Formulas and Percent Composition

Slides:



Advertisements
Similar presentations
Chapter 7- Molecular Formulas
Advertisements

Ch. 7 – The Mole Formula Calculations.
Percent Composition Empirical Formula Molecular Formula
Chemical Quantities or
HOLD UP YOUR BOARD! Chapter 7 Review game.
Chapter 7 Chemical Quantities
From percentage to formula
III. Formula Calculations
AP Chemistry Chapter 3 Chemical Quantities (The Mole)
Percent Composition Percent = part / whole Example: MgO
III. Formula Calculations (p )
Chemical Quantities or
Empirical and Molecular Formulas
85.7% C Percent Composition 24g C 28g CH4 mass of element x 100 x 100
The Mole Chapter 6.
Moles and Formula Mass.
12 Chemistry 2.3 gravimetric analysis CR 07 Empirical formulae The Empirical Formula (EF) is the ratio of all elements in a Compound It is the smallest/simplest.
Unit 11 Stoichiometry CP Chemistry.
What Could It Be? Finding Empirical and Molecular Formulas.
How Scientists Determine Formulas
Chapter 11 Empirical and Molecular Formulas
Empirical and Molecular Formulas
GRAB A CALCULATOR AND GET OUT A PIECE OF PAPER FOR NOTES. Empirical Formulas.
Calculating Empirical and Molecular Formulas
Section Percent Composition and Chemical Formulas
Percentage Composition and Empirical Formula
Section 5: Empirical and Molecular Formulas
Chemistry Notes Empirical & Molecular Formulas. Empirical Formula The empirical formula gives you the lowest, whole number ratio of elements in the compound.
Determining Chemical Formulas Experimentally % composition, empirical and molecular formula.
Percent Composition and Empirical Formula
Percent Composition, Empirical Formulas, Molecular Formulas
Molar Mass & Percent Composition
Percent Composition, Empirical Formulas, Molecular Formulas
  Proportion or ratio of each element in a chemical compound  Expressed as grams of element per 100 grams of chemical compound What is percent composition?
Empirical Formula The empirical formula indicates the ratio of the atoms of an element in a compound.
Formulas and Percent Composition Finding the Mystery Formulas.
Chemical Formulas and Molar Masses A few old ideas revisited and a few new.
Chemical Formulas and Molar Masses A few old ideas revisited and a few new.
The Mole and Chemical Composition
The Mole and Chemical Composition
Chemistry Dr. Davis. A. Percent Composition: the percentage by mass of each element in a compound 1. Looking at the formula for FeO, we might assume the.
Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Using Analytical Data The percentage composition is the percentage.
Using the MOLE. Percentage composition Percentage composition is the mass of individual elements in a compound expressed as a percentage of the mass of.
Percentage Percentage means ‘out of 100’
Molar Mass Practice Calculate the molar mass of the following: Hydrogen Fe Iron II sulfate.
What Could It Be? Empirical Formulas The empirical formula is the simplest whole number ratio of the atoms of each element in a compound. Note: it is.
3.10 Determining a Chemical Formula from Experimental Data
Mass % and % Composition Mass % = grams of element grams of compound X 100 % 8.20 g of Mg combines with 5.40 g of O to form a compound. What is the mass.
The Mole, part II Glencoe, Chapter 11 Sections 11.3 & 11.4.
Empirical & Molecular Formulas. Percent Composition Def – the percent by mass of each element in a compound Percent by mass = mass of element x 100 mass.
Mr. Chapman Chemistry 20. Converting from grams to moles Need: Moles and Mass worksheet.
Friday, April 26 th : “A” Day Monday, April 29 th : “B” Day Agenda  Begin 7.3: “Formulas & Percentage Composition”  In-Class Assignment: Practice.
Empirical & Molecular Formulas. Percent Composition Determine the elements present in a compound and their percent by mass. A 100g sample of a new compound.
Section 12.2: Using Moles (part 3). Mass Percent Steps: 1) Calculate mass of each element 2) Calculate total mass 3) Divide mass of element/ mass of compound.
Percent Composition, Empirical Formulas, Molecular Formulas
  Proportion or ratio of each element in a chemical compound  Percentage by mass of an element in a particular substance  Expressed as grams of element.
Calculating Empirical Formulas
Percent Composition What is the % mass composition (in grams) of the green markers compared to the all of the markers? % green markers = grams of green.
Formulas Empirical Formula – formula of a compound that expresses lowest whole number ratio of atoms. Molecular Formula – actual formula of a compound.
Empirical Formula EQ: How is the empirical formula calculated with experimental data?
IIIIII II. % Composition and Formula Calculations Ch. 3 – The Mole.
Percent Composition, Empirical and Molecular Formulas.
Percentage Composition
EMPIRICAL FORMULA The empirical formula represents the smallest ratio of atoms present in a compound. The molecular formula gives the total number of atoms.
Calculating Empirical and Molecular Formulas
Percent Composition Empirical Formula Molecular Formula
Empirical & Molecular Formulas
Chapter 11: More on the Mole
Final Review Day 2 Ch 11, 12 & 13.
Molecular Formula.
Presentation transcript:

