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Math with Chemical Formulas

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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? Back to Objectives

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? Back to Objectives

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