Transcript Ch 4 chemistry

 Defining the mole:
 A number equal to the number of atoms in exactly 12 gram
of 𝟏𝟐𝑪 atoms.
 1 mole of element X = gram atomic mass of X
Example:
 1 mole of sulfur = 32.6 g
Atomic mass of sulfur = 32.6 u
 The mole concept applied to compound:
Example:
 The molecular mass of water = 18.02 u
= 2 * atomic mass of H + 1 * atomic mass of O.
 1 mole of 𝑯𝟐 O = 16 + 2 = 18 g
 1 mole of ionic compound X = gram formula mass of X
Example:
1 mole of 𝑨𝒍𝟐 𝑶𝟑 =101.96 g
𝑨𝒍𝟐 𝑶𝟑 has 2Al with an atomic masses 26.98u and three
oxygen with mass of 16 u.
 Converting from gram to moles:
Example:
 In experiment to prepare of titanium(IV) oxide we start with
23.5g sample of titanium. How many moles of titanium do we
have?
 Solution:
1 mole of Ti = 47.867g Ti
23.5g Ti = ? Mole Ti
 Using conversion factor:
𝟏 𝒎𝒐𝒍𝒆 𝑻𝒊
23.5 g Ti × 𝟒𝟕,𝟖𝟔𝟕 𝒈𝑻𝒊 = 0.491 mole Ti
 Conversion from mole to grams:
Example:
 We need 0.254 moles of 𝑭𝒆𝑪𝒍𝟑 for certain experiment. How many
grams would you need weight?
 Solution:
1 mole of 𝑭𝒆𝑪𝒍𝟑 = 162.204 g
Molar mass 𝑭𝒆𝑪𝒍𝟑 = 55.845 g/mol + ( 3× 35.453) g/mol
= 162.204 g/mole
0.254 mole 𝑭𝒆𝑪𝑳𝟑 = ? g 𝑭𝒆𝑪𝑳𝟑
 By using Conversion factor :
So
0.254 ×
𝟏𝟔𝟐.𝟐𝟎𝟒 𝒈𝑭𝒆𝑪𝑳𝟑
𝟏 𝒎𝒐𝒍𝒆 𝑭𝒆𝑪𝑳𝟑
= 41.2 g 𝑭𝒆𝑪𝑳𝟑
Learning check
* How many moles of aluminum are there in 3.47 gram sheet of aluminum
foil.
 Avogadros number
 1 mole of X = 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 unit of X (Avogadros number)
 The unit can be atoms , molecules , formula unit.
Example :
1 mole of Xe = 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 atoms of Xe
1 mole of 𝑵𝑶𝟐 = 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 molecules of 𝑵𝑶𝟐
Example:
 In lightbulb the tungsten weight 0.653 g. how many atoms
of tungsten are there in such sample.
 Solution:
0.632 g W = ? Atoms of W
gram W to mole W to atom W
 1 mole of W = 183.84 g W
 1 mole W = 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 atoms W
 Conversion factors :
1.
𝟏 𝒎𝒐𝒍𝒆 𝑾
𝟏𝟖𝟑.𝟖𝟒 𝒈 𝑾
0.635 ×
𝟏 𝒎𝒐𝒍𝒆 𝑾
𝟏𝟖𝟑.𝟖𝟒 𝒈 𝑾
2.
×
𝟔.𝟎𝟐𝟐×𝟏𝟎𝟐𝟑 𝒂𝒕𝒐𝒎𝒔 𝑾
𝟏 𝒎𝒐𝒍𝒆 𝑾
𝟔.𝟎𝟐𝟐×𝟏𝟎𝟐𝟑 𝒂𝒕𝒐𝒎𝒔 𝑾
𝟏 𝒎𝒐𝒍𝒆 𝑾
𝐓𝐡𝐞𝐧 :-
= 2.08 × 𝟏𝟎𝟐𝟏 atoms W
Calculating the mass of molecules:
Example :What is the average mass of one molecule of carbon
tetrachloride.
 Solution:
1 molecule 𝑪𝑪𝒍𝟒 = ? g 𝑪𝑪𝒍𝟒
1 mole of 𝑪𝑪𝒍𝟒 = 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 molecules of 𝑪𝑪𝒍𝟒
= 153.823 g 𝑪𝑪𝒍𝟒
Molar mass of 𝑪𝑪𝒍𝟒 = 12 + (4× 35.45) = 153.823 g\mole
 Conversion factor :
1.
2.