Formulas and Percent Composition

Percent Composition The percent composition is the percentage by mass of each element in a compound This helps distinguish compounds made up of the same elements FeO or Fe2O3 In Iron (III) Oxide, Iron is 69.94% Fe and 30.06% O In Iron (II) Oxide, Iron is 77.73% Fe and 22.27% O

Percent Composition Calculate the percent composition of copper (I) sulfide. 1. Write the molecular formula Cu2S 2. Find the molar mass of each element in the compound (2 mol Cu)(63.55 g/mol Cu) = 127.1 g Cu (1 mol S)(32.07 g/mol S) = 32.07 g S Molar mass of Cu2S = 159.2 g/mol 3. Calculate the percent by mass of each element (127.1g Cu)/(159.2g Cu2S) x 100 = 79.85% Cu (32.07g S)/(159.2g Cu2S) x 100 = 20.15% S 4. Check to be sure it adds to 100%

Percent Composition Calculate the percent composition of aluminum nitrate. Al = 12.67% N = 19.73% O = 67.60%

Empirical Formulas The Empirical Formula is the simplest whole number ratio among the atoms in a compound Ammonium Nitrite NH4NO2 What is the simplest whole number ratio of each element in the compound? N2H4O2 becomes NH2O This is the empirical formula

Percent Composition and Empirical Formula One can determine the Empirical Formula from percent composition Chemical analysis of a liquid shows that it is 60.0% C, 13.4% H, and 26.6% O by mass. What is the empirical formula for the liquid? 1. Assume you have 100.0 g of the substance Then you have 60.0g of C 13.4g of H and 26.6g of O

Percent Composition and Empirical Formula 2. Convert from mass to moles (60.0g C)(1 mol C/12.011g C) = 5.00 mol C (13.4g H)(1 mol H/1.01g H) = 13.3 mol H (26.6g O)(1 mol O/16.00g O) = 1.66 mol O 3. Divide all moles by the smallest number to find the ratio 5.00 mol C/1.66 = 3.01 mol C 13.3 mol H/1.66 = 8.01 mol H 1.66 mol O/1.66 = 1.00 mol O 4. This is your Empirical Formula C3H8O

Empirical Formulas Benzoic Acid contains 68.8% Carbon, 4.95% Hydrogen, and 26.2% Oxygen. Find the empirical formula. (68.8gC)(1molC/12.01gC) = 5.73 mol C (4.95gH)(1molH/1.01gH) = 4.90 mol H (26.2gO)(1molO/16.00gO) = 1.64 mol O Divide by 1.64 (smallest number) 5.73molC/1.64 = 3.5 mol C 4.90molH/1.64 = 3.0 mol H 1.64molO/1.64 = 1.0 mol O

Empirical Formulas Can you have C3.5H3O? No! You must multiply all by 2 so you can have whole numbers 3.5 x 2, 3 x 2 and 1 x 2 Your final empirical formula is C7H6O2

Molecular Formulas Molecular formulas are multiples of empirical formulas CH2O C2H4O2 is acetic acid (2 x Empirical Formula) C6H12O6 is Glucose (6 x Empirical Formula) You can determine the molecular formula from the empirical formula and the mass of the compound

Molecular Formulas The empirical formula for a compound is P2O5. Its experimental molar mass is 284 g/mol. Determine the molecular formula for the compound. Determine the molar mass of the empirical formula 141.94 g/mol Find the ratio of the compound molar mass to the empirical formula molar mass (284 g/mol)/(141.94 g/mol) = 2.00 Multiply the empirical formula by this number 2(P2O5) = P4O10 Verify the results with a check of the molar mass of the new formula