unit 1 9 chapter 3 5 n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Finding Empirical Formulas PowerPoint Presentation
Download Presentation
Finding Empirical Formulas

Loading in 2 Seconds...

play fullscreen
1 / 28

Finding Empirical Formulas - PowerPoint PPT Presentation


  • 136 Views
  • Uploaded on

Unit 1-9 Chapter 3.5. Finding Empirical Formulas. Formulas. Empirical formula : the lowest whole number ratio of atoms in a compound. molecular formula = (empirical formula) n [ n = integer] molecular formula = C 6 H 6 = (CH) 6 empirical formula = CH.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Finding Empirical Formulas' - keilah


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
formulas
Formulas

Empirical formula: the lowest whole number ratio of atoms in a compound.

  • molecular formula = (empirical formula)n [n = integer]
  • molecular formula = C6H6 = (CH)6
  • empirical formula = CH

Molecular formula: the true number of atoms of each element in the formula of a compound.

formulas continued
Formulas (continued)

Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio).

Examples:

NaCl

MgCl2

Al2(SO4)3

K2CO3

formulas continued1
Formulas (continued)

Formulas for molecular compoundsMIGHTbe empirical (lowest whole number ratio).

Molecular:

C6H12O6

H2O

C12H22O11

Empirical:

H2O

CH2O

C12H22O11

calculating empirical formulas by mass
Calculating Empirical Formulas by mass %

One can calculate the empirical formula from the percent composition:

  • Base calculation on 100 grams of compound.
  • Determine moles of each element in 100 grams of compound.
  • Divide each value of moles by the smallest of the values.
  • Multiply each number by an integer to obtain all whole numbers.
calculating empirical formulas
Calculating Empirical Formulas

The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

calculating empirical formulas1

Assuming 100.00 g of para-aminobenzoic acid,

C: 61.31 g x = 5.105 mol C

H: 5.14 g x = 5.09 mol H

N: 10.21 g x = 0.7288 mol N

O: 23.33 g x = 1.456 mol O

1 mol

16.00 g

1 mol

14.01 g

1 mol

1.01 g

1 mol

12.01 g

Calculating Empirical Formulas
calculating empirical formulas2

Calculate the mole ratio by dividing by the smallest number of moles:

C: = 7.005  7

H: = 6.984  7

N: = 1.000

O: = 2.001  2

5.09 mol

0.7288 mol

0.7288 mol

0.7288 mol

1.458 mol

0.7288 mol

5.105 mol

0.7288 mol

Calculating Empirical Formulas
calculating empirical formulas3
Calculating Empirical Formulas

These are the subscripts for the empirical formula:

C7H7NO2

combustion analysis finding mass
Combustion Analysis (finding mass %)
  • Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this
    • C is determined from the mass of CO2 produced
    • H is determined from the mass of H2O produced
    • O is determined by difference after the C and H have been determined (law of conservation of mass)
elemental analyses
Elemental Analyses
  • Compounds containing other elements are analyzed using methods analogous to those used for C, H and O
  • Example: Ag2O
finding the molecular formula
Finding the Molecular Formula

The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?

1. Find the formula mass of C3H5O2

3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

finding the molecular formula1
Finding the Molecular Formula

The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?

2. Divide the molecular mass by the mass given by the emipirical formula.

3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

finding the molecular formula2
Finding the Molecular Formula

The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?

3. Multiply the empirical formula by this number to get the molecular formula.

3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

(C3H5O2) x 2 =

C6H10O4

homework
Homework
  • Page 98: 3.45, 47, 49, 51, 53, 55
stoichiometric calculations
Stoichiometric Calculations

The coefficients in the balanced equation give the ratio of moles of reactants and products

stoichiometric calculations1
Stoichiometric Calculations

From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)

stoichiometric calculations2
Stoichiometric Calculations

C6H12O6 + 6 O2 6 CO2 + 6 H2O

Starting with 1.00 g of C6H12O6…

we calculate the moles of C6H12O6…

use the coefficients to find the moles of H2O…

and then turn the moles of water to grams

homework1
Homework
  • Page 99: 57, 59, 61, 63
how many cookies can i make
How Many Cookies Can I Make?
  • You can make cookies until you run out of one of the ingredients
  • Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat)
how many cookies can i make1
How Many Cookies Can I Make?
  • In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make
limiting reactants reagents
Limiting Reactants/reagents
  • The limiting reactant is the reactant present in the smallest stoichiometric amount
    • In other words, it’s the reactant you’ll run out of first (in this case, the H2)
limiting reactants
Limiting Reactants

In the example below, the O2 would be the excess reactant/reagent

theoretical yield
Theoretical Yield
  • The theoretical yield is the amount of product that can be made
    • In other words it’s the amount of product possible as calculated through the stoichiometry problem
  • This is different from the actual yield, the amount one actually produces and measures
percent yield

Actual Yield

Theoretical Yield

Percent Yield = x 100

Percent Yield

A comparison of the amount actually obtained to the amount it was possible to make

homework2
Homework
  • Page 100: 67, 75, 77, 79