𝟏 𝒎𝒐𝒍𝒆 𝑪𝑪𝒍𝟒
𝟔.𝟎𝟐𝟐×𝟏𝟎𝟐𝟑 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆 𝑪𝑪𝒍𝟒
𝟏𝟓𝟑.𝟖𝟐𝟑 𝒈𝑪𝑪𝒍𝟒
𝟏 𝒎𝒐𝒍𝒆 𝑪𝑪𝒍𝟒
1 molecule 𝑪𝑪𝒍𝟒 ×
𝟏 𝒎𝒐𝒍𝒆 𝑪𝑪𝒍𝟒
𝟔.𝟎𝟐𝟐×𝟏𝟎𝟐𝟑 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆 𝑪𝑪𝒍𝟒
= 𝟐. 𝟓𝟓𝟓 × 𝟏𝟎−𝟐𝟐 g 𝑪𝑪𝒍𝟒 .
×
𝟏𝟓𝟑.𝟖𝟐𝟑 𝒈 𝑪𝑪𝒍𝟒
𝟏 𝒎𝒐𝒍𝒆 𝑪𝑪𝒍𝟒
 Percentage composition: Called percentage by mass of element: is the number of grams of the
element present in 100g of the compound.
 Percentage by mass of element =
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒆𝒍𝒆𝒎𝒆𝒏𝒕
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒘𝒉𝒐𝒍𝒆 𝒔𝒂𝒎𝒑𝒍𝒆
× 𝟏𝟎𝟎%
Example:
 A sample of liquid with a mass of 8.657 g was decomposed into its element
and gave 5.217g of carbon, 0.9620 g of hydrogen, and 2.478 g of oxyge.
What is the percentage composition of this compound?
 Solution:
 For C :
 For H :
 For O :
𝟓.𝟐𝟏𝟕 𝒈
× 𝟏𝟎𝟎% = 𝟔𝟎. 𝟐𝟔% C
𝟖.𝟔𝟓𝟕 𝒈
𝟎.𝟗𝟔𝟐 𝒈
× 𝟏𝟎𝟎% = 𝟏𝟏. 𝟏𝟏 %
𝟖.𝟔𝟓𝟕 𝒈
𝟐.𝟒𝟕𝟖 𝒈
× 𝟏𝟎𝟎 % = 𝟐𝟖. 𝟔𝟐 %
𝟖.𝟔𝟓𝟕 𝒈
Sum of percentage: 99.99%
 Learning check
An organic compound weighing 0.6672 g is decomposed , giving 0.3481 g of
carbon , 0,087 g hydrogen. What is the percentage of hydrogen and carbon in
this compound?
 Determining empirical and molecular formulas:
 Calculating an empirical formula from mass data:
Example:
 2.57 g sample of compound composed on only tin and chlorine was
found to contain 1.17 g of tin. What the compound empirical formula?
Solution: Mass of Cl = 2.57 g compound – 1.17 g Sn = 1.40 g Cl
 1 mol Sn = 118.7 g Sn
,
1 mol Cl = 35.45 g Cl
 Conversion factors : 1.
1 𝑚𝑜𝑙 𝑆𝑛
1 𝑚𝑜𝑙 𝑆𝑛
118.7 𝑔 𝑆𝑛
1.17g × 118.7 𝑔 𝑆𝑛 = 0.00986 mol Sn
1.40 g ×
2.
1 𝑚𝑜𝑙 𝐶𝑙
35.45 𝑔 𝐶𝑙
1 𝑚𝑜𝑙 𝐶𝐿
35.45 𝑔 𝐶𝐿
= 0.0395 mol Cl
Formula: 𝑆𝑛0.00986 𝐶𝑙0.0395
 To convert the decimal subscripts to integer by dividing each by the smallest number in
the set .
𝑆𝑛 0.00986 𝐶𝑙 0.0395 = 𝑆𝑛1.00 𝐶𝑙4.01 and the empirical formula is Sn𝐶𝑙4
0.00986
0.00986
Determining a molecular formula from an empirical formula and a
molecular mass:𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅
𝒆𝒎𝒑𝒊𝒓𝒊𝒄𝒂𝒍 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅
= integer
Example:Styrene has an empirical formula of CH it molecular mass is 104. what is
its molecular formula?
Solution:
The formula mass is:- 12.01 + 1.008 = 13.02
𝟏𝟎𝟒
𝟏𝟑.𝟎𝟐
= 7.99 = 8
Molecular formula of styrene is 𝑪𝟖 𝑯𝟖
learning check
The empirical formula of hydrazine is N𝑯𝟐 and its molecular mass is 32.0
what is its molecular formula?
 Calculate empirical formula from mass percent :
 Find the molecular formula of a compound has 20 % H, 80 % C, if its Mw = 30 g/mol ?
 Writing and balancing equations:
 Always the balancing of an equation as a two-step process:
 Step 1: write the unbalanced equation.
 Step 2: adjust the coefficient to get equal number of each kind of atoms on both
side.
