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Chemical Quantities

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How you measure how much?

You can measure mass, or volume,or you can count pieces.We measure mass in grams.We measure volume in liters.

We count pieces in MOLES.

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Moles

A unit of measureDefined as the number of carbon atoms

in exactly 12 grams of carbon-12. 1 mole is 6.02 x 1023 particles.Treat it like a very large dozen6.02 x 1023 is called Avagadro’s number.

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Representative particles

The smallest pieces of a substance.For a molecular compound it is a

molecule.For an ionic compound it is a formula

unit (simplest ratio).For an element it is an atom.

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Measuring Moles

Since the mole is the number of atoms in 12 grams of carbon-12,

the decimal number on the periodic table is also the mass of 1 mole of those atoms in grams (this goes for all elements).

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Molar Mass

The mass of 1 mole of an element in grams.

12.01 grams of carbon has the same number of pieces as 1.008 grams of hydrogen and 55.85 grams of iron.

We can right this as 12.01 g C = 1 mole

We can count things by weighing them.

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What about compounds?

in 1 mole of H2O molecules there are two

moles of H atoms and 1 mole of O atomsTo find the mass of one mole of a

compound determine the moles of the elements they

haveFind out how much they would weighadd them up

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What about compounds?What is the mass of one mole of CH4?

1 mole of C = 12.01 g4 mole of H x 1.01 g = 4.04g

1 mole CH4 = 12.01 + 4.04 = 16.05g

The molar mass of CH4 is 16.05g

The mass of one mole of a molecular compound.

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Molar MassThe mass of one mole of an ionic compound.Calculated the same way.

What is the molar mass of Fe2O3?

2 moles of Fe x 55.85 g = 111.70 g3 moles of O x 15.99 g = 47.97 gThe molar mass = 111.70 g + 47.97 g =

159.67g

YOU TRY #1

Calculate the molar mass of the followingA. Na2S

B. N2O4

C. C

D. Ca(NO3 )2

E. C6H12O6

F. (NH4 )3PO4

Using Molar Mass

6.02 x 1023

MolesMass Representative Particles

PT

Both

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Molar Mass

The number of grams of 1 mole of atoms, ions, or molecules.

We can make conversion factors from these.

To change grams of a compound to moles of a compound.

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For example

How many moles is 5.69 g of NaOH?

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For example

How many moles is 5.69 g of NaOH?

5 69. g

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For example

How many moles is 5.69 g of NaOH?

5 69. g mole

g

need to change grams to moles

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For example

How many moles is 5.69 g of NaOH?

5 69. g mole

g

need to change grams to moles for NaOH

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For example

How many moles is 5.69 g of NaOH?

5 69. g mole

g

need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g

1 mole of H = 1.01 g

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For example

How many moles is 5.69 g of NaOH?

5 69. g mole

g

need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g

1 mole of H = 1.01 g 1 mole NaOH = 40.00 g

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For example

How many moles is 5.69 g of NaOH?

5 69. g 1 mole

40.00 g

need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g

1 mole of H = 1.01 g 1 mole NaOH = 40.00 g

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For example

How many moles is 5.69 g of NaOH?

5 69. g 1 mole

40.00 = 0.142 mol NaOH

g

need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g

1 mole of H = 1.01 g 1 mole NaOH = 40.00 g

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More Examples

How much would 2.34 moles of carbon weigh?

How many atoms of lithium in 1.00 g of Li?

How many molecules in 6.8 g of CH4?

How many grams is 9.87 moles of H2O?

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YOU TRY #2

A. How many moles of magnesium in 24.31 g of Mg?

B. How much would 3.45 x 1022 atoms of U weigh?

C. How many moles is 4.56 g of CO2 ?

D. 49 molecules of C6H12O6 weighs how

much?

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All the things we can change

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Moles

Mass

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Moles

MassPT

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Moles

Mass

Representative Particles

PT

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6.02 x 1023

Moles

Mass

Representative Particles

PT

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Moles

Mass

Representative Particles

6.02 x 1023

PT

Atoms

Moles

Mass

Representative Particles

6.02 x 1023

PT

Atoms IonsMolecules

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Percent CompositionLike all percents Part x 100 % = %

wholeFind the mass of each component,divide by the total mass.

Molar mass of element (g) x 100 = % of element

Molar mass of formula (g)

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ExampleCalculate the percent composition of

Cu2S? Cu = 63.55 g x 2 = 127.10 g S = 32.06 g --- Cu2S = 159.16 g

% Cu 127.10g 159.16g

x 100 = 79.86 % Cu

% S32.06 g159.16g

x 100 = 20.14 % S

TIP - Each % should add up to equal100%!!!!!!!!!!!!

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Example 2Calculate the percent composition of

glucose C6H12O6Molar mass breakdownC = 72.06 gH = 12.12 gO = 96.00 g

Total = 180.18 g

% C = 72.06g 180.18 g

x 100 = 39.99 %

% H = 12.12g 180.18g

x 100 = 6.73 %

% O = 96 g 180.18g

x 100 = 53.28 %

You Try #3Find the percent composition of the following

A.Potassium nitrateB.H2SO4

C.C2H5OHD.C6H5NH2

Percent composition with WATER

Some compounds trap water moleculesSodium carbonate – it traps 10 water

molecules for every 1 sodium carbonateNa2CO3 10 H2O

To calculate the percent of water…1. Find the molar mass of the WHOLE formula

Sometimes this number is given

2. Find the molar mass of the water present3. Mm of water (g) x 100% = % water

mm of whole formula (g)

Example 1Calculate the percent of water in

Na2CO3 10 H2O. It has a molar mass of 286.14 g.

