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IIIIII Formula Calculations Ch. 7 – The Mole. A. Percentage Composition n the percentage by mass of each element in a compound EMPIRICAL and MOLECULAR.

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Presentation on theme: "IIIIII Formula Calculations Ch. 7 – The Mole. A. Percentage Composition n the percentage by mass of each element in a compound EMPIRICAL and MOLECULAR."— Presentation transcript:

1 IIIIII Formula Calculations Ch. 7 – The Mole

2 A. Percentage Composition n the percentage by mass of each element in a compound EMPIRICAL and MOLECULAR FORMULAS

3 100 = A. Percentage Composition %Cu = g Cu g Cu 2 S 100 = %S = g S g Cu 2 S % Cu 20.15% S n Find the % composition of Cu 2 S.

4 %Fe = 28 g 36 g 100 = 78% Fe %O = 8.0 g 36 g 100 = 22% O n Find the percentage composition of a sample that is 28 g Fe and 8.0 g O. A. Percentage Composition

5 n How many grams of copper are in a 38.0-gram sample of Cu 2 S? (38.0 g Cu 2 S)( ) = 30.3 g Cu Cu 2 S is % Cu A. Percentage Composition

6 100 = %H 2 O = g H 2 O g H 2 O 24.51% H 2 O n Find the mass percentage of water in calcium chloride dihydrate, CaCl 2 2H 2 O? A. Percentage Composition

7 B. Empirical Formula C2H6C2H6 CH 3 reduce subscripts n Smallest whole number ratio of atoms in a compound

8 B. Empirical Formula 1. Find mass (or %) of each element. 2. Find moles of each element. 3. Divide moles by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #s.

9 What ?????????? How can I remember? n There is a poem to help 1. Percent to Mass 2. Mass to Mole 3. Divide by Small 4. Multiply til Whole n Sometimes the mole values are still not whole numbers so multiply by the smallest factor that will make them whole numbers

10 B. Empirical Formula n Find the empirical formula for a sample of 25.9% N and 74.1% O g N 1 mol N g N = 1.85 mol N 74.1 g 1 mol g = 4.63 mol O 1.85 mol N 1.85 mol = 1 N = 2.5 O

11 B. Empirical Formula N 1 O 2.5 Need to make the subscripts whole numbers multiply by 2 N2O5N2O5

12 Example of Determining Empirical Formula Ex. What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% Oxygen? 25.9g N1 mol N = 1.85 mol N 14.0g N 74.1g O1 mol O = 4.63 mol O 16.0g O Divide each element by the lowest number of moles in the compound: 1.85 mol N = 1 mol N mol O = 2.50 mol O 1.85 Now find the lowest common denominator. 1 mol N x 2= N mol O x 2 = O 5 Therefore, N 2 O 5 is the empirical formula

13 Practice Problems n Calculate the empirical formula for a compound that is 94.1% O and 5.90%H n Calculate the empirical formula for a compound that is 79.8% C and 20.2% H

14 C. Molecular Formula n True Formula - the actual number of atoms in a compound CH 3 C2H6C2H6 empirical formula molecular formula ?

15 C. Molecular Formula 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empirical mass. 4. Multiply each subscript by the answer from step 3.

16 C. Molecular Formula n The empirical formula for ethylene is CH 2. Find the molecular formula if the molecular mass is 28.1 g/mol? 28.1 g/mol g/mol = 2.00 empirical mass = g/mol 2(CH 2 ) C 2 H 4

17 Example of Molecular Formula Calculate the molecular formula of the compound whose Molar mass is 60.0g and empirical formula is CH 4 N. Molar Mass of Compound Empirical formula Empirical formula mass (efm) Molar Mass efm Molecular Formula 60.0gCH 4 N = 30.0g CH 4 N 60.0/30.0 = 2 CH 4 N x 2 = C 2 H 8 N 2


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