1. Covalent Calculations
**Calculations mean the return of SIGNIFICANT FIGURES**

2. Representative Particles
The form in which a substance exists is its “representative particle”.
Covalent compounds: molecules
Substance
Representative Particle
Element
Atom
Ion
Ionic Compound
Formula Unit
Molecular Compound
Molecule

3. Solving Problems with Avagadro’s Number
1 mole = 6.02 x 1023 representative particles
Representative Particles: unit needs to
be applicable to what is being
calculated.

4. Packet #1. How many moles are 3.891 x 1024 molecules of CO2?
1 mol CO2
3.891 x 1024 molecules CO2
6.02 x 1023 molecules CO2
6.463 mol CO2

5. Packet #2. How many molecules are in 0.445 mol of sulfur hexafluoride?
0.445 mol SF6
6.02 x 1023 molecules SF6
1 mol SF6
2.68 x 1023 molecules SF6

6. Packet #3. How many atoms of Hydrogen are in 0.621 mol of ammonia?
0.621 mol NH3
6.02 x 1023
molecule NH3
3 atoms H
1 mole NH3
1 molecule
NH3
= x 1024 atoms H

7. Packet #4. How many atoms of nitrogen are in 2
Packet #4. How many atoms of nitrogen are in 2.03 mol of dinitrogen tetroxide?
2.03 mol N2O4
6.02 x 1023
molecule N2O4
2 atoms N
1 mole N2O4
1 molecule
N2O4
= x 1024 atoms N

8. Mole-Mass and Mole-Volume Relationships

9. #5. What is the molar mass of carbon tetrafluoride?
C 1 x = F 4 x = g
Finish b, c, and d for homework!

10. Terminology Summary Mass terminology Used For Example Gram atomic mass
Atoms
What is the mass of Zn?
Gram molecular mass
Molecules (2 or more non-metals together)
What is the mass of CO2?
Gram formula mass
Formula units (metal and nonmetal combinations)
What is the mass of MgF2?

11. Mole – Mass Conversions
Conversion factor
1 mol = molar mass (g)

12. #6. What is the mass of 0.667 moles of silicon dioxide?
0.667 mol SiO2
60.09 g SiO2
1 mol SiO2
= g SiO2

13. #7. How many moles is equivalent to 7.055 grams of dinitrogen monoxide?
N2O = (2 x 14.01) = g
7.055 g N2O
1 mol N2O
44.02 g N2O
= mol N2O

14. Multi-step Mole Problems
Using multiple conversion factors.

15. #8. What is the mass of 8.120 x 1023 molecules of tetranitrogen monoxide?
8.120 x 1023 mc N4O
1 mol N4O
72.04 g N4O
6.02x1023 mc N4O
1 mol N4O
= g N4O

16. #9. How many molecules are in 12.5 grams of chlorine gas?
12.5 g Cl mol Cl x 1023 mc Cl g Cl mol Cl2
= 1.06 x 1023 mc Cl2

17. #10. How many atoms of phosphorus are present in 20
#10.How many atoms of phosphorus are present in 20.0 g of diphosphorus tetrahydride?
20.0 g P2H4
1 mol P2H4
6.02x1023 mc P2H4
2 atoms P
65.98 g P2H4
1 mol P2H4
1 mc P2H4
= 3.65 x 1023 atoms P

18. Calculating Empirical Formulas
Empirical formula - lowest whole number ratio of the elements in a compound
It may or may not be the same as the molecular formula.
Ex. H2O is both the empirical & molecular formula
H2O2 is a molecular formula while HO is the empirical formula for H2O2.

19. Steps to calculate empirical formula:
Find moles of each element.
Set up mole ratio for each element in the compound.
Simplify mole ratio (divide by smallest). If your answers are not in whole numbers, you must multiply by 2,3,4,or 5 to get whole numbers.
Use mole ratio as subscripts in the formula. If given % composition, assume 100 g of compound.
Memory Trick: % to mass, mass to moles, divide by smallest and round ‘till whole!

20. #24. A compound is 79.8% C and 20.2% H. Find its empirical formula.
79.8gC
1mol C
= mol C
/6.64 = 1
12.01 g C
20.2 g H
1 mol H
= 20.0 mol H
/6.64 = 3
1.01 g H
CH3

21. #25. An oxide of aluminum is formed by the complete reaction of 4
#25. An oxide of aluminum is formed by the complete reaction of 4.151g of aluminum with g of oxygen. Calculate the empirical formula for this compound.
4.151g Al 1 mol Al = mol Al g Al 3.692g O 1 mol O = mol O g O
= 1
x 2
0.1539
= 2
= 1.5
x 2
0.1539
= 3
Al2O3

22. Calculating Molecular Formulas
Molecular formulas are the actual formulas. They may be the same as the empirical formula or a multiple of it.
To find the multiple (n), take the gram formula mass (gfm) and divide by the empirical formula mass (efm):
Multiply each subscript in the empirical formula by n to get the molecular formula.
n = gfm
efm

23. Ex. A white powder is analyzed and found to have the empirical formula P2O5. The compound has a molar mass of g. What is the compound’s molecular formula?
P2O5 = g/mol gfm/efm g/mol g/mol = 2 2(P2O5) = P4O10

24. 27. A compound used as an additive for gasoline to help prevent engine knock shows the following percentage composition: % Cl 24.27% C 4.07% H The molar mass is known to be g. Determine the empirical formula and the molecular formula for this compound.
/2.021 = 1
71.65 g Cl 1 mol Cl = mol Cl
35.45 g Cl
24.27g C 1 mol C = mol C
12.01 g C
4.07g H 1 mol H = mol H
1.01 g H
ClCH2 or CH2Cl
efm = =49.48 g/mol
98.96/49.48= C2H4Cl2
/2.021 = 1
/2.021 = 2