• One mole is the number of atoms in exactly 12.0 g of the pure isotope carbon-12

• Avogadro's number (N_A) is the number of atoms in one mole
  N_A = 6.022 x 10²³ atoms/mole
  Avogadro's number is also the number of molecules/mole; ions/mole; nuclei/mole; electrons/mole

• The mass of 1 mole of any naturally occurring element is the sum of the contributions from each of its isotopes:

  Elemental molar mass = (fractional abundance) X (isotopic molar mass)

• One mole of any chemical compound is 1 mole of its chemical formula unit

• The mass of one mole of a chemical compound is referred to as the molecular weight (or molar mass) of the compound

Example Problems: The Mole

• Determing number of moles in a specific amount
• Calculating the molar mass of a chemical compound
• Finding mass from moles/atoms/etc.
• Typical problems involving the mole

Calculate the number of moles in the following masses:
(a) 7.85 g of Fe (b) 6.55 x 10¹³ atoms of ¹⁴C

Approach to solving these problems:

1. Determine what it is we need to calculate
   
2. Determine what information we have or can use for this calculation
   
3. Set up equation to solve

   
   

1. Want to calculate number of moles of Fe
   
2. We are given the following information
   
   1. We have 7.85 g of Fe
      
   2. We know the molar mass of Fe: 55.847 g/mol
      
3. Set up the equation

   (1 mole Fe/55.847 g Fe) x 7.85 g Fe = 0.141 mole Fe

1. Want to calculate number of moles of ¹⁴C
   
2. We are given the following information
   
   1. We have6.55 x 10¹³ atoms of ¹⁴C
      
   2. We know N_A = 6.022 x 10²³ atoms/mole
      
3. Set up the equation

   (1 mole ¹⁴C/6.022 x 10²³ atoms ¹⁴C) x (6.55 x 10¹³ atoms ¹⁴C)= 1.09 x 10^-10 mole ¹⁴C

Calculate the molar mass of ammonium carbonate

1. First we have to know the chemical formula of ammonium carbonate
   The ammonium ion is NH₄⁺ and the carbonate anion is CO₃^2-. Thus, the chemical formula is (NH₄)₂CO₃

2. Next, we want to calculate the molar mass
   
3. We know that the molar mass of a chemical compound is the mass of 1 mole of a chemical compound and we know that 1 mole of a chemical compound is one mole of its chemical formula unit.
   
4. Therefore, the final step is to calculate the answer:

   molar mass = (2 x N) + (8 x H) + (1 x C) + (3 x O)
            = (2 x 14.007) + (8 x 1.008) + (1 x 12.011) + (3 x 15.999)
            = 28.014 + 8.064 + 12.011 + 47.997
            = 96.086 g/mol

    molar mass = 169.872 g/mol

Listing of Example Problems: The Mole

Calculate the mass of 375,000 molecules of methane

1. How many grams of methane (CH₄)
   
2. What do we know?
   
   1. 375,000 molecules of methane
      
   2. 6.022 x 10²³ molecules methane/mole methane
      
   3. We need to know the molar mass of methane

      molar mass CH₄ = (1 x C) + (4 x H)
          = (1 x 12.011) + (4 x 1.008)
          = 12.011 + 4.032
          = 16.043 g/mol

3. Setup equation and solve
   (16.043 g CH₄/1 mole CH₄) x (1 mole CH₄/6.022 x 10²³ molecules CH₄) x 375,000 molecules CH₄ = 9.990 x 10^-18 g CH₄

Calculate the mass of one molecule of chlorophyll (C₅₅H₇₂MgN₄O₅)

1. Question asks us to find grams (mass) of chlorophyll (C₅₅H₇₂MgN₄O₅)
   
2. We know the following:
   
   1. Avagadro's Number - 6.022 x 10²³ molecules chlorophyll/mole chlorophyll
      
   2. We can calculate the molar mass of chlorophyll: 893.509 g/mol

3. Set up equation and solve
   (893.509 g chlorophyll/1 mole chlorophyll) x (1 mole chlorophyll/6.022 x 10²³ molecules chlorophyll) x 1 molecule chlorophyll = 1.484 x 10^-21 g chlorophyll

In 1992 the United States manufactured 2.47 x 10⁸ pounds of phosphoric acid (H₃PO₄). How many moles is this? If 35% of this material was made by burning elemental phosphorus, how many moles and how many kilograms of phosphorus were consumed?

