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Chapter 10 “Chemical Quantities” You will need a calculator for this chapter!

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Page 1: Chapter 10 Chemical Quantities

Chapter 10 “Chemical Quantities”

You will need a

calculator for this

chapter!

Page 2: Chapter 10 Chemical Quantities

Section 10.1 p. 287 The Mole: A Measurement of

Matter

Page 3: Chapter 10 Chemical Quantities

How do we measure items?

You can measure mass,

volume,

or count pieces

We measure mass in grams

We measure volume in liters

We count pieces in MOLES

Page 4: Chapter 10 Chemical Quantities

Other Ways to Measure Amount

Pair: 1 pair of socks = 2 socks

Dozen: 1 dozen donuts = 12 donuts

Gross: 1 gross of pencils = 144 pencils (12 dozen)

Ream: 1 ream of paper = 500 sheets of paper

Guided Practice Problem p. 289

Page 5: Chapter 10 Chemical Quantities

Practice Problem #2 pg. 289

• Assume 2.0 kg of apples is 1 dozen and

that each apple has 8 seeds. How

many apple seeds are in 14 kg of

apples? (work INDEPENDENTLY to solve)

Page 7: Chapter 10 Chemical Quantities

Moles (abbreviated mol) Derived from German word

molekül (molecule)

SI measurement of an amount

1 mole = 6.02 x 1023 of representative particles, or…..

# of carbon atoms in exactly 12 g of Carbon-12 isotope

Called Avogadro’s number

Page 8: Chapter 10 Chemical Quantities

What are Representative Particles?

(Table 10.1 p. 290)

The smallest pieces of a substance:

1) molecular cmpd - molecule

2) ionic cmpd - formula unit (made

of ions)

3) element: is the

• Remember the 7 diatomic elements?

(made of molecules) BrINClHOF

H2 N2 O2 F2 Cl2 I2 Br2

atom

Guided Practice Problem #3 p. 291

Page 10: Chapter 10 Chemical Quantities

Quick Quiz • How big is a mole?

• If everyone in the world got a mole of pennies, how much $ would every person have?

• If you stacked a mole of paper how many times would it go from the Earth to the moon?

• How long would it take for every person in the world to eat through a mole of marshmellows?

6.02 x 1023

1 trillion bucks $1,000,000,000,000

80 billion times 80,000,000,000

40,000,000 years w/o a bathroom break!

Page 11: Chapter 10 Chemical Quantities

Consider these questions:

• How many oxygen atoms in the following?

CaCO3

Al2(SO4)3

• How many ions in the following?

CaCl2

NaOH

Al2(SO4)3

3 atoms of oxygen

12 (3 x 4) atoms of oxygen

3 total ions (1 Ca2+ ion and 2 Cl1- ions)

2 total ions (1 Na1+ ion and 1 OH1- ion)

5 total ions (2 Al3+ + 3 SO42- ions)

Page 12: Chapter 10 Chemical Quantities

Practice problems

Page 13: Chapter 10 Chemical Quantities

The Mass of a Mole of an Element

Atomic mass of element (mass

of 1 atom) expressed in amu

- atomic masses - relative

masses based on mass of C-12

(12.0 amu)

- 1 amu is 1/12 mass of C-12

atom

Page 14: Chapter 10 Chemical Quantities

Molar Mass…. = mass of 1 mol of element in grams

(periodic table)

12.01 grams C has same # particles

as 1.01 g H & 55.85 g Fe

12.01 g C = 1 mol C

1.01 g H = 1 mol H

55.85 g Fe = 1 mol Fe

All contain

6.02 x 1023

atoms

Page 16: Chapter 10 Chemical Quantities

What about compounds? 1 mol of H2O molecules has 2 mol

of H atoms & 1 mol of O atoms (think

of a compound as a molar ratio)

To find mass of 1 mol of a cmpd:

odetermine # moles of elements present

oMultiply # times their mass (from periodic table)

oadd up for total mass

Page 17: Chapter 10 Chemical Quantities

Calculating Molar Mass

Calculate molar mass of

magnesium carbonate, MgCO3.

24.3 g + 12.0 g + 3 x (16.00 g) = 84.3 g

So, 84.3 g = molar mass for MgCO3

Page 18: Chapter 10 Chemical Quantities

Section 10.2 p. 297

Mole-Mass and Mole-Volume Relationships

Page 19: Chapter 10 Chemical Quantities

Molar Mass Molar mass - generic term for

mass of 1 mol of any substance

(expressed in grams/mol)

Same as:

1) Gram Molecular Mass (for molecules)

2) Gram Formula Mass (ionic compounds)

3) Gram Atomic Mass (for elements)

o molar mass is more broad term than

these other specific masses

Page 20: Chapter 10 Chemical Quantities

Examples Calculate the molar mass of:

Na2S

N2O4

C

Ca(NO3)2

C6H12O6

(NH4)3PO4

= 78.05 g/mol

= 92.02 g/mol

= 12.01 g/mol

= 164.10 g/mol

= 180.12 g/mol

= 149.12 g/mol

Page 21: Chapter 10 Chemical Quantities

Molar Mass is… # of g in 1 mol of atoms,

formula units, or molecules

Make conversion factors

from these

- To change btwn g of

cmpd and mol of cmpd

Page 22: Chapter 10 Chemical Quantities

Using the Mole Roadmap

How many moles is 5.69 g

of NaOH?

0.142 mol NaOH

Page 23: Chapter 10 Chemical Quantities

The Mole-Volume Relationship

gases - hard to determine mass

how many moles of gas?

