CHM 1010 ‐ Sinex
Chemical Calculations
Chemical Calculations
• Moles and Molecules
• Moles and Chemical Reactions
• Moles, Chemical Reactions, and Molarity
• All done as UNIT CONVERSIONS!!!
• …and practice, practice, practice
CHM 1010 ‐ Sinex
1
Mass‐moles‐particles conversions
Mass of substance in grams
Molar mass of substance ‐ g/mole
Moles of substance
Conversion factors
Avogrado’s number ‐
6.02 x 1023 particles/mole
Number of particles of substance
(atoms, molecules, or ions)
Molar mass – sum of atomic masses from periodic table
Practice, practice, practice...
1
CHM 1010 ‐ Sinex
Chemical Calculations
How many molecules of aspirin (C9H8O4) in a dose of 1000 mg?
Find MM of aspirin:
MM = 9*12+8*1+4*16 = 180 g/mole
First find moles:
moles = 1000 mg *1g/1000mg * 1mole/180g = 0.0055 moles
Conversion factors in boxes
Now find molecules:
molecules = 0.0055 moles * 6.02x1023 molecules/mole = 3.3 x 1021 molecules
How many atoms of carbon?
atoms C = 9 atoms/molecule * 3.3 x 1021 molecules = 3.0 x 1022 atoms from the formula
3
For 12.6g NaHCO3, how many moles do you have?
Mass of substance A
MM
Moles of substance A
How many CO2 molecules in 0.17 moles CO2?
Moles of substance A
6.02 x 1023
Molecules of substance A
4
Practice, practice, practice...
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CHM 1010 ‐ Sinex
Chemical Calculations
Another way to find moles…
For a solution (homogeneous mixture), there are two components:
solute (substance being dissolved) and solvent (dissolving medium, usually water) A unit of concentration – amount solute per amount solution or solvent
Molarity = moles of solute/volume of solution in liters (moles/L)
Rearranging the definition of molarity:
moles = M x VL
Another way to find moles!
How many moles NaCl in 50 mL of 0.28 M NaCl?
50 mL x 1L/1000 mL x 0.28 mole/L = 0.014 mole NaCl
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What is the molarity if 35g KCl are dissolved to make 350 mL of solution?
Moles of substance A
Molarity
Volume of substance A
How many moles are in 300 mL of a 0.23 M NaOH solution?
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Practice, practice, practice...
3
CHM 1010 ‐ Sinex
Chemical Calculations
The ratio of moles in a reaction
2C4H10 + 13O2  8CO2 + 10H2O
Mole ratios
2
4
20
1
13
26
130
6.5
8
16
80
4
10
20
100
5
6
1.3
32
50
1.37
1.37
X 13/2
X 8/2
X 10/2
8.91
5.48
6.85
Click for answers
7
Mole‐to‐mole conversions
This comes from the balanced chemical reaction!
…and it relates substances in the reaction!
Moles of substance A
Mole ratio
Moles of substance B
2HCl + CaCO3  CaCl2 + H2O
How many moles of HCl react with a mole of CaCO3?
1 mole CaCO3 x 2 mole HCl/1 mole CaCO3 = 2 mole HCl
mole ratio
Practice, practice, practice...
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4
CHM 1010 ‐ Sinex
Chemical Calculations
Stoichiometry Mass of substance A
Mass of substance B
Molar mass of substance B
Molar mass of substance A
Moles of substance A
Mole ratio
Moles of substance B
Molarity of solution A
Molarity of solution B
Volume of substance A
Volume of substance B
Conversion factors
Molarity (M) = moles of solute/volume in liters of solution – moles/L
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Mass of substance A
How many grams of O2 are required to reaction with 10g C3H8? Mass of substance B
Molar mass of substance B
Molar mass of substance A
Mole ratio
Moles of substance A
Moles of substance B
Molarity of solution A
Molarity of solution B
Volume of substance A
Volume of substance B
C3H8 + 5O2  3CO2 + 4H2O
10g C3H8 x 1 mole/44g = 0.23 moles C3H8
Conversion factors in boxes
0.23 moles C3H8 x 5 moles O2/1mole C3H8 = 1.15 moles O2
1.15 moles O2 x 32.0g/mole = 37 g O2
This is a classic stoichiometry problem!
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Practice, practice, practice...
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CHM 1010 ‐ Sinex
Chemical Calculations
1
How many grams of O2 are required to reaction with 10g C3H8? 3
Mass of substance A
Molar mass of substance B
Molar mass of substance A
Mole ratio
Moles of substance A
2
Molarity of solution A
Moles of substance B
Molarity of solution B
Volume of substance A
Another view of solving
Mass of substance B
Volume of substance B
C3H8 + 5O2  3CO2 + 4H2O
Step 1
10g C3H8 x 1 mole/44g = 0.23 moles C3H8
Step 2
0.23 moles C3H8 x 5 moles O2/1mole C3H8 = 1.15 moles O2
Step 3
1.15 moles O2 x 32.0g/mole = 37 g O2
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Mass of substance A
If 74g O2 react, how many grams of CO2 will be produced? Mass of substance B
Molar mass of substance B
Molar mass of substance A
Mole ratio
Moles of substance A
Molarity of solution A
Volume of substance A
Moles of substance B
Molarity of solution B
Volume of substance B
C3H8 + 5O2  3CO2 + 4H2O
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Practice, practice, practice...