Note: The first thing to do when you see a problem like this is not to panic. Reread the question and see what it is looking for

1. First question: How many moles H₃PO₄?
   
2. We know:
   
   1. 2.47 x 10⁸ pounds H₃PO₄
      
   2. The conversion factor between kilogram and pounds is 0.453592 kg/pound
      
   3. We need to know the molar mass of H₃PO₄
      molar mass H₃PO₄ = (3 x H) + (1 x P) + (4 x O) = 97.994 g/mol

3. Now setup your equation and solve
   (1 mole H₃PO₄/97.994 g H₃PO₄) x (10³ g H₃PO₄/1 kg H₃PO₄) x (0.453592 kg H₃PO₄/1 pound H₃PO₄) x (2.47 x 10⁸ pounds H₃PO₄) = 1.14 x 10⁹ mole H₃PO₄

   ---

4. Second question: How many moles Phosphorus?
   
5. We know:
   1. 35% of this material comes from phosphorus
      (0.35) x (2.47 x 10⁸ pounds) = 8.64 x 10⁷ pounds H₃PO₄
      
   2. The conversion factor between kilogram and pounds is 0.453592 kg/pound
      
   3. The molar mass of H₃PO₄ was calculated above: 97.994 g H₃PO₄/mol H₃PO₄
      
   4. We also know one more thing: the mole ratio of phosphorus to phosphoric acid
      1 mole P/ 1 mol H₃PO₄
      Remember, the mole ratio comes from the chemical formula

6. Now setup your equation and solve
   (1 mole P,1 mole H₃PO₄) x (1 mole H₃PO₄/97.994 g H₃PO₄) x (10³ g H₃PO₄/1 kg H₃PO₄) x (0.453592 kg H₃PO₄/ 1 pound H₃PO₄) x (8.64 x 10⁷ pounds H₃PO₄) = 4.00 x 10⁸ mole P

   But wait, there is a quicker way to the answer:

   
   From part one of question we know that we have 1.14 x 10⁹ mole H₃PO₄ so,
   (1 mole P/1 mole H₃PO₄) x (0.35)(1.14 x 10⁹ mole H₃PO₄) = 4.00 x 10⁸ mole P

   ---

7. Third question: How many kilograms of phosphorus?
   
8. We know:
   1. We have 4.00 x 10⁸ mole P (We calculated that directly above)
      
   2. The molar mass of phosphorus is 30.974 g P/ mol P

9. Setup the equation and solve:
   (1 kg P/10³ g P) x (30.974 g P/ mole P) x (4.00 x 10⁸ mole P) = 1.24 x 10⁷ kg P

1. Find moles
   
2. We know the following:
   
   1. 0.035 mg C₂₀H₂₄O₂
      
   2. The molar mass of C₂₀H₂₄O₂ is calculated to be 296.41 g/mol
      
   3. 1 g = 10³ mg
      
3. Setup equation and solve
   (1 mole C₂₀H₂₄O₂/296.41 g C₂₀H₂₄O₂) x (1 g C₂₀H₂₄O₂/10³ mg C₂₀H₂₄O₂) x 0.035 mg C₂₀H₂₄O₂ = 1.2 x 10^-7 moles C₂₀H₂₄O₂

   Listing of Example Problems: The Mole

• Elemental Analysis ­ the percent by mass of each element present in a compound (also called the mass percent composition)
  
• Empirical Formula ­ the formula with the smallest set of whole numbers that match the elemental analysis
  
