Math with Chemical Formulas

Honors Chemistry Unit 5

Unit ObjectivesBe 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.

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.

Unit Objectives, cont.Understand the mole and

Avogadro’s number.Be able to convert to/from

atoms, ions, molecules, moles and grams

What is a “Mole” (mol)?

A mole is a counting unitjust like a dozenother examples…..

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

Avogadro’s NumberConstantthe number of particles in

exactly one mole of a pure substance

6.02 X 1023

Memorize this NumberMemorize this Number

Conversion Factors – 1 mol

1 mol = 6.02 X1023 of anything

Conversion Factors – 1 mol Examples…

1 mol 6.02 X 1023 atoms

1 mol 6.02 X 1023 ions

1 mol 6.02 X 1023 molecules

Molar Mass

Mass in g of 1 mole of anything

For elements, the molar mass is equal to the atomic mass

Molar Mass Examples…..

1 mol atomic wt. (g)

1 mol C 12.01 g C

1 mol Li 6.94 g Li

You try……

1 mol Ca ? g Ca

1 mol Fe ? g Fe

Now you have Two conversion Factors for a mole…..

1 mol 6.02 X 1023 atoms, ions or molecules

andand

1 mol ______(g)

Example 1How many g in 2.0 mol of He?

2.0 mol He 4.0g He = 8.0g He 1.0 mol He

Oct. 16-22, 2005

Example 2How many moles in 3.01 X 1023

atoms Ag?

3.01 X1023 atoms Ag 1mol Ag =

6.02 X 1023 atoms Ag

0.500 mol Ag

Example 3 What is the mass of 1.20 X 108 atoms of Cu?

1.20 X108 atoms Cu 1mol Cu 63.6g Cu =

6.02 X 1023 atoms Cu 1 mol Cu

1.27 X 10-14 g Cu

You try:Convert 11.5 g B to moles B

1.06 mol BConvert 8.0 X 1019 atoms of

Ag to g

0.014 g Ag

Formula Mass/Molecular Mass/Molar Mass

Sum of masses in a compoundMolar Mass of sodium chloride

NaClNa 1 mol X 22.99g/mol = 22.99g

Cl 1 mol X 35.45g/mol = 35.45g

Total: 58.44g

Example 2Molar Mass of magnesium chloride

MgCl2Mg 1 mol X 24.31g/mol = 24.31g

Cl 2 mol X 35.45g/mol = 70.90g

Total: 95.21g

Example 3 - Calcium Nitrate

Molar Mass of Ca(NO3)2

Ca 1 mol X 40.08 g/mol = 40.08g

N 2 mol X 14.01 g/mol = 28.02g

O 6 mol X 16.00 g/mol = 96.00g

Total: 164.10g

You try:Calculate the molar mass of

sodium phosphate

Na3PO4

Na 3 mol X 22.99 g/mol = 68.97g

P 1 mol X 30.97 g/mol = 30.97g

O 4 mol X 16.00 g/mol = 64.00g

Total: 163.94g

Conversion Factors using Formula Mass, Molecular Mass/Molar Mass

1 mol NaCl58.44g NaCl

1 mol MgCl295.21g MgCl2

1 mol Ca(NO3)2

164.1g Ca(NO3)2

can be used as a conversion factors

You try….How many mol in 127g barium

chloride?

(set up on board)

Answer: 0.610 mol BaCl2

Percent CompositionPercent composition is the percent by

mass of each element in a compound.

Percent composition is the same, regardless of the size of the sample.

% Composition Calculations

% comp = mass of element X 100% molar mass of cpd

= % element in the compound

ExamplesFind the % composition of Cu2SFirst, 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: 159.15g/mol

Example, cont. Next, find the % of each elementFor Cu:For Cu:% Cu = 127.1g X 100% = 79.86% Cu

159.15gFor S:For S:% S = 32.06g X 100% = 20.14% S

159.15gNext, check your work – do the %s add up to 100?

You try:barium choride

sodium phosphate

Answers:barium chloride

barium 65.90%

chloride 34.10%sodium phosphate

sodium 42.07%

phosphorus 18.89%

oxygen 39.04%

Back to Objectives

Determining Formulas

Empirical Formula = Simplest Formula

To find the empirical formula from data:

1. Assume 100% sample; change % to grams for each element

2. Find moles from the grams of each element

3. Find the smallest whole # ratio by dividing by the smallest number of moles

4. If necessary, multiply to get rid of fractions.

Example A compound is 78% B and 22% H. What is

the empirical formula?

FirstFirst, change % to grams and find moles:

78g B 1mol B = 7.22 mol10.81g B

22g H 1mol H = 21.78 mol1.01g H

Example, cont.NextNext, divide all mole numbers by the smallest

number of moles:

B: 7.22 mol = 17.22 mol

H: 21.78 mol = 3.02 = 37.22 mol

Example, cont. Finally, use these whole numbers as the

number of each individual element. They are the subscripts.

Empirical Formula = BH3

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:

FirstFirst (always!) assume 100g sample, convert % to g and then find moles of each element

Example 2 cont. Next,Next, Conversion to moles:

26.56g K 1mol K = 0.6793 mol K39.10g K

35.41g Cr 1mol Cr = 0.6810 mol Cr52.00g Cr

38.03g O 1mol O = 2.377 mol O16.00g O

Next,Next, divide all numbers by the smallest whole number to find the smallest whole number ratios:

0.6793 mol K = 1.00 mol K0.6793

0.6810 mol Cr = 1.003 mol Cr ~ 1.00 mol Cr0.6793

2.377 mol O = 3.499 mol O – can’t be rounded

0.6793

So, if you have: multiply all by:

.25 or .75 4

.33 or .66 3

.50 2

For our example:

2 X 1.00 mol K = 2 mol K

2 X 1.00 mol Cr = 2 mol Cr

2 X 3.499 mol O= 7 mol O

Empirical Formula = K2Cr2O7

You try:What is the empirical formula if we have

a sample containing 66.0% Ca and 34.0% P?

Answer: Ca3P2

You try:Find the empirical formula of a

compound with 32.38% Na; 22.65% S; and 44.99% O.

Answer: Na2SO4

Molecular FormulaMolecular Formula = Actual Formula

Example:

C2H6 CH3

molecular empirical

MF = (EF)x where X = Molecular mass Empirical mass

Example The empirical formula of a compound was

found to be P2O5. Experimentation shows that the molar mass of this compound is 283.89 g/mol. What is the compound’s molecular formula?

Back to Objectives

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)

Example What is the molarity of a 0.5L solution

containing 2 moles of NaCl?

M = 2moles NaCl = 4 M NaCl0.5 L

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 1000mL

L 86.8g LiBr 250mL 1L

= 0.59 M LiBr

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 86.841g = 13.894gLiBr in .500L of Water 1 mole

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 0.50 M CaCl2 solution?

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