Stoichiometry [Compatibility Mode]

8/8/2019 Stoichiometry [Compatibility Mode]

1/33

Stoichiometry


2/33

What happens to matter when itun ergoes c em ca c anges

The law of conservation of mass:

, ,any chemical reaction

Stoichiometry the "measurement of

"


3/33

Chemical equations

Example :

Steps involved in writing a 'balanced' equation for a chemical

reaction:

1. Experimentally determine reactants and products

2. Write 'un-balanced' equation using formulas of reactants and

products

3. Write 'balanced' equation by determining coefficients thatprov e equa num ers o eac type o atom on eac s e o

the equation (generally, whole number values)


4/33

indicated by using the symbols (g), (l), and (s)(for gas, liquidand solid, respectively):


5/33

Alkali rou


6/33

Combustion reactions are rapid reactions thatroduce a flame.


7/33

combination reactions

decomposition reactions


8/33

1 molecule of propane reacts with 5molecules of oxygen to

water.


9/33

Atoms of different elements have different masses.

grams a er . grams y rogen + .

grams Oxygen

water was com osed of two atoms of h dro en foreach atom of oxygen.

in the 11.1 grams of hydrogen there were twice asmany a oms as n e . grams o oxygen.


10/33

The atomic mass unit (amu) was standardized against the

12C isotope of carbon (amu = 12).

1 amu = 1.66054 x 10

-24

gramsconversely:

.

Average atomic mass

the mass of the hydrogen atom (1H) is 1.0080 amu, and themass of an oxygen atom (16O) is 15.995 amu.

12 13

. . .The mass of 12C is 12 amu, and that of 13C is 13.00335 amu.Therefore, the average atomic mass of carbon is:

(0.98892)*(12 amu) + (0.01108)*(13.00335 amu) = 12.011 amu


11/33

The formula wei ht of a substance is the sum of the

atomic weights of each atom in its chemical formula.

H2O has a formula weight of:

* *. . .

chemically bonded chemical formula is the molecular

formula, and the formula weight is the molecular weight.

Ionic substances are not chemically bonded and do notexist as discrete molecules no molecular wei ht

Table salt (NaCl) has a formula weight of:

23.0 amu + 35.5 amu = 58.5 amu


12/33

Percentage composition fromformulas

Formula and molecular weight:1*(12.011 amu) + 4*(1.008) = 16.043 amu

%C = 1*(12.011 amu)/16.043 amu = 0.749 = 74.9%

%H = 4*(1.008 amu)/16.043 amu = 0.251 = 25.1%


13/33

A mole is defined as the amount of matter that

contains as man ob ects as the number of

atoms in exactly 12 grams of12

C.

Various experiments have determined that this numberis 6.0221367 x 1023

23known as Avogadro's number.

us ow g s s num er ne mo e o mar esspread over the earth would result in a layer three milesthick.


14/33

The mass in rams of 1 mole mol of a

substance is called its molar mass.

always numerically equal to its formula weight

(in amu).

One H2O molecule weighs 18.0 amu; 1 mol of H2Owei hs 18.0 rams

One NaCl ion pair weighs 58.5 amu; 1 mol of NaCl.


15/33

Interconverting masses, moles, and

num ers o part c es

Ti or! brin me 1.5 moles of calcium chloride

Chemical formula of calcium chloride = CaCl2Molecular mass of Ca = 40.078 amu

Molecular mass of Cl = 35.453 amu

Therefore, the formula weight of CaCl2 = (40.078) +2(35.453) = 110.984 amu

Therefore one mole of CaCl2 would have a mass of

110.984 grams.

, .

(1.5 mole)(110.984 grams/mole) = 166.476 grams


16/33

Interconverting masses, moles, and

num ers o part c es

Tigor! I have 2.8 g of gold, how many atoms do I have?"

Molecular formula of gold is: Au= .

Therefore, 1 mole of gold weighs 196.9665 grams. So, in 2.8

(2.8 gram)(1 mole/196.9665 gram) = 0.0142 mole

rom voga ro s num er, we now a ere areapproximately 6.02 x 1023 atoms/mole. Therefore, in0.0142 moles we would have:. mo e . x a oms mo e = . x a oms


17/33

different atoms in a compound.= 2

if we know the molar amounts of eachelement in a compound we can determine

the empirical formula.


