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Percent Composition. 10.3. Percent Composition. I think you already know how to do this?. What did Bob score on his test? How do you figure it out?. 87 / 100. Bob. What did Bob score on his test? How do you figure it out? Part X 100 = 87 - PowerPoint PPT Presentation
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Percent Composition
10.3
Percent Composition
I think you already know how to do this?
What did Bob score on his test?
How do you figure it out? 87 / 100
Bob
What did Bob score on his test?
How do you figure it out?
Part X 100 = 87 Whole
87 / 100 Bob
What did Sue score on her paper?Sue
174 / 205
What did Sue score on her paper?
Part x 100 = 83 Whole
.8341 x 100 = 83
Sue
171 / 205
Percent Composition
= the relative amount of elements in a compound or substance.
The percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound, multiplied by 100%.
% mass of element = mass of element x 100% mass of compound
Sample Problem 10.9 on pg. 306
When 13.60 gram sample of a compound containing only oxygen and magnesium is decomposed, 5.40 g of oxygen is obtained. What is the percent composition of this compound? Knowns Unknowns Mass of compound= % Mg = Mass of oxygen= % O =Mass of magnezium=
Analyze Calculate Evaluate
% Composition from Chemical Formula
What is the percent composition of propane?C3H8
Knowns Unknowns
Calculate the percent composition of H2S?
Calculate the percent composition of Mg(OH)2?
10.2 HW
25. 22.4 L26. 567 g CaCO3
27. 11.0 mol C2H6O
28. 33.6 L Cl2
29. 39.9 g/mol30. Gas A: 28.0 g, nitrogen 31. The balloons have the same number of molecules. Each has 1 mole of gas or Avogadro’s number of particles. They will have different masses.
Percent Composition as a Conversion Factor
If you know the % composition you can use it to determine how much of a specific element you have.
Percent Composition as a Conversion Factor
If you know the % composition you can use it to determine how much of a specific element you have.
Propane is 82% C and 18% H. If you have 300 g of C3H8, how much C and H do you have?
Percent Composition as a Conversion Factor
Propane is 82% C and 18% H. If you have 300 g of C3H8, how much C and H do you have?
300 g C3H8 x 82 g C =
100 g C3H8
300 g C3H8 x 18 g H =
100 g C3H8
Analyze
Percent Composition as a Conversion Factor
Propane is 82% C and 18% H. If you have 300 g of C3H8, how much C and H do you have?
300 g C3H8 x 82 g C = 246 g C
100 g C3H8
300 g C3H8 x 18 g H = 54 g H
100 g C3H8
Calculate
Percent Composition as a Conversion Factor
Propane is 82% C and 18% H. If you have 300 g of C3H8, how much C and H do you have?
300 g C3H8 x 82 g C = 246 g C
100 g C3H8
300 g C3H8 x 18 g H = 54 g H
100 g C3H8
Evaluate
246
+ 54 _______ 300
Empirical Formula
= the empirical formula of a compound shows the smallest whole number ratio of the elements in the compound.
C2H2 Acetylene Empirical Formula
C8H8 Styrene
Empirical Formula and % Composition
A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula?
Empirical Formula and % Composition
A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula?
Empirical Formula and % Composition
A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? Know Unknown% N = 25.9 % N N?O?
% O = 74.1 % O
Because percent means parts per 100, you can assume the 100.0g of the compound contains 25.9g of N and 74.1g of O. Use these values and convert to moles.
Empirical Formula and % Composition
A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? Know Unknown% N = 25.9 % N N?O?
% O = 74.1 % O
25.9 g N X 1 mol N = 1.85 mol N 14.0 g N
74.1 g O X 1mol O = 4.63 mol O 16.0 g O
Empirical Formula and % Composition
A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? Know Unknown% N = 25.9 % N N?O?
% O = 74.1 % O
25.9 g N X 1 mol N = 1.85 mol N 14.0 g N
74.1 g O X 1mol O = 4.63 mol O 16.0 g O
Remember …. The
amount of N to O is a
ratio between the two
Empirical Formula and % Composition
A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? Know Unknown% N = 25.9 % N N?O?
% O = 74.1 % O
25.9 g N X 1 mol N = 1.85 mol N 14.0 g N
74.1 g O X 1mol O = 4.63 mol O 16.0 g O
1.85 mol N = 1 mol N : 4.63 mol O = 2.5 mol O 1.85 1.85
Empirical Formula and % Composition
A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? Know Unknown% N = 25.9 % N N?O?
% O = 74.1 % O
25.9 g N X 1 mol N = 1.85 mol N 14.0 g N74.1 g O X 1mol O = 4.63 mol O 16.0 g O
1.85 mol N = 1 mol N ; 4.63 mol O = 2.5 mol O 1.85 1.85
1 mol N X 2 = 2 mol N 2.5 mol O X 2 = 5 mol O
N2O5 is the empirical formula