1. Chapter 10 Chemical Quantities
10.1 The Mole: A Measurement of Matter
10.2 Mole-Mass and Mole-Volume Relationships
10.3 Percent Composition and Chemical Formulas
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2. Percent Composition of a Compound
percent composition: percent by mass of each element in a compound
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3. Percent Composition of a Compound
Percent Composition from the Chemical Formula
% by mass of element
mass of element in 1 mol compound
molar mass of compound
× 100% =
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4. Percent Composition of a Compound
Percent Composition from the Chemical Formula
Calculate the percent composition of
a) Al(OH)3
b) K2S
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5. Percent Composition of a Compound
Percent Composition from the Chemical Formula
Calculate the percent composition of
a) Al(OH)3
Al: 34.6%, O: 61.5%, H: 3.9%
b) K2S
K: 70.9%, S: 29.1%
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6. Percent Composition of a Compound
Percent Composition from Mass Data
% by mass of element =
mass of element
mass of compound
× 100%
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7. Calculate the mass of nitrogen and the mass of hydrogen in
Sample Problem 10.11
Calculating the Mass of an Element in a Compound Using Percent Composition
Calculate the mass of nitrogen and the mass of hydrogen in
125 g of NH3 fertilizer.
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8. 1. Calculate the % composition for each component.
Sample Problem 10.11
1. Calculate the % composition for each component.
14.0 g N
17.0 g NH3
3.0 g H
17.0 g NH3
% N = = 82.35%
% H = =17.65%
2. Multiply each % composition by the mass of NH3.
125 g NH3 ×
= 103 g N
125 g NH3 ×
= 22 g H
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9. Does the result make sense?
Sample Problem 10.11
Does the result make sense?
The sum of the two masses equals 125 g,
(103 g N + 22 g H = 125 g NH3).
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10. Calculate the masses of hydrogen, sulfur, and oxygen in 90
Calculate the masses of hydrogen, sulfur, and oxygen in 90.0 g of H2SO4.
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11. Calculate the masses of hydrogen, sulfur, and oxygen in 90
Calculate the masses of hydrogen, sulfur, and oxygen in 90.0 g of H2SO4.
Hydrogen: 1.8 g
Sulfur: g
Oxygen: g
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12. When a 14.2-g sample of HgO is decomposed into its elements by heating, 13.2 g Hg is obtained. What is the percent composition of the compound?
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13. mass of Hg × 100% % Hg = mass of compound 13.20 g = 14.20 g = 93.0% Hg
Sample Problem 10.9
% Hg =
mass of Hg
mass of compound
× 100%
= 93.0% Hg =
13.20 g
14.20 g
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14. mass of O × 100% % O = mass of compound 1.0 g = 14.2 g = 7.0% O
Sample Problem 10.9
% O =
mass of O
mass of compound
× 100%
1.0 g
= 7.0% O =
14.2 g
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15. Check that percents add up to 100%. 93.0% + 7.0% = 100%
Sample Problem 10.9
Does the result make sense?
Check that percents add up to 100%.
93.0% % = 100%
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16. Do Now: Calculate the percent composition of acetylene, C2H2.
C2H2 is used in welder’s torches.
2. Calculate the percent
composition of
styrene, C8H8.
Styrene is used to make polystyrene.
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17. Empirical Formulas
empirical formula: lowest whole-number ratio of atoms or moles of elements in a compound.
Compound
Molecular Formula
Empirical Formula
Hydrogen Peroxide
H2O2 HO
Water
H2O
Sucrose
C6H12O6
CH2O
Acetylene
C2H2 CH
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18. Determining the Empirical Formula of a Compound
Sample Problem 10.12
Determining the Empirical Formula of a Compound
A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula of the compound?
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19. Empirical Formula Determination Start with percent composition
Empirical Formulas
Empirical Formula Determination
Start with percent composition
Assume you have 100 g of compound. The % composition will equal the mass of each element.
Convert each mass to moles.
Determine lowest whole number mole ratio
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20. Analyze List the knowns and the unknown.
Sample Problem 10.12
Analyze List the knowns and the unknown. 1
KNOWNS percent by mass of N = 25.9% N percent by mass of O = 74.1% O
Assuming there is 100 g of compound, then you would have 25.9 g N and 74.1 g O
UNKNOWN empirical formula = N?O?
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21. Calculate Solve for the unknown.