6
CHM 1010 ‐ Sinex
Chemical Calculations
Mass of substance A
What volume of stomach acid, 0.16 M HCl, reacts with 1.0 g CaCO3?
Mass of substance B
Molar mass of substance B
Molar mass of substance A
Mole ratio
Moles of substance A
Moles of substance B
Molarity of solution A
Molarity of solution B
Volume of substance A
Volume of substance B
2HCl + CaCO3  CaCl2 + H2O
1.0 g CaCO3 x 1 mole/100g = 0.010 mole CaCO3
0.010 mole CaCO3 x 2 mole HCl/1 mole CaCO3 = 0.020 mole HCl
0.020 mole HCl x 1L/0.16 moles = 0.13 L = 130 mL stomach acid 13
Mass of substance A
What volume of 0.25 M CH3COOH will reaction with 25.0 mL of 0.37M NaOH?
Mass of substance B
Molar mass of substance B
Molar mass of substance A
Mole ratio
Moles of substance A
Molarity of solution A
Volume of substance A
Moles of substance B
Molarity of solution B
Volume of substance B
CH3COOH + NaOH  CH3COONa + H2O
25.0 mL x 1L/1000 mL x 0.37 mole/L = 0.0093 moles NaOH
0.0093 moles NaOH x 1 mole CH3COOH/1 mole NaOH = 0.0093 mole CH3COOH
0.0093 mole x 1L/0.25 moles = 0.037 L = 37 mL CH3COOH
This is a classic titration problem!
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7
CHM 1010 ‐ Sinex
Chemical Calculations
Could you find one of the conversion factors?
Suppose 2.0g CaCO3
reacted with 25.0 mL of HCl. What is the molarity of the HCl?
Mass of substance A
Mass of substance B
Molar mass of substance B
Molar mass of substance A
Mole ratio
Moles of substance A
Moles of substance B
Molarity of solution A
Molarity of solution B
Volume of substance A
Volume of substance B
2HCl + CaCO3  CaCl2 + H2O
2.0 g CaCO3 x 1 mole/100g = 0.020mole CaCO3
0.020 mole CaCO3 x 2 mole HCl/1 mole CaCO3 = 0.040 mole HCl
Molarity = moles/ VL = 0.040 mole HCl / (25.0 mL x 1L/1000mL) = 1.6 M HCl Moles of substance B
Volume of substance B
15
Mass of substance A
How many grams of Al(OH)3 will react with 100 mL of 0.15 M HCl?
Mass of substance B
Molar mass of substance B
Molar mass of substance A
Mole ratio
Moles of substance A
Molarity of solution A
Volume of substance A
Moles of substance B
Molarity of solution B
Volume of substance B
Al(OH)3 + 3HCl  AlCl3 + 3H2O
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Practice, practice, practice...
8
CHM 1010 ‐ Sinex
Chemical Calculations
For a 1.0 kg of Al2O3, how much aluminum metal, in grams, can be produced?
2Al2O3 (l) + 3C (s)  4Al (l) + 3CO2 (g)
How many atoms of carbon are consumed for the problem above?
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How many grams of CH3COOH are needed to react with 27.3 mL of 0.21 M NaOH?
CH3COOH + NaOH  CH3COONa + H2O
If 0.121g HX react with 26.2 mL of 0.229 M NaOH, what is the molar mass of HX?
HX + NaOH  NaX + H2O
What is element X?
Practice, practice, practice...
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9
CHM 1010 ‐ Sinex
Chemical Calculations
What if two reagents are added, how much product do you get?
Suppose 10 g Na and 10 g Cl2 are mixed together. How much NaCl can be produced?
2Na + Cl2  2NaCl
2Na + 1Cl2
First find the moles of each reactant:
10 g Na x 1 mole/23.0 g = 0.43 mole Na
10 g Cl2 x 1 mole/71.0 g = 0.14 mole Cl2
Find the moles of NaCl produced for each reactant:
2
1
Have 0.43
Need 0.22
Need 0.28
Have 0.14
0.43 mole Na x 2 mole NaCl/2 mole Na = 0.43 mole BrCl
0.14 mole Cl2 x 2 mole NaCl/1 mole Cl2 = 0.28 mole NaCl
 This is what can be produced! Cl2
is limiting reagent.
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A chemical analysis problem
A chemist decides to check the molarity of H2SO4 in her car battery. A 10.0 mL sample is reacted with 31.03 mL of 2.73M NaOH. What is the molarity of H2SO4?
H2SO4 + 2NaOH  Na2SO4 + 2H2O
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Practice, practice, practice...
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Practice, practice, practice...
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21
Answers to problems by slide number:
4‐ 0.15 mole NaHCO3; 1.0 x 1023 molecules CO2
6‐ 1.3 M KCl; 0.069 mole NaOH
12‐ 61g CO2
16‐ 0.39g Al(OH)3
17‐ 5.3 x 102g Al; 8.9 x 1024 atoms C
18‐ 0.34g CH3COOH; MM = 20.2 g/mole; X = F
20‐ 4.2M H2SO4
CHM 1010 ‐ Sinex
Chemical Calculations