stoichiometry hw

1. How many moles of O2 will be formed from 1.65 moles of KClO3?

1.65 mol KClO3 * (3 mol O2 / 2 mol KClO3) = 2.475 mol O2

2. How many moles of KClO3 are needed to make 3.50 moles of KCl?

3.50 mol KCl * (2 mol KClO3 / 2 mol KCl) = 3.50 mol KClO3

3. How many moles of KCl will be formed from 2.73 moles of KClO3?

2.73 mol KClO3 * (2 mol KCl / 2 mol KClO3) = 2.73 mol KCl  

4. How many moles of Fe2O3 are produced when 0.275 moles of Fe is reacted?

0.275 mol Fe * (2 mol Fe2O3 / 4 mol Fe) = 0.1375 mol Fe2O3  

5. How many moles of Fe2O3 are produced when 31.0 moles of O2 is reacted?

31.0 mol O2 * (2 mol Fe2O3 / 3 mol O2) = 20.67 mol Fe2O3

6. How many moles of O2 are needed to react with 8.9 moles of Fe?

8.9 mol Fe * (3 mol O2 / 4 mol Fe) = 6.675 mol O2

7. How many moles of O2 are produced when 1.26 moles of H2O is reacted?

1.26 mol H2O * (1 mol O2 / 2 mol H2O) = 0.63 mol O2

8. How many moles of H2O are needed to produce 55.7 moles of H2?

55.7 mol H2 * (2 mol H2O / 2 mol H2) = 55.7 mol H2O

9. If enough H2O is reacted to produce 3.40 moles of H2, then how many moles of O2 must have been made?

3.40 mol H2 * (1 mol O2 / 2 mol H2) = 1.70 mol O2

10. How many grams of O2 will be formed from 3.76 grams of KClO3?

3.76 g KClO3 * (1 mol KClO3 / 122.55 g KClO3) * (3 mol O2 / 2 mol KClO3) * (32 g O2 / 1 mol O2) = 1.47 g O2

11. How many grams of KClO3 are needed to make 30.0 grams of KCl?

30.0 g KCl * (1 mol KCl / 74.55 g KCl) * (2 mol KClO3 / 2 mol KCl) * (122.55 g KClO3 / 1 mol KClO3) = 49.3 g KClO3

12. How many grams of KCl will be formed from 2.73 g of KClO3?

2.73 g KClO3 * (1 mol KClO3 / 122.55 g KClO3) * (2 mol KCl / 2 mol KClO3) * (74.55 g KCl / 1 mol KCl) = 1.66 g KCl

13. How many grams of Fe2O3 are produced when 42.7 grams of Fe is reacted?

42.7 g Fe * (1 mol Fe / 55.85 g Fe) * (2 mol Fe2O3 / 4 mol Fe) * (159.7 g Fe2O3 / 1 mol Fe2O3) = 61.0 g Fe2O3

14. How many grams of Fe2O3 are produced when 17.0 grams of O2 is reacted?

17.0 g O2 * (1 mol O2 / 32 g O2) * (2 mol Fe2O3 / 3 mol O2) * (159.7 g Fe2O3 / 1 mol Fe2O3) = 56.6 g Fe2O3

15. How many grams of O2 are needed to react with 125 grams of Fe?

125 g Fe * (1 mol Fe / 55.85 g Fe) * (3 mol O2 / 4 mol Fe) * (32 g O2 / 1 mol O2) = 53.7 g O2

16. How many grams of CO2 are produced from the combustion of 100. grams of butane?

g C4H10 * (1 mol C4H10 / 58.12 g C4H10) * (8 mol CO2 / 2 mol C4H10) * (44.01 g CO2 / 1 mol CO2) = 303 g CO2

17. How many grams of O2 are needed to react with 100. grams of butane?

g C4H10 * (1 mol C4H10 / 58.12 g C4H10) * (13 mol O2 / 2 mol C4H10) * (32 g O2 / 1 mol O2) = 358 g O2

18. How many grams of H2O are produced when 5.38 g of O2 is reacted?

5.38 g O2 * (1 mol O2 / 32 g O2) * (10 mol H2O / 13 mol O2) * (18.02 g H2O / 1 mol H2O) = 2.34 g H2O

Empirical And Molecular Formulas

Empirical formula - (simplest formula) of a compound shows the lowest whole number ratio of the elements in the compound

Molecular formula - indicates the actual formula. It describes the number of each element that make up the a molecule or formula unit. The molecular formula is always a whole number multiple of the empirical formula

ex. C6H12O6 (molecular formula)
CH2O (empirical formula) glucose

ex. If the molecular formula is C4H10, what is the empirical formula? C2H5

ex. If the empirical formula is CH2, what are some possible molecular formulas? C2H4, C3H6, C4H8, C100H200

Determining the Empirical Formula:

ex. A compound consists of 85.6% carbon and 14.4% hydrogen. What is the empirical formula?

Assume a 100 g sample
Step 1: Convert % into mass (assume a 100g sample)

• 85.6 g C
• 14.4 g H

Step 2: Convert mass to moles

• 85.6 g C * (1 mol C / 12.01 g C) = 7.13 mol C
• 14.4 g H * (1 mol H / 1.01 g H) = 14.3 mol H

Step 3: Divide by the smallest number of moles. (If you do not get a whole number - multiply to make whole numbers.)

• 7.13 mol C / 7.13 = 1 mol C
• 14.3 mol H / 7.13 = 2 mol H

Therefore the empirical formula is CH2

ex. 2 What is the empirical formula of a compound containing 52.1% carbon, 34.7% oxygen and 13.2% hydrogen?