 Some guideline for balancing equation:
1. Start balancing with the most complicated formula first, ending with
element, particularly 𝑯𝟐 𝒂𝒏𝒅 𝑶𝟐 .
2. Balance atoms that appear in only two formula.
3. Balance as a group those polyatomic ions that appear unchanged on both
side of the arrow.
 Example: Sodium hydroxide and phosphoric acid 𝐇𝟑 𝐏𝐎𝟒 React to give sodium phosphate and
water. Write the balanced equation for this reaction?
 Solution:
NaOH + 𝐇𝟑 𝐏𝐎𝟒
𝐍𝐚𝟑 𝐏𝐎𝟒 + 𝐇𝟐 O (unbalanced)
1. Balance element ( Na and P)
3 NaOH
+ 𝐇𝟑 𝐏𝐎𝟒
𝐍𝐚𝟑 𝐏𝐎𝟒 + 𝐇𝟐 O
2. Balance particular ( 𝐇𝟐 𝐎)
3 NaOH + 𝐇𝟑 𝐏𝐎𝟒
𝐍𝐚𝟑 𝐏𝐎𝟒 + 𝟑 𝐇𝟐 O
LEARNING CHECK
Balance the following equation:
1. Ca𝐂𝐥𝟐 + 𝐊 𝟑 𝐏𝐎𝟒
𝐂𝐚𝟑 (𝐏𝐎𝟒 )2 + KCl
2. 𝐊 𝟐 𝐒𝐎𝟒 + NaOH
𝐍𝐚𝟐 𝐒𝐎𝟒 + KOH
Amounts of Reactants and Products
 To interpret a reaction quantitatively, we need to apply our knowledge of
molar masses and the mole concept.
 Stoichiometry is the quantitative study of reactants and products in a
chemical reaction.
 We use moles to calculate the amount of product formed in a reaction.
 This approach is called the mole method, which means simply that the
stoichiometric coefficients in a chemical equation can be interpreted as
the number of moles of each substance.
 For example, industrially ammonia is synthesized from hydrogen and
nitrogen as follows:
N2(g) + 3H2(g) →2NH3(g)
 Let’s consider a simple example in which 6.0 moles of H2 react
completely with N2 to form NH3. To calculate the amount of NH3
produced in moles, we use the conversion factor that has H2 in the
denominator and write
 Now suppose 16.0 g of H2 react completely with N2 to form
NH3. How many grams of NH3 will be formed?
The conversion steps are :
1. Convert 16.0 g of H2 to number of moles of H2, using the molar mass
of H2 as the conversion factor:
2. Calculate the number of moles of NH3 produced.
3. Calculate the mass of NH3 produced in grams using the molar mass of
NH3 as the conversion factor
• Similarly, we can calculate the mass in grams of N2 consumed in this
reaction. The conversion steps are :
 The general approach for solving stoichiometry problems is
summarized next.
1. Write a balanced equation for the reaction.
2. Convert the given amount of the reactant (in grams or other
units) to number of moles.
3. Use the mole ratio from the balanced equation to calculate the
number of moles of product formed.
4. Convert the moles of product to grams (or other units) of
product.
Limiting Reagents
 Reactants are usually not present in exact stoichiometric amounts, that is, in
the proportions indicated by the balanced equation.
 In stoichiometric calculations involving limiting reagents, the first step is to
decide which reactant is the limiting reagent.
 limiting reagent is the reactant used up first in a reaction .
 Excess reagents are the reactants present in quantities greater than necessary to
react with the quantity of the limiting reagent.
 Consider the industrial synthesis of methanol (CH3OH) from
carbon monoxide and hydrogen at high temperatures:
CO(g) + 2H2 (g)
→ CH3OH(g)
• Suppose initially we have 4 moles of CO and 6 moles of H2 .
Starting with 4 moles of CO, we find the number of moles of CH3OH
produced is :
and starting with 6 moles of H2, the number of moles of CH3OH
formed is
 Hydrogen produces a smaller amount of CH3OH, so it is called the
limiting reagent. Therefore, CO will be the excess reagent.
a)
b)
c)
Reaction Yield
 Theoretical yield of the reaction is the amount of product that would result
if all the limiting reagent reacted.
 The theoretical yield, then, is the maximum obtainable yield, predicted by
the balanced equation.
 In practice, the actual yield, or the amount of product actually obtained
from a reaction, is almost always less than the theoretical yield.
 The difference between actual and theoretical yields:
1. Many reactions are reversible, and so they do not proceed 100 percent
from left to right.
2. Difficult to recover all of the product from the reaction medium.
3. Some products react further among themselves or with the reactants to form
another new products.
 Percent yield, describes the proportion of the actual yield to the
theoretical yield. It is calculated as follows:
 Percent yields may range from a fraction of 1 percent to 100
percent.
a)
b)