• Chemical Formula ­ the formula with the correct set of whole numbers

Example Problems: Determining Chemical Formulas

• Mass percent calculations
• Determining empirical formulas
• Combustion Analysis

Given the chemical formula NH₄NO₃, determine the mass percent composition

1. First we need to determine the molar mass of NH₄NO₃
   Molar Mass of NH₄NO₃ = (2 x 14.007) + (4 x 1.008) + (3 x 15.999) = 80.05 g/mol
   
2. Next we find the amount of each element relative to the total molar mass:
   1. From the chemical formula we know there are 2 moles of nitrogen for every one mole of ammonium nitrate
      % N = (mass of N/mass of NH₄NO₃) x 100%
          =  ((2 x 14.007)/80.05) x 100%
          =  35%
      
   2. There are four moles of hydrogen for every one mole of ammonium nitrate
      % H = ((4 x 1.008)/80.05) x 100% = 5.04%
      
   3. And there are three moles of oxygen for every one mole of ammonium nitrate
      
   4. % O = ((3 x 15.999)/80.05) x 100% = 59.96%

Guidelines for solving elemental analysis problems

1. Base the formula determination on a 100 gram sample
   
2. Divide each mass percentage by the molar mass of the element to obtain the number of moles of the element in a 100 gram sample
   
3. Divide each molar amount by whichever amount is the smallest
   
4. If some results from Step 3 are far from integers, multiply through by a common factor that converts all molar amounts to integers or near-integers
   
5. Round off each molar number to the nearest integer

Given the following mass percent composition, determine the empirical formula.
   49.5 %C; 5.2 %H; 28.8 %N; 16.5 %O

1. Assume 100 g sample: 49.5 g C; 5.2 g H; 28.8 g N; 16.5 g O
   
2. Divide each mass number by the molar mass:
   (49.5 g C/12.011 g/mol C) = 4.121 mol C
   (5.2 g H/1.008 g/mol H) = 5.159 mol H
   (28.8 g N/14.007 g/mol N) = 2.056 mol N
   (16.5 g O/15.999 g/mol O) = 1.031 mol O

3. Divide each molar amount by smallest amount
   (4.121 mol C/1.031 mol O) = (4 mol C/mol O)
   (5.159 mol H/1.031 mol O) = (5 mol H/mol O)
   (2.056 mol N/1.031 mol O) = (2 mol N/mol O)
   (1.031 mol O/1.031 mol O) = (1 mol O/mol O)

4. Empirical Formula is then C₄H₅N₂O

What is the chemical formula for this compound if the molar mass is 194.2 g/mol?

1. First find molar mass of the empirical formula
   Molar Mass = (4 x 12.011) + (5 x 1.008) + (2 x 14.007) + (1 x 15.999) = 97.097 g/mol
   
2. Notice that, in this case, the empirical formula is not the chemical formula (because the molar masses are different). But the chemical formula must be some integer factor of the empirical formula (thus, the molar mass of the chemical formula must be some integer factor of the empirical formula).
   
3. (Factor) X Molar Mass Empirical Formula = Molar Mass Chemical Formula
   (Factor) X 97.097 g/mol = 194.2 g/mol
   (Factor) = 2
   
4. Therefore, the chemical formula is C₈H₁₀N₄O₂

Combustion Analysis is one of the most widely used methods for determining the empirical formula of an unknown compound. Most widely used for compounds which contain carbon. In combustion analysis, an accurately known mass of a compound is burned in a stream of oxygen gas. All of the carbon in the sample is converted to carbon dioxide and all of the hydrogen is converted to water.
The most important feature in this analysis is that atoms of each element involved in the reaction are conserved.

• Every carbon atom in the original sample ends up in a CO₂ molecule
  
• Every hydrogen atom in the original sample ends up in an H₂O molecule

Police officers confiscate a packet of white powder, which they believe contains heroin. Purification by a forensic chemist gave a 38.7 mg sample for combustion analysis. This sample gave 97.7 mg CO₂ and 20.81 mg H₂O. A second sample was analyzed for its nitrogen content, which was 3.8 %. Show by calculations whether or not these data are consistent with the formula for heroin C₂₁H₂₃O₅N.

