Bab 3 Stoikiometri

 

Mass Relationships in Chemical Reactions Chapter 3

 

Micro World atoms & molecules

Macro World grams

A t o m i c m as s    is the mass of an atom in  is atomic mass units (amu)  By definition: 1 atom 12C “weighs” 12 amu  amu  On this scale 1H

= 1.008 amu

16O

= 16.00 amu 3.1

 

Natural lithium is: 7.42% 6Li (6.015 amu) 92.58% 7Li (7.016 amu)

A v e rra ag e a att o m i c m a s s    of lithium:   of (7.42% x 6.015) + (92.58% x 7.016) = 6.941 amu 100

3.1

 

The m o l e (m (m o l )  is   is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12C 1 mol = N  A = 6.0221367 x 1023   Avogadro’s  Avogadro’ s number ((N  N  A )  )  

3.2

 

eggs Mola arr m a s s    is the mass of 1 mole of shoes in grams marbles atoms 1 mole 12C atoms = 6.022 x 10 23 atoms = 12.00 g 1 12C atom = 12.00 amu 1 mole

12C

atoms = 12.00 g

12C

1 mole lithium atoms = 6.941 g of Li For any element atomic mass (amu) = molar mass (grams) 3.2

 

One Mole of: S

C

Hg

Cu

Fe 3.2

 

1 12C atom 12.00 amu

12.00 g

x

6.022 x 1023 12C atoms

-24

1 amu = 1.66 x 10

M=

=

1.66 x 10-24 g 1 amu

23

g or 1 g = 6 6.022 .022 x 10  amu

molar mass in g/mol

N  A = Avogadro’s nu number  mber   3.2

 

Do You Understand Molar Mass? How many atoms are in 0.551 g of potassium (K) ? 1 mol K = 39.10 g K 1 mol K = 6.022 x 1023 atoms K 0.551 g K x

1 mol K 39.10 g K

x

6.022 x 1023 atoms K

=

1 mol K

8.49 x 1021 atoms K 3.2

 

M o l ec e c u l ar ar m a s s    (or molecular weight) is the sum of  (or the atomic masses (in amu) in a molecule. 

SO2 

1S

32.07 amu

2O

+ 2 x 16.00 amu

SO2 

64.07 amu

For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2  3.3

 

Do You You Understand Molecular Mass? How many H atoms are in 72.5 g of C3H8O ? 1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C 3H8O 1 mol C3H8O molecules = 8 mol H atoms 1 mol H = 6.022 x 1023 atoms H 23

72.5 g C3H8O x 1 mol C3H8O x 8 mol H atoms x6.022 x 10  H atoms = 60 g C3H8O 1 mol C3H8O 1 mol H atoms

5.82 x 1024 atoms H

3.3

 

   t    h   g    i

  y   v   a   e

   L

   H

   t    h   g    i    L

KE = 1/2 x m x v2 v = (2 x KE/m)1/2

  y   v   a   e    H

F=qxvxB 3.4

3.4  

P er e r c e n t c o m p o s i t io io n    of an element in a compound =  of

element  x 100% n x molar mass of element  molar mass of compound n is the number of moles of the element in 1 mole of the compound  compound  2 x (12.01 g) x 100% = 52.14% 46.07 g 6 x (1.008 g) 46.07 g x 100% = 13.13% %H = 1 x (16.00 g) %O = x 100% = 34.73% 46.07 g

%C =

C2H6O 52.14% + 13.13% + 34.73% = 100.0%

3.5  

Types of Formulas • Empirical Formula The formula of a compound that expresses the s m a l l e s t w h o l e n u m b e r ratio   of the atoms present.  of

Ionic formula are always empirical formula • Molecular Formula The formula that states the a c tu a l     number kind of atom found in o n e m o l e c u l eof   ofeach  of the compound.

 

To obtain an Empirical Formula   1. Determine the mass in grams of each element present, if necessary. 2. Calculate the number of m o l e s   of each element. 3. Divide ea each ch by the smallest number of es t w h o l e moles to obtain the s i m p l es n u m b e r r at at i o .

4. If w who hole le n num umbe bers rs a are re n not ot ob obta tain ined ed* in step 3), multiply through by the smallest number that will give all whole numbers * Be

careful! Do not round off numbers prematurely  prematurely 

 

A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine a formula for this substance. require m o l e   ratios so convert grams to moles moles of N = 2.34g of N = 0.167 moles of N 14.01 g/mole moles of O = 5.34 g

= 0.334 moles of O

16.00 g/mole 0.167

Formula:  N

0.334

O

 N 0.167 O 0.334 0.167

0.167

NO2

(HONORS only)  

Calculation of the Molecular Formula (HONORS only)

A compound has an empirical formula of NO2. The colourless liquid, used in rocket engines has a molar mass of 92.0 g/mole. e c u l ar a r f o r m u l a of this What is the m o l ec substance? empirical formula mass: 14.01+2 (16.00) = 46.01 g/mol n = molar mass emp. f. mass n = 2