2 things affect gas V:

a) Temp & b) Pressure

compare all gases at = temp &

pressure

Page 24: Chapter 10 Chemical Quantities

Standard Temperature and Pressure

0ºC & 1 atm pressure

- abbreviated “STP”

At STP, 1 mol of any gas has

V of 22.4 L

- Called molar volume

1 mol of any gas at STP = 22.4 L

Page 25: Chapter 10 Chemical Quantities

Practice Examples

Page 26: Chapter 10 Chemical Quantities

Mole Day

Celebrated on

October 23rd from 6:02

am until 6:02 pm

(6:02 on 10-23)

Page 27: Chapter 10 Chemical Quantities

Density of a gas D = m / V (density = mass/volume)

- for gas units are: g / L

find density of a gas at STP if

formula known

You need: 1) mass and 2) volume

Assume 1 mol, so mass is molar

mass (from periodic table)

At STP, V = 22.4 L

Page 29: Chapter 10 Chemical Quantities

Another way: If given density, find molar mass of gas

Assume 1 mol at STP, so V = 22.4 L

modify: D = m/V to show:

“m” will be mass of 1 mol, given 22.4 L

What is molar mass of a gas with

density of 1.964 g/L?

How about a density of 2.86 g/L?

= 44.0 g/mol

64.0 g/mol

m = D x V

Page 30: Chapter 10 Chemical Quantities

Summary • all equal:

a) 1 mole

b) molar mass (in grams/mol)

c) 6.02 x 1023 representative particles (atoms, molecules, or formula units)

d) 22.4 L of gas at STP

make conversion factors from these 4 values (p.303)

Page 31: Chapter 10 Chemical Quantities

Notice all conversions must go through the MOLE!

Copy this conversion map into your notes!

Page 32: Chapter 10 Chemical Quantities

Section 10.3 p. 305

Percent Composition and Chemical Formulas

Page 33: Chapter 10 Chemical Quantities

All percent problems:

part

whole 1) Find mass of each element,

2) Divide by total mass of cmpd; & x 100

x 100 % = percent

%mass of element = mass of element

x 100% mass of cmpd

Page 34: Chapter 10 Chemical Quantities

% composition from mass Calculate the percent composition

of a compound that is made of 29.0

grams of Ag with 4.30 grams of S.

29.0 g Ag

33.3 g total X 100 = 87.1 % Ag

4.30 g S

33.3 g total X 100 = 12.9 % S

Total = 100 %

Page 35: Chapter 10 Chemical Quantities

% comp from the chemical formula

If we know formula, assume

you have 1 mole,

Subscripts used to calculate

mass of each element in 1

mole of cmpd

sum of masses is molar mass

Page 36: Chapter 10 Chemical Quantities

% Composition Examples

Page 37: Chapter 10 Chemical Quantities

% composition as conversion factor

We can also use % as

conversion factor to calculate

# grams of element in cmpd

Calculate % C in C3H8

What is mass of C in 82.0 g sample

of propane (C3H8) 67.1 g C

Page 39: Chapter 10 Chemical Quantities

What is an Empirical Formula?

• Like ingredients for recipe –

double recipe, you double each

ingredient, but ratio of

ingredients stays same

• Empirical formula: lowest

whole number ratio of atoms

in cmpd

Page 40: Chapter 10 Chemical Quantities

Calculating Empirical Find lowest whole number ratio

C6H12O6

CH4N

A formula is not just ratio of atoms, it

is also ratio of moles

1 molecule of CO2 = 1 atom of C

and 2 atoms of O

1 mol of CO2 = 1 mol C and 2 mol O

= CH2O

= this is already the lowest ratio.

Page 41: Chapter 10 Chemical Quantities

Calculating Empirical get a ratio from % composition

1) Assume you have a 100 g sample - the percentage become grams (75.1% = 75.1 grams)

2) Convert grams to moles.

3) Find lowest whole number ratio by

dividing each # of moles by

smallest value

Page 42: Chapter 10 Chemical Quantities

Example Calculate empirical formula of

cmpd composed of 38.67 % C,

16.22 % H, and 45.11 %N. Assume 100 g sample, so

38.67 g C x 1mol C = 3.22 mole C 12.0 g C

16.22 g H x 1mol H = 16.22 mole H 1.0 g H

45.11 g N x 1mol N = 3.22 mole N 14.0 g N

Now divide each value by the smallest value

CH5N

Page 43: Chapter 10 Chemical Quantities

Example

The ratio is 3.22 mol C = 1 mol C

3.22 mol N 1 mol N

The ratio is 16.22 mol H = 5 mol H

3.22 mol N 1 mol N

= C1H5N1 which is = CH5N

Page 44: Chapter 10 Chemical Quantities

Practice Problem 36 p. 310

Page 45: Chapter 10 Chemical Quantities

What is a Molecular Formula?

• Molecular formula: true # of atoms of

each element in formula of cmpd

• molecular cmpds only

• Example: molecular formula for

benzene is C6H6 (note that

everything is divisible by 6)

• Therefore, empirical formula = CH (the

lowest whole number ratio)

Page 46: Chapter 10 Chemical Quantities

Formulas (continued)

ionic compounds ALWAYS

empirical (cannot be reduced).

Examples:

NaCl MgCl2 Al2(SO4)3 K2CO3

Page 47: Chapter 10 Chemical Quantities

Formulas (continued)

Formulas for molecular compounds

MIGHT be empirical (lowest whole

number ratio).

Molecular:

H2O

C6H12O6 C12H22O11

Empirical:

H2O

CH2O C12H22O11

(Correct formula)

(Lowest whole

number ratio)

Page 48: Chapter 10 Chemical Quantities

Empirical to molecular Since empirical formula is lowest

ratio, the actual molecule weighs more

Molar mass

Empirical formula mass

whole # to increase

each coefficient in

empirical formula

=

Page 49: Chapter 10 Chemical Quantities

Empirical to molecular practice problem