 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.

6

1 molecule of H2O - MW = 18.02 amu 1 mole of H2O - Mm = 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 MgSO4 . 7H2O

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 hydrates

6.022 x 1023 molecules

How is it possible that Mm 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.

Mass and Moles of a SubstanceMole 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

A"" of massmolar A"" of mass A"" of moles

8

     

Mm NaCl = 58.44 g/mol

Converting mass to mols or vice versa will require using molar mass.

Note: we can use Mm as a conversion factor between grams and mols with g in numerator or denominator.

Mass and Moles of a Substance

Mole calculationsSuppose 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.

Mass and Moles of a Substance

Mole calculationsSuppose we have 100.0 grams of H2O. How many moles does this represent?

10

  

 

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 #:

atomsFe1002.6Femole1moleculesHCl10 6.02 HCl mole 1

23

23

How many molecules are there in 56 mg HCN?

HW 20-2211

     

 

 

code: moles

Determining Chemical FormulasThe 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,

%100whole the of mass

whole in A"" of mass A"" % mass

A % 20

12

sample g 100A g 20

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

Mass Percentages from FormulasLet’s calculate the percent composition (%C, %H) of one molecule of butane, C4H10.

First, we need the molecular wt of C4H10.

amu 48.04 amu/atom 12.01 @ carbons 4 amu 10.10 amu/atom 1.01 @ hydrogens 10 amu 58.14 HC of molecule 1 104

Now, we can calculate the percents (basically, part/whole).

C%63.82%100 C % amu to tal 14.58Camu 48 .04

H%37.17%100 H % amu to tal 14.58Hamu 10 .10

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.

How many grams of carbon are there in 83.5 g of formaldehyde, CH2O, (40.0% C, 6.73% H, 53.3% O)?

14

  

 

rewrite % into ratio

= 33.4 g C

An unknown acid contains only C, H, O. A 4.24 mg sample of acid is completely burned. It gives 6.21 mg of CO2 and 2.54 mg H2O. What is mass% of each element in the unknown acid?

15

%C, H, O in sample unknown acid CO2 + H2O

O2

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 Omass 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.

Determining Chemical FormulasWe 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.

16

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)

Cx HyOz

17mols

Determining Chemical FormulasFor 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 HyOz 18

   

Determining Chemical Formulas

C )8(72.5g 12.01C mol 1 8.68 molCg

H )5(9.4g 1.01H mol 1 0.5 molHg

O )7(63.1g 16.00O mol 1 2.26 molOg

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:

19Note: 68.8g + 5.0 g + 26.2 g = assumed 100.0 g sample

Determining Chemical FormulasNext 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.

3.501.63(7)C 728.5 mol

3.01.63(7)H .954 mol

1.001.63(7)O )7(63.1 mol 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 C7H6O2 .

2 x C7H6O2

20

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.

Determining Chemical FormulasAn 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 C2H6O

To determine the molecular formula, we must know the molecular weight (molar mass) of the compound.

21

C2H4O2 is divisible by 2; CH2O would be its empirical formula

Determining Chemical FormulasDetermining the molecular formula from the empirical formula.

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

n = molar mass empirical mass

22

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 CH2O C2H4O2

Next, we get them on same scale by dividing each by the smallest mols, 3.33 mols

code: formula

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.

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:

)g(NH 2 (g)3H (g)N 322

322 NH mol 2 H mol 3 N mol 1 1 molecule N2 + 3 molecules H2 2 molecules NH3

28.02 g N2 + 3(2.02 g) H2 2 (17.04 g) NH3

28.02 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.

Molar Interpretation of a Chemical EquationSuppose we wished to determine the number of moles of NH3 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 H2 is gone, the reaction will stop.

)g(NH 2 (g)3H (g)N 322

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.

Mass Relationships in Chemical Equations

How many grams of HCl are required to react with 5.00 grams MnO2 according to this equation?

)s(MnO HCl(aq) 4 2 )g(Cl (aq)MnCl O(l)H 2 222

27

     

 

 

? g 5.00 g

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

Mass Relationships in Chemical Equations

How many grams of CO2 gas can be produced from 1.00 kg Fe2O3?

Fe2O3 (s) + 3 CO (g) 2 Fe (s) + 3 CO2 (g)

HW 24

28

     

 

   

1.00 kg assume excess ? g

code: stoich

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 bike

29

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.

Limiting ReagentIf 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, how many grams of H2 are produced?

)g(H (aq)ZnCl HCl(aq) 2 Zn(s) 22

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.

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

ZnS8 S Zn(s)8 8

31

7.36 g 6.45 g ? g

  

 

  

  

 

 

  

smaller mols; limiting reagent

This is known as the theoretical yield.

mols produced by each reactant:

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).

%100(calc) yield ltheoretica

(exp) yield actual Yield%

32

Once again, % is a unit.

Theoretical and Percent YieldTo 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?

%7.84%100 ZnSg 11.0 ZnSg 9.32 Yield%

33

experimental value

theoretical value;calculated value

% is unit, not button on calculator

If 11.0 g CH3OH 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 HC2H3O2?

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