= 92.0 g/mol 46.01 g/mol

2(NO2)

N2O4 

 

Empirical Formula from % Compositio Compositionn (HONORS only)

A substance has the following composition by mass: 60.80 % Na ; 28.60 % B ; 10.60 % H What is the empirical formula of the substance? Consider a sample size of 100 grams This will contain 28.60 grams of B and 10.60 grams H Determine the number of moles of each Determine the simplest whole number ratio

 

Combust 11.5 g ethanol Collect 22.0 g CO2 and 13.5 g H2O

g CO2 

mol CO2 

mol C

gC

6.0 g C = 0.5 mol C

g H2O

mol H2O

mol H

gH

1.5 g H = 1.5 mol H

g of O = g of sample – sample – (g  (g of C + g of H)

4.0 g O = 0.25 mol O

Empirical formula C0.5H1.5O0.25  Divide by smallest subscript (0.25) Empirical formula C H O

2

6

 

Mass Changes in Chemical Reactions

1. Wri rite te b bal alan ance ced d chemi chemica call equat equatio ion n 2. Conve Convert rt quant quantiti ities es of k know nown n substa substance nces s into into moles moles 3. Use coeffici coefficients ents in balance balanced d equati equation on to to calcu calculate late the number of moles of the sought quantity 4. Conve Convert rt mole moles s of soug sought ht quan quantit tity y into into desi desired red u unit nits s

3.6

3.8  

Methanol burns in air according to the equation 2CH3OH + 3O2 

2CO2 + 4H2O

If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH3OH

moles CH3OH

molar mass CH3OH

moles H2O

grams H2O

molar mass coefficients H2O chemical equation

209 g CH3OH x 1 mol CH3OH x 4 mol H2O x 18.0 g H2O 32.0 g CH3OH 1 mol H2O 2 mol CH3OH

235 g H2O

=

3.8  

Limiting Reagents

6 red green leftused overup

3.9  

Method 1 • Pick A Product • Try ALL the reactants • The lowest answer will be the correct answer • The reactant that gives the lowest answer will be the limiting reactant

 

Limiting Limiting Reactant: Method 1 Reactant • 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is  produced? 2 Al + 3 Cl2  2 AlCl3 • Start with Al: 10.0 g Al

1 mol Al 27.0 g Al

2 mol AlCl3 2 mol Al

133.5 g AlCl3 1 mol AlCl3 

= 49.4g 49.4 g AlCl3

•  Now Cl2: 35.0g Cl2

1 mol Cl2 71.0 g Cl2

2 mol AlCl3  133.5 g AlCl3  3 mol Cl2

1 mol AlCl3 

= 43.9g AlCl3 

 

Method 2 • Convert one of the reactants to the other REACTANT • See See if there there is enou enough gh react reactant ant “A” to use use up up the other reactants • If there is less than the GIVEN amount, it is the limiting reactant • Then, you can find the desired species

 

Do You You Understand Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe2O3  2Al + Fe2O3 

Al2O3 + 2Fe

Calculate the mass of Al2O3 formed. g Al

mol Al

mol Fe2O3 needed

g Fe2O3 needed

OR g Fe2O3 124 g Al x

mol Fe2O3 1 mol Al 27.0 g Al

x

mol Al needed 1 mol Fe2O3 

2 mol Al

Start with 124 g Al

x

160. g Fe2O3 

1 mol Fe2O3 

g Al needed =

367 g Fe2O3 

need 367 g Fe O 2

3

Have more Fe O (601 g) so Al Al is limiting reagent

2

3

3.9

 

Use limiting reagent (Al) to calculate amount of product that can be formed. g Al

mol Al

mol Al2O3 

2Al + Fe2O3  124 g Al x

1 mol Al 27.0 g Al

x

Al2O3 + 2Fe

1 mol Al2O3  2 mol Al

g Al2O3 

x

102. g Al2O3  1 mol Al2O3 

=

234 g Al2O3 

3.9  

Theor etical Y ield   is the amount of product that would  is

result if all the limiting reagent r eagent reacted.  A ct u al Y ield   is the amount of product actually obtained  is from a reaction. 

% Yield = Yield =

 Actual Yield Theoretical Yield

x 100

3.10  

Chemistry In Action: Chemical Fertilizers

Plants need: nee d: N, P, K, Ca, C a, S, & Mg

3H2 (  (g  g ) + N2 (  (g  g ) NH3 (aq  (aq)) + HNO3 (aq  (aq))

2NH3 (  (g  g ) NH4NO3 (  (aq aq))

fluorapatite

2Ca5(PO4)3F (s (s) + 7H2SO4 (  (aq aq)) 3Ca(H2PO4)2 (  (aq aq)) + 7CaSO4 (  (aq aq)) + 2HF (g  (g )