The Mole Molar Conversions

A. What is the Mole? A counting unit Similar to a dozen, except instead of 12, it’s 602 billion trillion 602,000,000,000,000,000,000,000 6.02 X 1023 (in scientific notation) This number is named in honor of Amedeo Avogadro (1776 – 1856),

A. What is the Mole? HOW LARGE IS IT??? 1 mole of hockey pucks would equal the mass of the moon! 1 mole of basketballs would fill a bag the size of the earth! 1 mole of pennies would cover the Earth 1/4 mile deep!

How does a Mole relate? 6.02 x 1023 cookies 6.02 x 1023 cars 1 Mol = 6.02 x 1023 particles If I had 1 mol of cookies, how many cookies would I have? 6.02 x 1023 cookies If I had 1 mol of cars, how many cars would I have? 6.02 x 1023 cars If I had 1 mol of Aluminum, how many atoms I have? 6.02 x 1023 atoms of Al Note that the NUMBER is always the same, but the MASS is very different!

A Mole of Anything Contains 6.02 x 1023 particles = 6.02 x 1023 C atoms = 6.02 x 1023 H2O molecules = 6.02 x 1023 NaCl “formula units” 6.02 x 1023 Na+ ions and 6.02 x 1023 Cl– ions 1 mole C 1 mole H2O 1 mole NaCl

Learning Check 1. Number of atoms in 0.500 mole of Al a) 500 Al atoms b) 6.02 x 1023 Al atoms c) 3.01 x 1023 Al atoms 2.Number of moles of S in 1.8 x 1024 S atoms a) 1.0 mole S atoms b) 3.0 mole S atoms c) 1.1 x 1048 mole S atoms

B. Molar Mass Describes the mass (how heavy) 1 mole of an element or compound is. Atomic mass tells the... molar mass (g/mol) the mass of one mole of an element or atom formula mass (g/mol) the mass of one mole of an ionic compound molecular mass (g/mol) the mass of one mole of a covalent compound

B. Molar Mass Examples carbon aluminum zinc 12.01 g/mol 26.98 g/mol

Learning Check! Find the molar mass – Round to the hundredths position. It is important for significant digits! = 79.90 g/mol 1 mole of Br atoms 1 mole of Sn atoms = 118.71 g/mol

B. Molar Mass Examples water H2O 2(1.01) + 16.00 = 18.02 g/mol sodium chloride H2O 2(1.01) + 16.00 = 18.02 g/mol NaCl 22.99 + 35.45 = 58.44 g/mol

B. Molar Mass Examples sodium bicarbonate NaHCO3 sucrose NaHCO3 22.99 + 1.01 + 12.01 + 3(16.00) = 84.01 g/mol C12H22O11 12(12.01) + 22(1.01) + 11(16.00) = 342.34 g/mol

B. Molar Mass of antacid Al(OH)3 = ? Learning Check! Molar Mass of K2O = ? g/mol B. Molar Mass of antacid Al(OH)3 = ?

C. Molar Conversions molar mass 6.02  1023 MOLES (g/mol) IN GRAMS MOLES NUMBER OF PARTICLES (g/mol) (particles/mol)

C. Molar Conversion Examples How many moles of carbon are in 26 g of carbon? 26 g C 1 mol C 12.01 g C = 2.2 mol C

C. Molar Conversion Examples How many moles of oxygen are in 85 g of Oxygen gas? 85 g O2 1 mol O2 32.00 g O2 = 2.6 mol O2

C. Reverse! How many grams of nitrogen are in 0.25 moles of nitrogen gas? 0.25 mol N2 28.02 g N2 1 mol N2 = 7.0 grams N2

C. Reverse! How many grams of water are in 3.45 moles of water? 3.45 mol H2O 18.02 g H2O 1 mol H2O = 62.2 g H2O

C. Moles to Molecules Ex How many molecules are in 2.50 moles of C12H22O11? 6.02  1023 molecules 1 mol 2.50 mol = 1.50  1024 molecules C12H22O11

C. Molecules to Moles Ex How many moles are in 1.505 x 1023 molecules of HCl? 1.505 1023 molecules 1 mol = .25 Moles HCl 6.02  1023 molecules

Learning Check! The artificial sweetener aspartame (Nutra-Sweet) formula C14H18N2O5 is used to sweeten diet foods, coffee and soft drinks. How many moles of aspartame are present in 225 g of aspartame?

Percent Composition

When you are asked to determine the percentage composition of a compound, it should be understood that this refers to the percentage by mass.  In other words, the percentage composition of water shows what % (by mass) is Hydrogen, and what % is Oxygen. Percentage composition by mass does not tell us the ratio by which the atoms combine. (Does not tell us the subscripts)

Example 1 Homer has a dozen assorted doughnuts.  3 of these doughnuts are chocolate.  What percentage of Homer's doughnuts are chocolate? (note, we are talking about total number, not mass, in this case.)

Example 2 Jim goes to a school that has 258 total students.  If 164 of the students are boys, what percentage of the student body is made up of girls?