Sample Problem 10.12
Calculate Solve for the unknown. 2
Convert mass to moles.
25.9 g N ×
1 mol N
14.0 g N
= 1.85 mol N
74.1 g O ×
1 mol O
16.0 g O
= 4.63 mol O
The mole ratio of N to O is N1.85O4.63.
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22. Calculate Solve for the unknown.
Sample Problem 10.12
Calculate Solve for the unknown. 2
Divide each molar quantity by the smaller number of moles to get 1 mol for the element with the smaller number of moles.
1.85 mol N
1.85
= 1 mol N
4.63 mol O
1.85
= 2.50 mol O
The mole ratio of N to O is N1O2.5.
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23. Calculate Solve for the unknown.
Sample Problem 10.12
Calculate Solve for the unknown. 2
Multiply each part of the ratio by the smallest whole number that will convert both subscripts to whole numbers.
1 mol N × 2 = 2 mol N
2.5 mol O × 2 = 5 mol O
The empirical formula is N2O5.
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24. A compound is analyzed and found to contain: 36.1% Ca 63.9% Cl
What is the empirical formula for the compound?
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25. A compound is analyzed and found to contain: 36.1% Ca 63.9% Cl
What is the empirical formula for the compound? CaCl2
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26. Finding the Molecular Formula of a Compound
Sample Problem 10.13
Finding the Molecular Formula of a Compound
Calculate the empirical formula mass.
Divide the molar mass by the empirical formula mass.
Multiply the empirical formula subscripts by the answer from step 2.
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27. Finding the Molecular Formula of a Compound
Sample Problem 10.13
Finding the Molecular Formula of a Compound
Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N.
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28. 1. Calculate the empirical formula mass (efm).
Sample Problem 10.13
1. Calculate the empirical formula mass (efm).
efm of CH4N = 12.0 g/mol + 4(1.0 g/mol) g/mol
= 30.0 g/mol
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29. 2. Divide the molar mass by the empirical formula mass.
Sample Problem 10.13
2. Divide the molar mass by the empirical formula mass.
molar mass
efm =
60.0 g/mol
30.0 g/mol
= 2
3. Multiply the formula subscripts by this value.
(CH4N) × 2 = C2H8N2
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30. Evaluate Does the result make sense?
Sample Problem 10.13
Evaluate Does the result make sense? 3
The molecular formula has the molar mass of the compound.
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31. A compound is composed of carbon and hydrogen and has a molar mass of 86.0 g/mol. It’s empirical formula is C3H7. What is the molecular formula?
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32. A compound is composed of carbon and hydrogen and has a molar mass of 86.0 g/mol. It’s empirical formula is C3H7. What is the molecular formula? C6H14
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33. Calculate the grams of nitrogen in 125 g of NH4NO3.
Do Now:
Calculate the grams of nitrogen in 125 g of NH4NO3.
Calculate the empirical formula for a compound that is 94.1% O and 5.9% H.
What is the empirical formula for C6H12O6?
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34. Calculate the grams of nitrogen in 125 g of NH4NO3. 43.8 g
Do Now:
Calculate the grams of nitrogen in 125 g of NH4NO g
Calculate the empirical formula for a compound that is 94.1% O and 5.9% H. OH
What is the empirical formula for C6H12O6?
CH2O
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35. A compound is analyzed and found to contain: 40% C 6.7% H 53.3% O
What is the empirical formula for the compound?
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36. A compound is analyzed and found to contain: 40% C 6.7% H 53.3% O
What is the empirical formula for the compound? CH2O
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37. A compound is analyzed and found to contain: 3.7% H 44.4% C 51.9% N
What is the empirical formula for the compound?
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38. A compound is analyzed and found to contain: 3.7% H 44.4% C 51.9% N
What is the empirical formula for the compound? HCN
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39. END OF 10.3
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40. What is the density of N2 gas at STP?
Do Now:
What is the density of N2 gas at STP?
A gas has a density of g/L. What is the molar mass of this gas?
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41. What is the density of N2 gas at STP? 1.25 g/L
Do Now:
What is the density of N2 gas at STP?
1.25 g/L
A gas has a density of g/L. What is the molar mass of this gas?
17.7 g/mol
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42. Take out the “OREO” lab handout.
Do Now:
Sit with your lab group.
Take out the “OREO” lab handout.
Work through the last 4 problems on the back of the lab handout.
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