1. Determining Chemical Formulas
empirical formula- consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole number mole ratio of the different atoms in the compound
CH3 = empirical formula (does not exist)
C2H6 = molecular formula (ethene)

2. Empirical Formulas
The formulas of ionic compounds are empirical formulas by the definition of ionic formulas.
The formulas of molecular compounds may or may not be the same as its empirical formula.

3. Calculating an Empirical Formula
If the elements are in % composition by mass form, covert the percentages to grams.
Convert the masses of each element to moles by dividing the mass of the element by its molar mass.
Select the element with the smallest number of moles and divide the number of moles of each element by that number which will give you a 1:---:--- ratio.
IF the ratio is very close to a whole number ratio, apply the numbers to each element. If one of the number is not close to a whole number, use a multiplier to convert the ratio to a whole number ratio.

4. 1- If the elements are in % composition by mass form, covert the percentages to grams.
e.g. C = 40.0%  40.0 g
H = 6.67%  6.67 g
O = 53.3%  53.3 g

5. 2- Convert the masses of each element to moles by dividing the mass of the element by its molar mass.
e.g. C = 40.0/12 = 3.33 mol
H = 6.67/1 = mol
O = 53.3/16 = 3.33 mol

6. 3- Select the element with the smallest number of moles and divide the number of moles of each element by that number which will give you a 1:---:--- ratio.
e.g. C = 3.33/3.33 = 1
H = 6.67/3.33 = 2
O = 3.33/3.33 = 1

7. 4- IF the ratio is very close to a whole number ratio, apply the numbers to each element. If one of the number is not close to a whole number, use a multiplier to convert the ratio to a whole number ratio.
e.g. 1:2:1 ratio  CH2O

8. Calculating an Empirical Formula
Sample Problem
32.38% Na, % S, & 44.99% O.
1- convert to g Na, g S, & g O
÷ = mol Na
22.65 ÷ = mol S
44.99 ÷ = mol O
÷ = mol Na  2
÷ = 1 mol S
2.812 ÷ = mol O  4
4- Rounding  2:1:4  Na2SO4

9. Calculating an Empirical Formula
Review sample problem M on page 247.
Do practice problems #1, 2, & 3 on page 247.

10. Practice problem #1 page 247 63.52% iron (Fe) 36.48% sulfur (S)
Convert % to grams: Fe = 63.52g S = 36.48g
Divide each element by its molar mass:
Fe = 63.52/55.8 = 1.14 mol
S = /32.1 = 1.14 mol
Divide each number of moles by the smallest number:
Fe = 1.14/1.14 = 1 S = 1.14/1.14 = 1
Ratio = 1:1 so FeS is the empirical formula

11. Practice problem #2 page 247
K = 26.56% Cr = 35.41% O = 38.03%
K = 26.56/39.1 = mol
Cr = 35.41/52.0 = mol
O = 38.03/16.0 = 2.38 mol
K = 0.679/0.679 = 1
Cr = 0.681/0.679 = 1.003
O = 2.38/0.679 = 3.51
1:1:3.5 ratio
Double the ratio to get whole numbers  2:2:7
Empirical formula is K2Cr2O7

12. Practice problem #3 page 247.
20.0 g calcium & bromine
4.00 g Ca so g Br
Already in grams so divide by molar mass:
4.00/ 40.1 = mol Ca
16.00/79.9 = mol Br
Ca = .0997/.0997 = 1
Br = .2003/.0997 =  2
Empirical formula is CaBr2

13. Ch 7 quiz #4 Empirical Formulas
1- A compound is 27.3% carbon and 72.7% oxygen by mass. What is the empirical formula of the compound? 2- A compound is 11.1% hydrogen and 88.9% oxygen. What is its empirical formula?

14. Calculating a Molecular Formula
molecular formula- the actual formula of a molecular compound (it may or may not be the same as the empirical formula of the compound)
The molar mass of a compound is determined by analytical means & is given.
Calculate the formula mass of the empirical formula. Divide the molar mass of the compound by its empirical mass.
“Multiply” the empirical formula by this factor.