Molar mass of waterbreakdown

H = 1.01g x 20 = 20.20gO = 16.00g x 10 = 160.00g

Total mm of water=180.20g

% H2O = 180.20g 286.14g

x 100 = 62.98 % H2O

Example 2Calculate the percent composition of

water in ZnSO4 7H2O

Molar mass breakdownZn = 65.39gS = 32.06gO = 16.00g x 4 = 64.00g

H = 1.01g x 14 = 14.14gO = 16.00g x 7 = 112.00g

% H2O = 126.14 g 287.59 g

x 100 = 43.86 %

Molar massOf WHOLEFormula =287.59 g Molar mass

Of water only =126.14 g

You Try #4

A.Calculate the percent composition of water in CuSO4 5H2O

B. Find the percentage of water in MgSO4 7H2O

Empirical Formula

From percentage to formula

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The Empirical FormulaThe lowest whole number ratio of elements in

a compound (reduced form).The molecular formula the actual ratio of

elements in a compound.The two can be the same. CH2 empirical formulaC2H4 molecular formulaC3H6 molecular formulaH2O bothJust find the lowest whole number ratio

Calculating EmpiricalMeans we can get ratio from percent

composition.Assume you have a 100 g.The percentages become grams.Can turn grams to moles.

Remember conversions!!!!!!!! Find lowest whole number ratio by

dividing by the smallest mole number.

Example 1Calculate the empirical formula of a compound

composed of 38.67 % C, 16.22 % H, and 45.11 %N.Assume 100 g so38.67 g C x 1mol C = 3.22 mole C

12.01 g C 16.22 g H x 1mol H = 16.06 mole H

1.01 g H45.11 g N x 1mol N = 3.22 mole N

14.01 g N

NOW divide by your smallest mole number to get a whole number ratio…

3.22/3.22 = 1 mole C

16.06/3.22 = 5 moles H

3.22/3.22 = 1 mole N

This is the ONLY time you can round !!!

Empirical formula = CH5N

Example 2Calculate the empirical formula of a

compound composed of 70 % Mn and 30% O.

Assume 100 g so70 g Mn x 1mol Mn = 1.27 mol Mn

54.94 g Mn

30 g O x 1mol O = 1.88 mole O 16.00 g O

NOW divide by your smallest mole number to get a whole number ratio…

1.27/1.27 = 1 mole Mn

1.88/1.27 = 1.5 moles O

This is the ONLY time you can round !!!

Empirical formula = Mn2O3

BUT onlyround to the nearest wholeor half number

You Try #5A.Determine the formula for a black, powdery compound made of 63% manganese and 37% oxygen.

B.Find the formula for an ingredient of rechargeable batteries that has the followingpercentage composition: 21.9% O, 1.4% H, and 76.7% Cd.

84 potatoes63 carrots15 onions3 heads of garlic27 turnips42 pieces of celery9 cans of green beans6 cans of tomatoes6 cans of corn6 cans of lima beans12 lbs. of beef6 bottles of hot sauce6 gallons of water

Empirical to molecular

Related by a whole number multiple.Divide the actual molar mass by the the mass of one mole of the empirical formula.

Example (easy type)

1. Determine the molecular formula of a compound having an empirical formula of CH and a molar mass of 78.11 g.

GIVEN:empirical = CHmm = 78.11 g

First step:calculate the molar mass of the empirical… 13.02 g

Next step:find the multiple

78.1113.02

= 6

Final step:run the multiple through the empirical to get your molecular

C6H6

Example 2 (a bit more challenging)

A compound has the following composition: 76.54% C, 12.13% H, 11.33% O. If its molar mass is 282.45 g, what is its molecular formula?

GIVEN:mm = 282.45 g% per element

First step:calculate the empirical…

76.54 g C x 1mol C = 6.37 mol C 12.01 g C

12.13 g H x 1mol H = 12.01 mol H

1.01 g H

11.33 g O x 1mol O = 0.71 mole O 16.00 g O

NOW divide by your smallest mole number to get a whole number ratio…

6.37/0.71 = 9 moles C

12.01/0.71 = 17 moles H

0.71/0.71 = 1 mole O

Empirical formula = C9H17O

Example 2 CONTINUEDA compound has the following

composition: 76.54% C, 12.13% H, 11.33% O. If its molar mass is 282.45 g, what is its molecular formula?

First step:calculate the molar mass of the empirical (C9H17O)…141.26 g

Next step:find the multiple

282.45 141.26

= 2

Final step:run the multiple through the empirical to get your molecular

C18H34O2

You Try #6A. The empirical formula for mercury (I) chloride is

found to be HgCl. The molar mass is found to be 472.08 g. Molecular formula is…

B. The empirical formula for trichloroisocyanuric acid, the active ingredient in dry household bleach, is OCNCl (that last letter is an L). The molar mass is 232.42 g. What is the molecular formula?

C. Calculate the molecular formula for a compound that has a molar mass of 86.17 g and contains 83.62% carbon and 16.38% hydrogen.