1. First determine how many grams of each element are present
   
   1. Carbon first-remember, every carbon atom in the sample ends up in a CO₂ molecule
      (12.001 g C/1 mol C) X (1 mol C/1 mol CO₂) X (1 mol CO₂/44.01 g CO₂) X (1 g CO₂/10₃ mg CO₂) X 96.7 mg CO₂ = .0267 g C
      
   2. Next hydrogen-remember, every hydrogen atom in the sample ends up in a H₂O molecule
      (1.008 g H/1 mol H) X (2 mol H/1 mol H₂O) X (1 mol H₂O/18.015 g H₂O) X (1 g H₂O/10³ mg H₂O) X 20.81 mg H₂O = .0023 g H
      
   3. Now nitrogen. This atom is not found from the combustion analysis. But the problem gives us additional information on the nitrogen content.
      (1 g N/10³ mg N) X (0.038)(38.7 mg) = 0.0015 g N
      
   4. Finally oxygen. You always find the oxygen content last in a combustion analysis problem, and the only way to find oxygen is by finding out how much of the original sample remains.
      
   5. Amount of O = amount sample - amount C - amount H - amount N
      = 38.7 mg - 26.7 mg - 2.3 mg - 1.5 mg
      = 8.2 mg O

2. Now we need to determine how many moles of each element we have:
   
   1. (0.0267 g C/12.011 g/mol C) = 0.00222 mol C
      
   2. (0.0023 g H/1.008 g/mol H) = 0.00231 mol H
      
   3. (0.0015 g N/14.007 g/mol N) = 0.0001 mol N
      
   4. (0.0082 g O/15.999 g/mol O) = 0.0005 mol O

3. Next, divide each number above by smallest molar amount
   
   1. 0.00222 mol C/0.0001 mol N = 21 mol C/mol N
      
   2. 0.00231 mol H/0.0001 mol N = 23 mol H/mol N
      
   3. 0.0001 mol N/0.0001 mol N = 1 mol N/mol N
      
   4. 0.0005 mol O/0.0001 mol N = 5 mol O/mol N

4. Finally we find that the empirical and chemical formula is C₂₁H₂₃NO₅

A chemical compound was found to contain only Fe, C, and H. When 5.00 g of this compound was completely burned in O₂, 9.1 g of CO₂ and 1.80 g of H₂O were produced. Find the percentage by mass of each element in the original compound and its empirical formula.

1. First find amounts of each element in the original 5.00 g sample.
   (12.011 g C/1 mol C) X (1 mol C/1 mol CO₂) X (1 mol CO₂/44.01 g CO₂) X 9.1 g CO₂ = 2.48 g C
   (1.008 g H/1 mol H) X (2 mol H/1 mol H₂O) X (1 mol H₂O/18.015 g H₂O) X 1.8 g H₂O = 0.20 g H
   5.00 g sample - 2.48 g C - 0.20 g H = 2.32 g Fe
   
2. Next find mass percentage composition.
   %C = 2.48/5.00 x 100% = 49.6%
   %H = 0.20/5.00 x 100% = 4.0%
   %Fe = 2.32/5.00 x 100% = 46.4%
   
3. Now find empirical formula
   (1 mol C/12.011 g C) X 2.48 g C = 0.206 mol C
   (1 mol H/1.008 g H) X 0.20 g H = 0.198 mol H
   (1 mol Fe/55.847 g Fe) X 2.32 g Fe = 0.042 mol Fe

   0.206 mol C/0.042 mol Fe = 5 mol C/mol Fe
   0.198 mol H/0.042 mol Fe = 5 mol H/mol Fe
   0.042 mol Fe/0.042 mol Fe = 1 mol Fe/mol Fe

4. Empirical Formula: C₅H₅Fe