Step 1: Assume a 100 g sample

• 52.1 g C
• 34.7 g O
• 13.2 g H

Step 2: Convert to moles

• 52.1 g C * (1 mol C / 12.01 g C) = 4.34 mol C
• 34.7 g O * (1 mol O / 16.00 g O) = 2.17 mol O
• 13.2 g H * (1 mol H / 1.01 g H) = 13.1 mol H

Step 3: Divide by smallest

• 4.34 mol C / 2.17 = 2 mol C
• 2.17 mol O / 2.17 = 1 mol O
• 13.1 mol H / 2.17 = 6 mol H

Therefore, the empirical formula is C2H6O

ex. 3. Phosphorus combines with oxygen to form an oxide containing 43.6% oxygen. What is the empirical formula of this oxide?

Step 1: Assume 100 g sample

• 43.6 g O
• 56.4 g P

Step 2: Convert to moles

• 43.6 g O * (1 mol O / 16.00 g O) = 2.73 mol O
• 56.4 g P * (1 mol P / 30.97 g P) = 1.82 mol P

Step 3: Divide by smallest

• 2.73 mol O / 1.82 = 1.5 mol O
• 1.82 mol P / 1.82 = 1 mol P

Since we have 1.5, multiply both by 2 to get whole numbers:

• 1.5 mol O * 2 = 3 mol O
• 1 mol P * 2 = 2 mol P

Therefore, the empirical formula is P2O3

Practice: Calculate the empirical formula of the compounds whose percent composition by mass are:

a. Fe 34.78% Cl 65.22% 
Step 1: 34.78 g Fe, 65.22 g Cl
Step 2: 0.623 mol Fe, 1.84 mol Cl
Step 3: 1 mol Fe, 3 mol Cl
Empirical Formula: FeCl3

b. As 60.98% rest S 
Step 1: 60.98 g As, 39.02 g S
Step 2: 0.813 mol As, 1.22 mol S
Step 3: 2 mol As, 3 mol S
Empirical Formula: As2S3

c. N 25.93% rest O 
Step 1: 25.93 g N, 74.07 g O
Step 2: 1.85 mol N, 4.63 mol O
Step 3: 2 mol N, 5 mol O
Empirical Formula: N2O5

2. 9.23 g calcium are heated in an excess of nitrogen. The final product has a mass of 11.38 g. What is the simplest formula of the compound?

   • Mass of Nitrogen = 11.38 g (product) - 9.23 g (Ca) = 2.15 g N
     Step 1: 9.23 g Ca, 2.15 g N
     Step 2: 0.230 mol Ca, 0.153 mol N
     Step 3: 1.5 mol Ca, 1 mol N
     Empirical Formula: Ca3N2

Determining the Molecular Formula:

Step 1: Calculate the molar mass of the empirical formula.

Step 2: Compare and multiply as needed

• Method 1 - Given the empirical formula and the molar mass
  ex. A compound has a molar mass of 30.00 g/mol and has an empirical formula of CH3. What is the molecular formula?

  Step 1: Molar mass of CH3 = 12.01 + (3 * 1.01) = 15.04 g/mol
  Step 2: 30.00 g/mol / 15.04 g/mol = 2
  Molecular Formula: (CH3) * 2 = C2H6

• Method 2 - Given % composition and molar mass
  ex. A combustion analysis shows butane to contain 82.5% carbon and 17.5 % hydrogen. The molar mass of butane is 58.0 g/mol. What is the molecular formula of butane?

  Empirical Formula: (using steps above) C2H5
  Step 1: Molar mass of C2H5 = (2 * 12.01) + (5 * 1.01) = 29.07 g/mol
  Step 2: 58.0 g/mol / 29.07 g/mol = 2
  Molecular Formula: (C2H5) * 2 = C4H10

1. The empirical formula of butane, the fuel used in disposable lighters, is C2H5. In an experiment, the molar mass of butane was determined to be 58 g/mol. What is the molecular formula of butane?  

   Step 1: Molar mass of C2H5 = 29.07 g/mol
   Step 2: 58 g/mol / 29.07 g/mol = 2
   Molecular Formula: (C2H5) * 2 = C4H10

2. Oxalic acid has the empirical formula CHO2. Its molar mass is 90 g/mol. What is the molecular formula of oxalic acid?

   Step 1: Molar mass of CHO2 = 12.01 + 1.01 + (2 * 16.00) = 45.02 g/mol
   Step 2: 90 g/mol / 45.02 g/mol = 2
   Molecular Formula: (CHO2) * 2 = C2H2O4

3. The empirical formula of codeine is C18H21NO3. If the molar mass of codeine is 299 g/mol, what is the molecular formula?

   Step 1: Molar mass of C18H21NO3 = (18 * 12.01) + (21 * 1.01) + 14.01 + (3 * 16.00) = 299.4 g/mol
   Step 2: 299 g/mol / 299.4 g/mol = 1 (approximately)
   Molecular Formula: C18H21NO3

4. A compound’s molar mass is 240.28 g/mol. Its percentage composition is 75.0% carbon, 5.05% hydrogen and 20.0% oxygen. What is the compound’s molecular formula?  

   Empirical Formula: (using steps above) CH2O
   Step 1: Molar mass of CH2O = 12.01 + (2 * 1.01) + 16.00 = 30.03 g/mol
   Step 2: 240.28 g/mol / 30.03 g/mol = 8
   Molecular Formula: (CH2O) * 8 = C8H16O8