In both cases, you found the percent composition by dividing one part by the total and multiplying by 100.                part % composition = -------------- x 100        whole      

Finding the percentage composition by mass works the same way Finding the percentage composition by mass works the same way.  You divide the mass that one element contributes to the compound by the mass of the entire compound.                                                                                                    mass from element % Composition = ------------------------------ x 100                          total mass of the compound

Example 3 What percentage of the mass of carbon dioxide (CO2) is made up by the carbon? Mass of Carbon = 12.01 x 1 = 12.01 Mass of Oxygen = 16.00 x 2 = 32.00

Continued part % comp. of Carbon = --------------- x 100 whole 12.01 44.01 Answer = 27.29%

Example 4 What is the percentage composition of glucose (C6H12O6) ? Start by determining the molar mass! C = 12.01 x 6 = 72.06 H = 1.01 x 12 = 12.12 O = 16.00 x 6 = 96.00 --------------- 180.18

Remember!!! part % comp. of Element = --------------- x 100 whole 72.06 % for Carbon = -------------- x 100                          180.18 40.00% C

12.12 % for Hydrogen = ------------- x 100 96.00 % for Oxygen = -------------- x 100                          180.18 53.28 % O

Check your Answer! One way to check your answer is to make sure that all of the percentages add up to approximately 100%.  (i.e. 40.0% + 6.7% + 53.3% = 100%)   Your total may be off by a few tenths of a percent, due to rounding.

Empirical and Molecular Formulas

Empirical Formula What are we talking about??? Empirical Formula represents the smallest ratio of atoms in a formula. In other words it represents the simplest chemical formula for a particular compound.

Steps to Solve! You need to follow these basic steps to determine your answer. 1) Make sure you are starting in grams 2) Convert all grams to moles (Molar Mass) 3) Divide each by the smallest number of moles. 4) Create whole number ratio of subscripts 5) Multiply if necessary

Example 1 In an unknown molecule, there are 4.15 grams of Carbon and 1.38 grams of Hydrogen. Determine the empirical formula for the substance! Step 1) Make sure you are starting in grams Step 2) Convert each mass in grams to moles.

Now What? Step 3) Divide each by the smallest number of moles. Set up the Ratio of elements

Solution! The 1 and the 4 represent the whole number ratio of the elements in the chemical formula. The 1 represents that there is only one Carbon atom The 4 represents that there are four Hydrogen atoms The Solution: The empirical formula is CH4

Phenol, a general disinfectant, is 76. 57 % Carbon, 6 Phenol, a general disinfectant, is 76.57 % Carbon, 6.43 % Hydrogen, and 17.00 % Oxygen. Determine it’s empirical formula. Step 1) Make sure you are starting in grams……. How can we go from percent to grams? Easy! Assume we have a 100 gram sample so there would be how much of each element?

Step 2) Convert each mass in grams to moles.

Try this! Element “A” is 78.1% abundant and has a molar mass of 10.81 g/mol. Element “B” is 21.9% abundant and has a molar mass of 1.01 g/mol. Determine the empirical formula. 78.1g A X 1mol/10.81g = 7.22 mol B 21.9g B X 1mol/1.01g = 21.7 mol H 7.22 mol ÷ 7.22mol =1 21.7 mol ÷ 7.22mol = 3.01 Ratio of 1:3 AB3

Molecular Formula The molecular formula is related, but different to the empirical formula. Remember that Empirical Formula represents the lowest ratio of the atoms in a compound. The Molecular Formula is the actual, or true ratio of the elements in the compound.

Needs and Steps! 1. In order to solve you need the molar mass of the molecular formula for a compound in g/mol. 2. Determine the empirical formula. 3. Calculate the molar mass of the empirical formula 4. Divide the molar mass of the molecular formula by the molar mass of the empirical formula. Apply ratio to all subscripts.

Example The empirical formula for hydroquinone, a chemical used in photography, is C3H3O. The molecular weight of the compound Is 110 g/mol. Determine the molecular formula. Step 1) Determine the molecular weight of the compound: Given at 110 g/mol.

Step 3) Determine the molar mass of the empirical formula. Step 2) Determine the empirical formula: Given C3H3O. Step 3) Determine the molar mass of the empirical formula.

Think…I Know it’s hard… Step 4) Think about the molar mass of the molecular formula compared to the molar mass of the empirical formula. Mol. Formula Emp. Formula 110 g/mol 55.06 g/mol Divide and get the ratio 110 g/mol / 55.06 g/mol = 2

So…. That 2 represents what you will multiply all the subscripts by to determine the correct molecular formula. C3H3O x 2 = C6H6O2 Molecular Formula

Try this The empirical formula for a compound is C3H7. If the molecular weight is 86 g/mol, then what is the molecular formula? 1. The molecular weight is given: 86 g/mol 2. The empirical formula is given: - C3H7

3. Calculate the molar mass for the empirical formula.

Compare the two…. The Molecular weight was 86 g/mol The Empirical weight was 43.1 g/mol Divide Molecular by Empirical to determine the ratio! 86 g/mol / 43.1 g/mol = 2

So 2 is the ratio, multiply all the subscripts by 2 C3H7 x 2 = C6H14 is the Mol. Formula

Practice! A) Empirical Formula is S Molecular weight is 256 grams per mol B) Empirical Formula is NO2 Molecular Weight is 46 g/mol