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
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
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