1. Calculating Percent Composition, (Empirical & Molecular Formulas)
Unit 6 Learning Target 2
Calculating Percent Composition, (Empirical & Molecular Formulas)

2. What you will need (so get it out!)
Periodic Table
Calculator
Writing Instrument
Note Sheet (from last week, the one with the definitions.)

3. Percent Composition
The percent by mass of each element in a compound. (pg. 341)
Application:
Percent composition= mass of element X (of element) mass of compound

4. Let’s Practice (on back of note sheet)
Start just like you would if calculating molar mass.
NaCl Na
1 x
23 = 23 Cl
35 =
+35 58

5. Next…
Divide the total mass of each element by the total mass of the compound. Na
23/58 = Cl
35/58 = .6034

6. Finally…
Multiply decimal by 100. Na
x 100
39.66 % Cl
x 100
60.34 %

7. One more…
Try this one on your own:
Mg3(PO4)2

8. Molar Mass Mg
3 x 24 = 72 P
2 x 31 = 62 O
8 x 16 =
+128
262

9. Divide Mg
72/262 =
0.2749 P
62/262 =
0.2366 O
128/262 =
0.4885

10. Multiply by 100 Mg 0.2749 x 100 = 27.46 % P 0.2366 x100 = 23.66 % O
48.85 %

11. Worksheet Time
Work independently or with a partner to complete the worksheet.
Formulas you may need:
6) Fe3(PO4)2
7) BeN
8) KCN
9) Mn(NO3)3
10) Li3P
11) Ni2(SO4)3

12. Empirical Formula
A formula that shows the smallest whole number ratio of the elements of a compound, and might be the same as the actual molecular formula. (pg. 344)
Application:
CH2O
CH3OH
H2O

13. How to solve an empirical formula problem
Convert masses of elements to moles.
If the given information is in percentages, assume a 100 gram sample so percents become grams.
Divide moles of each element by the smallest number of moles.
Round to nearest whole number unless it is close to .5
This ratio gives the subscripts for each element in the formula.

14. Sample Problem
Methyl Acetate is a solvent commonly used in paints, inks, and adhesives. Determine the empirical formula for Methyl Acetate, which has the following chemical analysis: 48.64% Carbon, 8.16% Hydrogen, and 43.20% Oxygen.

15. Convert to Moles 48.64% Carbon, 8.16% Hydrogen, and 43.20% Oxygen C
48.64 %
48.64 g C x 1 mol C =
g C
4.05 mol C H
8.16 %
8.16 g H x 1 mol H =
g H
8.16 mol H O
43.20 %
43.20 g O x 1 mol O =
g O
2.70 mol O

16. Divide by Smallest # of MolesC
4.05 moles C =
2.70 moles
1.5 moles C
1.5 x 2 = 3 H
8.16 moles H =
3 moles H
3 x 2 = 6 O
2.70 moles O =
1 mole O
1 x 2 = 2
The number of moles of each element was multiplied
by 2 to make 1.5 a whole number.

17. Write the formula
Methyl Acetate
C3H6O2

18. Let’s do one more!
Succinic Acid is a substance produced by lichens. Chemical analysis indicates that it is composed of g Carbon, 5.08 g Hydrogen, and g Oxygen. Determine the empirical formula.

19. Convert to Moles 40.68 g Carbon, 5.08 g Hydrogen, and 54.24 g Oxygen C
40.68 g C x 1 mol C =
g C
3.39 mol C H
5.08 g
5.08 g H x 1 mol H =
g H
5.08 mol H O
54.24 g
54.24 g O x 1 mol O =
g O
3.39 mol O

20. Divide by the smallest # of Moles
Succinic Acid= C2H3O2 C
3.39 moles C =
3.39 moles
1 moles C
1 x 2 = 2 H
5.08 moles H =
1.5 moles H
1.5 x 2 = 3 O
3.39 moles O =
1 mole O

21. Empirical  Molecular Formula
Solved just like empirical formula with three additional steps.
Find molar mass of the empirical formula.
Divide molar mass of compound molar mass of empirical formula
Multiply the coefficients of the empirical formula by the quotient from the previous step.

22. Practice
Succinic Acid is a substance produced by lichens. Chemical analysis indicates that it is composed of g Carbon, 5.08 g Hydrogen, and g Oxygen. The molar mass of the compound is g/mol. Determine the empirical and molecular formulas.

23. Sound familiar? We have already found the empirical formula.
C2H3O2
Calculate the molar mass:
2(12 g/mol)+3(1 g/mol)+2(16 g/mol)= 59.0 g/mol
Divide mass of molecular by mass of empirical:
118.1 g/mol = g/mol
2 (C2H3O2) = C4H6O4

24. Check yourself
4(12 g/mol)+6(1 g/mol)+4(16 g/mol)= g/mol

25. Worksheet time!