15. Calculating a Molecular Formula
empirical formula = P2O5
molecular mass is
empirical mass is
Dividing the molecular mass by the empirical mass gives a multiplication factor of : ÷ =  2
2 x (P2O5)  P4O10

16. Chapter 7 Problems Do practice problems #1 & 2 on page 249.
Do section review problems #1-4 on
page 249.

17. Practice problem #1 page 249
empirical formula = CH
formula mass = amu
empirical mass = ? = = 13.0 amu
molecular mass / empirical mass = /13.0 =  multiplication factor of 6
molecular formula = CH x 6  C6H6

18. Practice problem #2 page 249
formula mass = amu
0.44 g H & g O
1st find empirical formula:
H = 0.44/1.0 = 0.44
O = 6.92/16.0 = 0.43
0.44/0.43  1 H & 0.43/0.43  1 O
empirical formula = HO
empirical mass = 17.0
formula mass / empirical mass = 34.00/17.0 = 2
HO x 2  H2O2

19. Section review problem #1 page 249.
36.48% Na 25.41% S % O
36.48/23.0 = 1.58 mol Na
25.41/32.1 = mol S
38.11/16.0 = 2.38 mol O
1.58/0.792 =  2
0.792/0.792 = 1  1
2.38/0.792 =  3
2:1:3  Na2SO3

20. Section review problem #2 page 249.
53.70% Fe 46.30% S
53.70/55.8 = mol Fe
46.30/32.1 = 1.44 mol S
0.962/0.962 = 1 Fe
1.44/0.962 = 1.50 S
1:1.5 doubled  2:3  Fe2S3

21. Section review problem #3 page 249
1.04 g K 0.70 g Cr g O
1.04/39.1 = mol K
0.70/52.0 = mol Cr
0.86/16.0 = mol O
.0266/.0135 = 1.97  2
.0135/.0135 = 1
.0538/.0135 =  4
Empirical formula = K2CrO4

22. Section Review problem #4 page 249
4.04 g N g O f.m. = amu
4.04/14.0 = .289 mol N
11.46/16.0 =.716 mol O
.289/.289 = /.289 = 2.45
double ratio  2:5  N2O5
e.f.m. = 108
f.m./e.f.m. = 108/108 = 1
empirical formula is same as molecular formula N2O5

23. To find molar mass: add the masses of the elements in the formula of the compound.
To find number of grams (mass): multiply # of moles times the molar mass of the compound.
To find the number of moles: divide the number of grams by the molar mass of the compound.

24. To calculate % composition by mass:
1- find the molar mass of a compound
2- divide the mass of each element by the
molar mass of the compound
3- multiply by 100 to convert each ratio to a
percent

25. Calculating an Empirical Formula
If the elements are in % composition by mass form, convert the percentages to grams.
Convert the masses of each element to moles by dividing the mass of the element by its molar mass.
Select the element with the smallest number of moles and divide the number of moles of each element by that number which will give you a 1:---:--- ratio.
IF the ratio is very close to a whole number ratio, apply the numbers to each element. If one of the number is not close to a whole number, use a multiplier to convert the ratio to a whole number ratio.

26. Calculating a Molecular Formula
molecular formula- the actual formula of a molecular compound (it may or may not be the same as the empirical formula of the compound)
The molar mass of a compound is determined by analytical means & is given.
Calculate the formula mass of the empirical formula. Divide the molar mass of the compound by its empirical mass.
“Multiply” the empirical formula by this factor.

27. Final Practice- chapter 7 part 2
1- Determine the molar mass of the compound Na3PO How many moles of CO2 are in 198 g ? 3- What is the mass of 2.25 moles of H2O ? 4- What is the % composition of each element of the compound P4O10 ? 5- What is the empirical formulas of a compound that is 25.9% N and 74.1% O ? What is it molecular formula if its molecular mass is 216 ?

28. Final Practice- chapter 7 part 2
1- Determine the molar mass of the compound Na3PO4 . Na = 3 x 23.0 = 69.0 P = 1 x 31.0 = 31.0 O = 4 x 16.0 = g/mol 2- How many moles of CO2 are in 198 g ? C = 1 x 12.0 = 12.0 O = 2 x 16.0 = g/mol 198/44.0 = 4.5 mol CO2

29. 3- What is the mass of 2. 25 moles of H2O. H = 2 x 1. 0 = 2
3- What is the mass of 2.25 moles of H2O ? H = 2 x 1.0 = 2.0 O = 1 x 16.0 = g/mol 2.25 mol x 18.0 g/mol = 40.5 g H2O 4- What is the % composition of each element of the compound P4O10 ? P = 4 x 31.0 = O = 10 x 16.0 = g/mol P = 124/284(100) = 43.7% O = 160/284 (100) = 56.3%

30. 5- What is the empirical formulas of a compound that is 25.9% N and 74.1% O ? What is it molecular formula if its molecular mass is 216 ? N = 25.9/14.0 = 1.85 O = 74.1/16.0 = 4.63 N = 1.85/1.85 = 1 O = 4.63/1.85 = 2.5 1:2.5 doubled  2:5 so empirical formula = N2O5 e.f.m. = (2 x 14) + (5 x 16) = (mfm)/108 (efm) = 2 2 x 2:5  4:10  N4O10