18/33

Mercury forms a compound with chlorine that

is 73.9% mercur and 26.1% chlorine b mass.What is the empirical formula?

a cu a on ase: gram samp e o s compoun .

For Mercury:The sample contain 73.9 grams of mercury.Mole number (73.9 g)*(1 mol/200.59 g) = 0.368 moles

The sample contain 26.1 grams of chlorine.Mole number(26.1 g)*(1 mol/35.45 g) = 0.736 mol

Molar ratio : ( 0.736 mol Cl/0.368 mol Hg) = 2.0

The empirical formula would thus be (remember to list

cation first, anion last): HgCl2


19/33

Molecular formula from empirical

formula

the empirical formula if we know the.

e c em ca ormu a w a ways e someinteger multiple of the empirical formula

.e. n eger mu p es o e su scr p s othe empirical formula).


20/33

Vitamin C (ascorbic acid) contains 40.92 % C, 4.58 % H,and 54.50 % O, by mass. The experimentally

e erm ne mo ecu ar mass s amu. a s eempirical and chemical formula for ascorbic acid?

:

In 100 grams of ascorbic acid we would have:40.92 grams C. grams

54.50 grams O

This would give us how many moles of each element?


21/33

Determine the simplest whole number ratio by dividing by

-Oxygen):

C = (1.0)*3 = 3

H = (1.333)*3 = 4

O = (1.0)*3 = 3

or, C3H

4O

3


22/33

What about the chemical formula?

The experimentally determined molecular mass is 176 amu.The molecular mass of our empirical formula:

* * * =. . . .

What is the ratio between the two values?

amu . amu = .

Thus, the actual molecular formula is:2* C3H4O3 = C6H8O6


23/33

When a compound containing carbon and hydrogen is subjecto com us on w oxygen n a spec a com us on appara us

all the carbon is converted to CO2 and the hydrogen to H2O.


24/33

Consider the combustion of alcohol. The sample is knownto contain only C, H and O. Combustion of 0.255 grams of

e a co o pro uces . grams o 2 an . gramsof H2O. What is the empirical formula?

Answer:

alcohol + nO mCO + O

Determining the amount of C and H in the sample:The amount of C:


25/33

The amount of H O:

There are 2 H in H2O, thus:The number of H = 2 x 0.017 mols = 0.034 molsThe mass of H = 0.034 mol x (1 g/mol H) = 0.034 g

0.154 grams (C) + 0.034 grams (H) = 0.188 grams

u we now we com us e . grams o a co o . e'missing' mass must be from the oxygen atoms in theisopropyl alcohol:

0.255 grams - 0.188 grams = 0.067 grams oxygen


26/33

This much oxygen is how many moles?

Overall therfore, we have:.

0.0340 moles Hydrogen

0.0042 moles Oxygenv e y e sma es mo ar amoun o norma ze:C = 3.05 atomsH = 8.1 atoms

O = 1 atomWithin experimental error, the most likely empirical

C3H8O


27/33

Quantitative Information from

Balanced Equations

balanced equation is:

burning 1.00 gram of C4H10.


28/33

First of all we need to calculate how man moles of butanewe have in a 1 gram sample:

now, the stoichiometric relationship between C4H10 and CO2is: , therefore:

The question called for the determination of the mass of CO2produced, thus we have to convert moles of CO2 into grams

2


29/33

chemical reaction is called the limiting reactant(or limiting reagent) because it determines (orlimits) the amount of product formed.


30/33

Consider the following reaction:

Suppose that a solution containing 3.50 grams of Na3PO

4is

mixed with a solution containing 6.40 grams of Ba(NO3)2. How

3 4 2

Answer:

1. First we need to convert the grams of reactants into moles:


31/33

2. Now we need to define the stoichiometric ratios between

2 Na3PO4 Ba3(PO4)23 Ba (NO3)2 Ba3(PO4)2

3. We can now determine the moles of product that would be

during the course of the reaction:

. 3 2 make at most 0.0082 moles of the Ba3(PO4)2 product.


32/33

5. 0.0082 moles of the Ba3(PO4)2 product would be equal to:


33/33

The uantit of roduct that is calculated to form

when all of the limiting reactant is consumed in areaction is called thetheoretical yield.

the actual yield.

Actual yield < Theoretical yieldfor the following reasons:

for some reason not all the reactants may react

physical recovery of 100% of the sample may beimpossible (like getting all the peanut butter out of the

Documents

Stoichiometry [Compatibility Mode]