1. Math with Chemical Formulas
Honors Chemistry Unit 5

2. Unit Objectives
Be able to perform math functions with and without your calculator using correct scientific notation.
Be able to find molar/molecular/formula mass using the periodic table.

3. Unit Objectives, cont. Be able to calculate Molarity.
Be able to calculate percent composition.
Be able to determine empirical and molecular formulas using lab data.

4. Unit Objectives, cont. Understand the mole and Avogadro’s number.
Be able to convert to/from atoms, ions, molecules, moles and grams

5. What is a “Mole” (mol)? A mole is a counting unit just like a dozen
other examples…..

6. What is a Mole, cont.?
“Official Definition”
the amount of a substance that contains as many particles as there are atoms in exactly 12g of carbon-12

7. Avogadro’s Number Constant
the number of particles in exactly one mole of a pure substance
6.02 X 1023
Memorize this Number

8. Conversion Factors – 1 mol
1 mol = 6.02 X1023 of anything

9. Conversion Factors – 1 mol Examples…
1 mol X atoms
1 mol X ions
1 mol X molecules

10. Molar Mass Mass in g of 1 mole of anything
For elements, the molar mass is equal to the atomic mass

11. Molar Mass Examples….. 1 mol C 12.01 g C 1 mol atomic wt. (g)
1 mol Li g Li

12. You try……
1 mol Ca ? g Ca
1 mol Fe ? g Fe

13. Now you have Two conversion Factors for a mole…..
1 mol X atoms, ions or molecules
and
1 mol ______(g)

14. Example 1 How many g in 2.0 mol of He?
2.0 mol He g He = 8.0g He mol He

15. Example 2 How many moles in 3.01 X 1023 atoms Ag?
3.01 X1023 atoms Ag mol Ag = X 1023 atoms Ag
0.500 mol Ag
Oct , 2005

16. Example 3
What is the mass of 1.20 X 108 atoms of Cu?
1.20 X108 atoms Cu 1mol Cu 63.6g Cu = X 1023 atoms Cu mol Cu
1.27 X g Cu

17. You try: Convert 11.5 g B to moles B 1.06 mol B
Convert 8.0 X 1019 atoms of Ag to g
0.014 g Ag

18. Formula Mass/Molecular Mass/Molar Mass
Sum of masses in a compound
Molar Mass of sodium chloride
NaCl
Na mol X 22.99g/mol = 22.99g
Cl 1 mol X g/mol = g
Total: g

19. Example 2 MgCl2 Molar Mass of magnesium chloride
Mg mol X 24.31g/mol = 24.31g
Cl 2 mol X g/mol = g
Total: g

20. Example 3 - Calcium Nitrate
Molar Mass of Ca(NO3)2
Ca mol X g/mol = 40.08g
N 2 mol X g/mol = g
O 6 mol X g/mol = 96.00g
Total: g

21. You try: Calculate the molar mass of sodium phosphate Na3PO4
Na mol X g/mol = 68.97g
P 1 mol X g/mol = g
O 4 mol X g/mol = 64.00g
Total: g

22. Conversion Factors using Formula Mass, Molecular Mass/Molar Mass
1 mol NaCl g NaCl
1 mol MgCl g MgCl2
1 mol Ca(NO3) g Ca(NO3)2
can be used as a conversion factors

23. You try…. How many mol in 127g barium chloride? (set up on board)
Answer: mol BaCl2

24. Percent Composition
Percent composition is the percent by mass of each element in a compound.
Percent composition is the same, regardless of the size of the sample.

25. % Composition Calculations
% comp = mass of element X 100% molar mass of cpd
= % element in the compound

26. Examples Find the % composition of Cu2S First, find the molar mass:
2 mol Cu = 2 X 63.55g/mol = 127.1g Cu
1 mol S = 1 X 32.06g/mol = 32.06g S
Total: g/mol

27. Example, cont. Next, find the % of each element For Cu:
% Cu = 127.1g X 100% = % Cu
159.15g
For S:
% S = 32.06g X 100% = 20.14% S g
Next,
check your work – do the %s add up to 100?

28. You try:
barium choride
sodium phosphate

29. Answers: barium chloride sodium phosphate barium 65.90%
phosphorus 18.89%
oxygen 39.04%
Back to Objectives

30. Determining Formulas
Empirical Formula = Simplest Formula

31. To find the empirical formula from data:
Assume 100% sample; change % to grams for each element
Find moles from the grams of each element
Find the smallest whole # ratio by dividing by the smallest number of moles
If necessary, multiply to get rid of fractions.

32. Example A compound is 78% B and 22% H. What is the empirical formula?
First, change % to grams and find moles:
78g B 1mol B = mol g B
22g H 1mol H = mol g H

33. Example, cont.
Next, divide all mole numbers by the smallest number of moles:
B: 7.22 mol = mol
H: mol = 3.02 = mol

34. Example, cont.
Finally, use these whole numbers as the number of each individual element. They are the subscripts.
Empirical Formula = BH3

35. Example 2
Analysis shows a compound to contain 26.56% K, 35.41% Cr, and 38.03% O. Find the empirical formula of this compound:
First (always!) assume 100g sample, convert % to g and then find moles of each element

36. Example 2 cont. Next, Conversion to moles:
26.56g K 1mol K = mol K g K
35.41g Cr 1mol Cr = mol Cr g Cr
38.03g O 1mol O = mol O g O

37. 2.377 mol O = 3.499 mol O – can’t be rounded 0.6793
Next, divide all numbers by the smallest whole number to find the smallest whole number ratios:
mol K = 1.00 mol K
mol Cr = mol Cr ~ 1.00 mol Cr
2.377 mol O = mol O – can’t be rounded

38. So, if you have: multiply all by:
.25 or
.33 or

39. Empirical Formula = K2Cr2O7
For our example:
2 X 1.00 mol K = 2 mol K
2 X 1.00 mol Cr = 2 mol Cr
2 X mol O= 7 mol O
Empirical Formula = K2Cr2O7

40. You try:
What is the empirical formula if we have a sample containing 66.0% Ca and 34.0% P?
Answer: Ca3P2

41. You try:
Find the empirical formula of a compound with 32.38% Na; 22.65% S; and 44.99% O.
Answer: Na2SO4

42. Molecular Formula Molecular Formula = Actual Formula Example: C2H6 CH3
molecular empirical
MF = (EF)x where X = Molecular mass Empirical mass

43. Example
The empirical formula of a compound was found to be P2O5. Experimentation shows that the molar mass of this compound is g/mol. What is the compound’s molecular formula?
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44. Moles in Solution
Molarity is the term used for moles dissolved in solution
Symbol for Molarity = M
Definition – moles of solute per liter of solution
Formula M = moles solute (mol) liter solution (L)

45. Example
What is the molarity of a 0.5L solution containing 2 moles of NaCl?
M = 2moles NaCl = 4 M NaCl L

46. Example 2
What is the molarity of a 250 mL solution containing 12.7 g of lithium bromide?
M = moles = 12.7g LiBr 1mol LiBr mL
L g LiBr 250mL 1L
= 0.59 M LiBr

47. Example 3
How would you make 500mL of a 0.32M solution of LiBr and water?
Given: M=0.32M, L=0.500
M= #moles/liters
.32M= #moles
.500L
#moles= 0.16 molesLiBr
0.16 molesLiBr X g = gLiBr in .500L of Water
1 mole

48. You try:
Calculate the M of a 700. mL solution of 23.2g calcium chloride
How would you make a 0.2 L solution of M CaCl2 solution?
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