# Math with Chemical Formulas

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Math with Chemical Formulas
Honors Chemistry Unit 5

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.

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 unit just like a dozen
other 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

the number of particles in exactly one mole of a pure substance 6.02 X 1023 Memorize this Number

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

Conversion Factors – 1 mol Examples…
1 mol X atoms 1 mol X ions 1 mol X 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 C 12.01 g C 1 mol atomic wt. (g)
1 mol Li 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 X atoms, ions or molecules and 1 mol ______(g)

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

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

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

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

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

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

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

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

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

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

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.

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

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

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?

You try: barium choride sodium phosphate

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

Determining Formulas Empirical Formula = Simplest Formula

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.

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

Example, cont. Next, divide all mole numbers by the smallest number of moles: B: 7.22 mol = mol H: mol = 3.02 = 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: First (always!) assume 100g sample, convert % to g and then find moles of each element

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

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

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

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

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 Formula Molecular 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 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 NaCl 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 mL L g 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 g = gLiBr 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 M CaCl2 solution? Back to Objectives